ON THE FUNCTIONAL EQUATION OF THE ARTIN L–FUNCTIONS

TABLE OF CONTENTS

0. Introduction

1. Weil Groups

2. The Main Theorem

3. The Lemmas of Induction

4. The Lemma of Uniqueness

5. A Property of λ-Functions

6. A Filtration of the Weil Group

7. Consequences of Stickelberger’s Result

8. A Lemma of Lamprecht

9. A Lemma of Hasse

10. The First Main Lemma

11. Artin–Schreier Equations

12. The Second Main Lemma

13. The Third Main Lemma

14. The Fourth Main Lemma

15. Another Lemma

16. Definition of the λ-Functions

17. A Simplification

18. Nilpotent Groups

19. Proof of the Main Theorem

20. Artin L-Functions

21. Proof of the Functional Equation

22. Appendix

23. References O. Introduction

In this paper I want to consider not just the L-functions introduced by Artin [1] but the more general functions introduced by Weil [15]. To define these one needs the notion of a Weil group as described in [3]. This notion will be explained in the first paragraph. For now a rough idea will suffice. If E is a global field, that is an algebraic number field of finite degree over the rationals or a function field over a finite field, C_{E} will be the idéle class group of E. If E is a local field, that is the completion of a global field at some place [16], archimedean or nonarchimedean, C_{E} will be the multiplicative group of E. If K/E is a finite Galois extension the Weil group W_{K/E} is an extension of 𝔊(K/E), the Galois group of K/E, by C_{K}. It is a locally compact topological group.

If E ⊆ E′ ⊆ K and K/E is finite and Galois W_{K/E′} may be regarded as a subgroup of W_{K/E}. It is closed and of finite index. If E ⊆ K ⊆ L there is a continuous map of W_{L/E} onto W_{K/E}. Thus any representation of W_{K/E} may be regarded as a representation of W_{L/E}. In particular the representations ρ_{1} of W_{K_{1}/E} and ρ_{2} of W_{K_{2}/E} will be called equivalent if there is a Galois extension L/E containing K_{1}/E and K_{2}/E such that ρ_{1} and ρ_{2} determine equivalent representations of W_{L/E}. This allows us to refer to equivalence classes of representations of the Weil group of E without mentioning any particular extension field K.

In this paper a representation of W_{K/E} is understood to be a continuous representation ρ in the group of invertible linear transformations of a finite-dimensional complex vector space which is such that ρ(w) is diagonalizable, that is semisimple, for all w in W_{K/E}. Any one-dimensional representation of W_{K/E} can be obtained by inflating a one-dimensional representation of W_{E/E} = C_{E}. Thus equivalence classes of one-dimensional representations of the Weil group of E correspond to quasi-characters of C_{E}, that is, to continuous homomorphisms of C_{E} into ℂ^{×}.

Suppose E is a local field. There is a standard way of associating to each equivalence class ω of one-dimensional representations a meromorphic function L(s, ω). Suppose ω corresponds to the quasi-character χ_{E}. If E is nonarchimedean and ϖ_{E} is a generator of the prime ideal P_{E} of O_{E}, the ring of integers in E, we set

L(s, ω) = \frac{1}{1 − χ_{E}(ϖ_{E}) |ϖ_{E}|^{s}}

if χ_{E} is unramified. Otherwise we set L(s, ω) = 1. If E = ℝ and

χ_{E}(x) = (sgn x)^{m} |x|^{r}

with m equal to 0 or 1 we set

L(s, ω) = π^{− \frac{1}{2} (s+r+m)} Γ (\frac{s + r + m}{2}) .

If E = ℂ and z ∈ E then, for us, |z| will be the square of the ordinary absolute value. If

χ_{E}(z) = |z|^{r}z^{m}z̄^{n}

where m and n are integers such that m + n ≥ 0, mn = 0, then

L(s, ω) = 2 (2π)^{−(s+r+m+n)} Γ(s + r + m + n)

It is not difficult to verify, and we shall do so later, that it is possible, in just one way, to define L(s, ω) for all equivalence classes so that it has the given form when ω is one-dimensional, so that

L(s, ω_{1} ⊕ ω_{2}) = L(s, ω_{1}) L(s, ω_{2})

so that if E′ is a separable extension of E and ω is the equivalence class of the representation of the Weil group of E induced from a representation of the Weil group of E′ in the class Θ then L(s, ω) = L(s, Θ).

Now take E to be a global field and ω an equivalence class of representations of the Weil group of E. It will be seen later how, for each place v, ω determines an equivalence class ω_{v} of representations of the Weil group of the corresponding local field E_{v}. The product

∏_{v} L(s, ω_{v})

which is taken over all places, including the archimedean ones, will converge if the real part of s is sufficiently large. The function it defines can be continued to a function L(s, ω) meromorphic in the whole complex plane. This is the Artin L-function associated to ω. It is fairly well-known that if ω̃ is the class contragredient to ω there is a functional equation connecting L(s, ω) and L(1 − s, ω̃).

The factor appearing in the functional equation can be described in terms of the local data. To see how this is done we consider separable extensions E of the fixed local field F. If Ψ_{F} is a non-trivial additive character of F let ψ_{E/F} be the non-trivial additive character of E defined by

ψ_{E/F} (x) = ψ_{F} (S_{E/F} x)

where S_{E/F} x is the trace of x. We want to associate to every quasi-character χ_{E} of C_{E} and every non-trivial additive character ψ_{E} of E a non-zero complex number Δ(χ_{E}, ψ_{E}). If E is nonarchimedean, if 𝔓_{E}^{m} is the conductor of χ_{E}, and if 𝔓_{E}^{−n} is the largest ideal on which ψ_{E} is trivial choose any γ with O_{E}γ = 𝔓_{E}^{m+n} and set

Δ(χ_{E}, ψ_{E}) = χ_{E}(γ) \frac{∫_{U_{E}} ψ_{E} (\frac{α}{γ}) χ_{E}^{−1}(α)dα}{|∫_{U_{E}} ψ_{E} (\frac{α}{γ}) χ_{E}^{−1}(α)dα|}.

The right side does not depend on γ. If E = ℝ,

χ_{E}(x) = (sgnx)^{m} |x|^{r}

with m equal to 0 or 1, and ψ_{E}(x) = e^{2πiux} then

Δ(χ_{E}, ψ_{E}) = (i sgnu)^{m} |u|^{r}.

If E = ℂ, ψ_{ℂ}(z) = e^{4πi Re(wz)}, and

χ_{ℂ}(z) = |z|^{r}z^{m}z̄^{n}

with m + n ≥ 0, mn = 0 then

Δ(χ_{ℂ}, ψ_{ℂ}) = i^{m+n} χ_{ℂ}(w).

The bulk of this paper is taken up with a proof of the following theorem.

Theorem A

Suppose F is a given local field and ψ_{F} a given non-trivial additive character of F. It is possible in exactly one way to assign to each separable extension E of F a complex number λ(E/F, ψ_{F}) and to each equivalence class ω of representations of the Weil group of E a complex number ε(ω, Ψ_{E/F}) such that

(i) If ω corresponds to the quasi-character χ_{E} then

ε(ω, ψ_{E/F}) = Δ(χ_{E}, ψ_{E/F}).

(ii)

ε(ω_{1} ⊕ ω_{2}, ψ_{E/F}) = ε(ω_{1}, ψ_{E/F}) ε(ω_{2}, ψ_{E/F}).

(iii) If ω is the equivalence class of the representation of the Weil group of F induced from a representation of the Weil group of E in the class θ then

ε(ω, ψ_{F}) = λ(E/F, ψ_{F})^{dim θ} ε(θ, ψ_{E/F}).

α_{F}^{s} will denote the quasi-character x ⟶ |x|_{F}^{s} of C_{F} as well as the corresponding equivalence class of representations. Set

ε(s, ω, ψ_{F}) = ε (α_{F}^{s− \frac{1}{2}} ⊗ ω, ψ_{F} ) .

The left side will be the product of a non-zero constant and an exponential function.

Now take F to be a global field and ω to be an equivalence class of representations of the Weil group of F. Let 𝔸 be the adéle group of F and let ψ_{F} be a non-trivial character of 𝔸/F. For each place v let ψ_{v} be the restriction of ψ_{F} to F_{v}. ψ_{v} is non-trivial for each v and almost all the functions ε(s, ω_{v}, ψ_{v}) are identically 1 so that we can form the product

∏_{v} ε(s, ω_{v}, ψ_{v}).

Its value will be independent of ψ_{F} and will be written ε(s, ω).

Theorem B

The functional equation of the L-function associated to ω is

L(s, ω) = ε(s, ω) L(1 − s, ω̃).

This theorem is a rather easy consequence of the first theorem together with the functional equations of the Hecke L-functions.

For archimedean fields the first theorem says very little. For nonarchimedean fields it can be reformulated as a collection of identities for Gaussian sums. Four of these identities which we formulate as our four main lemmas are basic. All the others can be deduced from them by simple group-theoretic arguments. Unfortunately the only way at present that I can prove the four basic identities is by long and involved, although rather elementary, computations. However Theorem A promises to be of such importance for the theory of automorphic forms and group representations that we can hope that eventually a more conceptual proof of it will be found. The first and the second, which is the most difficult, of the four main lemmas are due to Dwork [6]. I am extremely grateful to him not only for sending me a copy of the dissertation of Lakkis [9] in which a proof of these two lemmas is given but also for the interest he has shown in this paper.

Chapter One.

Weil Groups

The Weil groups have many properties, most of which will be used at some point in the paper. It is impossible to describe all of them without some prolixity. To reduce the prolixity to a minimum I shall introduce these groups in the language of categories.

Consider the collection of sequences

S : C ⟶^{λ_{1}} G ⟶^{μ} 𝔊

of topogical groups where λ is a homeomorphism of C with the kernel of μ and μ induces a homeomorphism of G/λC with 𝔊. Suppose

S_{1} : C_{1} ⟶^{λ_{1}} G_{1} ⟶^{μ_{2}} 𝔊_{1}

is another such sequence. Two continuous homomorphisms φ and ψ from G to G_{1} which take C into C_{1} will be called equivalent if there is a c in C_{1} such that ψ(g) = cφ(g)c^{−1} for all g in G. S will be the category whose objects are the sequences S and Hom_{S_{0}}(S, S_{1}) will be the collection of these equivalence classes. S will be the category whose objects are the sequences S for which C is locally compact and abelian and 𝔊 is finite; if S and S_{1} belong to S

Hom_{S}(S, S_{1}) = Hom_{S_{0}}(S, S_{1}).

Let P_{1} be the functor from S to the category of locally compact abelian groups which takes S to C and let P_{2} be the functor from S to the category of finite groups which takes S to 𝔊. We have to introduce one more category S_{1,0}. The objects of S_{1} will be the sequences on S for which G^{c}, the commutator subgroup of G, is closed. Moreover the elements of Hom_{S_{1}}(S, S_{1}) will be the equivalence classes in Hom_{S}(S, S_{1}) all of whose members determine homeomorphisms of G with a closed subgroup fo finite index in G_{1}.

If S is in S_{1} let V (S) be the topological group G/G^{c}. If Φ ∈ Hom_{S_{1}}(S, S_{1}) let φ be a homeomorphism in the class Φ and let \overline{G} = φ(G). Composing the map G_{1}/G_{1}^{c} ⟶ \overline{G}/\overline{G}^{c} given by the transfer with the map \overline{G}/\overline{G}^{c} ⟶ G/G^{c} determined by the inverse of φ we obtain a map Φ_{v} : V (S_{1}) ⟶ V (S) which depends only on Φ. The map S ⟶ V (S) becomes a contravariant functor from S_{1} to the category of locally compact abelian groups. If S is the sequence

C ⟶ G ⟶ 𝔊

the transfer from G to C determines a homomorphism τ from G/G^{c} to the group of G-invariant elements in C. τ will sometimes be regarded as a map from G to this subgroup.

The category E will consist of all pairs K/F where F is a global or local field and K is a finite Galois extension of F. Hom(K/F, L/E) will be a certain collection of isomorphisms of K with a subfield of L under which F corresponds to a subfield of E. If the fields are of the same type, that is all global or all local we demand that E be finite and separable over the image of F. If F is global and E is local we demand that E be finite and separable over the closure of the image of F. I want to turn the map which associates to each K/F the group C_{K} into a contravariant functor which I will denote by C^{∗}. If φ : K/F ⟶ L/E and F and E are of the same type let K_{1} be the image of K in L and let φ_{C^{∗}} be the composition of N_{L/K_{1}} with the inverse of φ. If F is global and E is local let K_{1} be the closure in L of the image of K. As usual C_{K_{1}} may be considered a subgroup of the group of idèles of K. φ_{C^{∗}} is the composition of N_{L/K_{1}} with the projection of the group of idèles onto C_{K}.

If K is given let E^{K} be the subcategory of E whose objects are the extensions with the larger field equal to K and whose maps are equal to the identity on K. Let C_{∗} be the functor on E^{K} which takes K/F to C_{F}. If F is given let E_{F} have as objects the extensions with the smaller field equal to F. Its maps are to equal the identity on F.

A Weil group is a contravariant functor W from E to S with the following properties:

(i) P_{1} ∘ W is C^{∗}. (ii) P_{2} ∘ W is the functor 𝔊 : L/F ⟶ 𝔊(L/F). (iii) If φ ∈ 𝔊(L/F) ⊆ Hom(L/F, L/F) and g is any element of W_{L/F}, the middle group of the

sequence W (L/F), whose image in 𝔊(L/F) is φ then the map h ⟶ ghg^{−1} is in the class φ_{w}. (iv) The restriction of W to E^{K} takes values in S_{1}. Moreover, if K/F belongs to E^{K}

τ: W_{K/F} /W_{K/F}^{c} ⟶ C_{F}

is a homeomorphism. Finally, if φ : K/F ⟶ K/E is the identity on K and Φ = φ_{w} then the diagram

W_{K/F} /W_{K/F}^{c}  ⟶^{Φ_{v}}  W_{K/E}/W_{K/E}^{c}
τ ↓  ↓ τ
C_{F}  ⟶_{φ_{C^{∗}}}  C_{E}

is commutative and if ψ : F/F ⟶ K/F is the imbedding, ψ_{W} is τ.

Since the functorial properties of the Weil group are not all discussed by Artin and Tate, we should review their construction of the Weil group pointing out, when necessary, how the functorial properties arise. There is associated to each K/F a fundamental class α_{K/F} in H^{2}(𝔊(K/F), C_{K}). The group W (K/F) is any extension of 𝔊(K/F) by C_{K} associated to this element. We have to show, at least, that if φ : K/F ⟶ L/E the diagram

1 ⟶ C_{L} ⟶ W_{L/E} ⟶  𝔊(L/E) ⟶ 1
↓ φ_{C^{∗}}  ↓ φ_{𝔊}
1 ⟶ C_{K} ⟶ W_{K/F} ⟶  𝔊(K/F) ⟶ 1

can be completed to a commutative diagram by inserting φ̂ : W_{L/E} ⟶ W_{K/F}. The map φ_{C^{∗}} commutes with the action of 𝔊(L/E) on C_{L} and C_{K} so that φ̂ exists if and only if φ_{C^{∗}}(α_{L/E}) is the restriction φ_{𝔊}^{∗}(α_{K/F}) of φ_{K/F} to 𝔊(L/E). If this is so, the collection of equivalence classes to which φ̂ may belong is a principal homogeneous space of H^{1}(𝔊(L/E), C_{K}). In particular, if this group is trivial, as it is when φ_{𝔊} is an injection, the class of φ̂ is uniquely determined.

An examination of the definition of the fundamental class and shows that it is canonical. In other words, if φ is an isomorphism of K and L and of F and E, then φ_{𝔊}^{∗}(α_{K/F}) = φ^{−1}α_{L/E} = φ_{C^{∗}}(α_{L/E}). If K = L and φ is the identity on K, the relation φ_{𝔊}^{∗}(α_{K/F}) = α_{L/E} = φ_{C^{∗}}(α_{L/E}) is one of the basic properties of the fundamental class. Thus in these two cases φ̂ exists and its class is unique. Now take K to be global and L local. Suppose at first that K is contained in L, that its closure is L, and that F = K ∩ E. Then, by the very definition of α_{K/F} , φ_{𝔊}^{∗}(α_{K/F}) = φ_{C^{∗}}(α_{L/E}). More generally, if K_{1} is the image of K in L, and F_{1} the image of F in E, we can write φ as φ_{1}φ_{2}φ_{3} where φ_{3} : K/F ⟶ K_{1}/F_{1}, φ_{2} : K_{1}/F_{1} ⟶ K_{1}/K_{1} ∩ E, and φ_{1} : K_{1}/K_{1} ∩ E ⟶ L/E. φ̂_{3} and φ̂_{2} exist. If the closure of K_{1} is L then φ̂_{1} and therefore φ̂ = φ̂_{3}φ̂_{2}φ̂_{1} also exist. The class of φ̂ is uniquely determined.

Artin and Tate show that W_{K/F}^{c} is a closed subgroup of W_{K/F} and that τ is a homeomorphism of W_{K/F} /W_{K/F}^{c} and C_{F}. Granted this, it is easy to see that the restriction of W to ξ^{K} takes values in S_{1}. Suppose we have the collection of fields in the diagram with L and K normal over F and L and K′ normal over F′. Let α, β, and ν be the imbeddings α : L/F ⟶ L/K, β : L/F′ ⟶ L/K′, ν : L/F ⟶ L/F′.

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We have shown the existence of α̂, β̂, and ν̂. It is clear that ν̂β̂(W_{L/K′}) is contained in α̂(W_{L/K}). Thus we have a natural map

π : ν̂β̂(W_{L/K′})/ν̂β̂(W_{L/K′}^{c}) ⟶ α̂(W_{L/K})/α̂(W_{L/K}^{c}).

Let us verify that the diagram

W_{L/K}/W_{L/K′}^{c} ⟶ ν̂β̂(W_{L/K′})/ν̂β̂(W_{L/K′}^{C}) ⟶^{π} α̂(W_{L/K})/α̂(W_{K/K}^{c}) ⟶ W_{L/K}/W_{L/K}^{c}
↓ τ ↓ τ
C_{k′}  ⟶^{N_{K′/K}}  C_{k}  (A)

is commutative. To see this let W_{L/K′} be the disjoint union

U_{i=1}^{r} C_{K}h_{i}.

Then we can choose h′_{i}, g′_{j}, 1 ≤ i ≤ r, 1 ≤ j ≤ s so that W_{L/K} is the disjoint union

⋃_{i=1}^{r} ⋃_{j=1}^{s} C_{K} g′_{j}h′_{i}

and ν̂β̂(h′_{i}) = α̂(h_{i}). Using these coset representatives to compute the transfer one immediately verifies the assertion. We should also observe that the transitivity of the transfer implies the commutativity of the diagram

W_{K/F} /W_{K/F}^{c}  ⟶^{Φ_{v}}  W_{K/F′}/W_{K/F′}^{c}
τ ↓  ↓
C_{F}  ⟶_{φ_{C^{∗}}}  C_{F′}

if Φ is the class of an imbedding φ̂ where φ is an imbedding K/F ⟶ K/F′.

We have still not defined φ_{W} for all φ. However we have defined it when φ is an isomorphism of the two larger fields or when the second large field is the closure of the first. Moreover the definition is such that the third condition and all parts of the fourth condition except the last are satisfied. The last condition of (iv) can be made a definition without violating (i) and (ii). What we do now is show that there is one and only one way of extending the definition of φ_{W} to all φ without violating conditions (i) or (ii) and the functorial property.

Suppose F ⊆ K ⊆ L, K/F and L/F are Galois, and ψ is the imbedding L/F ⟶ L/K. It is observed in Artin and Tate that there is one and only class of maps {θ} which make the following diagram commutative

1 ⟶ W_{L/K}/W_{L/K}^{c} ⟶ ψ̂W_{L/K}/ψ̂W_{L/K}^{c} ⟶ W_{L/F} /ψ̂W_{L/K}^{c} ⟶ W_{L/F} /ψ̂W_{L/K} ⟶ 1
τ ↓ ↓ θ ↓
1 ⟶ C_{K} ⟶ W_{K/F} ⟶ 𝔊(K/F) ⟶ 1.

The homomorphism on the right is that deduced from

W_{L/F} /W_{L/K} = 𝔊(L/F)/𝔊(L/K) ≃ 𝔊(K/F).

Let φ, μ, and ν be imbeddings φ : K/F ⟶ L/F, μ : K/K ⟶ L/K, ν : K/F ⟶ K/K. Then ψ ∘ φ = μ ∘ ν, so that ν̂∘ μ̂ = φ̂∘ ψ̂. Moreover ν̂∘ μ̂ is the composition of the map τ : W_{L/K} ⟶ C_{K} and the imbedding of C_{K} in W_{K/F}. Thus the kernel of φ̂ contains ψ̂W_{L/K}^{c} so that φ̂ ∘ ψ̂ restricted to W_{L/K}/W_{L/K}^{c} must be τ and the only possible choice for φ̂ is, apart from equivalence, θ. To see that this choice does not violate the second condition observe that the restriction of τ to C_{L} will be N_{L/K} and that ψ̂ is the identity on C_{L}.

Denote the map θ : W_{L/F} ⟶ W_{K/F} by θ_{L/K} and the map τ : W_{K/F} → C_{K} by τ_{K/F}. It is clear that τ_{K/F} ∘θ_{L/K} is the transfer from W_{L/F} /W_{L/F}^{c} to ψ̂W_{L/K}/ψ̂W_{L/K}^{c} followed by the transfer

from ψ̂W_{L/K}/ψ̂W_{L/K}^{c} to ψ̂C_{L} = C_{F}. By the transitivity of the transfer τ_{K/F} ∘ θ_{L/K} = τ_{L/F}. It follows immediately that if F ⊆ K ⊆ L ⊆ L′ and all extensions are Galois the map θ_{L′/K} and θ_{L/K}θ_{L′/L} are in the same class.

Suppose that φ is an imbedding K/F ⟶ K′/F′ and choose L so that K′ ⊆ L and L/F is Galois. Let ψ : K′/F′ ⟶ L/F′, μ : K/F ⟶ L/F, ν : L/F ⟶ L/F′ be imbeddings. Then ψ ∘ φ = ν ∘ μ so that μ̂ ∘ ν̂ = φ̂ ∘ ψ̂. If α : L/F ⟶ L/K, β : L/F′ ⟶ L/K′ are the imbeddings then the kernel of ψ̂ is ν̂β̂W_{L/K′}^{c} which is contained in α̂W_{L/K}^{c} the kernel of μ̂. Thus there is only

one way to define φ̂ so that μ̂ ∘ ν̂ = φ̂ ∘ ψ̂. The diagram

W_{L/K′}/W_{L/K′}^{c} \underrightarrow{_{β}} W_{L/F′}/β̂W_{L/K′}^{c} \underrightarrow{_{ψ}} W_{K′/F′}
↓ ν̂  ↓ φ̂
W_{L/F} /ν̂β̂W_{L/K′}^{c} ⟶ W_{L/F} /α̂W_{L/K}^{c} \underrightarrow{_{μ}} W_{K/F}

will be commutative. Since ψ̂ ∘ β̂ = τ_{L/K′} and μ̂ ∘ α̂ = τ_{K/F} diagram (A) shows that φ̂ has the required effect on C_{K}.

To define φ_{W} in general, we observe that every φ is the composition of isomorphisms, imbeddings of fields of the same type, and a map K/F ⟶ K′/F′ where K is global, K′ is local, K′ is the closure of K, and F = F′ ∩ K. Of course the identity

(φ ∘ ψ)_{W} = ψ_{W} φ_{W}

must be verified. I omit the verification which is easy enough. The uniqueness of the Weil groups in the sense of Artin and Tate implies that the functor W is unique up to isomorphism.

The sequence

S(n, ℂ) : GL(n, ℂ) ⟶^{id} GL(n, ℂ) ⟶ 1

belongs to S_{1}. If S : C ⟶ G ⟶ 𝔊 belongs to S_{1} then

Hom_{S_{0}}(S, S(n, ℂ))

is the set of equivalence classes of n-dimensional complex representations of G. Let Ω_{n}(S) be the set of all Φ in Hom_{S_{0}}(S, S(n, ℂ)) such that, for each φ ∈ Φ, φ(g) is semi-simple for all g in G. Ω_{n}(S) is a contravariant functor of S and so is Ω(S) = ⋃_{n=1}^{∞} Ω_{n}(S). On the category S_{1}, it can be turned into a covariant functor. If ψ : S ⟶ S_{1}, if Φ ∈ Ω(S), and if φ ∈ Φ, let ψ associate to Φ the matrix representations corresponding to the induced representation Ind(G_{1}, ψ(G), φ ∘ ψ^{−1}). It follows from the transitivity of the induction process that Ω is a covariant functor of S_{1}.

To be complete a further observation must be made.

Lemma 1.1 Suppose H is a subgroup of finite index in G and ρ is a finite-dimensional complex representation of H such that ρ(L) is semi-simple for all h in H. If

σ = Ind(G, H, ρ)

then σ(g) is semi-simple for all g.

H contains a subgroup H_{1} which is normal and of finite index in G, namely, the group of elements acting trivially on H\G. To show that a non-singular matrix is semi-simple one has only to show that some power of it is semi-simple. Since σ^{n}(g) = σ(g^{n}) and g^{n} belongs to H_{1} for some n we need only show that σ(g) is semi-simple for g in H_{1}. In that case σ(g) is equivalent to ∑_{i=1}^{r} ⊕ρ(g_{i}gg_{i}^{−1}) if G is the disjoint union

⋃_{i=1}^{r} Hg_{i}

Suppose L/F and K/F belong to E_{F} and φ ∈ Hom_{E_{F}} (L/F, K/F). Since the maps of the class φ_{W} all take W_{K/F} onto W_{L/F} the associated map Ω(W (L/F)) ⟶ Ω(W (K/F)) is injective. Moreover it is independent of φ. If L_{1}/F and L_{2}/F belong to E_{F} there is an extension K/F and maps φ_{1} ∈ Hom_{E_{F}} (L_{1}/F, K/F), φ_{2} ∈ Hom_{E_{F}} (L_{2}/F, K/F). ω_{1} in Ω(W (L_{1}/F)) and ω_{2} in Ω(W (L_{2}/F)) have the same image in Ω(W (K/F)) for one such K if and only if they have the same image for all such K. If this is so we say that ω_{1} and ω_{2} are equivalent. The collection of equivalence classes will be denoted by Ω(F). Its members are referred to as equivalence classes of representations of the Weil group of F.

Let F be the category whose objects are local and global fields. If F and E are of the same type Hom_{F}(F, E) consists of all isomorphisms of F with a subfield of E over which E is separable. If F is global and E is local Hom_{F}(F, E) consists of all isomorphisms of F with a subfield of E over whose closure E is separable. Ω(F) is clearly a covariant functor on F. Let F_{gℓ}, and F_{loc} be the subcategories consisting of the global and local fields respectively. Suppose F and E are of the same type and φ ∈ Hom_{F}(F, E). If ω ∈ Ω(E) choose K so that ω belongs to Ω(W (K/E)). We may assume that there is an L/F and an isomorphism ψ from L onto K which agrees with φ on F. Then ψ_{W} : W_{K/E} ⟶ W_{L/F} is an injection. Let θ be the equivalence class of the representation

σ = Ind(W_{L/F}, ψ_{W}(W_{K/E}), ρ ∘ ψ_{W}^{−1})

with ρ in ω. I claim that θ is independent of K and depends only on ω and φ. To see this it is enough to show that if L ⊆ L′, L′/F is Galois, ψ′ is an isomorphism from L′ to K′ which agrees with ψ on L, and ρ′ is a representation of W_{K′/F} in ω the class of

σ′ = Ind(W_{L′/F}, ψ′_{W}(W_{K′/E}), ρ′ ∘ ψ_{w}^{′ −1})

is also Θ. Suppose μ is a map from W_{K′/E} to W_{K/E} associated to the imbedding K/E ⟶ K′/E and ν is a map from W_{L′/F} to W_{L/F} associated to the imbedding L/F ⟶ L′/F. We may suppose that ψ_{W} ∘ μ = ν ∘ ψ′_{W}. The kernel of μ is W_{K′/K}^{c} if, for simplicity of notation, W_{K′/K} is regarded as a subgroup of W_{K′/E} and that of ν is W_{L′/L}^{c}. Moreover ψ′_{W} (W_{K′/K}^{c}) = W_{L′/L}^{c}. Take ρ′ = ρ ∘ μ. Then σ acts on the space V of functions f on W_{K/F} satisfying f(vw) ≡ ρ(ψ_{w}^{−1}(h))f(w) for v in ψ_{w}(W_{K/E}). Let V′ be the analogous space on which σ′ acts. Then

V′ = {f ∘ ν | f ∈ V }.

The assertion follows. Thus Ω(F) is a contravariant functor on F_{gℓ} and F_{loc}.

After this laborious and clumsy introduction we can set to work and prove the two theorems. The first step is to reformulate Theorem A.

Chapter Two.

The Main Theorem

It will be convenient in this paragraph and at various later times to regard W_{K/E} as a subgroup of W_{K/F} if F ⊆ E ⊆ K. If F ⊆ E ⊆ L ⊆ K we shall also occasionally take W_{L/E} to be W_{K/E}/W_{K/L}^{c}.

If K/F is finite and Galois, P(K/F) will be the set of extensions E′/E with F ⊆ E ⊆ E′ ⊆ K and P_{∘}(K/F) will be the set of extensions in P(K/F) with the lower field equal to F.

Theorem 2.1

Suppose K is a Galois extension of the local field F and ψ_{F} is a given non-trivial additive character of F. There is exactly one function λ(E/F, ψ_{F}) defined on P_{∘}(K/F) with the following two properties

(i)

λ(F/F, ψ_{F}) = 1.

(ii) If E_{1}, …, E_{r}, E′_{1}, …, E′_{s} are fields lying between F and K, if χ_{E_{i}}, 1 ≤ i ≤ r, is a quasicharacter of C_{E_{i}}, if χ_{E′_{j}}, 1 ≤ j ≤ s, is a quasi-character of C_{E′_{j}}, and if

⊕_{i=1}^{r} Ind(W_{k/F}, W_{K/E_{i}}, χ_{E_{i}})

is equivalent to

⊕_{j=1}^{s} Ind(W_{K/F}, W_{K/E′_{j}}, χ_{E′_{j}})

then

∏_{i=1}^{r} Δ(χ_{E_{i}}, ψ_{E_{i}/F})λ(E_{i}/F, ψ_{F})

is equal to

∏_{j=1}^{s} Δ(χ_{E′_{j}}, ψ_{E′_{j}/F})λ(E′_{j}/F, ψ_{F}).

A function satisfying the conditions of this theorem will be called a λ-function. It is clear that the function λ(E/F, ψ_{F}) of Theorem A when restricted to P_{∘}(K/F) becomes a λ-function. Thus the uniqueness in this theorem implies at least part of the uniqueness of Theorem A. To show how this theorem implies all of Theorem A we have to anticipate some simple results which will be proved in paragraph 4.

First of all a λ-function can never take on the value 0. Moreover, if F ⊆ K ⊆ L the λ-function on P_{∘}(K/F) is just the restriction to P_{∘}(K/F) of the λ-function on P_{∘}(L/F). Thus λ(E/F, ψ_{F}) is defined independently of K. Finally if E ⊆ E′ ⊆ E′′

λ(E′′/E, ψ_{E}) = λ(E′′/E′, ψ_{E′/E})λ(E′/E, ψ_{E})^{[E′′:E′]}.

We also have to use a form of Brauer’s theorem [4]. If G is a finite group there are nilpotent subgroups N_{1}, …, N_{m}, one-dimensional representations χ_{1}, …, χ_{m} of N_{1}, …, N_{m} respectively, and integers n_{1}, …, n_{m} such that the trivial representation of G is equivalent to

⊕_{i=1}^{m} n_{i}Ind(G, N_{i}, χ_{i}).

The meaning of this when some of the n_{i} are negative is clear

Lemma 2.2

Suppose F is a global or local field and ρ is a representation of W_{K/F}. There are intermediate fields E_{1}, …, E_{m} such that 𝔊(K/E_{i}) is nilpotent for 1 ≤ i ≤ m, one-dimensional representations χ_{E_{i}} of W_{K/E_{i}}, and integers n_{1}, …, n_{m} such that ρ is equivalent to

⊕_{i=1}^{m} n_{i}Ind(W_{K/F}, W_{K/E_{i}}, χ_{E_{i}}).

Theorem 2.1 and Lemma 2.2 together imply the uniqueness of Theorem A. Before proving the lemma we must establish a simple and well-known fact.

Lemma 2.3

Suppose H is a subgroup of finite index in the group G. Suppose τ is a representation of G, σ a representation of H, and ρ the restriction of τ to H. Then

τ ⊗ Ind(G, H, σ) ≃ Ind(G, H, ρ ⊗ σ).

Let τ act on V and σ on W. Then Ind(G, H, σ) acts on X, the space of all functions f on G with values in W satisfying

f(hg) = σ(h) f(g)

while Ind(G, H, ρ ⊗ σ) acts on Y, the space of all functions f on G with values in V ⊗ W satisfying

f(hg) = (ρ(h) ⊗ σ(h)) f(g).

Clearly, V ⊗ X and Y have the same dimension. The map of V ⊗ X to Y which sends v ⊗ f to the function

f′(g) = τ(g)v ⊗ f(g)

is G-invariant. If it were not an isomorphism there would be a basis v_{1}, …, v_{n} of V and functions f_{1}, …, f_{n} which are not all zero such that

Σ_{i=1}^{n} τ(g)v_{i} ⊗ f_{i}(g) ≡ 0.

This is clearly impossible.

To prove Lemma 2.2 we take the group G of Brauer’s theorem to be 𝔊(K/F). Let F_{i} be the fixed field of N_{i} and let ρ_{i} be the tensor product of χ_{i}, which we may regard as a representation of W_{K/F_{i}} and the restriction of ρ to W_{K/F_{i}}. Then

ρ ≃ ρ ⊗ 1 ≃ ⊕_{i=1}^{m}n_{i}Ind(W_{K/F_{i}}, ρ_{i}).

This together with the transitivity of the induction process shows that in proving the lemma we may suppose that 𝔊(K/F) is nilpotent.

We prove the lemma, with this extra condition, by induction on [K : F]. We use the symbol ω to denote an orbit in the set of quasi-characters of C_{K} under the action of 𝔊(K/F). The restriction of ρ to C_{K} is the direct sum of one-dimensional representations. If ρ acts on V let V_{ω} be the space spanned by the vectors transforming under C_{K} according to a quasi-character in ω. V is the direct sum of the spaces V_{ω} which are each invariant under W_{K/F}. For our purposes we may suppose that V = V_{ω} for some ω. Choose χ_{K} in this ω and let V_{0} be the space of vectors transforming under C_{K} according to χ_{K}. Let E be the fixed field of the isotrophy group of χ_{K}. V_{0} is invariant under W_{K/E}. Let σ be the representation of W_{K/E} in V_{0}. It is well-known that

ρ ≃ Ind(W_{K/F}, W_{K/E}, σ).

To see this one has only to verify that the space X on which the representation on the right acts and V have the same dimension and that the map

f ⟶ ∑_{W_{K/E}\W_{K/F}} ρ(g^{−1})f(g)

of X into V which is clearly W_{K/F} -invariant has no kernel. It is easy enough to do this.

If E ≠ F the assertion of the lemma follows by induction. If E = F choose L containing F so that K/L is cyclic of prime degree and L/F is Galois. Then ρ(W_{K/L}) is an abelian group and W_{K/L}^{c} is contained in the kernel of ρ. Thus ρ may be regarded as a representation of W_{L/F}. The assertion now follows from the induction assumption and the concluding remarks of the previous paragraph.

Now take a local field E and a representation ρ of W_{K/E}. Choose intermediate fields E_{1}, …, E_{m}, one-dimensional representations χ_{E_{i}} of W_{K/E_{i}}, and integers n_{1}, …, n_{m} so that

ρ ≃ ⊕_{i=1}^{m} n_{i} Ind(W_{K/E}, W_{K/E_{i}}, χ_{E_{i}}).

If ω is the class of ρ set

ε(ω, ψ_{E}) = ∏_{i=1}^{m} {Δ(χ_{E_{i}}, Ψ_{E_{i}/E})λ(E_{i}/E, Ψ_{E})}^{n_{i}}.

Theorem 2.1 shows that the right side is independent of the way in which ρ is written as a sum of induced representations. The first and second conditions of Theorem A are clearly satisfied. If ρ is the representation above and σ the representation

Ind(W_{K/F}, W_{K/E}, ρ)

then

σ ≃ ⊕_{i=1}^{m} n_{i}Ind(W_{K/F}, W_{K/E_{i}}, χ_{E_{i}}).

Thus if ω′ is the class of σ

ε(ω′, ψ_{F}) = ∏_{i=1}^{m} {Δ(χ_{E_{i}}, ψ_{E_{i}/F})λ(E_{i}/F, ψ_{F})}^{n_{i}}

while

ε(ω, ψ_{E/F}) = ∏_{i=1}^{m} {Δ(χ_{E_{i}}, ψ_{E_{i}/F}) λ(E_{i}/E, ψ_{E/F})}^{n_{i}}.

The third property follows from the relations

dim ω = Σ_{i=1}^{m} n_{i} [E_{i} : E]

and

λ(E_{i}/F, ψ_{F}) = λ(E_{i}/E, ψ_{E/F}) λ(E/F, ψ_{F})^{[E_{i}:E]}

Chapter Three.

The Lemmas of Induction

In this paragraph we prove two simple but very useful lemmas.

Lemma 3.1

Suppose K is a Galois extension of the local field F. Suppose the subset 𝔄 of P(K/F) has the following four properties.

(i) For all E, with F ⊆ E ⊆ K, E/E ∈ 𝔄. (ii) If E′′/E′ and E′/E belong to 𝔄 so does E′′/E. (iii) If L/E belongs to P(K/F) and L/E is cyclic of prime degree then L/E belongs to 𝔄. (iv) Suppose that L/E in P(K/F) is a Galois extension. Let G = 𝔊(L/E). Suppose G = H · C where H ≠ {1}, H ∩ C = {1}, and C is a non-trivial abelian normal subgroup of G which is contained in every non-trivial normal subgroup of G. If E′ is the fixed field of H and if every E′′/E in P_{∘}(L/E) for which [E′′ : E] < [E′ : E′] is in 𝔄 so is E′/E. Then 𝔄 is all of P(K/F).

It is convenient to prove another lemma first.

Lemma 3.2

Suppose K is a Galois extension of the local field F and F ⊂_{≠} E ⊆ K. Suppose that the only

normal subfield of K containing E is K itself and that there are no fields between F and E. Let G = 𝔊(K/F) and let E be the fixed field of H. Let C be a minimal non-trivial abelian normal subgroup of G. Then G = HC, H ∩ C = {1} and C is contained in every non-trivial normal subgroup of G. In particular if H = {1}, G = C is abelian of prime order.

H is contained in no subgroup besides itself and G contains no normal subgroup but {1}. Thus if H is normal it is {1} and G has no proper subgroups and is consequently cyclic of prime order. Suppose H is not normal. Since G is solvable it does contain a minimal non-trivial abelian normal subgroup C. Since C is not contained in H, H ⊂_{≠} HC and G = HC. Since H ∩ C is a

normal subgroup of G it is {1}. If D is a non-trivial normal subgroup of G which does not contain C then D ∩ C = {1} and D is contained in the centralizer Z of C. Then DC is also and Z must meet H non-trivially. But Z ∩H is a normal subgroup of G. This is a contradiction and the lemma is proved.

The first lemma is certainly true if [K : F] = 1. Suppose [K : F] > 1 and the lemma is valid for all pairs [K′ : F′] with [K′ : F′] < [K : F]. If the Galois extension L/E belongs to P(K/F) then 𝔄 ∩ P(L/E) satisfies the condition of the lemma with K replaced by L and F by E. Thus, by induction, if [L : E] < [K : F], P(L/E) ⊆ 𝔄. In particular if E′/E is not in 𝔊 then E = F and the only normal subfield of K containing E′ is K itself. If 𝔄 is not P(K/F) then amongst all extensions which are not in 𝔊 choose one E/F for which [E : F] is minimal. Because of (ii) there are no fields between F and E. Lemma 3.2, together with (iii) and (iv), show that E/F is in 𝔄. This is a contradiction.

There is a variant of Lemma 3.1 which we shall have occasion to use.

Lemma 3.3

Suppose K is a Galois extension of the local field F. Suppose the subset 𝔄 of P_{∘}(K/F) has the following properties.

(i) F/F ∈ 𝔄. (ii) If L/F is normal and L ⊂_{≠} K then P_{∘}(L/F) ⊆ 𝔄.

(iii) If F ⊂ E ⊆ E′ ⊆ K and E/F belong to 𝔊 then E′/F belong to 𝔄. (iv) If L/F in P_{∘}(K/F) is cyclic of prime degree then L/F ∈ 𝔄.

(v) Suppose that L/F in P_{∘}(K/F) is Galois and G = 𝔊(L/F). Suppose G = HC where H ≠ {1}, H ∩C = {1}, and C is a non-trivial abelian normal subgroup of G which is contained in every non-trivial normal subgroup. If E is the fixed field of H and if every E′/F in P_{∘}(L/F) for which [E′ : F] < [E : F] is in 𝔄 so is E/F.

Then 𝔄 is P_{∘}(K/F).

Again if 𝔄 is not P_{∘}(K/F) there is an E/F not in 𝔄 for which [E : F] is minimal. Certainly [E : F] > 1. By (ii) and (iii), E is contained in no proper normal subfield of K and there are no fields between E and F. Lemma 3.2 together with (iv) and (v) lead to the contradiction that E/F is in 𝔄.

Chapter Four.

The Lemma of Uniqueness

Suppose K/F is a finite Galois extension of the local field F and ψ_{F} is a non-trivial additive character of F. A function E/F ⟶ λ(E/F, ψ_{F}) on P_{∘}(K/F) will be called a weak λ-function if the following two conditions are satisfied.

(i) λ(F/F, Ψ_{F}) = 1. (ii) If E_{1}, …, E_{r}, E′_{1}, …, E′_{s} are fields lying between F and K, if μ_{i}, 1 ≤ i ≤ r, is a one-

dimensional representation of 𝔊(K/E_{i}), if ν_{j}, 1 ≤ j ≤ s, is a one-dimensional representation of 𝔊(K/E′_{j}), and if

⨁_{i=1}^{r} Ind(𝔊(K/F), 𝔊(K/E_{i}), μ_{i})

is equivalent to

⨁_{j=1}^{s} Ind(𝔊(K/F), 𝔊(K/E_{j}), ν_{j})

then

∏_{i=1}^{r} Δ(χ_{E_{i}}, ψ_{E_{i}/F})λ(E_{i}/F, ψ_{F})

is equal to

∏_{j=1}^{s} Δ(χ_{E′_{j}} ψ_{E′_{j}/F})λ(E′_{j}/F, ψ_{F})

if χ_{E_{i}} is the character of C_{E_{i}} corresponding to μ_{i} and χ_{E′_{j}} is the character of C_{E′_{j}} corresponding to ν_{j}.

Supposing that a weak λ-function is given on P_{∘}(K/F), we shall establish some of its properties.

Lemma 4.1

(i) If L/F in P_{∘}(K/F) is normal the restriction of λ(·, ψ_{F}) to P_{∘}(L/F) is a weak λ-function. (ii) If E/F belongs to P_{∘}(K/F) and λ(E/F, ψ_{F}) ≠ 0 the function on P_{∘}(K/E) defined by

λ(E′/E, ψ_{E/F}) = λ(E′/F, ψ_{F}) λ(E/F, ψ_{F})^{−[E′:E]}

is a weak λ-function.

Any one-dimensional representation μ of 𝔊(L/E) may be inflated to a one-dimensional representation, again called μ, of 𝔊(K/E) and

Ind(𝔊(K/F), 𝔊(K/E), μ)

is just the inflation to 𝔊(K/F) of

Ind(𝔊(L/F), 𝔊(L/E), μ).

The first part of the lemma follows immediately from this observation.

As for the second part, the relation

λ(E/E, ψ_{E/F}) = 1

is clear. If fields E_{i}, 1 ≤ i ≤ r, E′_{j}, 1 ≤ j ≤ s, lying between E and K and representations μ_{i} and ν_{j} are given as prescribed and if

⨁_{i=1}^{r} Ind(𝔊(K/E), 𝔊(K/E_{i}), μ_{i}) = ρ

is equivalent to

⨁_{j=1}^{s} Ind(𝔊(K/E), 𝔊(K/E′_{j}), ν_{j}) = σ

then

⨁_{i=1}^{r} Ind(𝔊(K/F), 𝔊(K/E_{i}, μ_{i})

is equivalent to

⨁_{j=1}^{s} Ind(𝔊(K/F), 𝔊(K/E′_{j}), ν_{j})

so that

∏_{i=1}^{r} Δ(χ_{E_{i}}, ψ_{E_{i}/F}) λ(E_{1}/F, ψ_{F})  (A)

is equal to

∏_{j=1}^{s} Δ(χ_{E′_{j}}, ψ_{E′_{j}/F}) λ(E′_{j}/F, ψ_{F}).  (B)

Since ρ and σ have the same dimension

Σ_{i=1}^{r} [E_{i} : E] = Σ_{j=1}^{s} [E′_{j} : E]

so that

∏_{i=1}^{r} λ(E/F, Ψ_{F})^{[E_{i}:E]} = ∏_{j=1}^{s} λ(E/F, ψ_{F})^{[E′_{j}:F]}.

Dividing (A) by the left side of this equation and (B) by the right and observing that the results are equal we obtain the relation needed to prove the lemma.

If K/F is abelian S(K/F) will be the set of characters of C_{F} which are 1 on N_{K/F} C_{K}.

Lemma 4.2

If K/F is abelian

λ(K/F, Ψ_{F}) = ∏_{μ_{F} ∈S(K/F)} Δ(μ_{F} , ψ_{F}).

μ_{F} determines a one-dimensional representation of 𝔊(K/F) which we also denote by μ_{F}. The lemma is an immediate consequence of the equivalence of

Ind(𝔊(K/F), 𝔊(K/K), 1)

and

⨁_{μ_{F} ∈S(K/F)} Ind(𝔊(K/F), 𝔊(K/F), μ_{F}).

Lemma 4.3

Suppose K/F is normal and G = 𝔊(K/F). Suppose G = HC where H ∩ C = {1} and C is a non-trivial abelian normal subgroup. Let E be the fixed field of H and L that of C. Let T be a set of representatives of the orbits of S(K/L) under the action of G. If μ ∈ T let B_{μ} be the isotropy group of μ and let B_{μ} = 𝔊(K/L_{μ}). Then [L_{μ} : F] < [E : F] and

λ(E/F, ψ_{F}) = ∏_{μ∈T} Δ(μ′, ψ_{L_{μ}/F}) λ(L_{μ}/F, ψ_{F}).

Here 𝔊(K/L_{μ}) = 𝔊(K/L) · (𝔊(K/L_{μ}) ∩ 𝔊(K/E)) and μ′ is the character of C_{L_{μ}} associated to the character of 𝔊(K/L_{μ}) : g ⟶ μ(g_{1}) if

g = g_{1}g_{2}, g_{1} ∈ 𝔊(K/L), g_{2} ∈ 𝔊(K/L_{μ}) ∩ 𝔊(K/E),

We may as well denote the given character of 𝔊(K/L_{μ}) by μ′ also. To prove the lemma we show that

Ind(𝔊(K/F), 𝔊(K/E), 1) = σ

is equivalent to

⨁_{μ∈T} Ind(𝔊(K/F), 𝔊(K/L_{μ}), μ′).

Since T has at least two elements it will follow that

[E : F] = dim Ind(𝔊(K/F), 𝔊(K/E), 1)

is greater than

[L_{μ} : F] = dim Ind(𝔊(K/F), 𝔊(K/L_{μ}), μ′).

The representation σ acts on the space of functions on H\G. If ν ∈ S(K/L), that is, is a character of C, let ψ_{ν}(hc) = ν(c) if h ∈ H, c ∈ C. The set

{ψ_{ν} | ν ∈ S(K/L)}

is a basis for the functions on H\G. If μ ∈ T let S_{μ} be its orbit; then

V_{μ} = Σ_{ν∈S_{μ}}ℂψ_{ν}

is invariant and irreducible under G. Moreover, if g belongs to 𝔊(K/L_{μ})

σ(g)ψ_{μ} = μ′(g)ψ_{μ}.

Since

dim V_{μ} = [𝔊(K/F), 𝔊(K/L_{μ})]

the Frobenius reciprocity theorem implies that the restriction of σ to V_{μ} is equivalent to

Ind(𝔊(K/F), 𝔊(K/L_{μ}), μ′).

Lemma 4.2 is of course a special case of Lemma 4.3.

Lemma 4.4 λ(E/F, Ψ_{F}) is different from 0 for all E/F in P_{∘}(K/F).

The lemma is clear if [K : F] = 1. We prove it by induction on [K : F]. Let 𝔊 be the set of E/F in P_{∘}(K/F) for which λ(E/F, ψ_{F}) ≠ 0. We may apply Lemma 3.3. The first condition of that lemma is clearly satisfied. The second follows from the induction assumption and the first part of Lemma 4.1; the third from the induction assumption and the second part of Lemma 4.1. The fourth and fifth follow from Lemmas 4.2 and 4.3 respectively. We of course use the fact that Δ(χ_{E}, Ψ_{E}), which is basically a Gaussian sum when E is non-archimedean, is never zero.

For every E′/E in P(K/F) we can define λ(E′/E, ψ_{E/F}) to be
λ(E′/F, ψ_{F}) λ(E/F, ψ_{F})^{−[E′:E]}.

Lemma 4.5

If E′′/E′ and E′/E belong to P(K/F) then

λ(E′′/E, ψ_{E/F}) = λ(E′′/E′, ψ_{E′/F})λ(E′/E, ψ_{E/F})^{E′′:E′]}.

Indeed

λ(E′′/E, ψ_{E/F}) = λ(E′′/F, ψ_{F}) λ(E/F, ψ_{F})^{−[E′′:E]}

which equals

{λ(E′′/F, ψ_{F}) λ(E′/F, ψ_{F})^{−[E′′:E′]}} {λ(E′/F, ψ_{F})^{[E′′:E′]} λ(E/F, ψ_{F})^{−[E′′:E]}}

and this in turn equals

λ(E′′/E′, ψ_{E′/F}) λ(E′/E, ψ_{E/F})^{[E′′:E′]}.

Lemma 4.6

If λ_{1}(·, Ψ_{F}) and λ_{2}(·, Ψ_{F}) are two weak λ-functions on P_{∘}(K/F) then
λ_{1}(E′/E, ψ_{E/F}) = λ_{2}(E′/E, ψ_{E/F})

for all E′/E in P(K/F).

We apply Lemma 3.1 to the collection 𝔊 of all pairs E′/E in P(K/F) for which the equality is valid. The first condition of that lemma is clearly satisfied. The second is a consequence of the previous lemma. The third and fourth are consequences of Lemmas 4.2 and 4.3 respectively.

Since a λ-function is also a weak λ-function the uniqueness of Theorem 2.1 is now proved.

Chapter Five.

A Property of λ-Functions

It follows immediately from the definition that if ψ′_{E}(x) = ψ_{E}(βx) then

Δ(χ_{E}, ψ′_{E}) = χ_{E}(β)Δ(χ_{E}, ψ_{E}).

Associated to any equivalence class ω of representations of the Weil group of the field F is a onedimensional representation or, what is the same, a quasi-character of C_{F}. It is denoted detω and is obtained by taking the determinant of any representation in ω. Suppose ρ is in the class ω and ρ is a representation of W_{K/F}. To find the value of the quasi-character detω at β choose w in W_{K/F} so that τ_{K/F} w = β. Then calculate det(ρ(w)) which equals detω(β).

If F ⊆ E ⊆ K the map τ = τ_{K/F} can be effected in two stages. We first transfer W_{K/F} /W_{K/F}^{c} into W_{K/E}/W_{K/E}^{c}; then we transfer W_{K/E}/W_{K/E}^{c} into C_{K}. If W_{K/F} is the disjoint union

⋃_{i=1}^{r} W_{K/E}w_{i}

and if w_{i}w = u_{i}(w)w_{j}(i) then the transfer of w in W_{K/E}/W_{K/E}^{c} is the coset to which w′ = ∏_{i=1}^{r} u_{i}(w) belongs.

Suppose σ is a representation of W_{K/E} and

ρ = Ind(W_{K/F}, W_{K/E}, σ).

ρ acts on a certain space V of functions on W_{K/F} and if V_{i} is the collection of functions in V which vanish outside of W_{K/E}w_{i} then

V = ⨁_{i=1}^{r} V_{i}.

We decompose the matrix of ρ(w) into corresponding blocks ρ_{ji}(w). ρ_{ji}(w) is 0 unless j = j(i) when ρ_{ji}(w) = σ(u_{i}, (w)). This makes it clear that if ι_{E/F} is the representation of W_{K/F} induced from the trivial representation of W_{K/E}

det(ρ(w)) = det(ι_{E/F} (w))^{dim σ}det(σ(w′))

or, if θ is the class of σ,

detω(β) = {det ι_{E/F} (β)}^{dim θ}{det θ(β)}.

Lemma 5.1

Suppose F is a local field and E/F ⟶ λ(E/F, ψ_{F}) and ω ⟶ ε(ω, ψ_{E/F}) satisfy the conditions of Theorem A for the character ψ_{F}. Let ψ′_{F} (x) = ψ_{F} (βx) with β in C_{F}. If E/F ⟶ λ(E/F, ψ′_{F}) and ω ⟶ ε(ω, ψ′_{E/F}) satisfy the conditions of Theorem A for ψ′_{F} then

λ(E/F, ψ′_{F}) = det ι_{E/F} (β) λ(E/F, ψ_{F})

and

ε(ω, ψ′_{E/F}) = det ω(β) ε(ω, ψ_{E/F}).

Because of the uniqueness all one has to do is verify that the expressions on the right satisfy the conditions of the theorem for the character ψ′_{F}. This can now be done immediately.

Chapter Six.

A Filtration of the Weil Group

In this paragraph I want to reformulate various facts found in Serre’s book [12] as assertions about a filtration of the Weil group. Although some of the lemmas to follow will be used to prove the four main lemmas, the introduction of the filtration itself is not really necessary. It serves merely to unite in a form which is easily remembered the separate lemmas of which we will actually be in need.

Let K be a finite Galois extension of the non-archimedean local field F and let G = 𝔊(K/F). Let O_{F} be the ring of integers in F and let 𝔭_{F} be the maximal ideal of O_{F}. If i ≥ −1 is an integer let G_{i} be the subgroup of G consisting of those elements which act trivially on O_{F} /𝔭_{F}^{i+1}. If u ≥ −1 is a real number and i is the smallest integer greater than or equal to u set G_{u} = G_{i}. Finally if u ≥ −1 set

φ_{K/F} (u) = ∫_{0}^{u} \frac{1}{[G_{0} : G_{t}]} dt.

The integrand is not defined at -1 but that is of no consequence. φ_{K/F} is clearly a piecewise linear, continuous, and increasing map of [−1, ∞) onto itself. The inverse function* ψ_{K/F} will have the same properties.

We take from Serre’s book the following lemma.

Lemma 6.1

If F ⊆ L ⊆ K and L/F is normal then φ_{K/F} = φ_{L/F} ∘ φ_{K/L} and ψ_{K/F} = ψ_{K/L} ∘ ψ_{L/F}.

The circle denotes composition not multiplication. This lemma allows us to define φ_{E/F} and ψ_{E/F} for any finite separable extension E/F by choosing a Galois extension L of F which contains E and setting

φ_{E/F} = φ_{L/F} ∘ ψ_{L/E}
ψ_{E/F} = φ_{L/E} ∘ ψ_{L/F}

because if L′ is another such extension we can choose a Galois extension K containing both L and L′ and

φ_{L/F} ∘ ψ_{L/E} = φ_{L/F} ∘ φ_{K/L} ∘ ψ_{K/L} ∘ ψ_{L/E} = φ_{K/F} ∘ ψ_{K/E} = φ_{L′/F} ∘ ψ_{L′/E}
φ_{L/E} ∘ ψ_{L/F} = φ_{L/E} ∘ φ_{K/L} ∘ ψ_{K/L} ∘ ψ_{L/F} = φ_{K/E} ∘ ψ_{K/F} = φ_{L′/E} ∘ ψ_{L′/F} .

Of course ψ_{E/F} is the inverse of φ_{E/F}.

Lemma 6.2

If E ⊆ E′ ⊆ E′′ and E′′/E is finite and separable, φ_{E′′/E} = φ_{E′/E} ∘ φ_{E′′/E′} and ψ_{E′′/E} = ψ_{E′′/E′} ∘ ψ_{E′/E}.

* In this chapter ψ_{K/F} does not appear as an additive character. Nonetheless, there is a regrettable conflict of notation.

Each of these relations can be obtained from the other by taking inverses; we verify the second

ψ_{E′′/E′} ∘ ψ_{E′/E} = φ_{L/E′′} ∘ ψ_{L/E′} ∘ φ_{L/E′} ∘ ψ_{L/E} = φ_{L/E′′} ∘ ψ_{L/E} = ψ_{E′′/E}.

It will be necessary for us to know the values of these functions in a few special cases.

Lemma 6.3

(i) If K/F is Galois and unramified ψ_{K|F} (u) ≡ u. (ii) If K/F is cyclic of prime degree ℓ and if G = G_{t} while G_{t+1} = {1} where t is a non-negative

integer then

ψ_{K/F} (u) = u  u ≤ t
= t + ℓ(u − t) u ≥ t.

These assertions follow immediately from the definitions.

Lemma 6.4

Suppose K/F is Galois and G = 𝔊(K/F) is a product HC where H ≠ {1}, H ∩ C = {1}, and C is a non-trivial abelian normal subgroup of G which is contained in every non-trivial normal subgroup.

(i) If K/F is tamely ramified so that G_{1} = {1} then G_{0} = C is a cyclic group of prime order ℓ and [G : G_{0}] = [H : 1] divides ℓ − 1. If E is the fixed field of H, ψ_{E/F} (u) = u for u ≤ 0 and ψ_{E/F} (u) = ℓu for u ≥ 0.

(ii) If K/F is wildly ramified there is an integer t ≥ 1 such that C = G_{1} = … = G_{t} while G_{t+1} = {1}. [G_{0} : G_{1}] divides [G_{1} : 1] − 1 and every element of C has order p or 1. If E is the fixed field of H and L that of C

ψ_{L/F} (u) = u  u ≤ 0
= [G_{0} : G_{1}]u u ≥ 0

while

ψ_{E/F} (u) = u  u ≤ \frac{t}{[G_{0} : G_{1}]}

= \frac{t}{[G_{0} : G_{1}]} + [G_{1} : 1] (u − \frac{t}{[G_{0} : G_{1}]}) u ≥ \frac{t}{[G_{0} : G_{1}]}

We observed in the third paragraph that C must be its own centralizer. G_{0} cannot be {1}. Thus C ⊆ G_{0}. In case (i) G_{0} is abelian and thus G_{0} = C. In both cases if ℓ is a prime dividing the order of C the set of elements in C of order ℓ or 1 is a non-trivial normal subgroup of G and thus C itself. In case (i) C is cyclic and thus of prime order ℓ. Moreover, H which is isomorphic to G/G_{0} is abelian and, if h ∈ H, {c ∈ C|hc = ch} is a normal subgroup of G and hence {1} or C. If h ≠ 1 it must be 1. Consequently each orbit of H in C − {1} has [H : 1] elements and [H : 1] divides ℓ − 1.

In case (ii) G_{1} is a non-trivial normal subgroup and hence contains C. G_{1} and C are both p-groups. The centralizer of G_{1} in C is not trivial. As a normal subgroup of G it contains C.

Therefore it is C and G_{1} is contained in C which is its own centralizer. Since each G_{1}, i ≥ 1, is a normal subgroup of G, it is either C or {1}. Thus there is an integer t ≥ 1 such that G_{1} = G_{t} = C while G_{t+1} = {1}. If i ≥ 0 is an integer let U_{K}^{i} be the group of units of O_{K} which are congruent to 1 modulo p_{K}^{i+1}; let U_{K}^{(−1)} = C_{K}, and if U ≥ −1 is any real number let i be the smallest integer greater than or equal to u and set U_{K}^{u} = U_{K}^{i}. If θ_{t} is the map of G_{t}/G_{t+1} into p_{K}^{t}/p_{K}^{t+1} and θ_{0} the map of G_{0}/G_{1} into U_{K}^{0}/U_{K}^{1} introduced in Serre then, for g in G_{0} and h in C,

θ_{t}(ghg^{−1}) = θ_{0}(g)^{t}Θ_{t}(h).

If h ≠ 1, ghg^{−1} = h if and only if θ_{0}(g)^{t} = 1 and then g belongs to the centralizer of C, that is to G_{1}. Again C − {1} is broken up into orbits, each with [G_{0} : G_{1}] elements and [G_{0} : G_{1}] divides [G_{i} : 1] − 1. Observe that t must be prime to [G_{0} : G_{1}].

It follows immediately from the definitions that H_{u} = H ∩ G_{u}. In case (i) H_{0} will be {1} and φ_{K/E}(u) will be identically u. Thus ψ_{E/F} = ψ_{K/F} and, from the definition, ψ_{K/F} (u) = u if u ≤ 0 while ψ_{K/F} (u) = [G_{0} : 1]u if u ≥ 0. In case (ii), φ_{K/E}(u) = u if u ≤ 0 and

φ_{K/E}(u) = \frac{u}{[H_{0} : 1]} = \frac{u}{[G_{0} : G_{1}]}

if u ≥ 0 while ψ_{K/F} (u) = u if u ≤ 0 and

ψ_{K/F} (u) = [G_{0} : G_{1}]u  0 ≤ u ≤ \frac{t}{[G_{0} : G_{1}]}

= t + [G_{0} : 1] (u − \frac{t}{[G_{0} : G_{1}]}) \frac{t}{[G_{0} : G_{1}]} ≤ u.

The lemma follows.

Lemma 6.5

For every separable extension E′/E the function ψ_{E′/E} is convex, and if u is an integer so is ψ_{E′/E}(u).

All we have to do is prove that the assertion is true for all E′/E in P(K/F) if F is an arbitrary non-archimedean local field and K an arbitrary Galois extension of it. To do this we just combine the previous three lemmas with Lemma 3.1. We are going to use the same method to prove the following lemma.

Lemma 6.6

For every separable extension E′/E and any u ≥ −1

N_{E′/E} (U_{E′}^{ψ_{E′/E}(u)}) ⊆ U_{E}^{u}.

We have to verify that the set 𝔊 of all E′/E in P(K/F) for which the assertion is true satisfies the conditions of Lemma 3.1. There is no problem with the first two.

Lemma 6.7 E′/E belongs to 𝔊 if and only if for every integer n ≥ −1

N_{E′/E} (U_{E′}^{ψ_{E′/E}(n)}) ⊆ U_{E}^{n}

and

N_{E′/E} (U_{E′}^{ψ_{E′/E}(n)+1}) ⊆ U_{E}^{n+1}.

If E′/E belongs to 𝔊 choose ε > 0 so that ψ_{E′/E}(n + ε) = ψ_{E′/E}(n) + 1. The smallest integer greater than or equal to n + ε is at least n + 1 so

N_{E′/E} (U_{E′}^{ψ_{E′/E}(n)+1}) ⊆ U_{E}^{n+ε} ⊆ U_{E}^{n+1}.

Conversely suppose the conditions of the lemma are satisfied and n < u < n + 1. Since ψ_{E′/E}(n) is an integer the smallest integer greater than or equal to ψ_{E′/E}(u) is at least ψ_{E′/E}(n) + 1. Thus

N_{E′/E} (U_{E′}^{ψ_{E′/E}(u)}) ⊆ N_{E′/E} (U_{E′}^{ψ_{E′/E}(n)+1}) ⊆ U_{E}^{n+1} = U_{E}^{u}.

Lemma 6.8

If L/E is Galois then, for every integer n ≥ −1,

N_{L/E} (U_{L}^{ψ_{L/E}(n)}) ⊆ U_{E}^{n}

and

N_{L/E} (U_{L}^{ψ_{L/E}(n)+1}) ⊆ U_{E}^{n+1}.

The assertion is clear if n = −1. A proof for the case n ≥ 0 and L/E totally ramified is given in Serre’s book. Since that proof works equally well for all L/E we take the lemma as proved.

Lemma 6.9

Suppose K/F is Galois and G = 𝔊(K/F). Suppose G = HC where H ≠ {1}, H ∩ C = {1}, and C is a non-trivial abelian normal subgroup of G which is contained in every non-trivial normal subgroup of G. If E is the fixed field of H

N_{E/F} (U_{E}^{ψ_{E/F} (u)}) ⊆ U_{F}^{u}

for all u ≥ −1.

Let L be the fixed field of C. If K/F is tamely ramified K/E and L/F are unramified so that ψ_{E/F} = ψ_{K/L} and U_{E}^{v} = C_{E} ∩ U_{K}^{v}, U_{F}^{v} = C_{F} ∩ U_{L}^{v} for every v ≥ −1. If α belongs to C_{E}, then delete N_{K/L}α = N_{E/F} α. Since K/L is Galois

N_{E/F} (U_{E}^{ψ_{E/F} (u)}) ⊆ C_{F} ∩ N_{K/L} (U_{L}^{ψ_{K/L}(u)}) ⊆ C_{F} ∩ U_{L}^{u} = U_{F}^{u}.

If K/F is not tamely ramified

𝔭_{E}^{n} = E ∩ p_{k}^{[G_{0}:G_{1}]n−m}

if n ≥ 1 and 0 ≤ m < [G_{0} : G_{1}]. Thus

U_{E}^{v} = G_{E} ∩ U_{K}^{v}

if −1 ≤ v ≤ 0 and

U_{E}^{v} = C_{E} ∩ U_{K}^{[G_{0}:G_{1}]v}

if v ≥ 0 or, more briefly,

U_{E}^{v} = C_{E} ∩ U_{K}^{ψ_{K/E}(v)}

for all v ≥ −1. In the same way we find

U_{F}^{v} = C_{F} ∩ U_{L}^{ψ_{L/F} (v)}

for all v ≥ −1. Since K/L is normal

N_{E/F} (U_{E}^{ψ_{E/F} (u)}) ⊆ C_{F} ∩ N_{K/L} (U_{K}^{ψ_{K/F} (u)}) ⊆ C_{F} ∩ U_{L}^{ψ_{L/F} (u)} = U_{F}^{u}.

Lemma 6.6 now follows immediately.

Lemma 6.10

(a) Suppose K/F is Galois and G = 𝔊(K/F). Suppose t ≥ −1 is an integer such that G = G_{t} ≠ G_{t+1}. Then ψ_{K/F} (u) = u for u ≤ t. Moreover N_{K/F} defines an isomorphism of C_{K}/U_{K}^{t} with C_{F} /U_{F}^{t} and if −1 ≤ u ≤ t the inverse image of U_{F}^{u}/U_{F}^{t} is U_{K}^{u}/U_{K}^{t}. However the map of C_{K}/U_{K}^{t+1} into C_{F} /U_{F}^{t+1} defined by the norm is not surjective.

(b) Suppose K/F is Galois and G = 𝔊(K/F). Suppose s ≥ −1 is an integer and G = G_{s}. If F ⊆ E ⊆ K, ψ_{E/F} (u) = u for u ≤ s and N_{E/F} defines an isomorphism of C_{E}/U_{E}^{s} and C_{F} /U_{F}^{s}. If −1 ≤ u ≤ s the inverse image of U_{F}^{u}/U_{F}^{s} is U_{E}^{u}/U_{E}^{s}.

If t = −1 the assertions of part (a) are clear. If t ≥ 0, K/F is totally ramified. The relation ψ_{K/F} (u) = u for u ≤ t is an immediate consequence of the definition. Since the extension is totally ramified N_{K/F} defines an isomorphism of U_{K}^{−1}/U_{K}^{0} and U_{F}^{−1}/U_{F}^{0} . It follows from Proposition V.9 of Serre’s book that if 0 ≤ n < t the associated map U_{K}^{n}/U_{K}^{n+1} ⟶ U_{F}^{n}/U_{F}^{n+1} is an isomorphism but that the map U_{K}^{t}/U_{K}^{t+1} ⟶ U_{F}^{t} /U_{F}^{t+1} has a non-trivial cokernel. The first part of the lemma is an immediate consequence of these facts.

To prove part (b) we first observe that there is a t ≥ s such that G = G_{t} ≠ G_{t+1}. It then follows from part (a) that the map N_{K/F} determines an isomorphism of C_{K}/U_{K}^{s} and C_{F} /U_{F}^{s} under which U_{K}^{u}/U_{K}^{s} and U_{F}^{u}/U_{F}^{s} correspond if −1 ≤ u ≤ s. Let E be the fixed field of H. We have H_{s} = H ∩ G_{s} = H, so that N_{K/E} determines on isomorphism of C_{K}/U_{K}^{s} and C_{E}/U_{E}^{s} under which U_{K}^{u}/U_{K}^{s} and U_{E}^{u}/U_{E}^{s} correspond if −1 ≤ u ≤ s. Moreover if u ≤ s, ψ_{K/F} (u) = ψ_{K/E}(u) = u so that ψ_{E/F} (u) = u. Part (b) follows from these observations and the relation N_{K/F} = N_{E/F} N_{K/E}.

If E is any non-archimedean local field and u > −1

U_{E}^{u} = ∩_{v<u}U_{E}^{v}.

If α belongs to C_{E} set

v_{E}(α) = sup{u|α ∈ U_{E}^{u}}.

Then v_{E}(1) = ∞, but v_{E}(α) is finite if α ≠ 1 and α belongs to U_{E}^{v_{E}(α)}.

If F ⊆ L ⊆ K, τ_{K/F,L/F} will be any of the maps W_{K/F} ⟶ W_{L/F} associated to the imbedding L/F ⟶ K/F. We abbreviate τ_{K/F,F/F} to τ_{K/F}. If w belongs to W_{K/F} , σ(w) is the image of w in 𝔊(K/F), and E is the fixed field of σ(w), we set

v_{K/F} (w) = φ_{E/F} (v_{E}(τ_{K/E}(w)).

Note that we regard W_{K/E} as a subgroup of W_{K/F}. If v ≥ −1 let

W_{K/F}^{v} = {w | v_{K/F} (w) ≥ v}.

We shall show that W_{K/F}^{v} is a normal subgroup of W_{K/F}. These groups provide a filtration of the Weil group, some of whose properties are established in the following lemmas.

Lemma 6.11

If σ ∈ 𝔊(K/F) and t = sup{u | σ ∈ G_{u}}, set v_{K/F} (σ) = φ_{K/F} (t). Then

v_{K/F} (σ) = max{v_{K/F} (w) | σ(w) = σ}.

If σ = 1 both sides are infinite and the assertion is clear. If σ ≠ 1 let E be the fixed field of σ. If σ(w) = σ, w belongs to W_{K/E} and v_{K/F} (w) = φ_{E/F} (v_{K/E}(w)). Also v_{K/F} (σ) = φ_{E/F} (v_{K/E}(σ)). Consequently it is sufficient to prove the lemma when F = E. The set

S = {τ_{K/F} (w) | σ(w) = σ}

is a coset of N_{K/F} (C_{K}) in C_{F} and C_{F} is generated by N_{K/F} (C_{K}) together with any element of S. Moreover s = max{v_{F} (β) | β ∈ S} is the largest integer such that S ∩ U_{F}^{s} is not empty. Since G = G_{t} ≠ G_{t+1} the preceding lemma shows that s = t = φ_{K/F} (t).

Lemma 6.12

(a) For all w and w_{1} in W_{K/F} , v_{K/F} (w) = v_{K/F} (w^{−1}) and v_{K/F} (w_{1}w w_{1}^{−1}) = v_{K/F} (w).

(b) If F ⊆ E ⊆ K and w belong to W_{K/E} then

v_{K/F} (w) = φ_{E/F} (v_{K/E}(w)).

(c) For all w in W_{K/F} , τ_{K/F} (w) ⊂ U_{F}^{v_{K/F} (w)}.

The first two assertions follow immediately from the definitions and the basic properties of the Weil group. I prove only the third. Let me first observe that if F ⊆ E ⊆ K and w ⊂ W_{K/E}, then

τ_{K/F} (w) = N_{E/F} (τ_{K/E}(w)).

To see this, choose a set of representatives w_{1}, …, w_{r} for the cosets of C_{K} in W_{K/E} and then a set of representatives v_{1}, …, v_{s} for the cosets of W_{K/E} in W_{K/F}. Let w_{i}w = a_{i}w_{j(i)} with a_{i} in C_{K}; then

τ_{K/E}(w) = ∏_{i=1}^{r} a_{i}.

However v_{j}w_{i}w = v_{j}a_{i}v_{j}^{−1}v_{j}w_{j(i)} so that

τ_{K/F} (w) = ∏_{j=1}^{s} ∏_{i=1}^{r} v_{j}a_{i}v_{j}^{−1} = ∏_{j=1}^{s} v_{j}τ_{K/E}(w)v_{j}^{−1} = N_{E/F} (τ_{K/E}(w)).

In particular, if E is the fixed field of σ(w), τ_{K/E}(w) is contained U_{E}^{ψ_{E/F} (v_{K/F} (w))} and τ_{K/F} (w) is contained in

N_{E/F} (U_{E}^{ψ_{E/F} (v_{K/F} (w))}) ⊆ U_{F}^{v_{K/F} (w)}.

Lemma 6.13

If u and v belong to W_{K/F} then

v_{K/F} (uv) ≥ min{v_{K/F} (u), v_{K/F} (v)}.

Let σ = σ(u) and let τ = σ(v). Because of the second assertion of the previous lemma we may assume that σ and τ generate 𝔊(K/F). Let E be the fixed field of στ. If

t = {min (ψ_{K/F} (v_{K/F} (σ)), ψ_{K/F} (v_{K/F} (τ))}

and G = 𝔊(K/F) then G = G_{t} ≠ G_{t+1}. According to Lemma 6.11, if

s = min{v_{K/F} (u), v_{K/F} (v)},

then t ≥ ψ_{K/F} (s) which, by Lemma 6.10, is therefore equal to s. Since τ_{K/F} (uv) = τ_{K/F} (u)τ_{K/F} (v), τ_{K/F} (uv) lies in U_{F}^{s}. On the other hand

τ_{K/F} (uv) = N_{E/F} (τ_{K/E}(uv))

so that, by Lemma 6.10 again, τ_{K/E}(uv) belongs to U_{E}^{s} and

v_{K/F} (uv) ≥ φ_{E/F} (s) = s.

Thus the sets W_{K/F}^{x} , x ≥ −1, give a filtration of W_{K/F} by a collection of normal subgroups. The next sequence of lemmas show that the filtration is quite analogous to the upper filtration of the Galois groups.

Lemma 6.14

For each x ≥ −1 the map τ_{K/F,L/F} takes G_{K/F}^{x} into G_{L/F}^{x}.
If w belongs to W_{K/F} let w̄ = τ_{K/F,L/F} (w). We must show that
v_{L/F} (w̄) ≥ v_{K/F} (w).

Let σ = σ(w) and let σ̄ = σ(w̄). If E is the fixed field of σ then \overline{E} = E ∩ L is the fixed field of σ̄. Since

v_{L/F} (w̄) = φ_{\overline{E}/F} (v_{L/\overline{E}}(w̄))

and

v_{K/F} (w) = φ_{\overline{E}/F} (v_{K/\overline{E}}(w))

we may suppose \overline{E} = F. Since τ_{K/F} (w) = τ_{L/F} (w̄), Lemma 6.12 implies that τ_{L/F} (w̄) lies in

U_{F}^{v_{K/F} (w)}. Thus

v_{L/F} (w̄) = v_{F} (τ_{L/F} (w̄)) ≥ v_{K/F} (w).
Of course W_{F/F} is C_{F} and, if v ≥ −1, W_{F/F}^{v} = U_{F}^{v}.

Lemma 6.15

For each v ≥ −1, τ_{K/F} maps W_{K/F}^{v} onto U_{F}^{v}.

Since v_{1} ≤ v_{2} implies W_{K/F}^{v_{2}} ⊆ W_{K/F}^{v_{1}} it is enough to prove the lemma when v = n is an integer. The lemma is clear if [K : F] = 1; so we proceed by induction on [K : F]. If [K : F] > 1, choose an intermediate normal extension L so that [L : F] = ℓ is a prime. Let \overline{G} = 𝔊(L/F). Lemma 6.12 implies that

W_{K/L}^{ψ_{L/F} (v)} = W_{K/L} ∩ W_{K/F}^{v} .

There is an integer t ≥ −1 such that \overline{G} = \overline{G}_{t} and \overline{G}_{t+1} = \{1\}. It is shown in Chapter V of Serre’s book that if n > t

N_{L/F} (U_{L}^{ψ_{L/F} (n)}) = U_{F}^{n}.

By induction

τ_{K/L} (W_{K/L}^{ψ_{L/F} (n)}) = U_{L}^{ψ_{L/F} (n)}.

Since τ_{K/F} (w) = N_{L/F} (τ_{K/L}(w)) if w is in W_{K/L},

τ_{K/F} (W_{K/F}^{n}) = U_{F}^{n}

if n > t. Suppose \overline{σ} generates \overline{G}. Then V_{L/F} (\overline{σ}) = t. By Herbrand’s theorem there is a σ in 𝔊(K/F) with v_{K/F} (σ) = t whose restriction to L is \overline{σ}. By Lemma 6.11 there is a w in W_{K/F} such that σ = σ(w) and v_{K/F} (w) = t. Then τ_{K/F} (w) lies in U_{F}^{t} but not in N_{L/F} (C_{L}). From Serre’s book again

[U_{F}^{t} : N_{L/F} U_{F}^{ψ_{L/F} (t)}] = ℓ

so that U_{F}^{t} is generated by τ_{K/F} (w) and N_{L/F} (U_{L}^{ψ_{L/F} (t)}) and hence is contained in the image of W_{K/F}^{t}. To complete the proof of the lemma we have only to observe that Lemma 6.10 implies that

U_{F}^{n} = U_{F}^{t} N_{L/F} (U_{L}^{ψ_{L/F} (n)})

if n ≤ t.

Lemma 6.16

Suppose F ⊆ L ⊆ K and L/F and K/F are Galois. Then, for each v ≥ −1, τ_{K/F,L/F} maps W_{K/F}^{v} onto W_{L/F}^{v}.

Chapter 6 30

If [L : F] = 1 this is just the previous lemma so we proceed by induction on [L : F]. We have to show that if \overline{w} belongs to W_{L/F} there is a w in W_{K/F} such that \overline{w} = τ_{K/F,L/F} (w) and v_{K/F} (w) ≥ v_{L/F} (\overline{w}). Let \overline{σ} = σ(\overline{w}) and let \overline{E} be the fixed field of \overline{σ}. If \overline{E} ≠ F then, by the induction assumption, there is a w in W_{K/E} such that τ_{K/\overline{E},L/\overline{E}}(w) = \overline{w} and v_{K/\overline{E}}(w) ≥ v_{L/\overline{E}}(\overline{w}). By Lemma 6.12, v_{K/F} (w) ≥ v_{L/F} (\overline{w}). Moreover, we may assume that τ_{K/\overline{E},L/\overline{E}} is the restriction to W_{K/\overline{E}} of τ_{K/F,L/F}.

Suppose \overline{E} = F. Then v_{L/F} (\overline{w}) = v_{F} (τ_{L/F} (\overline{w})). Choose w_{1} in W_{K/F} so that τ_{K/F} (w_{1}) = τ_{L/F} (\overline{w}) and v_{K/F} (w_{1}) ≥ v_{F} (τ_{L/F} (\overline{w})). Let \overline{w}_{1} = τ_{K/F,L/F} (w_{1}) and set \overline{u} = \overline{w}_{1}^{−1}\overline{w}. Certainly v_{L/F} (\overline{u}) ≥ v_{L/F} (\overline{w}). Moreover, τ_{L/F} (\overline{u}) = 1. Let F ⊆ L_{1} ⊆ L where L_{1}/F is cyclic of prime order. If \overline{u} does not belong to W_{L/L_{1}} the group C_{F} is generated by N_{L_{1}/F} (C_{L_{1}}) and τ_{L/F} (\overline{u}), which is impossible since τ_{L/F} (\overline{u}) = 1. Thus \overline{u} belongs to W_{L/L_{1}} and, as observed, there is a u in W_{K/L_{1}} such that τ_{K/F,L/F} (u) = τ_{K/L_{1},L/L_{1}}(u) = \overline{u}. Then τ_{K/F,L/F} (uw_{1}) = \overline{w}.

Chapter Seven.

Consequences of Stickelberger’s Result

Davenport and Hasse [5] have shown that Stickelberger’s arithmetic characterization of Gaussian sums over a finite field can be used to establish identities between these Gaussian sums. After reviewing Stickelberger’s result we shall prove the identities of Davenport and Hasse together with some more complicated identities. However for the proof of Stickelberger’s result itself, I refer to Davenport and Hasse.

If Z = e^{\frac{2πi}{p}} and α belongs to GF (p) the meaning of Z^{α} is clear. If κ is any finite field and S is the absolute trace of κ let ψ_{κ}^{0} be the character of κ defined by ψ_{κ}^{0}(α) = Z^{S(α)}. If χ_{κ} is any character of κ^{∗} and ψ_{κ} is any non-trivial additive character of κ we will take the Gaussian sum τ(χ_{κ}, ψ_{κ}) to be

− ∑_{α∈κ^{∗}} χ_{κ}^{−1}(α)ψ_{κ}(α).

We abbreviate τ(χ_{κ}, ψ_{κ}^{0}) to τ(χ_{κ}).

Let 𝔨_{n} be the field obtained by adjoining the n^{th} roots of unity to the rational numbers. If ϖ = Z − 1 then in 𝔨_{p} the ideal (p) equals (ϖ^{p−1}). If q = p^{f} and κ has q elements then in 𝔨_{q−1} the ideal (p) is a product 𝔭𝔭′ … where the residue fields of 𝔭𝔭′, … are isomorphic to κ. In 𝔨_{p(q−1)}

(p) = (𝔭 𝔭′ …)^{p−1}

with 𝔓 = (𝔭, ϖ), 𝔓′ = (𝔭′, ϖ), and so on. The residue fields of 𝔓, 𝔓′, … are also isomorphic to κ. Choose one of these prime ideals, say 𝔓. Once an isomorphism of the residue field with κ is chosen the map of the (q − 1)^{th} roots of unity to the residue field defines an isomorphism of κ^{∗} and the group of (q − 1)^{th} roots of unity. Then χ_{κ} can be regarded as a character of the latter group. Choose α = α(χ_{κ}, 𝔓) with 0 ≤ α < q − 1 so that χ_{κ}(ζ) = ζ^{α} for all (q − 1)^{th} roots of unity. Write

α = α_{0} + α_{1}p + … + α_{f−1}p^{f−1} 0 ≤ α_{i} < p.

Not all of the α_{i} can be equal to p − 1. Set

σ(α) = α_{0} + α_{1} + … + α_{f−1}
γ(α = α_{0}!α_{1}! … α_{f−1}!

The following lemma is Stickelberger’s arithmetical characterization of τ(χ_{κ}).

Lemma 7.1

(a) τ(χ_{κ}) lies in 𝔨_{p(q−1)} and is an algebraic integer.

(b) If χ_{κ} = 1 then τ(χ_{κ}) = 1 but if χ_{κ} ≠ 1 the absolute value of τ(χ_{κ}) and all its conjugates is \sqrt{q}.

(c) Every prime divisor of τ(χ_{κ}) in 𝔨_{p(q−1)} is a divisor of p.

(d) If β is a non-zero element of the prime field then the automorphism Z ⟶ Z^{β} of 𝔨_{p(q−1)} over 𝔨_{q−1} sends τ(χ_{κ}) to χ_{κ}(β) τ(χ_{κ}).

(e) If 𝔓 is a prime divisor of p in 𝔨_{p(q−1)} and α = α(χ_{κ}, p) the multiplicative congruence

τ(χ_{κ}) ≡ \frac{ϖ^{σ}(α)}{γ(α)} (mod^{∗}𝔓)

is valid.

(f) Suppose ℓ is a prime dividing q − 1 and χ_{κ} = χ′_{κ}χ′′_{κ} where the order of χ′_{κ} is a power of ℓ and that of χ′′_{κ} is prime to ℓ. If ℓ^{a} is the exact power of ℓ dividing q − 1 and λ = ζ_{0} − 1 where ζ_{0} is a primitive ℓ^{a}-th root of unity then

τ(χ_{κ}) ≡ τ(χ′′_{κ}) (mod λ).

Before stating the identities for Gaussian sums which are implied by this lemma, I shall prove a few elementary lemmas.

Lemma 7.2

Suppose 0 ≤ α < p^{f} − 1 and
α = α_{0} + α_{1}p + … + α_{f−1}p^{f−1} 0 ≤ α_{i} < p.

Suppose also that 0 ≤ j_{0} < j_{1} < … < j_{r} = f and set

β_{s} = α_{j_{s}} + α_{j_{s}+1} p + … + α_{j_{s+1^{−1}}} p^{j_{s+1}−j_{s}−1}.

If σ = ∑_{s=0}^{r−1} β_{s} and γ = ∏_{s=0}^{r−1} β_{s}! then

\frac{ϖ^{σ}}{γ} = \frac{ϖ^{σ(α)}}{γ(α)} (mod^{∗}𝔓).

First of all, I remark once and for all that if n ≥ 1, 0 < u ≤ p^{n} − 1, and v = u(mod p^{n}) then v ≡ u(mod^{∗}p). Thus if 0 ≤ u ≤ p^{n} − 1 and v ≥ 0

(u + vp^{n})! = (vp^{n})! ∏_{w=1}^{u}(w + vp^{n}) ≡ u!(vp^{n})! (mod^{∗}p).

Also if v ≥ 0

∏_{w=1}^{p^{n}}(vp^{n} + w) ≡ (v + 1) p^{n}! (mod^{∗}p)

and, by induction,

(vp^{n})! ≡ v!(p^{n}!)^{v} (mod^{∗}p).

In particular p^{(n+1)}! ≡ p!(p^{n}!)^{p} ≡ (−p) (p^{n}!)^{p}. Apply induction to obtain

p^{n}! ≡ (−p)^{\frac{p^{n}−1}{p−1}} (mod^{∗}p).

From the relations

p = ∏_{i=1}^{p−1}(1 − Z^{i}) = (−ϖ)^{p−1} ∏_{i=1}^{p−1} \frac{Z^{i} − 1}{Z − 1}

and

\frac{Z^{i} − 1}{Z − 1} = 1 + Z + … + Z^{i−1} ≡ i (mod^{∗}p).

We conclude that

p = (p − 1)!(−ϖ)^{p−1} ≡ −ϖ^{p−1} (mod^{∗}p).

The lemma itself is clear if r = f so we proceed by induction downward from f. Suppose r < f, j_{s+2} − j_{s} = t > 1, and the lemma is valid for the sequence j_{0}, j_{1}, …, j_{s+1} − 1, j_{s+1}, … j_{r}. To prove it for the given sequence we have only to show that if

x = α_{j_{s}} + α_{j_{s}+1} p + … + α_{j_{s+1^{−2}}} p^{t−2}

and y = α_{j_{s+1}−1} then

\frac{ϖ^{x+y}}{x!y!} ≡ \frac{ϖ^{x+yp^{t−1}}}{(x + yp^{t−1})!} (mod^{∗}p).

But

ϖ^{y(p^{t−1}−1)} ≡ (−p)^{y \frac{p^{t−1}−1}{p−1}} (mod^{∗}p)

and

(x + yp^{t−1})! = x!y!(p^{t−1}!)^{y} ≡ x!y!(−p)^{\frac{y^{t−1}−1}{p−1}} (mod)^{∗}p).

Lemma 7.3

Suppose β_{0}, …, β_{r−1} and γ_{0}, …, γ_{r−1} are non-negative integers all of which are less than or equal to q − 1. Suppose that q = p^{f} is a prime power and

∑_{i=0}^{r−1} (β_{i} + γ_{i})q^{i} < 2(q^{r} − 1).

Suppose also that δ_{i}, 0 ≤ i ≤ r − 1, are given such that 0 ≤ δ_{i} ≤ q − 1,

∑_{i=0}^{r−1} δ_{i}q^{i} < q^{r} − 1

and

∑_{i=0}^{r−1}(β_{i} + γ_{i})q^{i} = ∑_{i=0}^{r−1} δ_{i}q^{i} (mod q^{r} − 1).
(a) If ∑_{i=0}^{r−1}(β_{i} + γ_{i})q^{i} < q^{r−1} and if ν is the number of k, 1 ≤ k ≤ r, for which
∑_{i=0}^{k−1}(β_{i} + γ_{i}) ≥ q^{k} then
∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = ν(q − 1).
(b) If ∑_{i=0}^{r−1}(β_{i} + γ_{i})q^{i} ≥ q^{r} − 1 and if ν is the number of k, 1 ≤ k ≤ r, for which
1 ≠ ∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≥ q^{k} then
∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = ν(q − 1).

Observe immediately that if 1 ≤ k ≤ r, then 0 ≤ β_{k−1} + γ_{k−1} ≤ 2(q − 1) and
∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≤ 2(q − 1) ∑_{i=0}^{k−1} q^{i} = 2(q^{k} − 1).

If r = 1 then β_{0} + γ_{0} = δ_{0} + ε(q − 1) with ε equal to 0 or 1. If ε = 0 we are in case (a) and ν = 0 while β_{0} + γ_{0} − δ_{0} = 0. If ε = 1 we are in case (b); here ν = 1 and β_{0} + γ_{0} − δ_{0} = q − 1. Suppose then that r ≥ 2 and that if β′_{0}, …, β′_{r−1}, γ′_{0}, …, γ′_{r−2}, δ′_{0}, …, δ′_{r−2}, and ν′ are given as in the lemma (with r replaced by r − 1) then

∑_{i=0}^{r−2}(β′_{i} + γ′_{i} − δ′_{i}) = ν′(q − 1).

We establish part (a) first. In this case

∑_{i=0}^{r−1}(β_{i} + γ_{i})q^{i} = ∑_{i=0}^{r−1} δ_{i}q^{i}

and

∑_{i=0}^{r−2}(β_{i} + γ_{i})q^{i} = ∑_{i=0}^{r−2} δ_{i}q^{i} + εq^{r−1}

with ε = δ_{r−1} − β_{r−1} − γ_{r−1}. If ε were negative the left side of the equation would be negative; if ε were greater than 1 the left side would be greater than 2(q^{r−1} − 1). Since neither of these possibilities occur ε is 0 or 1.

Suppose first that ε = 0. If ∑_{i=0}^{r−2} δ_{i}q^{i} < q^{r−1} − 1 choose β′_{i} = β_{i}, γ′_{i} = γ_{i}, 0 ≤ i ≤ r − 2. Then δ′_{i} = δ_{i}, 0 ≤ i ≤ r − 2, and ν′ = ν. The assertion of the lemma follows in this case. If ∑_{1=0}^{r−2} δ_{i}q^{i} = q^{r−1} − 1 then δ_{i} = q − 1, 0 ≤ i ≤ r − 2. Then β_{0} + γ_{0} ≡ q − 1 (mod q) and, as a consequence, β_{0} + γ_{0} = q − 1. We show by induction that β_{i} + γ_{i} = q − 1, 0 ≤ i ≤ r − 2. If this is so for i < j then

∑_{i=j}^{r−2} (β_{i} + γ_{i})q^{i} = ∑_{i=j}^{r−2}(q − 1)q^{i}.

Hence β_{j} + γ_{j} ≡ q − 1 (mod q) and β_{j} + γ_{j} = q − 1. It follows immediately that ν = 0 and ∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = 0.

Now suppose that ε = 1. If

∑_{i=0}^{r−2}(β_{i} + γ_{i})q^{i} = 2(q^{r−1} − 1)

then β_{i} = γ_{i} = q − 1, 0 ≤ i ≤ r − 2, δ_{0} = q − 2, and δ_{i} = q − 1, 1 ≤ i ≤ r − 2. Thus ν = r − 1 and

∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = 1 + (r − 1) (q − 1) − 1 = (r − 1) (q − 1).

Suppose then that

∑_{i=0}^{r−2}(β_{i} + γ_{i})q^{i} < 2(q^{r−1} − 1).

From the relation

∑_{i=0}^{r−2}(β_{i} + γ_{i})q^{i} = ∑_{i=0}^{r−2} δ_{i}q^{i} + 1 + (q^{r−1} − 1).

We conclude that ∑_{i=0}^{r−2} δ_{i}q^{i} < q^{r−1} − 1. Then for some m, with 0 ≤ m ≤ r − 2, δ_{m} < q − 1. We choose the minimal value for m.

∑_{i=0}^{r−2}(β_{i} + γ_{i})q^{i} = (δ_{m} + 1)q^{m} + ∑_{i=m+1}^{r−2} δ_{i}q^{i} + (q^{r−1} − 1).

Thus if β′_{i} = β_{i}, γ′_{i} = γ_{i}, 0 ≤ i ≤ r−2, then δ′_{i} = 0, i < m, δ′_{m} = δ_{m}+1, and δ′_{i} = δ_{i}, m < i ≤ r−2. Arguing by congruences as before we see that β_{i} + γ_{i} = q − 1 for i < m. Thus

∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} = q^{k} − 1

for k ≤ m. However β_{m} +γ_{m} ≠ q −1 and thus β_{m} +γ_{m} +1 is prime to q. Moreover if r−1 ≥ k > m

1 + ∑_{i=1}^{k−1}(β_{i} + γ_{i})q^{i} ≡ (β_{m} + γ_{m} + 1)q^{m} (mod q^{m+1}).

Thus it is greater than or equal to q^{k} if and only if it is greater than or equal to q^{k} + 1. It follows that ν′ = ν + m and that

∑_{i=0}^{r−2}(β_{i} + γ_{i} − δ_{i}) = −m(q − 1) + ∑_{i=0}^{r−2}(β′_{i} + γ′_{i} − δ′_{i}) = ν(q − 1) + 1.

Since β_{r−1} + γ_{r−1} − δ_{r−1} = −1 the assertion of the lemma follows.

Now let us treat part (b). In this case

∑_{i=0}^{r−1}(β_{i} + γ_{i})q^{i} = ∑_{i=0}^{r−1} δ_{i}q^{i} + (q^{r} − 1)

and

1 + ∑_{i=0}^{r−2}(β_{i} + γ_{i})q^{i} = ∑_{i=0}^{r−2} δ_{i}q^{i} + εq^{r−1}

with ε = δ_{r−1} − β_{r−1} − γ_{r−1} + q. Again ε is 0 or 1. If β_{i} = γ_{i} = q − 1 for 0 ≤ i ≤ r − 2 then ε = 1 and δ_{i} = q − 1 for 0 ≤ i ≤ r − 2. Also ν = r and

∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = (r − 1) (q − 1) + β_{r−1} + γ_{r−1} − δ_{r−1} = r(q − 1).

Having taken care of this case, we suppose that

∑_{i=0}^{r−2}(β_{i} + γ_{i})q^{i} < 2(q^{r−1} − 1).

First take ε = 0. If δ_{0} = 0 then 1 + β_{0} + γ_{0} ≡ 0 (mod q) and β_{0} + γ_{0} = q − 1. Thus one of them is less than q − 1. By symmetry we may suppose it is β_{0}. Let β′_{0} = β_{0} + 1, β′_{i} = β_{i}, 1 ≤ i ≤ r − 2, and γ′_{i} = γ_{i}, 0 ≤ i ≤ r − 2. Since δ_{0} = 0

∑_{i=0}^{r−2} δ_{i}q^{i} ≤ q^{r−1} − q < q^{r−1} − 1

and δ′_{i} = δ_{i}, 0 ≤ i ≤ r − 1. Also ν = ν′ + 1 so that

∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = ∑_{i=0}^{r−2}(β′_{i} + γ′_{i} − δ′_{i}) − 1 + q = ν(q − 1)

as required. If δ_{0} > 0 take β′_{i} = β_{i} and γ′_{i} = γ_{i}, 0 ≤ i ≤ r − 2. Then δ′_{0} = δ_{0} − 1, δ′_{i} = δ_{i}, 1 ≤ i ≤ r − 2. Also if k ≤ r − 1

∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≡ δ_{0} − 1 ≠ −1(mod q)

and the left-hand side is greater than or equal to q^{k} if and only if it is greater than or equal to q^{k} − 1. It follows that ν = ν′ + 1. Consequently

∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = ∑_{i=0}^{r−2}(β′_{i} + γ′_{i} − δ′_{i}) − 1 + q = ν(q − 1).

If ε = 1 take γ′_{i} = γ_{i} and β′_{i} = β_{i}, 0 ≤ i ≤ r − 2. Then δ′_{i} = δ_{i}, 0 ≤ i ≤ r − 2, and ν = ν′ + 1 so that

∑_{1=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = ν′(q − 1) + (β_{r−1} + γ_{r−1} − δ_{r−1}) = ν(q − 1).

Lemma 7.4

Suppose β_{i} and γ_{i} are two periodic sequences of integers with period r. That is β_{i+r} = β_{i} and γ_{i+r} = γ_{i} for all i in ℤ. Suppose 0 ≤ β_{i} ≤ q − 1, 0 ≤ γ_{i} ≤ q − 1 for all i and that none of the numbers

ε_{k} = ∑_{i=0}^{r−1}(β_{i+k} + γ_{i+k})q^{i}

is divisible by q^{r} − 1. Let

∑_{i=0}^{r−1}(β_{i} + γ_{i})q^{i} ≡ ∑_{i=0}^{r−1} δ_{i}q^{i} (mod q^{r} − 1)

with 0 ≤ δ_{i} ≤ q − 1 and ∑_{i=0}^{r−1} δ_{i}q^{i} < q^{r} − 1. If μ is the number of ε_{k}, 1 ≤ k ≤ r, which are greater than or equal to q^{r} − 1 then

∑_{i=0}^{r−1}(β_{i} + γ_{i} − δ_{i}) = μ(q − 1).

Since ε_{0} ≤ 2(q^{r} − 1) and is not divisible by q^{r} − 1 it is less than 2(q^{r} − 1). Thus all we need do is show that the μ of this lemma is equal to the ν of the preceding lemma. Observe first of all that ε_{j} ≥ q^{r} − 1 if and only if ε_{j} ≥ q^{r}.

Suppose ε_{0} < q^{r}. If 1 ≤ k < r
∑_{i=k}^{r−1}(β_{i} + γ_{i})q^{i} < q^{r}

so that

∑_{i=k}^{r−1}(β_{i} + γ_{i})q^{i−k} < q^{r−k}.

Thus, if ε_{k} ≥ q^{r},

q^{r} ≤ ∑_{i=r−k}^{r−1}(β_{i+k} + γ_{i+k}) q^{i} + ∑_{i=0}^{r−k−1}(β_{i+k} + γ_{i+k})q^{i}
< q^{r−k} ∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} + q^{r−k}

and

∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≥ q^{k}.

Conversely if 1 ≤ k < r and

∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≥ q^{k},

then

∑_{i=0}^{r−1}(β_{i+k} + γ_{i+k})q^{i} ≥ ∑_{i=r−k}^{r−1}(β_{i+k} + γ_{i+k})q^{i}
= q^{r−k} ∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≥ q^{r}.

Thus μ = ν in this case.

Now suppose ε_{0} ≥ q^{r}. If 1 ≤ k < r

∑_{i=k}^{r−1}(β_{i} + γ_{i})q^{i} ≥ q^{r} − ∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≥ q^{r} − 2(q^{k} − 1).

If

∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≥ q^{k} − 1

then

∑_{i=0}^{r−1}(β_{i+k} + γ_{i+k})q^{i} ≥ ∑_{i=r−k}^{r−1}(β_{i+k} + γ_{i+k})q^{i} + ∑_{i=0}^{r−k−1}(β_{i+k} + γ_{i+k})q^{i}
= q^{r−k} ∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} + q^{−k} ∑_{i=k}^{r−1}(β_{i} + γ_{i})q^{i}
≥ q^{r−k}(q^{k} − 1) + q^{r−k} − 2 + 2q^{−k}.

Thus ε_{k} ≥ q^{r} − 1 and hence ε_{k} ≥ q^{r}. Conversely if ε_{k} ≥ q^{r},

q^{r−k} ∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} = ∑_{i=r−k}^{r−1}(β_{i+k} + γ_{i+k})q^{i}
≥ q^{r} − ∑_{i=0}^{r−k−1}(β_{i+k} + γ_{i+k})q^{i}
≥ q^{r} − 2(q^{r−k} − 1)
= q^{r} − 2q^{r−k} + 2.

Thus

∑_{i=0}^{k−1}(β_{i} + γ_{i})q^{i} ≥ q^{k} − 1

and again μ = ν. Lemma 7.5

Suppose q = p^{f} is a prime power, ℓ is a positive integer, and (ℓ, q) = 1. Let ℓm ≡ 1 (mod q) and if x is any integer let φ(x), with 0 ≤ φ(x) < q, be the remainder of x upon division by q. If 0 ≤ β < q and if ψ(x) = φ(x)!

\frac{ℓβ}{β!} ∏_{k=0}^{ℓ−1} \frac{ψ((β − k)m)}{ψ(−km)} ≡ 1(mod^{∗}p).

If ℓ = ℓ_{1} + uq with ℓ_{1} > 0 and u ≥ 0 then ℓ^{β} ≡ ℓ_{1}^{β}(mod^{∗}p). Moreover

∏_{k=0}^{ℓ−1} \frac{ψ((β − k)m)}{ψ(−km)} = \{∏_{k=0}^{ℓ_{1}−1} \frac{ψ((β − k)m)}{ψ(−km)}\} \{∏_{k=ℓ_{1}}^{ℓ−1} \frac{ψ((β − k)m)}{ψ(−km)}\}

and

∏_{k=ℓ_{1}}^{ℓ−1} ψ(−km) = {∏_{j=0}^{q−1} j!}^{u} = ∏_{k=ℓ_{1}}^{ℓ−1} ψ((β − k)m).

Thus it is enough to prove the lemma with ℓ replaced by ℓ_{1}. In other words we may suppose that 0 < ℓ < q. The case ℓ = 1 is trivial and we exclude it from the following discussion. Finally we suppose that 0 < m < q.

Let q − 1 = rℓ + s with 0 < r and 0 ≤ s < ℓ. Arrange the integers from 0 to q − 1 into the following array.

0  1  2 .  .  .  .  .  . . l − 1
l  l + 1  l + 2 .  .  .  .  .  . . 2l − 1
.  .  .
.  .  .
.  .  β − l + 1  .
.  .  .  β  .
.  .  .
.  .  .
(r − 1)l (r − 1)l + 1  q − l . rl − 1
rl  rl + 1  . .  .  .  . rl + s

Since ℓ does not divide q, rℓ + s = q − 1 does not lie in the last column. Also q − ℓ lies in the column following that in which rℓ + s lies.

We replace each number j in the above array by φ(jm). The resulting array, which is written out below, has some special features which must be explained. The first column is explained by the observation xℓm ≡ x(mod q). The other entries, apart from those at the foot of each column, are explained by the observation that, when 1 ≤ j and xℓ + j lies in the first array, φ(x + mj) > r while 0 < φ(mj) + x < q + r so that φ(x + mj) = φ(mj) + x. The position of q − 1 is explained by the relation m(q − ℓ) ≡ −mℓ ≡ q − 1(mod q). The other entries at the feet of the columns are explained by the observation that if 1 ≤ j ≤ ℓ − 1 then φ(jm) > r ≥ 1 while m(q − k) = m(q − ℓ) + m(ℓ − k) ≡ φ((ℓ − k)m) − 1(mod q) if 1 ≤ k ≤ ℓ − 1.

0  m  .  .  .  .  .  . .  φ((l − 1)m)
l  m + 1 .  .  .  .  .  . . φ((l − 1)m) + 1
.  .  .
.  .  .
.  .  φ((β − l + 1)m)  .
.  .  .  φ(βm)  .
.  .  .
.  .  .
r − 1 m + r − 1  q − 1 . φ((l − 2 − s)m) − 1
r  m + r .  .  .  . φ((l − 1)m) − 1

Suppose first of all that β < ℓ − 1. Then the numbers φ((β − k)m), 0 ≤ k ≤ ℓ − 1 constitute the first β + 1 together with the last ℓ − β − 1 numbers in the array. (The order of the numbers in the array is the order in which they appear when the array is read as though it were a printed page.) The numbers φ(−km), 1 ≤ k ≤ ℓ − 1, are the last ℓ − 1 numbers of the array, that is, the numbers after q − 1. Cancelling in the product of the lemma the terms in numerator and denominator corresponding to the last ℓ − β − 1 terms of the array, we obtain

∏_{j=1}^{β} \frac{φ(jm)!}{(φ(jm) − 1)!} = ∏_{j=1}^{β} φ(jm) ≡ m^{β}β! (mod^{∗}p)

as required.

Now take β ≥ ℓ − 1. Then the numbers β, β − 1, …, β − (ℓ − 1) occur as indicated in the first array. In particular there is exactly one in each column. The numerator in the product of the lemma is the product of the factorials of the corresponding elements of the second array. The denominator is the product of the factorials of the elements appearing after q − 1. As indicated t is the element lying above q − 1. Thus t is larger than any element appearing in a column other than that of t. The product of the lemma is t! times the product of the factorials of the other elements on the broken line divided by the factorials of the elements at the foot of the column in which they lie. Thus it equals t! divided by the product of all the elements below the broken line except those which lie directly below t. But t! is the product of all numbers in the second array except those which lie below t. Thus the quotient is the product of all numbers which lie above or on the broken line, that is,

∏_{j=1}^{β} φ(jm) ≡ ∏_{j=1}^{β} jm = m^{β} β! (mod^{∗}p)

as required.

Lemma 7.6

Suppose that q = p^{f} is a prime power, that ℓ is a positive integer dividing q − 1, that 0 ≤ α_{1} < q − 1, that (ℓ, α_{1}) = 1, and that

α_{2} = \frac{α_{1}}{ℓ} \frac{q^{ℓ} − 1}{q − 1}.

Then α_{2} is an integer and 0 ≤ α_{2} < q − 1. Moreover if

α_{2} = γ_{0} + γ_{1}q + … + γ_{ℓ−1} q^{ℓ−1}

with 0 ≤ γ_{i} ≤ q − 1 for 0 ≤ i ≤ ℓ − 1 then

∑_{j=0}^{ℓ−1} γ_{j} = α_{1} + ∑_{j=1}^{ℓ−1} j · \frac{q − 1}{ℓ}

and

ℓ^{α_{1}} ∏_{j=0}^{ℓ−1} γ_{j}! ≡ α_{1}! ∏_{j=1}^{ℓ−1} (\frac{jq − 1!}{ℓ}) (mod^{∗}p).

Certainly 0 ≤ α_{2} < q^{ℓ} − 1; moreover

\frac{q^{ℓ} − 1}{q − 1} = 1 + q + … + q^{ℓ−1} ≡ ℓ (mod ℓ)

so that α_{2} is an integer. Let α_{1} = mℓ + k with m ≥ 0 and 0 ≤ k < ℓ and for 0 ≤ j < ℓ let

(ℓ − 1 − j)k = i_{j} + δ_{j}ℓ

with δ_{j} ≥ 0 and 0 ≤ i_{j} < ℓ. Clearly i_{ℓ−1} = δ_{ℓ−1} = 0. Also (ℓ − 1)k = ℓ − k + ℓ(k − 1) so that i_{0} = ℓ − k and δ_{0} = k − 1. If j ≥ 1 then (δ_{j−1} − δ_{j})ℓ = k + (i_{j} − i_{j−1}). Since −ℓ < i_{j} − i_{j−1} < ℓ and 0 < k < ℓ the right-hand side is greater than −ℓ and less than 2ℓ so that δ_{j−1} − δ_{j} is 0 or 1. If it is 1 then k + i_{j} ≥ ℓ and i_{j} ≥ ℓ − k. If it is 0 then i_{j} = i_{j−1} − k < ℓ − k. Recalling that i_{0} = ℓ − k we see that

S = {j|1 ≤ j ≤ ℓ − 1 and δ_{j−1} − δ_{j} = 1} = {j|0 ≤ j < ℓ and i_{j} > ℓ − k}.

We shall prove that

γ_{0} = m + i_{0} \frac{(q − 1)}{ℓ} + k − δ_{0}

and

γ_{j} = m + i_{j} \frac{(q − 1)}{ℓ} + δ_{j−1} − δ_{j} 1 ≤ j < ℓ.

Since (k, ℓ) = 1 the numbers i_{j} are distinct and it will follow immediately that

∑_{j=0}^{ℓ−1} γ_{j} = (mℓ + k) + ∑_{j=0}^{ℓ−1} j · \frac{q − 1}{ℓ} = α_{1} + ∑_{j=1}^{ℓ−1} j · \frac{q − 1}{ℓ}.

Moreover we will have

∏_{j=0}^{ℓ−1} γ_{j}! = \{∏_{j=0}^{ℓ−1} (m + j \frac{(q − 1)}{ℓ})!\} \{m+i_{0} \frac{q − 1}{ℓ} +k −δ_{0}\}\{∏_{j∈S} (m + 1 + i_{j} \frac{(q − 1)}{ℓ})\}.

Recall that k − δ_{0} = 1. Dividing the first term by

∏_{j=0}^{ℓ−1} (j · \frac{(q − 1)}{ℓ})!

we obtain

∏_{j=0}^{ℓ−1} ∏_{n=1}^{m} (n + j \frac{(q − 1)}{ℓ}) .

The product of the last two terms is

∏_{j=ℓ−k}^{ℓ−1} (m + 1 + j \frac{(q − 1)}{ℓ}) .

If 1 ≤ n ≤ m and 0 ≤ j ≤ ℓ − 1 then nℓ − j < α_{1} < q so that the product of ℓ^{mk} and the first of these two expressions is multiplicatively congruent to

∏_{j=0}^{ℓ−1} ∏_{n=1}^{m} (nℓ − j) = (mℓ)!

Moreover, if ℓ − k ≤ j ≤ ℓ − 1, then 0 ≤ (m + 1)ℓ − j ≤ (m + 1)ℓ − (ℓ − k) = α_{1} < q and the second of these expressions upon multiplication by ℓ^{k} becomes multiplicatively congruent to

∏_{j=ℓ−k}^{ℓ−1} ((m + 1)ℓ − j) = ∏_{j=1}^{k} (mℓ + j).

The relations together imply the second identity of the lemma.

To verify that the γ_{j}, 0 ≤ j < ℓ, have the form asserted, we start with the relation

α_{2} = \frac{q^{ℓ} − 1}{q − 1} · \frac{mℓ + k}{ℓ} = ∑_{j=0}^{ℓ−1} q^{j}m + ∑_{j=0}^{ℓ−1} \frac{q^{j}k}{ℓ} .

The second term is equal to

∑_{j=0}^{ℓ−1} \{(∑_{i=0}^{j−1} q^{i}) \frac{q − 1}{ℓ} k\} + k = k + ∑_{j=0}^{ℓ−2} (ℓ − 1 − j)q^{j} · \frac{q − 1}{ℓ} · k.

Thus

α_{2} = (m + (ℓ − 1)k · \frac{q − 1}{ℓ} + k) + ∑_{j=1}^{ℓ−1} (m + (ℓ − 1 − j)k · \frac{q − 1}{ℓ}) q^{j}

= (m + i_{0} · \frac{q − 1}{ℓ} + k − δ_{0}) + ∑_{j=1}^{ℓ−1} (m + i_{j} · \frac{q − 1}{ℓ} + δ_{j−1} − δ_{j}) .

Moreover m < \frac{q−1}{ℓ} so that

0 ≤ m + i_{0} · \frac{q − 1}{ℓ} + k − δ_{0} < ℓ · \frac{q − 1}{ℓ} + 1 = q

and

0 ≤ m + i_{j} · \frac{q − 1}{ℓ} + δ_{j−1} − δ_{j} < ℓ · \frac{q − 1}{ℓ} + 1 = q.

The required relations follow immediately.

Now we can state and prove the promised identities for Gaussian sums. Each of these will amount to an assertion that a certain number in 𝔨_{p(q−1)} is 1. To prove this we will show first that the number is invariant under all automorphisms of 𝔨_{p(q−1)} over 𝔨_{(q−1)} and thus lies in 𝔨_{q−1}. The only prime ideals occurring in the factorization of the number, which is not a priori an algebraic integer, into prime ideals will be divisors of p. We show that every conjugate of the number has absolute value 1 and that it is multiplicatively congruent to 1 modulo every divisor of p. It will follow that it is a root of unity in 𝔨_{q−1} and hence a (q−1)th root of unity if q is odd and a 2(q−1)th root of unity if q is even. If q is odd the multiplicative congruences imply that the number is 1. If q is even they imply that the number is ±1. To show that it is actually 1 some supplementary discussion will be necessary.

Stickelberger’s result is directly applicable only to the normalized Gaussian sum τ(χ_{κ}). We shall have to use the obvious relation τ(χ_{κ}, Ψ_{κ}) = χ_{κ}(β)τ(χ_{κ}) if ψ_{κ}(α) = ψ_{κ}^{0}(βα). If κ is an extension of λ and ψ_{λ} is given, we set

ψ_{κ/λ}(α) = ψ_{λ}(S_{κ/λ}(α))

for α in κ. If χ_{λ} is given χ_{κ/λ} is the character defined by

χ_{κ/λ}(α) = χ_{λ}(N_{κ/λ}(α)).

Lemma 7.7

If κ is a finite extension of the finite field and χ_{λ} and ψ_{λ} are given then

τ(χ_{κ/λ}, ψ_{κ/λ}) = {τ(χ_{λ}, ψ_{λ})}^{[κ:λ]}.

Since χ_{κ/λ}(β) = χ_{λ}(β)^{[κ:λ]} it will be enough to show that

τ(χ_{κ/λ}) = {τ(χ_{λ})}^{[κ:λ]}.

Set

μ = \frac{\{τ(χ_{λ})\}^{[χ:λ]}}{τ(χ_{κ/λ})} .

Let λ have q = p^{ℓ} elements and let κ have p^{k} = q^{f}. It follows immediately from Lemma 7.1 that the absolute value of μ and all its conjugates is 1, that it lies in 𝔨_{p(q^{f} −1)}, that it is invariant under all automorphisms of 𝔨_{p(q^{f} −1)} over 𝔨_{q^{f} −1}, and that its only prime factors are divisors of p. The

mapping β ⟶ N_{κ/λ}β sends β to β^{\frac{q^{f} −1}{q−1}} . Thus if α = α(χ_{λ}, 𝔭) and 𝔓 divides 𝔭

α(χ_{κ/λ}, p) = \frac{q^{f} − 1}{q − 1} α = α + αq + … + αq^{f−1}.

Applying Lemmas 7.1 and 7.2, we see that

τ(χ_{λ}) ≡ \frac{ϖ^{α}}{α!} (mod^{∗}𝔓)

and

τ(χ_{κ/λ}) ≡ \frac{ϖ^{fα}}{(α!)^{f}} (mod^{∗}𝔓).

Consequently

μ ≡ 1 (mod^{∗}𝔓).

Thus μ = 1 if q is odd and μ = ±1 if q is even. If χ_{λ} = 1 then χ_{κ/λ} = 1 and, from part (b) of Lemma 7.1, μ is 1. If q = 2 then χ_{λ} = 1. Suppose then q is even and greater than 2. If χ_{λ} is not identically 1 choose a prime r dividing the order of χ_{λ}. Set χ_{λ} = χ′_{λ}χ′′_{λ} where the order of χ′_{λ} is a power of r and the order of χ′′_{λ} is prime to r. The analogous decomposition of χ_{κ/λ} is χ′_{κ/λ}χ′′_{κ/λ}. Of course χ_{λ} and χ_{κ/λ} have the same order. Define μ′ and μ′′ in the obvious way. According to part (f) of Lemma 7.1

μ ≡ μ′′ (mod r).

Since r does not divide 2 this implies that μ = μ′′. Thus one can show by induction on the number of primes dividing the order of χ_{λ} that μ = 1.

Lemma 7.8

Suppose λ is a finite field with q elements, κ is a finite extension of λ, and [κ : λ] = f. Suppose ℓ is a prime and the order of q modulo ℓ is f. Let T be a set of representatives for the orbits of the non-trivial characters of κ^{∗} of order ℓ under the action of 𝔊(κ/λ) and let χ_{λ} be a character of λ^{∗}. If ψ_{λ} is any non-trivial character of λ

χ_{λ}(ℓ^{ℓ})τ(χ_{λ}^{ℓ}, ψ_{λ}) ∏_{μ_{κ}∈T} τ(μ_{κ}, ψ_{κ/λ}) = τ(χ_{λ}, ψ_{λ}) ∏_{μ_{κ}∈T} τ(χ_{κ/λ} μ_{κ}, ψ_{κ/λ}).

Since the isotropy group of each point in T is trivial

χ_{λ}^{ℓ}(β) ∏_{μ_{κ}∈T} μ_{κ}(β) = χ_{λ}(β) ∏_{μ_{κ}∈T} χ_{κ/λ}(β) μ_{κ}(β)

and we may content ourselves with showing that

χ_{λ}(ℓ^{ℓ}) τ(χ_{λ}^{ℓ}) ∏_{μ_{κ}∈T} τ(μ_{κ}) = τ(χ_{λ}) ∏_{μ_{κ}∈T} τ(χ_{κ/λ} μ_{κ}).

Of course χ_{λ}(ℓ^{ℓ}) is the value of χ_{λ} at the element of the prime field corresponding to ℓ^{ℓ}. Let μ be the quotient of the right side by the left. The characters of κ^{∗} of order ℓ are the characters μ_{κ}^{k}, 0 ≤ k < ℓ, defined by

α(μ_{κ}^{k}, 𝔓) = k · \frac{q^{f} − 1}{ℓ}.

Since the order of q modulo ℓ is f, if T = {μ_{κ}^{k}|k ∈ A} every non-trivial character of order ℓ is representable as μ_{κ}^{η(q^{i}k)} with 0 ≤ i < f and k ∈ A. η(q^{i}k) is the remainder of q^{i}k upon division by ℓ. Thus as we already saw, T has \frac{ℓ−1}{f} elements. Lemma 7.1 again shows that μ and all its conjugates have absolute value 1 and that μ is invariant under all automorphisms of 𝔨_{p(q^{f} −1)} over 𝔨_{q^{f} −1}.

Let α = α(χ_{λ}, 𝔭) and let β = α(χ_{λ}^{ℓ}, 𝔭). Then ℓα = β + ν(q − 1) with ν ≥ 0. If 0 ≤ k < ℓ let

α(μ_{κ}^{k}, 𝔓) = k · \frac{q^{f} − 1}{ℓ} = ∑_{j=0}^{f−1} γ_{j}^{k} q^{j}

with 0 ≤ γ_{j}^{k} ≤ q − 1. In particular, γ_{0}^{k} is the residue of k · \frac{q^{f} −1}{ℓ} modulo q. Moreover if k_{1} ≡ q^{i}k (modulo ℓ) then

α(μ_{κ}^{k}, 𝔓) = ∑_{j=0}^{f−1} γ_{j+i}^{k_{1}} q^{j}.

It is understood that if j + i ≥ ℓ then γ_{j+i}^{k_{1}} = γ_{j+i−ℓ}^{k_{1}}. Thus if φ(x) is the remainder of x upon division by q,

\{γ_{j}^{k} | k ∈ A, 0 ≤ j < f\} = \{φ (k · \frac{q^{f} − 1}{ℓ}) | 0 < k < ℓ\}.

Certainly

α(χ_{κ/λ} μ_{κ}^{k}, 𝔓) ≡ \frac{q^{f} − 1}{q − 1} α + k · \frac{q^{f} − 1}{ℓ} (mod (q^{f} − 1)).

Let 0 ≤ k′ < ℓ and let ν + k ≡ k′ (mod ℓ). Since, by definition, ℓα = β + ν(q − 1)

\frac{q^{f} − 1}{q − 1} α + k \frac{q^{f} − 1}{ℓ} ≡ \frac{q^{f} − 1}{ℓ} · \frac{β}{q − 1} + k′ \frac{q^{f} − 1}{ℓ} (mod (q^{f} − 1)).

Since 0 ≤ β < q − 1 the right side is non-negative and at most q^{f} − 2. Thus it is α(χ_{κ/λ} μ_{κ}^{k}, 𝔓). Let

α(χ_{κ/λ} μ_{κ}^{k}, 𝔓) = ∑_{j=0}^{f−1} δ_{j}^{k} q^{j}

with 0 ≤ δ_{j}^{k} ≤ q − 1. Thus δ_{0}^{k} is the residue of

\frac{q^{f} − 1}{ℓ} · \frac{β}{q − 1} + k′ · \frac{q^{f} − 1}{ℓ}

modulo q. Since χ_{κ/λ} is invariant under automorphisms of κ/λ

α(χ_{κ/λ} μ_{κ}^{k}, 𝔓) = ∑_{j=0}^{f−1} δ_{j+i}^{k_{1}} q^{j}

if k_{1} ≡ q^{i}k (mod ℓ). Since the residue of \frac{q^{f} −1}{q−1}α modulo q is α,

\{α\} ∪ \{δ_{j}^{k} | 0 ≤ j < ℓ, k ∈ A\} = \{φ (\frac{q^{f} − 1}{ℓ} · \frac{β}{q − 1} + k \frac{q^{f} − 1}{ℓ}) | 0 ≤ k < ℓ\}.

Since χ_{λ}(ℓ^{ℓ}) ≡ ℓ^{αℓ} (mod^{∗}𝔓) the number μ is multiplicatively congruent modulo 𝔓 to the quotient of

\frac{ϖ^{α}ϖ^{ε}}{α! ∏_{k∈A} ∏_{j=0}^{ℓ−1} δ_{j}^{k}!}, ε = ∑_{k∈A} ∑_{j=0}^{f−1} δ_{j}^{k}

by

\frac{ℓ^{β} ϖ^{β} ϖ^{ε′}}{β! ∏_{k∈A} ∏_{j=0}^{ℓ−1} γ_{j}^{k}!}, ε′ = ∑_{k∈A} ∑_{j−0}^{f−1} γ_{j}^{k}.

Since

∑_{j=0}^{f−1} (α + γ_{j}^{k}) q^{j} ≡ ∑_{j=0}^{f−1} δ_{j}^{k} q^{j},

we conclude from Lemma 7.4 that

fα + ∑_{j=0}^{f−1} (γ_{j}^{k} − δ_{j}^{k}) = ρ(q − 1),

if ρ is the number of i, 0 ≤ i < f, such that

\frac{q^{f} − 1}{q − 1}α + α (μ_{κ}^{η(q^{i}k)}, 𝔓) ≥ q^{f}.

Since

\frac{q^{f} − 1}{q − 1} · α + α(μ_{κ}^{0}, 𝔓) = \frac{q^{f} − 1}{q − 1}α,

the number

(ℓ − 1)α + ∑_{k∈A} ∑_{j=0}^{f−1} (γ_{j}^{k} − δ_{j}^{k})

is (q − 1) times the number of k, 0 ≤ k < ℓ, such that

\frac{q^{f} − 1}{q − 1}α + k · \frac{q^{f} − 1}{ℓ} = \frac{q^{f} − 1}{q − 1} \frac{β}{ℓ} + (k + ν) \frac{q^{f} − 1}{ℓ} ≥ q^{f}.

The number of such k is ν because ν < ℓ and

\frac{q^{f} − 1}{q − 1} \frac{β}{ℓ} + (ℓ − ν + ν) \frac{q^{f} − 1}{ℓ} ≥ 1 + q^{f} − 1 = q^{f}

while

\frac{q^{f} − 1}{q − 1} \frac{β}{ℓ} + (ℓ − 1) \frac{q^{f} − 1}{ℓ} < \frac{q^{f} − 1}{ℓ} + (ℓ − 1) \frac{q^{f} − 1}{ℓ} = q^{f} − 1.

Thus

∑_{k∈A} ∑_{j=0}^{f−1} (γ_{j}^{k} − δ_{j}^{k}) = ν(q − 1) − (ℓ − 1)α = α − β.

If ℓm ≡ 1 (mod q)

φ (k · \frac{q^{f} − 1}{ℓ}) = φ(−km)

and

φ (\frac{q^{f} − 1}{ℓ} · \frac{β}{q − 1} + k \frac{q^{f} − 1}{ℓ}) = φ((β − k)m).

It follows immediately from Lemma 7.5 that

ℓ^{β} α! ∏_{k∈A} ∏_{j=0}^{f−1} δ_{j}^{k}! ≡ β! ∏_{k∈A} ∏_{j=0}^{f−1} γ_{j}^{k}!

Thus μ = 1 if q is odd and μ = ±1 if q is even. If χ_{λ} = 1 the number μ is clearly 1. This time too, one can apply part (f) of Lemma 7.1 and induction on the number of primes dividing the order of χ_{λ} to show that μ = 1 if q is even.

Lemma 7.9

Let λ be a finite field with q elements and let κ be a finite extension of λ with [κ : λ] = ℓ where ℓ is a prime dividing q − 1. Suppose χ_{λ} is a character of λ^{∗} whose restriction to the ℓth roots of unity is not trivial and χ_{κ} is a character of κ^{∗} such that χ_{κ}^{ℓ} = χ_{κ/λ}. If T is the set of non-trivial characters of λ^{∗} of order ℓ

χ_{λ}(ℓ)τ(χ_{λ}, ψ_{λ}) ∏_{μ_{λ}∈T} τ(μ_{λ}, ψ_{λ}) = τ(χ_{κ}, ψ_{κ/λ}).

If σ ∈ 𝔊(κ/λ) define χ_{κ}^{σ} by χ_{κ}^{σ}(α) = χ_{κ}(α^{σ^{−1}}). Since χ_{κ/λ}^{σ} = χ_{κ/λ}, χ_{κ}^{σℓ} = χ_{κ/λ} and χ_{κ}^{σ−1} is a character of order ℓ. If χ_{κ}^{σ−1} = 1 for some σ ≠ 1 then it is 1 for all σ and χ_{κ}(α) = 1 if α is a (q − 1)th power, that is, if N_{κ/λ}(α) = 1. Consequently there is a character ν_{λ} of λ^{∗} such that χ_{κ} = ν_{κ/λ}. Then ν_{λ}^{ℓ} = χ_{λ} and χ_{λ} is trivial on the ℓth roots of unity, contrary to assumption. Thus

{χ_{κ}^{σ−1} | σ ≠ 1} = {μ_{κ/λ} | μ_{λ} ∈ T }.

If β ∈ λ^{∗} and β = N_{κ/λ}(γ) then

χ_{κ}(β) = ∏_{σ} χ_{κ}(γ^{σ−1}) = χ_{κ}(γ^{ℓ}) ∏_{σ≠1} χ_{κ}^{σ−1}(γ) = χ_{λ}(β) ∏_{σ≠1} μ_{λ}(β),

because μ_{λ}(β) = μ_{κ/λ}(γ), and it will be enough to show that

χ_{λ}(ℓ) τ(χ_{λ}) ∏_{μ_{λ}∈T} τ(μ_{λ}) = τ(χ_{κ}).

Let μ be the quotient of the left side by the right. Thus μ is a number in 𝔎_{p(q−1)} and the only primes appearing in the factorization of μ are divisors of p. Since χ_{κ/λ} is not identically 1 neither is χ_{κ}. Thus the absolute value of μ and all its conjugates is 1. Let α = α(χ_{λ}, 𝔭) and let β = α(χ_{κ}, 𝔓) where 𝔓 divides 𝔭. Then

ℓβ ≡ α\frac{q^{ℓ} − 1}{q − 1} (mod(q^{ℓ} − 1)).

Since ℓ divides \frac{q^{ℓ}−1}{q−1} we can write

β = \frac{q^{ℓ} − 1}{q − 1} · \frac{α}{ℓ} − j · \frac{q^{ℓ} − 1}{ℓ}.

Since the restriction of χ_{λ} to the ℓth roots of unity is not trivial, α · \frac{q−1}{ℓ} ≢ 0 (mod (q − 1)). Thus ℓ does not divide α. For all i ≥ 0

τ (χ_{κ}^{q^{i}}) = τ(χ_{κ}).

Moreover

α (χ_{κ}^{q^{i}}, 𝔓) ≡ \frac{α^{ℓ} − 1}{q − 1} \frac{α}{ℓ} − j \frac{α^{ℓ} − 1}{ℓ} + (q^{i} − 1) \frac{q^{ℓ} − 1}{q − 1} \frac{α}{ℓ} − j(q^{i} − 1) \frac{q^{ℓ} − 1}{ℓ}

≡ \frac{q^{ℓ} − 1}{q − 1} \frac{α}{ℓ} + \{\frac{q^{i} − 1}{q − 1} α − j\} \frac{q^{ℓ} − 1}{ℓ}.

Since \frac{q^{i}−1}{q−1} ≡ i (mod ℓ) we choose i so that iα ≡ j (mod ℓ); then

α(χ_{κ}^{q^{i}}, 𝔓) ≡ \frac{q^{ℓ} − 1}{q − 1} \frac{α}{ℓ} (mod(q^{ℓ} − 1)).

Both sides of this congruence are non-negative and less than q^{ℓ} − 1. Thus it is an equality and we can assume that β = \frac{q^{ℓ}−1}{q−1} · \frac{α}{ℓ}. The set T consists of the characters μ_{λ}^{j}, 1 ≤ j ≤ ℓ − 1, defined by

α(μ_{λ}^{j}, 𝔓) = \frac{j}{ℓ} (q − 1).

Under the automorphism z ⟶ z^{m} of k_{p(q^{ℓ}−1)} over k_{q^{ℓ}−1} the number μ is multiplied by

χ_{κ}^{−1}(m) χ_{λ}(m) ∏_{μ_{λ}∈T} μ_{λ}(m) which is 1 because m belongs to λ. Let

β = γ_{0} + γ_{1}q + … + γ_{ℓ−1} q^{ℓ−1}

with 0 ≤ γ_{i} ≤ q − 1. Then

τ(χ_{κ}) ≡ \frac{ϖ^{ε}}{∏_{j=0}^{ℓ−1} γ_{j}!} (mod^{∗}𝔓), ε = ∑_{j=0}^{ℓ−1} γ_{j},

and

χ_{λ}(ℓ)τ(χ_{λ}) ∏_{j=1}^{ℓ−1} τ(μ_{λ}^{j}) ≡ \frac{ℓ^{α}ϖ^{ε′}}{α! ∏_{j=1}^{ℓ−1} (j · \frac{q−1}{ℓ})!} (mod^{∗}𝔓), ε′ = α + ∑_{j=0}^{ℓ−1} j · \frac{q − 1}{ℓ}.

Lemma 7.6 implies immediately that μ ≡ 1 (mod^{∗}𝔓). Thus μ = 1 if q is odd and μ = ±1 if q is even. If ℓ′ is a prime divisor of q − 1 different from ℓ, we write χ_{λ} as χ′_{λ}χ′′_{λ} where the order of χ′_{λ} is a power of ℓ′ and the order of χ′′_{λ} is prime to ℓ. In a similar fashion we write χ_{κ} as χ′_{κ}χ′′_{κ}. The pair χ′′_{λ} and χ′′_{κ} also satisfy the conditions of the lemma. The final assertion of Lemma 7.1 shows that, if μ′′ is defined in the same way as μ, μ = μ′′. Arguing by induction we see that it is enough to verify that μ = 1 when the order of χ_{λ} is a power of ℓ. Applying the last part of Lemma 7.1, again we see that there is a prime 𝔮 dividing ℓ such that

τ(χ_{λ}) ≡ τ(χ_{κ}) ≡ τ(μ_{λ}^{j}) ≡ 1 (mod 𝔮).

Since χ_{λ}(ℓ) is an ℓ^{ω}th root of unity for some ω,

χ_{λ}(ℓ) ≡ 1 (mod𝔮).

Thus μ ≡ 1 (modℓ) and μ = 1.

Chapter Eight.

A Lemma of Lamprecht

Let F be a non-archimedean local field and let ψ_{F} be a non-trivial character of F. n = n(ψ_{F}) is the largest integer such that ψ_{F} is trivial on 𝔓_{F}^{−n}. If χ_{F} is a quasi-character of C_{F} , m = m(χ_{F}) is the smallest non-negative integer such that χ_{F} is trivial on U_{F}^{m}. If γ in C_{F} is such that γO_{F} = 𝔓_{F}^{m+n} set

Δ_{1}(χ_{F} , ψ_{F}; γ) = \frac{∫_{U_{F}} ψ_{F} (\frac{α}{γ}) χ_{F}^{−1}(α)dα}{|∫_{U_{F}} ψ_{F} (\frac{α}{γ}) χ_{F}^{−1}(α)dα|}.

Then

Δ(χ_{F} , ψ_{F}) = χ_{F} (γ)Δ_{1}(χ_{F} , ψ_{F} , γ).

As suggested by Hasse [8], we shall, in the proofs, of the main lemmas, make extensive use of the following lemma which is central to the paper [10] of Lamprecht.

Lemma 8.1

(a) If m = m(χ_{F}) = 2d with d integral and positive there is a unit β in O_{F} such that

ψ_{F} (\frac{βx}{γ}) = χ_{F} (1 + x)

for all x in 𝔓_{F}^{d}. For any such β

Δ_{1}(χ_{F} , ψ_{F}; γ) = ψ_{F} (\frac{β}{γ} ) χ_{F}^{−1}(β).

(b) If m = m(χ_{F}) = 2d + 1 with d integral and positive there is a unit β in O_{F} such that, for all x in 𝔓_{F}^{d+1},

ψ_{F} (\frac{βx}{γ}) = χ_{F} (1 + x).

For any such β, Δ_{1}(χ_{F} , ψ_{F}; γ) is equal to

ψ_{F} (\frac{β}{γ} ) χ_{F}^{−1}(β) \frac{∫_{O_{F} /P_{F}} ψ_{F} (\frac{δβx}{γ}) χ_{F}^{−1}(1 + δx)dx}{|∫_{O_{F} /P_{F}} ψ_{F} (\frac{δβx}{γ}) χ_{F}^{−1}(1 + δx)dx|}

if δO_{F} = 𝔓_{F}^{d}.

Let m = 2d + ε with ε = 0 in case (a) and ε = 1 in case (b). The function ψ_{F} (\frac{xy}{γ} ) , x ∈

O_{F} , y ∈ 𝔓_{F}^{d+ε} can be regarded as a function on

O_{F} /𝔓_{F}^{d} × 𝔓_{F}^{d+ε}/𝔓_{F}^{m}.

For fixed x it is a character of 𝔓_{F}^{d+ε}/𝔓_{F}^{m} which is trivial if and only if x ∈ 𝔓_{F}^{d} and for fixed y it is a character of O_{F} /𝔓_{F}^{d} which is trivial if and only if y ∈ 𝔓_{F}^{m}. Thus it defines a duality of O_{F} /𝔓_{F}^{d} and 𝔓_{F}^{d+ε}/𝔓_{F}^{m}. The existence of a β such that

χ_{F} (1 + x) = ψ_{F} (\frac{βx}{γ})

for x in 𝔓_{F}^{d+ε} follows immediately from the relation

χ_{F} (1 + x) χ_{F} (1 + y) = χ_{F} (1 + x + y)

which is valid for x in 𝔓_{F}^{d+ε}. The number β must be a unit because χ_{F} (1 + x) is different from 1 for some x in 𝔓_{F}^{m−1}.

In case (a)

∫_{U_{F}} ψ_{F} (\frac{α}{γ}) χ_{F}^{−1}(α)dα

is equal to

∫_{U_{F} /U_{F}^{d}} ψ_{F} (\frac{α}{γ}) χ_{F}^{−1}(α) \{∫_{𝔓_{F}^{d}} ψ_{F} (\frac{(α − β)x}{γ}) dx\}dα.

The main integral is 1 or 0 according as α − β does or does not lie in 𝔓_{F}^{d}. Thus this expression is equal to

ψ_{F} (\frac{β}{γ} ) χ_{F}^{−1}(β) [U_{F} : U_{F}^{d}]^{−1}.

The first part of the lemma follows.

In case (b)

∫_{U_{F}} ψ_{F} (\frac{α}{γ}) χ_{F}^{−1}(α)dα

is equal to

∫_{U_{F} /U_{F}^{d+1}} ψ_{F} (\frac{α}{γ}) χ_{F}^{−1}(α) \{∫_{𝔓_{F}^{d+1}} ψ_{F} (\frac{(α − β)x}{γ}) dx\}dα.

The inner integral is 0 unless α − β lies in P_{F}^{d} when it is 1. Thus this expression is equal to

ψ_{F} (\frac{β}{γ} ) χ_{F}^{−1}(β) [U_{F} : U_{F}^{d}]^{−1} ∫_{O_{F} /𝔓_{F}} ψ_{F} (\frac{δβx}{γ}) χ_{F}^{−1}(1 + δx)dx.

The second part of the lemma follows.

The number β is only determined modulo 𝔓_{F}^{d}. When applying the lemma we shall, after choosing β, set

Δ_{2}(χ_{F} , ψ_{F}; γ) = ψ_{F} (\frac{β}{γ} ) χ_{F}^{−1}(β)

and then define Δ_{3}(χ_{F} , ψ_{F}; γ), which will be 1 when m is even, by the equation,

Δ_{1}(χ_{F} , ψ_{F}; γ) = Δ_{2}(χ_{F} , ψ_{F}; γ) Δ_{3}(χ_{F} , ψ_{F}; γ).

When we need to make the relation between β and χ_{F} explicit we write β as β(χ_{F}). To be of any use to us this lemma must be supplemented by some other observations.

If K is a finite Galois extension of F any quasi-character χ_{F} of C_{F} determines a onedimensional representation of W_{K/F} whose restriction to C_{K} is a quasi-character χ_{K/F} of C_{K}. The character χ_{K/F} may be defined directly by

χ_{K/F} (α) = χ_{F} (N_{K/F} α).

More generally, if E is any finite separable extension of F we define χ_{E/F} by

χ_{E/F} (α) = χ_{F} (N_{E/F} α).

To apply the lemma of Lamprecht we shall need to know, in some special cases, the relation between β(χ_{F}) and β(χ_{E/F}).

Suppose m is a positive integer and m = 2d + ε where ε is 0 or 1 and d is a positive integer. Let m′ = ψ_{E/F} (m − 1) + 1 and let m′ = 2d′ + ε′ where ε′ is 0 or 1 and d′ is a positive integer.* Since ψ_{E/F} is convex

ψ_{E/F} (\frac{m − 1}{2}) ≤ \frac{1}{2} ψ_{E/F} (m − 1) + \frac{1}{2} ψ_{E/F} (0) = \frac{1}{2} (m′ − 1) < d′ + ε′

and

d′ + ε′ = ψ_{E/F} (u)

with u > \frac{m−1}{2}. Since the least integer greater than \frac{m−1}{2} is d + ε, Lemma 6.6 implies that

N_{E/F} (U_{E}^{d′+ε′}) ≤ U_{F}^{u} ≤ U_{F}^{d+ε}.

In other words, if x ∈ 𝔓_{E}^{d′+ε′} then

N_{E/F} (1 + x) − 1 ∈ 𝔓_{F}^{d+ε}.

Lemma 6.6 also implies that

N_{E/F} (1 + x) − 1 ∈ 𝔓_{F}^{m}

if x ∈ 𝔓_{E}^{m′}. If x ∈ 𝔓_{E}^{d′+ε′} and y ∈ 𝔓_{E}^{m′} then

N_{E/F} (1 + x + y) − 1 = N_{E/F} (1 + x)N_{E/F} (1 + \frac{y}{1 + x}) − 1

is congruent to

N_{E/F} (1 + x) − 1

modulo 𝔓_{F}^{m}. Thus if x ∈ 𝔓_{E}^{d′+ε′} and y ∈ 𝔓_{E}^{d′+ε”} so that xy ∈ 𝔓_{E}^{m′}, then

N_{E/F} (1 + x + y) − 1 ≡ N_{E/F} (1 + x + y + xy) − 1 (mod 𝔓_{F}^{m}).

* We are here dealing not with an additive character, but with the function of Chapter 6!

The right side is

N_{E/F} (1 + x)N_{E/F} (1 + y) − 1,

which equals

{N_{E/F} (1 + x) − 1} + {N_{E/F} (1 + y) − 1} + {(N_{E/F} (1 + x) − 1) (N_{E/F} (1 + y) − 1)}

and this is congruent to

{N_{E/F} (1 + x) − 1} + {N_{E/F} (1 + y) − 1}

modulo 𝔓_{F}^{m}. Thus the map

P_{E/F} : x ⟶ N_{E/F} (1 + x) − 1

is a homomorphism from 𝔓_{E}^{d′+ε′}/𝔓_{E}^{m′} to 𝔓_{F}^{d+ε}/𝔓_{F}^{m}. If E ⊆ E′ we can replace F by E, E by E′, m by m′, and m′ by ψ_{E′/E}(m′ − 1) + 1, and define P_{E′/E}. Since ψ_{E′/F} = ψ_{E′/E} ∘ ψ_{E/F} and

N_{E′/F} (1 + x) − 1 = N_{E/F} (1 + (N_{E′/E}(1 + x) − 1)) − 1,

the relation

P_{E′/F} = P_{E/F} ∘ P_{E′/E}

is valid.

If n = n(ψ_{F}) and n′ = n(ψ_{E/F}), choose γ_{F} in C_{F} so that γ_{F} O_{F} = 𝔓_{F}^{m+n} and γ_{E} in C_{E} so that γ_{E}O_{E} = 𝔓_{E}^{m′+n′}. I apologize again for the unfortunate conflict of notation. ψ_{E/F} is on the one hand a function on {u ∈ ℝ | u ≥ −1} and on the other a character of E. However, warned one again, the reader should not be too inconvenienced by the conflict. Define

P_{E/F}^{∗} : O_{F} /𝔓_{F}^{d} ⟶ O_{E}/𝔓_{E}^{d′}

by the relation

ψ_{F} (\frac{x P_{E/F} (y)}{γ_{F}}) = ψ_{E/F} (\frac{P_{E/F}^{∗} (x)y}{γ_{E}} ) .

It will often be necessary to keep in mind the dependence of P_{E/F}^{∗} on γ_{F} and γ_{E}. Then we shall write

P_{E/F}^{∗} (x) = P_{E/F}^{∗} (x; γ_{E}, γ_{F}).

It is clear that

P_{E′/F}^{∗} (x; γ_{E′}, γ_{F}) = P_{E′/E}^{∗}(P_{E/F}^{∗} (x; γ_{E}, γ_{F}); γ_{E′}, γ_{E}).

Lemma 8.2

Let K/F be abelian and let G = 𝔊(K/F). Suppose there is an integer t such that G = G_{t} while G_{t+1} = {1}. Suppose m ≥ t + 1 and m > 1 and γ_{F} is chosen. If μ_{F} belongs to S(K/F), the set of characters of C_{F} /N_{K/F} C_{K}, then m ≥ m(μ_{F}) so that for some α(μ_{F}) in O_{F}

μ_{F} (1 + x) = ψ_{F} (\frac{α(μ_{F})x}{γ_{F}})

for all x in 𝔓_{F}^{d+ε}. The element γ_{K} may be taken equal to γ_{F} and if P_{K/F}^{∗} (β) = P_{K/F}^{∗} (β; γ_{F} , γ_{F}) then

N_{K/F} (P_{K/F}^{∗} (β)) ≡ ∏_{μ_{F}} (β + α(μ_{F})) (mod 𝔓_{F}^{d})

for all β in O_{F}.

If t = −1 then n(ψ_{F}) = n(ψ_{K/F}) and m′ = m so that γ_{K} may be taken equal to γ_{F}. If t ≥ 0 the extension is ramified. Let 𝔓_{K}^{δ_{K/F}} be the different of K/F. Then

n(ψ_{K/F}) = [K : F]n(ψ_{F}) + δ_{K/F} .

By definition

ψ_{K/F} (m − 1) = t + [K : F](m − t − 1).

By Proposition 4 of paragraph IV.2 of Serre’s book δ_{K/F} = ([K : F] − 1)(t + 1). Thus

m^{∗} + n(ψ_{K/F}) = [K : F] (m + n(ψ_{F}))

and we can again take γ_{K} = γ_{F}.

Since m(μ_{F}) = t + 1 we have m ≥ m(μ_{F}) and
μ_{F} (1 + x + y) = μ_{F} (1 + x) μ_{F} (1 + y)

for x and y in 𝔓_{F}^{d+ε}. Thus the existence of α(μ_{F}) is assured. The last assertion of the lemma will be proved by induction. We will need to know that if x ≡ y(mod 𝔓_{K}^{d′}) then

N_{K/F} x ≡ N_{K/F} y(mod 𝔓_{F}^{d}).

When proving this we may suppose that xO_{K} = 𝔓_{K}^{r} with r ≤ d′ and that \frac{y}{x} belongs to O_{K}. Then

N_{K/F} x − N_{K/F} y = N_{K/F} x\{1 − N_{K/F} (1 + \frac{y − x}{x})\}.

If r ≥ d there is nothing to prove. Suppose r ≤ d. If d′ − r = ψ_{K/F} (u) and s is the smallest integer greater than or equal to u the right side belongs to 𝔓_{F}^{s+r}. Since the derivative of ψ_{K/F} is at least one everywhere ψ_{K/F} (u + r) ≥ d′. But

d′ ≥ \frac{m′ − 1}{2} = \frac{1}{2} ψ_{K/F} (m − 1) + \frac{1}{2} ψ_{K/F} (0) ≥ ψ_{K/F} (\frac{m − 1}{2}) .

Thus u + r ≥ \frac{m−1}{2} and s + r ≥ d.

Suppose F ⊆ L ⊆ K and L/F is cyclic of prime order. Let H = 𝔊(K/L) and let \overline{G} = 𝔊(L/F). Certainly H = H_{t} while H_{t+1} = {1}. Since ψ_{K/F} (t) = ψ_{K/L}(t) = t, we have ψ_{L/F} (t) = t and, by Herbrand’s theorem,

\overline{G}_{t} = \overline{G}^{t} = HG^{t}/H = G/H = \overline{G}.

Moreover t + 1 = ψ_{L/F} (t + δ) with δ > 0 so that

\overline{G}_{t+1} = \overline{G}^{t+δ} = HG^{t+δ}/H = H/H = \{1\}.

Finally, ψ_{L/F} (m − 1) + 1 ≥ t + 1 so that L/F and K/L, with m replaced by ψ_{L/F} (m − 1) + 1, satisfy the conditions of the lemma. S(L/F) is a subgroup of S(K/F). If μ_{F} and ν_{F} belong to S(K/F) then μ_{L/F} = ν_{L/F} if and only if μ_{F} and ν_{F} belong to the same coset of S(L/F). Take S to be a set of representatives for these cosets; then

S(K/L) = {μ_{L/F} | μ_{F} | μ_{F} ∈ S}.

We take α(μ_{F} ν_{F}) = α(μ_{F}) + α(ν_{F}) if μ_{F} belongs to S and ν_{F} belongs to S(L/F). If μ_{F} belongs to S we take α(μ_{L/F}) to be P_{L/F}^{∗} (α(μ_{F})). If the lemma is valid for K/L and L/F then

N_{K/F} (P_{K/F}^{∗} (β)) = N_{L/F} (N_{K/L}(P_{K/L}^{∗}(P_{L/F}^{∗} (β)))

which is congruent modulo P_{F}^{d} to

N_{L/F} (∏_{μ_{F} ∈S}(P_{L/F}^{∗} (β) + P_{L/F}^{∗} (α(μ_{F}))))

or

∏_{μ_{F} ∈S}{N_{L/F} (P_{L/F}^{∗} (β + α(μ_{F})))}.

This is congruent modulo P_{F}^{d} to

∏_{μ_{F} ∈S} ∏_{ν_{F} ∈S(L/F)}{β + α(μ_{F}) + α(ν_{F})}

which equals

∏_{μ_{F} ∈S(K/F)}{β + α(μ_{F})}.

Thus it is enough to prove the lemma when K/F is cyclic of prime order. In this case more precise information is needed and the assertion of the lemma will follow immediately from it.

Lemma 8.3

If K/F is unramified and m ≥ 1 we may take P_{K/F}^{∗} (β) = β.

According to paragraph V.2 of Serre’s book

N_{K/F} (1 + y) − 1 ≡ S_{K/F} (y) (mod P_{F}^{m})

if y ∈ 𝔓_{K}^{d′+ε′}. Thus P_{K/F} (y) = S_{K/F} (y) and

ψ_{F} (\frac{x P_{K/F} y}{γ_{F}} ) = ψ_{K/F} (\frac{xy}{γ_{F}} ) .

Lemma 8.4

Suppose K/F is abelian, totally ramified, and [K : F] = ℓ is an odd prime. If d ≥ t + 1 we may take P_{K/L}^{∗}(β) = β.

The relation

m′ = t + 1 + ℓ(m − 1 − t) = ℓm − (t + 1) (ℓ − 1)

implies that m′ ≡ m (mod 2), ε′ = ε, and

d′ = ℓd + \frac{ℓ − 1}{2}(ε − t − 1) = d + \frac{ℓ − 1}{2}(m − t − 1).

Since

\frac{ℓ − 1}{2}(m − t − 1) ≥ m − t − 1 ≥ d + ε

we have

d′ + ε′ ≥ 2(d + ε) ≥ m.

Moreover

\frac{2(d′ + ε′) + δ_{K/F}}{ℓ} ≥ \frac{m′ + δ_{K/F}}{ℓ} = m

so that by Lemma 5 of paragraph V.3 of Serre’s book

N_{K/F} (1 + x) − 1 ≡ S_{K/F} (x) (mod 𝔓_{F}^{m})

if x ∈ 𝔓_{K}^{d′+ε′}. The lemma follows.

Let p be the characteristic of O_{F} /𝔓_{F}.

Lemma 8.5

Suppose K/F is abelian, totally ramified, and [K : F] = ℓ is an odd prime. Suppose t + 1 ≤ m ≤ 2t + 1. Choose a non-trivial character μ_{F} in S(K/F). We may choose α = α(μ_{F}) so that αO_{F} = 𝔓_{F}^{v}, if m = t + 1 + v, so that α = N_{K/F} α_{1} for some α_{1} in O_{K}, and so that

μ_{F} (1 + x) = ψ_{F} (\frac{αx}{γ_{F}})

for x in 𝔓_{F}^{s}. Here s is the least integer greater than or equal to \frac{t}{2}. If ζ is a (p−1)th root of unity in F there is a unique integer j with 1 ≤ j ≤ p − 1 such that ζ − j lies in 𝔓_{F}. Set μ_{F}^{ζ} = μ_{F}^{j}. We may take α(μ_{F}^{ζ}) to be ζα. If β belongs to O_{F} we can find a β_{1} in O_{K} such that β ≡ N_{K/F} β_{1}(mod 𝔓_{F}^{d}). Then

P_{K/F}^{∗} (β) ≡ β − β_{1} \frac{α}{α_{1}} (mod 𝔓_{K}^{d′}).

If

μ_{F} (1 + x) = ψ_{F} (\frac{αx}{γ_{F}})

for x in 𝔓_{F}^{s} then, necessarily, αO_{F} = 𝔓_{F}^{v}. Choose δ_{1} in O_{K} such that δ_{1}O_{K} = 𝔓_{K}^{v} and set δ = N_{K/F} δ_{1}. Set α = ωδ where ω is yet to be chosen. We must have

μ_{F} (1 + x) = ψ_{F} (\frac{ωδx}{γ_{F}})

if x ∈ 𝔓_{F}^{s}. This equation determines the unit w modulo 𝔓_{F}^{r} if r = t − s. Since any unit is a norm modulo 𝔓_{F}^{t} we may suppose ω = N_{K/F} ω_{1}. Take α_{1} = ω_{2}δ_{1}. β_{1} exists for a similar reason.

The number ζ −j must lie in pO_{F}. But K/F is wildly ramified, because 2t+1 ≥ m > 1, ℓ = p and p = S_{K/F} (1) so that, by paragraph V.3 of Serre’s book, p belongs to 𝔓_{F}^{u} if u is the greatest integer in

\frac{(ℓ − 1)}{ℓ} (t + 1) ≥ \frac{t + 1}{2}.

However d + ε ≥ s so that d + ε + u ≥ t + 1 and, if x belongs to 𝔓_{F}^{d+ε}, (ζ − j)x lies in 𝔓_{F}^{t+1}. Thus

ψ_{F} (\frac{α(ζ − j)x}{γ_{F}}) = μ_{F} (1 + (ζ − j)x) = 1

and

μ_{F}^{j} (1 + x) = μ_{F}^{j} (1 + x) = ψ_{F} (\frac{jαx}{γ_{F}}) = ψ_{F} (\frac{ζαx}{γ_{F}}) .

Since

\frac{2s + δ_{K/F}}{ℓ} ≥ \frac{t + 1 + δ_{K/F}}{ℓ} = t + 1.

The lemmas of paragraph V.3 of Serre’s book imply that

N_{K/F} (1 + x) ≡ 1 + S_{K/F} (x) + N_{K/F} (x) (mod 𝔓_{F}^{t+1})

if x belongs to 𝔓_{K}^{s} and then

1 = μ_{F} (N_{K/F} (1 + x)) = μ_{F} (1 + S_{K/F} (x) + N_{K/F} (x)).

As we observed d + ε ≥ s. Moreover d + ε ≤ t + 1 so that

\frac{d + ε + δ_{K/F}}{ℓ} ≥ d + ε

and S_{K/F} (x) and N_{K/F} (x) belong to 𝔓_{F}^{d+ε} if x belongs to 𝔓_{K}^{d+ε}. Thus, for such x,

ψ_{F} (\frac{αN_{K/F} (x)}{γ_{F}}) = ψ_{F} (\frac{−αS_{K/F} (x)}{γ_{F}}) .

Again

\frac{2(d′ + ε′) + δ_{K/F}}{ℓ} ≥ m

so that

N_{K/F} (1 + x) − 1 = S_{K/F} (x) + N_{K/F} (x) (mod 𝔓_{F}^{m})

if x ∈ 𝔓_{K}^{d′+ε′}. Moreover

d′ + ε′ = d + ε + \frac{ℓ − 1}{2} (m − t − 1) ≥ d + ε

so that N_{K/F} (x) and hence S_{K/F} (x) belong to 𝔓_{F}^{d+ε}. Thus

βN_{K/F} x ≡ αN_{K/F} (\frac{β_{1}x}{α_{1}}) (mod 𝔓_{F}^{m}).

But β_{1}x/α_{1} belongs to 𝔓_{K}^{d′+ε′−v} and

d′ + ε′ − v = d + ε + \frac{ℓ − 1}{2} v − v ≥ d + ε

so that

ψ_{F} (\frac{β P_{K/F} (x)}{γ_{F}}) = ψ_{F} (\frac{β S_{K/F} (x) + β N_{K/F} (x)}{γ_{F}})

which equals

ψ_{F} (\frac{β S_{K/F} (x) − αS_{K/F} (\frac{β_{1}x}{α_{1}})}{γ_{F}} ) = ψ_{K/F} ((β − \frac{αβ_{1}}{α_{1}} ) \frac{x}{γ_{F}} )

as required.

Lemma 8.6

Suppose K/F is a wildly ramified quadratic extension, m ≥ t + 1, and m = t + 1 + v. Let μ_{F} be the non-trivial character in S(K/F). If β belongs to O_{F} there is a β_{1} in O_{K} and a δ in U_{F}^{t} such that β ≡ δN_{K/F} β_{1}(mod 𝔓_{F}^{d}). We can choose α = α(μ_{F}) so that

μ_{F} (1 + x) = ψ_{F} (\frac{αδx}{γ_{F}})

if x is in 𝔓_{F}^{s} and so that α = N_{K/F} α_{1} for some α_{1} in O_{K}. Here s has the same meaning as before. Thus, if r is the integral part of \frac{t+1}{2}, t + 1 = r + s. With these choices

P_{K/F}^{∗} (β) ≡ β − \frac{β_{1}αδ}{α_{1}} (mod P_{K}^{d′}) .

If β = 0 the existence of δ and β_{1} is clear. Otherwise we can find a β_{1} such that N_{K/F} β_{1}/β is in U_{F}^{t}. We choose δ accordingly. If m = t + 1 + v and

μ_{F} (1 + x) = ψ_{F} (\frac{αδx}{γ_{F}})

for x in 𝔓_{F}^{s} then O_{F} α = 𝔓_{F}^{v}. Choose η_{1} in O_{K} so that O_{K}η_{1} = 𝔓_{K}^{v} and set η = N_{K/F} η_{1}. Set α = ωη where ω is yet to be chosen. We must have

μ_{F} (1 + x) = ψ_{F} (\frac{ωηδx}{γ_{F}})

if x ∈ 𝔓_{F}^{s}. This equation determines the unit ω modulo 𝔓_{F}^{r}. Since any unit is a norm modulo 𝔓_{F}^{t} we may suppose ω = N_{K/F} ω_{1}. Take α_{1} = ω_{1}η_{1}.

Since the extension is quadratic

N_{K/F} (1 + x) = 1 + S_{K/F} (x) + N_{K/F} (x).

Since

\frac{s + δ_{K/F}}{2} = \frac{s + t + 1}{2} ≥ s

both S_{K/F} (x) and N_{K/F} (x) are in 𝔓_{F}^{s} if x belongs to 𝔓_{K}^{s} and

ψ_{F} (\frac{αδN_{K/F} (x)}{γ_{F}}) = ψ_{F} (\frac{−αδS_{K/F} (x)}{γ_{F}}) .

We have m′ = 2m − (t + 1) and d′ = m − s, so that d′ + ε′ = m − r and d′ + ε′ − v = s. Thus if x belongs to 𝔓_{K}^{d′+ε′}

βN_{K/F} (x) ≡ αδN_{K/F} (\frac{β_{1}x}{α_{1}}) (mod 𝔓_{F}^{m})

and β_{1}x/α_{1} lies in 𝔓_{K}^{s}. Consequently

ψ_{F} (\frac{β P_{K/F} x}{γ_{F}}) = ψ_{K/F} ((β − β_{1} \frac{αδ}{α_{1}} ) \frac{x}{γ_{F}} )

as required.

Lemma 8.7

If K/F is a tamely ramified quadratic extension and m ≥ 2 we may take P_{K/F}^{∗} (β) = β.

Notice that t+1 = 1 so that m ≥ t+1. In this case m′ = 2m−1, d′ = m−1, and d′ +ε′ = m. If x ∈ 𝔓_{K}^{d′+ε′}

N_{K/F} (1 + x) = 1 + S_{K/F} (x) + N_{K/F} (x)

is congruent to

1 + S_{K/F} (x)

modulo 𝔓_{F}^{m}. The lemma follows.

To complete the proof of Lemma 8.2 we have to show that if K/F is cyclic of prime order

N_{K/F} (P_{K/F}^{∗} (β)) ≡ ∏_{μ_{F} ∈S(K/F)}(β + α(μ_{F})) (mod 𝔓_{F}^{d}).

We consider the cases discussed in the previous lemmas one by one. If the extension is unramified we may take all the numbers α(μ_{F}) to be 0. The congruences then reduce to the identity β^{n} = β^{n}. The same is true if K/F is cyclic of odd order and d ≥ t + 1 or K/F is quadratic and t = 0. If K/F is cyclic of odd order ℓ and t + 1 ≤ m ≤ 2t + 1 the right side becomes

β^{ℓ} − βα^{ℓ−1}.

If β ≡ 0 (mod 𝔓_{F}^{d}) both sides are congruent to 0 modulo 𝔓_{F}^{d}. Suppose β does not belong to 𝔓_{F}^{d} and βO_{F} = 𝔓_{F}^{u}. Then β_{1}O_{K} = 𝔓_{K}^{u} and

N_{K/F} (β − β_{1} \frac{α}{α_{2}} )

is congruent to

β^{ℓ} − βα^{ℓ−1} + ∑_{i=1}^{ℓ−1}(−1)^{i}β^{ℓ}E^{i} (\frac{β_{1}α}{βα_{1}} )

modulo 𝔓_{F}^{d}. If x ∈ K then E^{i}(x) is the ith elementary symmetric function of x and its conjugates. Moreover β_{1}α/βα_{1} belongs to 𝔓_{K}^{(ℓ−1) (v−u)}. If ℓ − 1 ≥ i ≥ 1

\frac{i(ℓ − 1) (v − u) + (ℓ − 1) (t + 1)}{ℓ} ≥ \frac{(ℓ − 1) (v + t + 1)}{ℓ} − ℓu.

The right side is

\frac{(ℓ − 1)}{ℓ}m − pu ≥ d − ℓu.

The argument of paragraph V.3 of Serre’s book shows that

N_{K/F} (β − β_{1} \frac{α}{α_{1}} ) ≡ β^{ℓ} − βα^{ℓ−1} (mod 𝔓_{F}^{d}).

For a wildly ramified quadratic extension we use the notation of Lemma 8.6. The right side of the congruence may be taken to be β^{2} + βαδ. The identity is again non-trivial only if βO_{F} = 𝔓_{F}^{u} with u < d. Then the left side may be taken to be

β^{2} − β^{2}δS_{K/F} (\frac{β_{1}α}{βα_{1}} ) + δ^{2}αN_{K/F} β_{1}

which is congruent to

β^{2} + αβδ − β^{2}δS_{K/F} (\frac{β_{1}α}{βα_{1}} )

modulo P_{F}^{d}. Since

\frac{v − u + t + 1}{2} ≥ \frac{m}{2} − u ≥ d − u

we have

β^{2}S_{K/F} (\frac{β_{1}α}{βα_{1}} ) ≡ 0 (mod 𝔓_{F}^{d}).

Suppose χ_{F} is a quasi-character of C_{F} , m = m(χ_{F}), and β = β(χ_{F}). If, as sometimes happens, m′ = m(χ_{K/F}) we can take β(χ_{K/F}) = P_{K/F}^{∗} (β).

Lemma 8.8

Suppose K/F is Galois and G = 𝔊(K/F). Suppose s ≥ 0 is an integer and G^{s} = {1}. If m = m(χ_{F}) and m > s then

m′ = ψ_{K/F} (m − 1) + 1 = m(ψ_{K/F}).

It follows from paragraph V.6 of Serre’s book that

N_{K/F} (U_{K}^{ψ_{K/F} (v)}) = U_{F}^{v}

if v ≥ s. Thus χ_{K/F} is trivial on U_{K}^{u} if u > ψ_{K/F} (m−1) but is not trivial on U_{K}^{i} if u = ψ_{K/F} (m−1).

We can now collect together, with one or two additional comments, the previous results in a form which will be useful in the proof of the first main lemma. We use the same notation.

Lemma 8.9

Suppose K/F is a cyclic extension of prime order ℓ, χ_{F} is a quasi-character of C_{F} , m(χ_{F}) ≥ t + 1, m(χ_{F}) > 1, and m(χ_{K/F}) − 1 = ψ_{K/F} (m(χ_{F}) − 1).

(a) If K/F is unramified we may take β(χ_{K/F}) = β(χ_{F}) and β(μ_{F} χ_{F}) = β(χ_{F}) for all μ_{F} in S(K/F).

(b) If ℓ is odd and d ≥ t + 1 we may take β(χ_{K/F}) = β(χ_{F}) and β(μ_{F} χ_{F}) = β(χ_{F}) for all μ_{F} in S(K/F).

(c) If ℓ is odd and t + 1 ≤ m ≤ 2t + 1 and μ_{F} is a given non-trivial character in S(K/F) we may choose α = α(μ_{F}) = N_{K/F} α_{1} as in Lemma 8.5 and β = β(χ_{F}) = N_{K/F} β_{1} for some β_{1} in U_{K}. Then we may choose

β(χ_{K/F}) = β − β_{1} \frac{α}{α_{1}}

and

β(μ_{F}^{ζ} χ_{F}) = N_{K/F} (β_{1} + ζα_{1}).

(d) If ℓ is 2 and K/F is wildly ramified we choose α = α(μ_{F}) as in Lemma 8.6. We may choose β = β(χ_{F}) in the form δN_{K/F} β_{1} with δ in U_{F}^{t}. Then we may choose

β(χ_{K/F}) = β − β_{1} \frac{αδ}{α_{1}}

and

β(μ_{F} χ_{F}) = β + αδ.

(e) If ℓ is 2 and K/F is tamely ramified we may take β(χ_{K/F}) = β(μ_{F} χ_{F}) = β(χ_{F}).

Only part (c) requires any further verification. It must be shown that

N_{K/F} (β_{1} + ζα_{1}) ≡ β + ζα (mod 𝔓_{F}^{d}).

The left side is congruent to

βN_{K/F} (1 + \frac{ζα_{1}}{β_{1}} ) .

All we need do is show that

N_{K/F} (1 + \frac{ζα_{1}}{β_{1}} ) ≡ 1 + \frac{ζα}{β} (mod 𝔓_{F}^{d}).

The right side is

1 + N_{K/F} (\frac{ζα_{1}}{β_{1}} ) .

According to paragraph V.3 of Serre’s book the congruence will be satisfied if

\frac{v + (ℓ − 1) (t + 1)}{ℓ} ≥ d.

But t + 1 = d + x with x ≥ 0 so that d + x + v = 2d + ε and v = d + ε − x. Thus

\frac{v + (ℓ − 1) (t + 1)}{ℓ} = \frac{d + ε − x + (ℓ − 1) (d + x)}{ℓ} = d + \frac{ε + (ℓ − 2)x}{ℓ} ≥ d.

The preceding discussion has now to be repeated with different hypotheses and different, but similar, conclusions.

Lemma 8.10

Suppose K/F is abelian and G = 𝔊(K/F). Suppose there is a t ≥ 0 such that G = G_{t} while G_{t+1} = {1}. If 2 ≤ m ≤ t + 1 then m′ = ψ_{K/F} (m − 1) + 1 is just m. Let t + 1 = m + v, let δ be such that δO_{F} = 𝔓_{F}^{t+1+n(ψ_{F})}, let ε_{1} in O_{K} be such that ε_{1}O_{K} = 𝔓_{K}^{v}, and let ε = N_{K/F} ε_{1}. We may choose γ_{F} = δ/ε and γ_{K} = δ/ε_{1}. Let r be the greatest integer in \frac{t+1}{2} and let s = t + 1 − r. If μ_{F} is a non-trivial character in S(K/F) then m(μ_{F}) = t + 1. Let

μ_{F} (1 + x) = ψ_{F} (\frac{β(μ_{F})x}{δ})

for x in 𝔓_{F}^{s}. Then

β ∏_{μ_{F} ≠1}(βε + β(μ_{F})) ≡ N_{K/F} (P_{K/F}^{∗} (β)) (mod 𝔓_{F}^{d}).

The relation m′ = ψ_{K/F} (m − 1) + 1 = m is an immediate consequence of the definitions. Since the extension is totally ramified

n(ψ_{K/F}) = [K : F]n + ([K : F] − 1)(t + 1)

if n = n(ψ_{F}). Thus

m + n = (t + 1 + n) − v

and

m′ + n(ψ_{K/F}) = [K : F](t + 1 + n) + (m − t − 1) = [K : F](t + 1 + n) − v.

Consequently γ_{F} and γ_{K} can be chosen as asserted. The results of chapter V of Serre’s book imply that m(μ_{F}) = t + 1 if μ_{F} is not trivial.

We saw when proving Lemma 8.2 that if x ≡ y(mod 𝔓_{K}^{d′}) then N_{K/F} x ≡ N_{K/F} y(mod 𝔓_{F}^{d}) and that if F ⊆ L ⊆ K both L/F and K/L satisfy the conditions of the lemma. For L/F, ε_{1} is replaced by N_{K/L}ε_{1} and, for K/L, ε is replaced by N_{K/L}ε_{1}. Take Q_{L/F}^{∗} to be P_{L/F}^{∗} in the special case that m = t + 1 and ε_{1} = 1. Then

ψ_{L/F} (\frac{N_{K/L}(ε_{1})x P_{L/F}^{∗} (β)}{δ}) = ψ_{F} (\frac{εP_{L/F} (x)β}{δ} )

by definition. The right side is equal to

ψ_{L/F} (\frac{x Q_{L/F}^{∗} (εβ)}{δ}) .

Thus

Q_{L/F}^{∗} (εβ) ≡ N_{K/L}(ε_{1})P_{L/F}^{∗} (β) (mod 𝔓_{L}^{s}).

If μ_{F} belongs to S(K/F) but not to S(L/F) then m(μ_{L/F}) = m(μ_{F}) and β(μ_{L/F}) may be taken to be Q_{L/F}^{∗} (β(μ_{F})). Let S′ be a set of representatives for the cosets of S(L/F) in S(K/F) − S(L/F) and suppose the lemma is true for K/L and L/F. Then

N_{K/F} (P_{K/F}^{∗} (β)) = N_{L/F} (N_{K/L}(P_{K/L}^{∗}(P_{L/F}^{∗} (β))))

is congruent to

N_{L/F} (P_{L/F}^{∗} (β) ∏_{μ_{F} ∈S′}{N_{K/L}(ε_{1})P_{L/F}^{∗} (β) + Q_{L/F}^{∗} (β(μ_{F}))})

modulo 𝔓_{F}^{d}. This in turn is congruent to

N_{L/F} (P_{L/F}^{∗} (β)) ∏_{μ_{F} ∈S′} N_{L/F} (Q_{L/F}^{∗} (εβ + β(μ_{F}))).

Applying the induction hypothesis to the first part and Lemma 8.2 to the second, we see that the whole expression is congruent to

β\{∏_{{ν_{F} ∈S(L/F) \atop ν_{F} ≠1}} (εβ + β(ν_{F}))\} \{∏_{μ_{F} ∈S′_{ν_{F} ∈S(L/F)}} (εβ + β(μ_{F}) + β(ν_{F}))\}

modulo 𝔓_{F}^{d} as required.

Once again we devote a lemma to cyclic extensions of prime order.

Lemma 8.11

Suppose K/F is cyclic of prime order ℓ and 2 ≤ m ≤ t + 1. Choose a non-trivial character μ_{F} in S(K/F). There is an α_{1} in U_{K} such that if α = N_{K/F} α_{1}

μ_{F} (1 + x) = ψ_{F} (\frac{αx}{δ})

for x in P_{F}^{s}. If β belongs to O_{F} there is a β_{1} in O_{K} such that β ≡ N_{K/F} (β_{1}) (mod 𝔓_{F}^{t}). Then

P_{K/F}^{∗} (β) ≡ β \frac{ε}{ε_{1}} − β_{1} \frac{α}{α_{1}} (mod 𝔓_{K}^{d′}).

Since β(μ_{F}) is determined only modulo 𝔓_{F}^{s} and s ≤ t we can take β(μ_{F}) = N_{K/F} α_{1} for some α_{1} in U_{K}. The existence of β_{1} also follows as before. Since t + 1 ≥ m

\frac{2(d′ + ε′) + (ℓ − 1) (t + 1)}{ℓ} ≥ \frac{m + (ℓ − 1) (t + 1)}{ℓ} ≥ m

and

N_{K/F} (1 + x) ≡ 1 + S_{K/F} (x) + N_{K/F} (x) (mod 𝔓_{F}^{m})

if x belongs to 𝔓_{K}^{d′+ε′}. Thus

ψ_{F} (\frac{εP_{K/F} (x)β}{δ} ) = ψ_{K/F} (\frac{εxβ}{δ} ) ψ_{F} (\frac{N_{K/F} (ε_{1}x)β}{δ} ) .

But d′ + ε′ + t ≥ t + 1 so that

N_{K/F} (ε_{1}x)β ≡ αN_{K/F} (\frac{ε_{1}β_{1}}{α_{1}} · x) (mod 𝔓_{F}^{t+1})

Since t + 1 = m + v, d′ + ε′ + v ≥ s and if y = \frac{ε_{1}β_{1}}{α_{1}} · x then y which lies in 𝔓_{K}^{d′+ε′+v} also lies in 𝔓_{K}^{s}. But

\frac{2s + (t + 1)(ℓ − 1)}{ℓ} ≥ t + 1

so that

N_{K/F} (1 + y) ≡ 1 + S_{K/F} (y) + N_{K/F} (y) (mod 𝔓_{F}^{t+1}).

Consequently

ψ_{F} (\frac{−αN_{K/F} (y)}{δ}) = ψ_{F} (\frac{αS_{K/F} (y)}{δ}) .

In conclusion

ψ_{F} (\frac{ε P_{K/F} (x)β}{δ} ) = ψ_{K/F} (\frac{ε_{1}}{δ} (\frac{ε}{ε_{1}} · β − \frac{α}{α_{1}} · β_{1}) x)

as required.

To complete the proof of Lemma 8.10 we have to show that when K/F is cyclic of prime order, ℓ

β ∏_{μ_{F} ≠1}(βε + β(μ_{F})) ≡ N_{K/F} (P_{K/F}^{∗} (β)) (mod 𝔓_{F}^{d})

Since 2 ≤ m ≤ t + 1 the extension is wildly ramified, ℓ = p, and once the character μ_{F} is chosen as in the previous lemma we can define μ_{F}^{ζ} as in Lemma 8.5. The left side is congruent to

β (β^{ℓ−1}ε^{ℓ−1} + (−1)^{ℓ}α^{ℓ−1}) .

If β ∈ 𝔓_{F}^{d} this is congruent to 0 and so is the right side. Suppose βO_{F} = 𝔓_{F}^{u} with u < d. The right side is congruent to

N_{K/F} (β \frac{ε}{ε_{1}} − β_{1} \frac{α}{α_{1}} ) ≡ βα^{ℓ−1}N_{K/F} (\frac{β}{β_{1}} \frac{α_{1}}{α} \frac{ε}{ε_{1}} − 1) .

Since

\frac{(ℓ − 1)(u + v) + (ℓ − 1)(t + 1)}{ℓ} ≥ \frac{ℓ − 1}{ℓ} · (t + 1) ≥ \frac{t + 1}{2} ≥ d

this is congruent to

βα^{ℓ−1}\{N_{K/F} (\frac{β}{β_{1}} \frac{α_{1}}{α} \frac{ε}{ε_{1}} ) + (−1)^{ℓ}\}.  (8.1)

Since

βN_{K/F} (β_{1}^{−1}) ≡ 1 (mod 𝔓_{F}^{t−u}).

We see that

β^{ℓ+1}N_{K/F} β_{1}^{−1} ≡ β^{ℓ}(mod 𝔓_{F}^{t})

and that the expression (8.1) is congruent to

β^{ℓ}ε^{ℓ−1} + (−1)^{ℓ}βα^{ℓ−1}

modulo 𝔓_{F}^{d}.

Lemma 8.12

Suppose K/F is abelian and G = 𝔊(K/F). Suppose there is an integer t such that G = G_{t} while G_{t+1} = {1}. Let χ_{F} be a quasi-character of C_{F} and suppose 2 ≤ m(χ_{F}) ≤ t + 1. If m(χ_{F}) < t + 1 then m(χ_{K/F}) = m(χ_{F}). If m(χ_{F}) = t + 1 then m(μ_{F} χ_{F}) < t + 1 for some μ_{F} in S(K/F) if and only if m(χ_{K/F}) < m(χ_{F}).

It follows immediately from Lemma 6.7 that if χ_{F} is any quasi-character of C_{F} and E any finite separable extension of F then

m(χ_{E/F}) − 1 ≤ ψ_{E/F} (m(χ_{F}) − 1).

In the particular case under consideration Lemma 6.10 shows that if m = m(χ_{F}) ≤ t then

N_{K/F} : U_{K}^{m−1}/U_{K}^{t} ⟶ U_{F}^{m−1}/U_{F}^{t}

is an isomorphism. Thus χ_{K/F} (α) will be different from 1 for some α in U_{K}^{m−1} and m(χ_{K/F}) will be at least m. If m(χ_{F}) = t + 1 then m(μ_{F} χ_{F}) is less than t + 1 for some μ_{F} in S(K/F) if and only if χ_{F} is trivial on the image of U_{K}^{t}/U_{K}^{t+1} in U_{F}^{t} /U_{F}^{t+1}. This is so if and only if m(χ_{K/F}) ≤ t.

We shall need the following lemma in the proof of the first main lemma.

Lemma 8.13

Suppose K/F is cyclic of prime order, χ_{F} is a quasi-character of C_{F} with m(χ_{F}) ≤ t + 1, and m(χ_{K/F}) = m(χ_{F}). Choose α, α_{1}, ε, ε_{1} in Lemma 8.11. We may choose β = β(χ_{F}) = N_{K/F} β_{1} with β_{1} in U_{K} and we may choose

β(χ_{K/F}) = β \frac{ε}{ε_{1}} − β_{1} \frac{α}{α_{1}} .

Moreover m(μ_{F}^{ζ} χ_{F}) = t + 1 and we may take

β(μ_{F}^{ζ} χ_{F}) = N_{K/F} (ζα_{1} + ε_{1}β_{1}).

Since β(χ_{F}) is determined only modulo 𝔓_{F}^{d} and d ≤ t the existence of β_{1} is clear. It is also clear that m(μ_{F}^{ζ} χ_{F}) = t + 1. The elements β(χ_{F}), β(χ_{K/F}), and β(μ_{F}^{ζ} χ_{F}) are to satisfy the following conditions:

Chapter 8 64 (i) If x is in 𝔓_{F}^{d}

χ_{F} (1 + x) = ψ_{F} (\frac{εβ(χ_{F})x}{δ}) .

(ii) If x is in 𝔓_{K}^{d′}

χ_{K/F} (1 + x) = ψ_{K/F} (\frac{ε_{1}β(χ_{K/F})x}{δ}) .

(iii) If x is in 𝔓_{F}^{s}

μ_{F}^{ζ} (1 + x)χ_{F} (1 + x) = ψ_{F} (\frac{β(μ_{F}^{ζ} χ_{F})x}{δ}) .

We have already shown that β(χ_{K/F}) may be taken to be

β \frac{ε}{ε_{1}} − β_{1} \frac{α}{α_{1}}

β(μ_{F}^{ζ} χ_{F}) must be congruent to ζα + εβ modulo 𝔓_{F}^{r}

N_{K/F} (ζα_{1} + ε_{1}β_{1}) = ζαN_{K/F} (1 + \frac{ε_{1}β_{1}}{ζα_{1}} ) .

Since

\frac{ν + (ℓ − 1)(t + 1)}{ℓ} ≥ \frac{ℓ − 1}{ℓ}(t + 1) ≥ r.

The right side is congruent to

ζα\{1 + N_{K/F} (\frac{ε_{1}β_{1}}{ζα_{1}} )\} = ζα + εβ

modulo 𝔓_{F}^{r}.^{∗}

∗(1998) The manuscript of Chapter 8 ends here.

Chapter Nine.

A Lemma of Hasse

Let λ ⊆ κ be two finite fields and let G = 𝔊(κ/λ). If x ∈ κ set

ω_{κ/λ}(x) = ∑ x^{σ_{1}}x^{σ_{2}}

where the sum is taken over all unordered pairs of distinct elements of G. It is clear that

ω_{κ/λ}(x + y) = ω_{κ/λ}(x) + ω_{κ/λ}(y) + S_{κ/λ}(x)S_{κ/λ}(y) − S_{κ/λ}(xy).

One readily verifies also that if λ ≤ η ≤ κ then

ω_{κ/λ}(x) = ω_{η/λ}(S_{κ/η}(x)) + S_{η/λ} (ω_{κ/η}(x)).

Suppose ψ_{λ} is a non-trivial character of λ and φ_{λ} is a nowhere vanishing function on λ satisfying the identity

φ_{λ}(x + y) = φ_{λ}(x)φ_{λ}(y) ψ_{λ}(xy).

Define φ_{κ/λ} on κ by

φ_{κ/λ}(x) = φ_{λ}(S_{κ/λ}(x)) ψ_{λ}(−ω_{κ/λ}(x)).

Then φ_{κ/λ}(x + y) is equal to

φ_{λ}(S_{κ/λ}(x + y)) ψ_{λ} (−ω_{κ/λ}(x) − ω_{κ/λ}(y) − S_{κ/λ}(x)S_{κ/λ}(y) + S_{κ/λ}(xy))

which is

φ_{κ/λ}(x)φ_{κ/λ}(y)ψ_{κ/λ}(xy).

If the fields have odd characteristic the following lemma is, basically, a special case of Lemma 7.7. That lemma has been proven in a simple and direct manner by Weil [14]. We shall use his method to prove the following lemma which in characteristic two, when it cannot be reduced to the previous lemma, is due to Hasse [8].

Lemma 9.1

Let

σ(φ_{λ}) = − ∑_{x∈λ} φ_{λ}(x)

and let

σ(φ_{κ/λ}) = − ∑_{x∈κ} φ_{κ/λ}(x).

Then

σ(φ_{κ/λ}) = σ(φ_{λ})^{[κ:λ]}.

If

P (X) = X^{m} − aX^{m−1} + bX^{m−2} − + …

is any monic polynomial with coefficients in λ set m(P) = m and

χ_{λ}(P) = φ_{λ}(a) ψ_{λ}(−b).

If the degree of the polynomial is 1, b is taken to be 0; if the degree is 0 both a and b are taken to be 0. If

P′(X) = X^{m′} − a′ X^{m′−1} + b′ X^{m′−2} − + …

then

P P′(X) = X^{m+m′} − (a + a′) X^{m+m′−1} + (b + b′ + aa′)X^{m+m′−2} − + …

and

χ_{λ}(P P′) = φ_{λ}(a + a′) ψ_{λ}(−b − b′ − aa′) = χ_{λ}(P)χ_{λ}(P′).

If t is an indeterminate we introduce the formal series

F_{λ}(t) = ∑ χ_{λ}(P)t^{m(P)} = ∏(1 − χ_{λ}(P)t^{m(P)})^{−1}.

The sum is overall monic polynomials with coefficients in λ and the product is overall irreducible polynomials of positive degree with coefficients in λ. If r ≥ 2

∑_{m(P)=r} χ_{λ}(P) = 0

so that

F_{λ}(t) = 1 − σ(φ_{λ})t.

If we replace λ by κ, φ_{λ} by φ_{κ/λ}, and ψ_{λ} by ψ_{κ/λ}, we can define F_{κ/λ}(t) in a similar way. If k = [κ : λ] and T is the set of kth roots of unity the problem is to show that

∏_{ζ∈T} F_{λ}(ζt) = F_{κ/λ}(t^{k}).

Suppose P is an irreducible monic polynomial with coefficients in λ and P′ is one of its monic irreducible factors over κ. Let m = m(P) and let r be the greatest common divisor of m and k. The field obtained by adjoining the roots of P to κ has degree \frac{mk}{r} over λ and is the same as the field obtained by adjoining the roots of P′ to κ. Thus m(P′) = \frac{m}{r} and P splits into r irreducible factors over κ. We shall show that

χ_{κ/λ}(P′) = \{χ_{λ}(P)\}^{\frac{k}{r}} .

Thus if P′_{1}, …, P′_{r} are the factors of P and ℓ = \frac{k}{r}

∏_{i=1}^{r}{1 − χ_{κ/λ}(P′_{i})t^{km(P′_{i})}} = {1 − χ_{λ}(P)^{ℓ}t^{ℓm}}^{r}

which equals

∏_{ζ∈T} {1 − χ_{λ}(P)ζ^{m}t^{m}}.

The necessary identity follows.

Let ν be the field obtained by adjoining a root x of P′ to κ and let μ be the field obtained by adjoining x to λ. If

P (X) = X^{m} − aX^{m−1} + bX^{m−2} …

then

a = S_{μ/λ}(x)

and

b = ω_{μ/λ}(x).

Thus

χ_{λ}(P) = φ_{λ}(S_{μ/λ}(x)) ψ_{λ}(−ω_{μ/λ}(x)) = φ_{μ/λ}(x).

Since φ_{ν/λ}(x) is equal to

φ_{λ}(S_{κ/λ}(S_{ν/κ}(x))) ψ_{λ}(−ω_{κ/λ}(S_{ν/κ}(x)) + S_{κ/λ}(ω_{ν/κ}(x)))

which in turn equals

φ_{κ/λ}(S_{ν/κ}(x)) ψ_{κ/λ}(−ω_{ν/κ}(x)).

We conclude that

χ_{κ/λ}(P′) = φ_{ν/λ}(x).

Replacing κ by μ we see that φ_{ν/λ}(x) equals

φ_{μ/λ}(S_{ν/μ}(x)) ψ_{μ/λ}(−ω_{ν/μ}(x)) = φ_{μ/λ}(ℓx) ψ_{μ/λ} (−ℓ \frac{(ℓ − 1)}{2} x^{2}) .

One easily shows by induction that for every integer ℓ

\{φ_{μ/λ}(x)\}^{ℓ} = φ_{μ/λ}(ℓx) Ψ_{μ/λ} (−ℓ \frac{(ℓ − 1)}{2} x^{2}) .  (9.1)

The relation

χ_{κ/λ}(P′) = {χ_{λ}(P)}^{ℓ}

follows.

Taking μ = λ in the identity (9.1) we see that

\{φ_{λ}(x)\}^{ℓ} = φ_{λ}(ℓx) ψ_{λ} (−ℓ \frac{(ℓ − 1)}{2} x^{2})

for every integer ℓ. Moreover {φ_{λ}(0)}^{2} = φ_{λ}(0) so that φ_{λ}(0) = 1. If the characteristic p of λ is odd take ℓ = p to see that {φ_{λ}(x)}^{p} = 1. If the characteristic is 2, take ℓ = 4 to see that {φ_{λ}(x)}^{4} = 1. Suppose φ′_{λ} is another function on λ which vanishes nowhere and satisfies

φ′_{λ}(x + y) = φ′_{λ}(x) φ′_{λ}(y) ψ_{λ}(xy).

Then φ′_{λ}φ_{λ}^{−1} is a character and for some α in λ

φ′_{λ}(x) ≡ φ_{λ}(x) ψ_{λ}(αx).

Of course

φ_{λ}(x)ψ_{λ}(αx) = φ_{λ}(x + α)φ_{λ}^{−1}(α).

Thus

σ(φ′_{λ}) = φ_{λ}^{−1}(α) σ(φ_{λ}).

If a and b are two non-zero complex numbers and m is a positive integer we write a ∼_{m} b if, for some integer r ≥ 0, (\frac{a}{b})^{m^{r}} = 1. Lemma 9.2

If α ∈ λ^{×}, the multiplicative group of λ, let ν(α) be 1 or -1 according as α is or is not a square in λ. Suppose ψ′_{λ}(x) = ψ_{λ}(αx), φ_{λ} and φ′_{λ} are nowhere vanishing, and

φ_{λ}(x + y) = φ_{λ}(x)φ_{λ}(y)ψ_{λ}(xy)

while

φ′_{λ}(x + y) = φ′_{λ}(x)φ′_{λ}(y)ψ′_{λ}(xy).

Then

σ(φ′_{λ}) ∼_{p} ν(α)σ(φ_{λ}).

Moreover

σ(φ_{λ}) ∼_{2p} |σ(φ_{λ})|.

Suppose first that p is odd. By the remarks preceding the statement of the lemma it is enough to prove the assertions for one choice of φ_{λ} and φ′_{λ}. For example we could take φ_{λ}(x) = Ψ_{λ} (\frac{(x^{2})}{2}) and if α = β^{2} we could take φ′_{λ}(x) = ψ_{λ} (\frac{(βx)^{2}}{2} ). In this case it is clear that σ(φ_{λ}) = σ(φ′_{λ}). However if α is not a square, we take φ′_{λ}(x) = ψ_{λ} (\frac{αx^{2}}{2} ). Then

σ(φ_{λ}) + σ(φ′_{λ}) = 2 ∑_{x∈λ} ψ_{λ} (\frac{x}{2}) = 0.

With this choice of φ_{λ},

\overline{φ_{λ}(x)} = ψ_{λ} (−\frac{x^{2}}{2} )

so that \overline{σ(φ_{λ})} = ν(−1) σ(φ_{λ}). Moreover it is well known and easily verified that σ(φ_{λ}) ≠ 0. Since

{σ(φ_{λ})}^{4} = {ν(−1)}^{2} |σ(φ_{λ})|^{4} = |σ(φ_{λ})|^{4}

we have

σ(φ_{λ}) ∼_{2p} |σ(φ_{λ})|.

The absolute value on the right is of course the ordinary absolute value.

Suppose p is 2. Again any choice of φ_{λ} and φ′_{λ} will do. In this case α is necessarily a square. Let α = β^{2}. We can take φ′_{λ}(x) = φ_{λ}(βx). Then σ(φ′_{λ}) = σ(φ_{λ}). It is enough to prove the second assertion for any ψ_{λ} and any φ_{λ}. Let ϕ be the prime field and let ψ_{ϕ} be the unique non-trivial additive character of ϕ. Take ψ_{λ} = ψ_{λ/ϕ}. Let φ_{ϕ}(0) = 1, φ_{ϕ}(1) = i. One verifies by inspection that

φ_{ϕ}(x + y) = φ_{ϕ}(x)φ_{ϕ}(y)ψ_{ϕ}(xy).

Take φ_{λ} = φ_{λ/ϕ}. Since

σ(φ_{λ/ϕ}) = {σ(φ_{ϕ})}^{[λ:ϕ]},

it is enough to verify that

σ(φ_{ϕ}) ∼_{2} |σ(φ_{ϕ})|.

Since σ(φ_{ϕ}) = −1 + i, this is no problem.

If a is a non-zero complex number set

A [a] = \frac{a}{|a|}.

The following lemma explains our interest in the numbers σ(φ_{λ}).

Lemma 9.3

Suppose L is a non-archimedean local field and χ_{L} is a quasi-character of C_{L} with m = m(χ_{L}) = 2d + 1, where d is a positive integer. Let ψ_{L} be a non-trivial additive character of L and let n = n(ψ_{L}). Let γ be such that γO_{L} = P_{L}^{m+n} and let β be a unit such that

χ_{L}(1 + x) = ψ_{L} (\frac{βx}{γ})

for x in 𝔓_{L}^{d+1}. Choose δ so that δO_{L} = 𝔓_{L}^{d} and let ψ_{λ} be the character of λ = O_{L}/𝔓_{L} defined by

ψ_{λ}(x) = ψ_{L} (\frac{βδ^{2}x}{γ}) .

If φ_{λ} is defined by

φ_{λ}(x) = ψ_{L} (\frac{βδx}{γ}) χ_{L}^{−1}(1 + δx)

then

φ_{λ}(x + y) = φ_{λ}(x)φ_{λ}(y)ψ_{λ}(xy)

and

Δ_{3}(χ_{L}, ψ_{L}, γ) = A [−σ(φ_{λ})].

In the statement of this lemma we have not distinguished, in the notation, between an element of O_{L} and its image in λ. This is convenient and not too ambiguous. It will be done again. The only questionable part of the lemma is the relation

φ_{λ}(x + y) = φ_{λ}(x)φ_{λ}(y)ψ_{λ}(xy).

Since

(1 + δx) (1 + δy) ≡ (1 + δx + δy) (1 + δ^{2}xy) (mod 𝔓_{L}^{m})

we have

χ_{L}^{−1}(1 + δx) χ_{L}^{−1}(1 + δy) = χ_{L}^{−1}(1 + δx + δy) ψ_{L} (−\frac{βδ^{2}xy}{γ} ) .

The required relation follows immediately.

There are a few remarks which we shall need later. It is convenient to formulate them explicitly now. We retain the notation of the previous lemma. Lemma 9.4

If m(μ_{L}) < m(χ_{L}) then

Δ_{3}(μ_{L} χ_{L}, Ψ_{L}; γ) ∼_{p} Δ_{3} (χ_{L}, Ψ_{L}; γ)

and if m(μ_{L}) ≤ d we may take β(μ_{L}χ_{L}) = β(χ_{L}) and then

Δ_{3}(μ_{L}χ_{L}, ψ_{L}; γ) = Δ_{3}(χ_{L}, ψ_{L}; γ).
In both cases m(μ_{L}χ_{L}) = m(χ_{L}). Moreover if x ∈ 𝔓_{L}^{2d}

ψ_{L} (\frac{β(μ_{L}χ_{L})x}{γ}) = μ_{L}(1 + x) χ_{L}(1 + x) = χ_{L}(1 + x)

which in turn equals

ψ_{L} (\frac{β(χ_{L})x}{γ}) .

Thus

β(μ_{L}χ_{L}) ≡ β(χ_{L}) (mod 𝔓_{L})

and if

ψ_{λ}(x) = ψ_{L} (\frac{β(χ_{L})δ^{2}x}{γ})

while

ψ′_{λ}(x) = ψ_{L} (\frac{β(μ_{L}χ_{L})δ^{2}x}{γ})

then ψ_{λ} = ψ′_{λ}. The first assertion of the lemma now follows from the previous two lemmas. It is clear that we can take β(μ_{L}χ_{L}) = β(χ_{L}) if m(μ_{L}) ≤ d. Let the common value of the two numbers be β. Then

ψ_{L} (\frac{βδx}{γ}) μ_{L}^{−1}(1 + δx)χ_{L}^{−1}(1 + δx)

is equal to

ψ_{L} (\frac{βδx}{γ}) χ_{L}^{−1}(1 + δx).

We see now that the second assertion is completely trivial.

There is a corollary of this lemma which it is convenient to observe. Lemma 9.5

Suppose m(χ_{L}) = 2d + ε where d is a positive integer and ε is 0 or 1. If m(μ_{L}) ≤ d and μ_{L} is of order r then

Δ(μ_{L}χ_{L}, ψ_{L}) ∼_{r} Δ(χ_{L}, ψ_{L}).

Choose γ in the usual way so that

Δ(χ_{L}, ψ_{L}) = χ_{L}(γ) Δ_{1}(χ_{L}, ψ_{L}; γ)

and

Δ(μ_{L}χ_{L}, ψ_{L}) = χ_{L}(γ)μ_{L}(γ) Δ_{1}(μ_{L}χ_{L}, ψ_{L}; γ).

It is clear that

μ_{L}(γ) ∼_{r} 1.

If we take

β(μ_{L}χ_{L}) = β(χ_{L})

then, clearly,

Δ_{2}(μ_{L}χ_{L}, ψ_{L}; γ) ∼_{r} Δ_{2}(χ_{L}, ψ_{L}, γ).

To complete the proof of Lemma 9.5 we have only to appeal to Lemma 9.4.

Lemma 9.6

Suppose K is an unramified extension of L and χ_{L} is a quasi-character of C_{L} with

m = m(χ_{L}) = 2d + 1

where d is a positive integer. Let ψ_{L} be a non-trivial additive character of F and let n = n(ψ_{L}). Suppose

χ_{L}(1 + x) = ψ_{L} (\frac{βx}{ϖ_{L}^{m+n}} )

for x in 𝔓_{L}^{d+1}. Take

β(χ_{L}) = β(χ_{K/L}) = β.

If

φ_{λ}(x) = ψ_{L} (\frac{βϖ_{L}^{d}x}{ϖ_{L}^{m+n}} ) χ_{L}^{−1}(1 + ϖ_{L}^{d}x)

and if

φ_{κ}(x) = ψ_{K/L} (\frac{βϖ_{L}^{d}x}{ϖ_{L}^{m+n}} ) χ_{K/L}^{−1}(1 + ϖ_{L}^{d}x)

for x in κ = 0_{K}/𝔓_{K} then φ_{κ} = φ_{κ/λ}. Moreover if [K : L] = ℓ then

Δ_{3}(χ_{K/L}, ψ_{K/L}, ϖ_{L}^{m+n}) = (−1)^{ℓ−1}{Δ_{3}(χ_{L}, ψ_{L}, ϖ_{L}^{m+n})}^{ℓ}.

Once we prove that φ_{κ} = φ_{κ/λ} this lemma will follow from Lemmas 9.1 and 9.3. If x belongs to K let E^{2}(x) be the second elementary symmetric function of x and its conjugates over L. If x belongs to O_{K}

N_{K/F} (1 + ϖ_{L}^{d}x) ≡ (1 + ϖ_{L}^{d}S_{K/L}x) (1 + ϖ_{L}^{2d}E^{2}(x)) (mod 𝔓_{L}^{m}).

Since

E^{2}(x) ≡ ω_{κ/λ}(x) (mod 𝔓_{F})

we have

φ_{κ}(x) = φ_{λ}(S_{K/L}x)ψ_{λ}(−ω_{κ/λ}(x)) = φ_{κ/λ}(x).

Now suppose K is a ramified abelian extension of L and [K : L] = ℓ is an odd prime. Let G = 𝔊(K/L) and suppose G = G_{t} while G_{t+1} = {1}. Suppose

m = m(χ_{L}) = 2d + 1

is greater than or equal to t + 1 and

χ_{L}(1 + x) = ψ_{L} (\frac{βx}{ϖ_{L}^{m+n}} )

for x in P_{L}^{d+1}. Let

d′ = ℓd − (\frac{ℓ − 1}{2}) t

and if x belongs to O_{K} set

φ′_{λ}(x) = ψ_{K/L} (\frac{βϖ_{K}^{d′}x}{ϖ_{L}^{m+n}}) χ_{L}^{−1}(1 + S_{K/L}(ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}x)).

Suppose also that

ϖ_{L} = N_{K/L}ϖ_{K}.

The assumptions listed, we may now state the next lemma. Lemma 9.7

If

ε = S_{K/L} (\frac{ϖ_{K}^{2d′}}{ϖ_{L}^{2d}} )

then ε is a unit. Moreover φ′_{λ} is a function on λ = O_{L}/P_{L} + O_{K}/P_{K} which satisfies

φ′_{λ}(x + y) = φ′_{λ}(x)φ′_{λ}(y)ψ_{λ}(εxy)

if

ψ_{λ}(u) = ψ_{L} (\frac{βu}{ϖ_{L}^{1+n}} ) .

If

φ_{λ}(x) = ψ_{L} (\frac{βx}{ϖ_{L}^{d+1+n}} ) χ_{L}^{−1} (1 + ϖ_{L}^{d}x)

then

A[σ(φ_{λ})]^{ℓ} = A[σ(φ′_{λ})].

Chapter 9 73

Since

\frac{d′ + (ℓ − 1) (t + 1)}{ℓ} ≥ d

the number

S_{K/L} (ϖ_{K}^{d′}x)

lies in 𝔓_{L}^{d}. Moreover E^{2}(ϖ_{K}^{d′}x) is a sum of traces of elements in 𝔓_{K}^{2d′}. Since

\frac{2d′ + (ℓ − 1) (t + 1)}{ℓ} = 2d + \frac{ℓ − 1}{ℓ} ≥ 2d

it lies in 𝔓_{L}^{2d}. If x lies in 𝔓_{K} it lies in 𝔓_{L}^{m} and

S_{K/L}(ϖ_{K}^{d′}x)

lies in 𝔓_{L}^{d+1} because

\frac{d′ + 1 + (ℓ − 1) (t + 1)}{ℓ} = d + 1 + \frac{t(ℓ − 1)}{2ℓ} ≥ d + 1.

Thus if x belongs to P_{K}

φ′_{λ}(x) = ψ_{L} (\frac{β}{ϖ_{L}^{m+n}} S_{K/L} (ϖ_{K}^{d′}x)) χ_{L}^{−1} (1 + S_{K/L}(ϖ_{K}^{d′}x)) = 1.

Since

E^{2}(ϖ_{K}^{d′}(x + y))

is equal to

E^{2}(ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}y) + S_{K/L}(ϖ_{K}^{d′}x) S_{K/L}(ϖ_{K}^{d′}y) − S_{K/L}(ϖ_{K}^{2d′}xy),

the expression

1 + S_{K/L}(ϖ_{K}^{d′}(x + y)) + E^{2}(ϖ_{K}^{d′}(x + y))

is congruent modulo 𝔓_{L}^{m} to the product of

1 + S_{K/L}(ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}x)

and

1 + S_{K/L}(ϖ_{K}^{d′}y) + E^{2}(ϖ_{K}^{d′}y)

and

1 − S_{K/L}(ϖ_{K}^{2d′}xy).

Thus

φ′_{λ}(x + y) = φ′_{λ}(x)φ′_{λ}(y)ψ_{λ}(εxy).

Since

\frac{2d′ + (ℓ − 1) (t + 1)}{ℓ} ≥ 2d

the number ε is in O_{L}. We conclude in particular that if y belongs to 𝔓_{K} then

φ′_{λ}(x + y) = φ′_{λ}(x).

If t = 0 let σ be a generator of G and let ϖ_{K}^{1−σ} = ν. In this case 2d′ = 2dℓ and

\frac{ϖ_{K}^{2d′}}{ϖ_{L}^{2d}} = \{∏_{τ∈G} ϖ_{K}^{1−τ} \}^{2d} ≡ ν^{dℓ(ℓ−1)} (mod 𝔓_{L})

and

ε ≡ ℓν^{dℓ(ℓ−1)} (mod 𝔓_{L})

is a unit. If t > 0

ϖ_{K}^{2d′} ≡ ϖ_{L}^{2d−t}ϖ_{K}^{t} (mod 𝔓_{K})

so that

ε ≡ S_{K/L} (\frac{ϖ_{K}^{t}}{ϖ_{L}^{t}} ) (mod 𝔓_{L}).

It is shown in paragraph V.3 of Serre’s book that the right side of this congruence is a unit. First take p odd and let

φ_{λ}(x) = ψ_{λ} (\frac{x^{2}}{2} + αx) = ψ_{λ} (\frac{(x + α)^{2}}{2} ) ψ_{λ} (\frac{−α^{2}}{2} ) .

Then

σ(φ_{λ}) = −ψ_{λ} (\frac{−α^{2}}{2} ) ∑ ψ_{λ} (\frac{(x + α)^{2}}{2} ) = −ψ_{λ} (\frac{−α^{2}}{2} ) ∑ ψ_{λ} (\frac{x^{2}}{2} ) .

Making use of the calculations in the proof of Lemma 9.2 we see that

A[σ(φ_{λ})]^{ℓ} = ν_{λ}(−1)^{\frac{ℓ−1}{2}} ψ_{λ} (\frac{−ℓα^{2}}{2} ) A[− ∑_{λ} ψ_{λ} (\frac{x^{2}}{2} )]

of ν_{λ} is the non-trivial quadratic character of λ^{×}.

Since

1 + S_{K/L} (ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}x)

is congruent to

{1 + S_{K/L}(ϖ_{K}^{d′}x)} {1 + E^{2}(ϖ_{K}^{d′}x)}

modulo 𝔓_{L}^{m} the value of φ′_{λ}(x) is

ψ_{λ} (\frac{(S_{K/L}y)^{2} + 2α S_{K/L}y − 2E^{2}(y)}{2})

if

y = \frac{ϖ_{K}^{d′}}{ϖ_{L}^{d}} x.

Thus

φ′_{λ}(x) = ψ_{λ} (\frac{(S_{K/L}(y + α)^{2}) − ℓα^{2}}{2} ) .

Replacing x by

x − \frac{ϖ_{L}^{d}}{ϖ_{K}^{d′}} α

and summing we find that

σ(φ′_{λ}) = ψ_{λ} (\frac{−ℓα^{2}}{2} ) \{− ∑_{x} ψ_{λ} (\frac{εx^{2}}{2} )\}.

Collecting this information together we see that to prove the lemma when the residual characteristic p is odd we must show that

ν_{λ}(−1)^{\frac{ℓ−1}{2}} = ν_{λ}(ε).

Since ν^{dℓ(ℓ−1)} is certainly a square we have to show that

ν_{λ}(−1)^{\frac{ℓ−1}{2}} = ν_{λ}(ℓ)

when t = 0. If the field λ is of even degree over the prime field both sides are 1. If not, an odd power of p is congruent to 1 modulo ℓ and the relation follows from the law of quadratic reciprocity. If t > 0 then

ε ≡ S_{K/L} (\frac{ϖ_{K}^{t}}{ϖ_{L}^{t}} ) (mod 𝔓_{L})

and we can appeal to paragraph V.3 for a proof that

ε + u^{p−1} ≡ 0

has a solution in λ. Thus ν_{λ}(ε) = ν_{λ}(−1) and we have to show that

ν_{λ}(−1)^{\frac{p−3}{2}} = 1.

If p ≡ 1 (mod 4) then ν_{λ}(−1) = 1 and if p ≡ 3 (mod 4) the exponent is even.

Before considering the case p = 2, we remark a simple consequence of the preceding discussion.

Chapter 9 76 Lemma 9.8

If p is odd let

φ_{λ}(x) = ψ_{λ} (\frac{x^{2}}{2} + αx) .

If t = 0 and

μ = ν^{dℓ \frac{(ℓ−1)}{2}}

then

φ′_{λ} (\frac{x}{μ}) = ψ_{λ} (ℓ (\frac{x^{2}}{2} + αx))

and if t > 0

φ′_{λ}(x) = ψ_{λ} (\frac{εx^{2}}{x} ) .

In both cases

φ′_{λ}(x) = ψ_{λ} (\frac{(S_{K/L} (y + α)^{2}) − ℓα^{2}}{2} )

with

y = \frac{ϖ_{K}^{d′}}{ϖ_{L}^{d}} x.

If t = 0 then y = μx. Thus if x belongs to O_{L}, as we may assume,

φ′_{λ} (\frac{x}{μ}) = ψ_{λ} (\frac{ℓ(x + α)^{2} − ℓα^{2}}{2} ) = ψ_{λ} (ℓ (\frac{x^{2}}{2} + αx)) .

If t > 0 then ℓ = p is odd and

\frac{d′ℓd + (ℓ − 1) (t + 1)}{ℓ} = \frac{1}{ℓ} \{(ℓ − 1) (t − 1) − \frac{(ℓ − 1)}{2} t\} = \frac{1}{ℓ} \{(ℓ − 1) + \frac{(ℓ − 1)}{2} t\} ≥ 1

so that, if x ∈ O_{F},

S_{K/L}(y + α)^{2} ≡ εx^{2} (mod P_{L}).

Now take p = 2 so that t is necessarily 0 and again let

μ = ν^{dℓ \frac{(ℓ−1)}{2}}

so that

\frac{ϖ_{K}^{d′}}{ϖ_{L}^{d}} ≡ μ (mod P_{K}).

If x is in O_{L} and y = \frac{x}{μ} then

φ′_{λ} (\frac{x}{μ}) = φ′_{λ}(y)

is equal to

ψ_{L} (\frac{ℓ x}{ϖ_{L}^{d+1+n}} ) χ_{L}^{−1} (1 + ℓ ϖ_{L}^{d}x + \frac{ℓ(ℓ − 1)}{2} ϖ_{L}^{2d}x^{2}) .

Chapter 9 77 Since

1 + ℓϖ_{L}^{d}x + \frac{ℓ(ℓ − 1)}{2} ϖ_{L}^{2d}x^{2} ≡ (1 + ℓϖ_{L}^{d}x) (1 + \frac{ℓ(ℓ − 1)}{2} ϖ_{L}^{2d}x^{2})

modulo 𝔓_{L}^{m} we have

φ′_{λ} (\frac{x}{μ}) = φ_{λ}(ℓ x) ψ_{λ} (\frac{−ℓ(ℓ − 1)}{2} x^{2})

which equals

{φ_{λ}(x)}^{ℓ}.

Moreover

{φ_{λ}(x)}^{2} = φ_{λ}(2x) ψ_{λ}(−x^{2}) = ψ_{λ}(−x^{2}).

Since the characteristic is 2 there is an α ≠ 0 such that

ψ_{λ}(x^{2}) = ψ_{λ}(αx).

Then the complex conjugate of φ_{λ}(x) is

φ_{λ}(x)ψ_{λ}(αx)

and

σ(φ_{λ}) = − ∑_{x} φ_{λ}(x + α)

which equals

− ∑_{x} φ_{λ}(x)φ_{λ}(α)ψ_{λ}(αx)

is equal to

φ_{λ}(α) \overline{σ(φ_{λ})}.

Consequently

A[σ(φ_{λ})]^{ℓ} = φ_{λ}(α)^{\frac{ℓ−1}{2}} A[σ(φ_{λ})].

Since

{φ_{λ}(x)}^{4} = 1

we have

A[σ(φ′_{λ})] = A[σ(φ_{λ})]

if ℓ ≡ 1 (mod 4) and

A[σ(φ′_{λ})] = A[\overline{σ(φ_{λ})}] = φ_{λ}^{−1}(α) A[σ(φ_{λ})]

if ℓ ≡ 3 (mod 4).

We have to show that

φ_{λ}(α)^{\frac{ℓ−1}{2}} = 1

if ℓ ≡ 1 (mod 4) and that

φ_{λ}(α)^{\frac{ℓ+1}{2}} = 1

if ℓ ≡ 3 (mod 4). These relations are clear if ℓ is congruent to 1 or 7 modulo 8. In general if ℓ ≡ 1 (mod 4)

φ_{λ}(α)^{\frac{ℓ−1}{2}} = ψ_{λ} (− \frac{(ℓ − 1) (ℓ − 3)}{8}α^{2})

and if ℓ ≡ 3 (mod 4)

φ_{λ}(α)^{\frac{ℓ+1}{2}} = ψ_{λ} (− \frac{(ℓ + 1) (ℓ − 1)}{8}α^{2}) .

Let ϕ be the prime field and let ψ_{ϕ} be its non-trivial additive character. Choose α_{1} such that

ψ_{λ/ϕ}(x) = ψ_{λ} (α_{1}^{2}x).

Then

ψ_{λ}(x^{2}) = ψ_{λ/ϕ} (\frac{x^{2}}{α_{1}^{2}} ) = ψ_{λ/ϕ} (\frac{x}{α_{1}} ) = ψ_{λ} (α_{1}x)

and α = α_{1}. Thus

ψ_{λ}(α^{2}) = ψ_{λ/ϕ}(1).

The right side is +1 or –1 according as f = [λ : ϕ] is even or odd. But ℓ divides 2^{f} − 1 so that, by the second supplement to the law of quadratic reciprocity, f is even if ℓ is congruent to 3 or 5 modulo 8.

There is a complement to Lemma 9.7.

Lemma 9.9

If m(χ_{L}) ≥ 2(t + 1) choose β(χ_{K/L}) = β(χ_{L}) = β in O_{L}. If t + 1 < m(χ_{L}) < 2(t + 1) choose β(χ_{L}) = β and

β(χ_{K/L}) = β − β_{1} \frac{α}{α_{1}}

as in Lemma 8.9. Then m(χ_{K/L}) = 2d′ + 1 and

ψ_{K/L} (\frac{β(χ_{K/L}) ϖ_{K}^{d′}x}{ϖ_{L}^{m+n}}) χ_{K/L}^{−1} (1 + ϖ_{K}^{d′}x)

is equal to

ψ_{K/L} (\frac{β ϖ_{K}^{d′}x}{ϖ_{L}^{m+n}}) χ_{L}^{−1} (1 + S_{K/L} (ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}x)).

From Lemma 8.8 we have

m(χ_{K/L}) = 1 + t + ℓ(m − 1 − t) = 2 (ℓ d − \frac{(ℓ − 1)}{2} t) + 1

as required. If d ≥ t + 1 then

d′ ≥ \frac{(ℓ + 1)}{2} d + \frac{(ℓ − 1)}{2} (d − t) ≥ m

because ℓ is odd. Moreover,

\frac{3d′ + (ℓ − 1) (t + 1)}{ℓ} ≥ \frac{m′ + (ℓ − 1) (t + 1)}{ℓ} = m.

Chapter 9 79 Consequently

N_{K/L}(1 + ϖ_{K}^{d′}x) ≡ 1 + S_{K/L}(ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}x) (mod 𝔓_{L}^{m})

and the lemma is valid if m ≥ 2(t + 1).

If t + 1 < m < 2(t + 1) we still have

\frac{3d′ + (ℓ − 1) (t + 1)}{ℓ} ≥ m

so that

N_{K/L}(1 + ϖ_{K}^{d′}x)

is congruent to

1 + S_{K/L}(ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}x) + N_{K/L}(ϖ_{K}^{d′}x)

modulo 𝔓_{L}^{m}. Since d′ ≥ d + 1 this is congruent to

{1 + S_{K/L}(ϖ_{K}^{d′}x) + E^{2}(ϖ_{K}^{d′}x)} {1 + N_{K/L}(ϖ_{K}^{d′}x)}

modulo 𝔓_{L}^{m}. Certainly

χ_{L}(1 + N_{K/L}(ϖ_{K}^{d′}x)) = ψ_{L} (\frac{β N_{K/L}(ϖ_{K}^{d′}x)}{ϖ_{L}^{m+n}}) .

Moreover, if m = t + 1 + v

d′ − v = d + \frac{ℓ − 3}{2} v ≥ d ≥ s

if s is the least integer greater than or equal to \frac{t}{2}. Thus, just as in the proof of Lemma 8.5,

ψ_{L} (\frac{β N_{K/L}(ϖ_{K}^{d′}x)}{ϖ_{L}^{m+n}}) = ψ_{L} (\frac{α N_{K/L} (\frac{β_{1}}{α_{1}} ϖ_{K}^{d′}x)}{ϖ_{L}^{m+n}} )

is equal to

ψ_{L} (\frac{−S_{K/L} (\frac{αβ_{1}}{α_{1}} ϖ_{K}^{d′}x)}{ϖ_{L}^{m+n}} ) .

Multiplying the inverse of this with

ψ_{K/L} ( (β − \frac{αβ_{1}}{α_{1}} ) \frac{ϖ_{K}^{d′}x}{ϖ_{L}^{m+n}} )

we obtain

ψ_{K/L} (\frac{β ϖ_{K}^{d′}x}{ϖ_{L}^{m+n}}) .

The lemma follows.

If m = t + 1 we may still choose

β(χ_{K/L}) = β − β_{1} \frac{α}{α_{1}}

as in Lemma 8.9. However the relation between φ_{λ}(x) and

φ_{K}(x) = ψ_{K/L} (\frac{β(χ_{K/L}) ϖ_{K}^{d′}x}{ϖ_{L}^{m+n}}) χ_{K/L}^{−1} (1 + ϖ_{K}^{d′}x)

will be more complicated. Here x = O_{K}/𝔓_{K} is the same field as λ = O_{L}/𝔓_{L}. We introduce it only for notational purposes.

Because m = t + 1 the number t is at least 2 and

d = d′ = \frac{t}{2}.

Since

\frac{3d + (p − 1) (t + 1)}{p} ≥ t + 1

and

\frac{d + (p − 1) (t + 1)}{p} ≥ d + 1

the expression

N_{K/L}(1 + ϖ_{K}^{d}x)

is congruent to

{1 + S_{K/L}(ϖ_{K}^{d}x) + E^{2}(ϖ_{K}^{d}x)} {1 + ϖ_{L}^{d} N_{K/L}x}

modulo 𝔓_{L}^{m} and

χ_{L}(1 + S_{K/L}(ϖ_{K}^{d}x) + E^{2}(ϖ_{K}^{d}x))

is equal to

Ψ_{L} (\frac{β S_{K/L}(ϖ_{K}^{d}x) + β E^{2}(ϖ_{K}^{d}x)}{ϖ_{F}^{m+n}}) .

According to Newton’s formulae

S_{K/L}(ϖ_{K}^{2d}x^{2}) − S_{K/L}(ϖ_{K}^{d}x)^{2} + 2E^{2}(ϖ_{K}^{d}x) = 0.

Thus

E^{2}(ϖ_{K}^{d}x) ≡ − \frac{1}{2} S_{K/L}(ϖ_{K}^{2d}x^{2}) (mod P_{L}^{m}).

Observe that p is equal to ℓ and therefore, in the present circumstances, odd.

Let μ_{L} be a character in S(K/L) as in Lemma 8.9(c) and let

ψ_{L} (\frac{α x}{ϖ_{L}^{d+1+n}} ) μ_{L}^{−1}(1 + ϖ_{L}^{d}x) = ψ_{λ} (ρ \frac{x^{2}}{2} + τ x)

Chapter 9 81 with

ρ = \frac{α}{β}.

Certainly

μ_{L}(N_{K/L}(1 + ϖ_{K}^{d}x)) = 1

if x belongs to O_{K}. Replacing x by \frac{β_{1}}{α_{1}} x we see that

ψ_{L} (\frac{1}{ϖ_{L}^{m+n}} \{α S_{K/L} (\frac{β_{1}}{α_{1}} ϖ_{K}^{d}x) − \frac{α}{2} S_{K/L} (\frac{β_{1}^{2}}{α_{1}^{2}} ϖ_{K}^{2d}x^{2}) + N_{K/L} (β_{1} ϖ_{K}^{d}x)\})

is equal to

ψ_{λ} (ρ \frac{z^{2}}{2} + τ z)

if

z = \frac{1}{ϖ_{L}^{d}} \{S_{K/L} (\frac{β_{1}}{α_{1}} ϖ_{K}^{d}x) − \frac{1}{2} S_{K/L} (\frac{β_{1}^{2}}{α_{1}^{2}} ϖ_{K}^{2d}x^{2}) + N_{K/L} (\frac{β_{1}}{α_{1}} ϖ_{K}^{d}x) \}

which is congruent to

\frac{β}{α} N_{K/L}x ≡ \frac{β}{α} x^{p}

modulo 𝔓_{L}.

Let

φ_{λ}(x) = ψ_{L} (\frac{β x}{ϖ_{L}^{d+1+n}} ) χ_{L}^{−1} (1 + ϖ_{L}^{d}x)

equal

ψ_{λ} (\frac{x^{2}}{2} + σ x) .

If x belongs to O_{K}

χ_{L}^{−1} (1 + ϖ_{L}^{d} N_{K/L}x) = ψ_{λ} (\frac{x^{2p}}{2} + σ x^{p}) ψ_{L} (\frac{−β N_{K/L}x}{ϖ_{L}^{d+1+n}}) .

We now put these facts together to find a suitable expression for φ_{κ}(x). We may as well take x in O_{L}. Then φ_{κ}(x) is the product of

ψ_{K/L} (\frac{β ϖ_{K}^{d′}x}{ϖ_{L}^{m+n}})

and

ψ_{K/L} (− \frac{β_{1}α}{α_{1}} \frac{ϖ_{K}^{d}x}{ϖ_{L}^{m+n}} ) χ_{L}^{−1} (1 + ϖ_{L}^{d} N_{K/L}x)

and

ψ_{L} (− \frac{β}{ϖ_{L}^{m+n}} \{S_{K/L}(ϖ_{K}^{d}x) − \frac{1}{2} S_{K/L}(ϖ_{K}^{2d}x^{2})\}) .

Chapter 9 82

The second of these three expressions is equal to the product of

ψ_{λ} (\frac{x^{2p}}{2} + σ x^{p}) ψ_{λ} (−ρ^{−1} \frac{x^{2p}}{2} − ρ^{−1}τ x^{p})

and

ψ_{K/L} (− \frac{αβ_{1}^{2} ϖ_{K}^{2d}x^{2}}{2α_{1}^{2} ϖ_{L}^{m+n}} ) = ψ_{λ} (− \frac{ερβ_{1}^{2}}{2α_{1}^{2}} x^{2})

if

ε = S_{K/L} (\frac{ϖ_{K}^{2d}}{ϖ_{L}^{2d}} ) .

The product of the first and third is equal to

ψ_{λ} (\frac{ε x^{2}}{2} ) .

As proven in paragraph V.3 of Serre’s book the elements of U_{L} congruent to

1 + (ε x + x^{p}) ϖ_{L}^{t}

modulo 𝔓_{L}^{t+1} are all norms, so that

ψ_{λ}(ρ x^{p}) = ψ_{λ}(−ρε x).

In particular

ψ_{λ} (− \frac{ερβ_{1}^{2}}{2α_{1}^{2}} x^{2}) = ψ_{λ} (ρ^{−1} \frac{x^{2p}}{2} ) .^{∗}

∗(1998) The manuscript of Chapter 9 ends with this formula.

Chapter Eleven.

Artin–Schreier Equations

The theory of Artin–Schreier equations is central to Dwork’s proof of the second main lemma. We first review the basic theory, which we take from Mackenzie and Whaples [11], and then review Dwork’s rather amazing calculations. These we take from Lakkis [9].

We start with an exercise from Serre’s book [12]. Suppose F is a non-archimedean local field and K/F is Galois. Let p be the residual characteristic. With the convention (0) = 𝔓_{F}^{∞} we let

p O_{F} = 𝔓_{F}^{e} .

Suppose G = 𝔊(K/F) and σ ∈ G_{i} with i ≥ 1. Let
ϖ_{K}^{σ} = ϖ_{K}(1 + a)

with a in 𝔓_{K}^{i}. Let

φ(x) = x^{σ} − x.

φ is an F -linear operator on K. If x = αϖ_{K}^{j} belongs to 𝔓_{K}^{j} then

φ(x) = x^{σ} − x = (α^{σ} − α) ϖ_{K}^{jσ} + α(ϖ_{K}^{jσ} − ϖ_{K})

is congruent to

αϖ_{K}^{j} (ϖ_{K}^{j(σ−1)} − 1) = αϖ_{K}^{j} {(1 + a)^{j} − 1}

modulo 𝔓_{K}^{i+j+1}. This in turn is congruent to

(α ϖ_{K}^{j}) (ja) = jax

modulo 𝔓_{K}^{i+j+1}.

If*

ψ(x) = x^{σ^{p}} − 1

then, as an operator,

ψ = (1 + φ)^{p} − 1 = ∑_{k=1}^{p} \binom{p}{k} φ^{k}.

If x belongs to 𝔓_{K}^{j} then

φ^{k}(x) ≡ j(j + i) … (j + (k − 1)i)a^{k}x (mod 𝔓_{K}^{j+ki+1})

and ψ(x) is congruent to

pjax + j(j + i) … (j + (p − 1)i)a^{p}x

or to

pjax + j(j^{p−1} − i^{p−1})a^{p}x

* We seem to be dealing with yet another use of the symbol ψ! modulo 𝔓_{K}^{i+j+e′+1} if p O_{K} = 𝔓_{K}^{e′}. We deduce the following congruences:

(i) If (p − 1)i > e′ then

ψ(x) ≡ pjax (mod 𝔓_{K}^{i+j+e′+1}).

(ii) If (p − 1)i = e′ then

ψ(x) ≡ pjax + j(1 − i^{p−1})a^{p}x (mod 𝔓_{K}^{i+j+e′+1}).

(iii) If (p − 1)i < e′ then

ψ(x) ≡ j(1 − i^{p−1})a^{p}x (mod 𝔓_{K}^{pi+j+1}).

Observe that if (j, p) = 1 and σ belongs to G_{i}, with i ≥ 1, then

φ(x) ≡ 0 (mod 𝔓_{K}^{i+j+1})

for all x in 𝔓_{K}^{j} if and only if σ belongs to G_{i+1}. It follows immediately that if σ belongs to G_{1} and i ≥ 1 then

φ(x) ≡ 0 (mod 𝔓_{K}^{i+j})

for all x in 𝔓_{K}^{j} only if σ belongs to G_{i}.

If σ is replaced by σP then φ is replaced by ψ. If k > \frac{e′}{p−1} and G_{k} ≠ \{1\} then, for some i ≥ k, G_{i} ≠ {1} and G_{i+1} = {1}. Taking (j, p) = 1 we infer from (i) that if σ belongs to G_{i} but not to G_{i+1} then σ^{p} is in G_{i+e′} but not in G_{i+e′+1}. This is impossible. Thus G_{k} = {1} if k > \frac{e′}{p−1}. If G_{1} ≠ \{1\} then p divides e′ so that if (p − 1)i = e′ the number i is also divisible by p. The congruence (ii) reduces to

ψ(x) ≡ j (pa + a^{p}) (mod 𝔓_{K}^{i+j+e′+1}).

Thus if σ belongs to G_{i} its pth power σ^{p} lies in G_{i+e} and is therefore 1. Consequently

pa + a^{p} ≡ 0 (mod 𝔓_{K}^{pi+1}).

Letting a = αϖ_{K}^{i} and p = βϖ_{K}^{e} we find that

α^{p} + βα ≡ 0 (mod 𝔓_{K}).

Since this congruence has only p roots the image of Θ_{i} lies in a subset of U_{K}^{i}/U_{K}^{i+1} with p elements and G_{i} is either {1} or cyclic of order p.

If (p − 1) < e′ and (i, p) = 1 the congruence (iii) implies that σ^{p} belongs to G_{pi+1} if σ belongs to G_{i}. However if (p − 1)i < e′ and p divides i it shows that σ^{p} belongs to G_{pi} but not to G_{pi+1} if σ belongs to G_{i} but not to G_{i+1}. Thus σ ⟶ σ^{p} defines an injection of G_{i}/G_{i+1} into G_{pi}/G_{pi+1}. If G_{i}/G_{i+1} is not trivial neither is G_{pi}/G_{pi+1} and (p−1)pi ≤ e′. If (p−1)pi < e′ we can repeat the process. Thus, for some positive integer h, (p − 1)p^{h}i = e′ and G_{p^{h}i} is not trivial. It is then cyclic of order p. According to Proposition IV.10 of Serre’s book those k ≥ 1 for which G_{k}/G_{k+1} ≠ {1} are all congruent modulo p. In particular if G_{k}/G_{k+1} is not trivial for some k ≥ 1 divisible by p it is not trivial only when k is divisible by p. The preceding discussion shows that if i is the smallest value of k ≥ 1 for which G_{k}/G_{k+1} is non-trivial then any σ in G_{i} but not in G_{i+1} generates G_{i} = G_{1}. In other words:

Lemma 11.1

If G_{1} is not cyclic then (i, p) = 1 if i ≥ 1 and G_{i}/G_{i+1} ≠ {1}.

Lemma 11.2

Suppose K/L is cyclic of prime degree and G = 𝔊(K/L) is equal to G_{L} with t ≥ 1 and (t, p) = 1. Then there is a Δ in K and an a in L such that aO_{L} = 𝔓_{L}^{−t} and

Δ^{p} − Δ = a.

We observe first of all that [K : L] must be p and that if pO_{K} = P_{K}^{e′} then (p − 1)t < e′. If x belongs to K the symbol O(x) will stand for an element in xO_{K} and the symbol o(x) will stand for an element in x𝔓_{K}. If

x = ∑_{i=0}^{p−1} a_{i} ϖ_{K}^{i}

with a_{i} in F then

|x| = max_{0≤i<p} |a_{i}| |ϖ_{K}^{i}|.

Moreover if σ is a generator of G

x^{σ} − 1 = ∑_{i=1}^{p−1} a_{i} ϖ_{K}^{i} (ϖ_{K}^{i(σ−1)} − 1)

and if ϖ_{K}^{σ−1} = (1 + aϖ_{K}^{t})

|ϖ_{K}^{i(σ−1)} − 1| = | (1 + aϖ_{K}^{t})^{i} − 1| = |ϖ_{K}^{t}|

for 1 ≤ i < p. Thus

|x^{σ} − x| = |ϖ_{K}^{t}|{ max_{1≤i<p} |a_{i}| |ϖ_{K}^{i}| } ≤ |ϖ_{K}^{t}| |x|.

There is equality if a_{0} = 0. In particular if

y = ∑_{i=1}^{p−1} a_{i} ϖ_{K}^{i}

then

x^{σ} − 1 = y^{σ} − 1

and

|y^{σ} − x| = |ϖ_{K}^{t}| |y|.

If x belongs to K let

𝔭(x) = x^{p} − x.

Then

𝔭(x + y) − 𝔭(x) − 𝔭(y) = ∑_{i=1}^{p−1} \binom{p}{i} x^{i}y^{p−i}  (11.1)

Chapter 11 86 Since e′ − (p − 1)t > 0 the right side is o(y) if

v_{K}(x) ≥ −t

and

v_{K}(y) ≥ −t.

We define v_{K}(x) by the equation

|x| = |ϖ_{K}|^{v_{K}(x)}.

To prove the lemma we construct a sequence Λ_{0}, Λ_{1}, Λ_{2}, … and a sequence Θ_{0}, Θ_{1}, … with the following properties:

(i) v_{K}(Λ_{n}) = −t for all n ≥ 0. (ii) If σ is a given generator of G and ζ is a given (p − 1)th root of unity

Λ_{n}^{σ} − Λ_{n} = ζ + o(1).

(iii)

𝔭(Λ_{n}^{σ}) − 𝔭(Λ_{n}) = Θ_{n}^{σ} − Θ_{n}

and

|Θ_{n}^{σ} − Θ_{n}| = |ϖ_{K}^{t}| |Θ_{n}|.

(iv)

Λ_{n+1} = Λ_{n} + Θ_{n}.

(v)

𝔭(Λ_{n+1}^{σ}) − 𝔭(Λ_{n+1}) = o(𝔭(Λ_{n}^{σ}) − 𝔭(Λ_{n})).

It will follow from (iii) and (v) that {Θ_{n}} is converging to 0. Then (iv) implies that {Λ_{n}} has a limit Δ. (i) implies that v_{K}(Δ) = −t and (v) implies that Δ^{p} − Δ = a belongs to F. From (ii)

Δ^{σ} − Δ = ζ + o(1).

To construct Λ_{0} let α belong to U_{K}^{i} and consider

\frac{α^{σ}}{ϖ_{K}^{σ t}} − \frac{α}{ϖ_{K}^{t}} = \frac{α^{σ} − α}{ϖ_{K}^{σ t}} + \frac{α}{ϖ_{K}^{t}} (ϖ_{K}^{t(1−σ)} − 1) = −taα + o(1)

if

ϖ_{K}^{σ} = ϖ_{K}(1 + aϖ_{K}^{t}).

We can choose α so that

−taα = ζ + o(1).

Then we set

Λ_{0} = \frac{α}{ϖ_{K}^{t}} .

We observe in passing that conditions (i) and (ii) determine Λ_{n} modulo 𝔓_{K}^{−t+1}.

Chapter 11 87

Suppose Λ_{0}, …, Λ_{n} have been defined. Then

𝔭(Λ_{n}^{σ}) = 𝔭(Λ_{n}) + 𝔭(ζ + o(1)) + o(1)

which equals

𝔭(Λ_{n}) + 𝔭(ζ) + o(1) = 𝔭(Λ_{n}) + o(1).

Choose Θ_{n} so that

Θ_{n}^{σ} − Θ_{n} = 𝔭(Λ_{n}^{σ}) − 𝔭(Λ_{n})

and

|Θ_{n}^{σ} − Θ_{n}| = |ϖ_{K}^{t}| |Θ_{n}|.

Then v_{K}(Θ_{n}) > −t and if

Λ_{n+1} = Λ_{n} + Θ_{n}

v_{K}(Λ_{n+1}) = −t. Moreover

Λ_{n+1}^{σ} − Λ_{n+1} = Λ_{n}^{σ} − Λ_{n} + o(1) = ζ + o(1).

and

𝔭(Λ_{n+1}) = 𝔭(Λ_{n}) + 𝔭(Θ_{n}) + x

with x = o(Θ_{n}). Then

x^{σ} − x = o(Θ_{n}^{σ} − Θ_{n}) = o(𝔭(Λ_{n}^{σ}) − 𝔭(Λ_{n})).

Also

𝔭(Θ_{n}^{σ}) − 𝔭(Θ_{n}) = 𝔭(Θ_{n}^{σ} − Θ_{n}) + o(Θ_{n}^{σ} − Θ_{n}).

Since v_{K}(Θ_{n}^{σ} − Θ_{n}) is positive the right side is

−(Θ_{n}^{σ} − Θ_{n}) + o(Θ_{n}^{σ} − Θ_{n}).

Thus

𝔭(Λ_{n+1}^{σ}) − 𝔭(Λ_{n+1})

which equals

𝔭(Λ_{n}^{σ}) − 𝔭(Λ_{n}) − (Θ_{n}^{σ} − Θ_{n}) + o(𝔭(Λ_{n}^{σ}) − 𝔭(Λ_{n}))

is

o(𝔭(Λ_{n}^{σ}) − 𝔭(Λ_{n})).

Lemma 11.3

Supposer Δ_{1} belongs to K, a belongs to L, v_{L}(a) = −t and

Δ_{1}^{p} − Δ_{1} = a + O(ϖ_{K}^{r})

with r ≥ 1. Define Δ_{n} inductively by

Δ_{n+1} = Δ_{n}^{p} − a.

Then

Δ_{n+1} − Δ_{n} = o(Δ_{n} − Δ_{n−1})

if n ≥ 2 and if r ≥ (e′ − (p − 1)t)

Δ_{n+1} − Δ_{n} = O(ϖ_{K}^{r+(n−1) (e′−(p−1)t)}).

Moreover

lim_{n→∞} Δ_{n} = Δ

exists and Δ^{p} − Δ = a.

The last assertion is a consequence of the first. It is clear that

Δ_{2} − Δ_{1} = O(ϖ_{K}^{r}).

Suppose n ≥ 2, and

Δ_{n} − Δ_{n−1} = x = o(1).

Then

Δ_{n+1} − Δ_{n} = Δ_{n}^{p} − Δ_{n−1}^{p} = (Δ_{n−1} + x)^{p} − Δ_{n−1}

is equal to

\{∑_{k=1}^{p−1} \binom{p}{k} Δ_{n−1}^{k} x^{p−k}\} + x^{p}

which is o(x) because e′ − (p − 1)t > 0. If

x = O(ϖ_{K}^{r+(n−2) (e′−(p−1)t)})

and r ≥ e′ − (p − 1)t it is

O(ϖ_{K}^{r+(n−1) (e′−(p−1)t)}).

The lemma has a couple of corollaries which should be remarked. Lemma 11.4

If a is in L, v_{L}(a) = −t, Δ^{p} − Δ = a, and ξ is a (p − 1)th root of unity there is a number Δ_{ξ} such that

Δ_{ξ} = Δ + ξ + O(ϖ_{K}^{e′−(p−1)t})

and

Δ_{ξ}^{p} − Δ_{ξ} = a.

Relation (11.1) shows that Δ + ξ satisfies the conditions of the previous lemma with r = e′ − (p − 1)t. Lemma 11.5

Suppose Δ belongs to K, b belongs to L, v_{L}(b) = −t and

Δ^{p} − Δ = b.

Then for any u in U_{L}^{t+1} the equation

Λ^{p} − Λ = bu

has a solution in K.

Take, in Lemma 11.3, a = bu and Δ_{1} = Δ. Lemma 11.5 shows that if S is the set of all in L with v_{L}(a) = −t for which the equation

Δ^{p} − Δ = a

has a solution in K then S = SU_{L}^{t+1}. Lemma 11.6

If ℓ is the integral part of \frac{t}{p} the number of cosets of U_{L}^{t+1} in S is

\frac{p − 1}{p} [O_{L} : 𝔓_{L}]^{1+ℓ}.

Fix a generator σ of G = 𝔊(K/L). If a belongs to S, Δ^{p} − Δ = a, and ξ is a (p − 1)th root of unity

(ξΔ)^{p} − ξΔ = ξ a.

By Lemma 11.4 there is a (p − 1)th root of unity ζ such that

Δ^{σ} = Δ + ζ + o(1).

Then

(ξΔ)^{σ} = ξΔ + ξζ + o(1).

Thus if S′ is the set of a in L with v_{L}(a) = −t for which

a = Δ^{p} − Δ

with

Δ^{σ} = Δ + 1 + o(1)

the number of cosets if U_{L}^{t+1} in S is p − 1 times the number of cosets of U_{L}^{t+1} in S′. Choose Δ_{0}, with v_{K}(Δ_{0}) = −t, for which Δ_{0}^{p} − Δ_{0} = a_{0} is in F and

Δ_{0}^{σ} = Δ_{0} + 1 + o(1).

If v_{K}(Δ) = −t, Δ^{p} − Δ is in F, and

Δ^{σ} = Δ + 1 + o(1)

then, according to an earlier remark,

Δ = Δ_{0} + Ω_{0}

with Ω_{0} = o(Δ_{0}).

Chapter 11 90

Choose any Ω_{0} = o(Δ_{0}) and set Λ_{0} = Δ_{0} + Ω_{0}. According to the relation (A)

𝔭(Λ_{0}) = 𝔭(Δ_{0}) + 𝔭(Ω_{0}) + o(Ω_{0}).

Since

Ω_{0}^{σ} − Ω_{0} = O(ϖ_{K}^{t}Ω_{0}) = o(1)

we have

Ω_{0}^{σ p} − Ω_{0}^{p} = ∑_{i=1}^{p−1} \binom{p}{i} Ω_{0}^{p−i}(Ω_{0}^{σ} − Ω_{0})^{i} = o(Ω_{0}^{σ} − Ω_{0}).

Thus

𝔭(Λ_{0})^{σ} − 𝔭(Λ_{0}) = Ω_{0}^{σ} − Ω_{0} + o(ϖ_{K}^{t}Ω_{0})

and 𝔭(Λ_{0}) is in L only if

Ω_{0}^{σ} − Ω_{0} = o(ϖ_{K}^{t}Ω_{0}),

that is, only if Ω_{0} = α_{0} + o(Ω_{0}) with α_{0} in L. On the other hand if

Ω_{0}^{σ} − Ω_{0} = o(ϖ_{K}^{t}Ω_{0})

and we construct the sequence Λ_{0}, Λ_{1}, Λ_{2}, … as before and let

Δ = lim_{n→∞} Λ_{n}

then

Δ = Λ_{0} + o(Ω_{0}).

We conclude that the number of cosets in 𝔓_{K}^{s}/𝔓_{K}^{s+1}, s > −t, containing an Ω_{0} such that

(Δ_{0} + Ω_{0})^{p} − (Δ_{0} + Ω_{0})

is in L is 1 if p does not divide s and is [O_{L} : 𝔓_{L}] if it does.

Choose Δ so that

Δ^{p} − Δ = a

is in S′. If Ω belongs to 𝔓_{K}^{s}, s > −t, but not to 𝔓_{K}^{s+1} and

(Δ + Ω)^{p} − Δ − Ω = b

is also in S′ then a and b belong to the same coset of U_{L}^{t+1} if and only if

b = a + o(1).

If s > 0

𝔭(Δ + Ω) = 𝔭(Δ) + o(1)

but if s ≤ 0

𝔭(Δ + Ω) = 𝔭(Δ) + Ω^{p} − Ω + o(Ω)

and

Ω^{p} − Ω + o(Ω) = o(1)

if and only if s = 0 and

Ω = ξ + o(1)

where ξ is some (p − 1)th root of unity. The lemma follows.

If x = {x_{1}, …, x_{n}} let E^{i}(x) be the ith elementary symmetric function of x_{1}, …, x_{n} and let

S^{i}(x) = ∑_{k=1}^{n} x_{k}^{i}.

If Z is an indeterminate and

Q(Z) = ∑_{i=0}^{n}(−1)^{i} E^{i}(x)Z^{i} = ∏_{i=1}^{n}(1 − x_{i}Z)

then

∑_{i=1}^{∞} S^{i}(x)Z^{i}

is clearly −Z times the logarithmic derivative of Q(Z). Thus

(∑_{i=1}^{∞} S^{i}(x) Z^{i}) (∑_{i=0}^{n}(−1)^{i}E^{i}(x)Z^{i}) = − ∑_{i=0}^{n}(−1)^{i} iE^{i}(x)Z^{i}.

This identity which we refer to as Newton’s identity is equivalent to the formulae of Newton. It implies in particular that

∑_{j=0}^{i−1}(−1)^{j} S^{i−j}(x) E^{j}(x) = (−1)^{i+1} iE^{i}(x)  (11.2)

if 1 ≤ i ≤ n. We may divide Newton’s identity by Q(Z) and then expand the right-hand side to obtain expressions for the S^{i}(x) as polynominals in E^{1}(x), …, E^{n}(x). The coefficients are necessarily integers. To calculate them we suppose that x_{1}, …, x_{n} lie in a field of characteristic zero. Let

Q(Z) = 1 + P (Z).

Then

log Q(Z) = − ∑_{k=1}^{∞} \frac{(−1)^{k}}{k} (P (Z))^{k}.

The coefficient of Z^{i−1} in the derivative of the right side is

− ∑_{k} ∑_{{α_{1}+...+α_{n}=k \atop α_{1}+2α_{2}+...+nα_{n}=i}} \frac{i(k − 1)!}{α_{1}! … α_{n}!} ∏_{j=1}^{n} \{E^{j}(x)\}^{α_{j}}.

This expression is therefore equal to −S^{i}(x).

Suppose K/L is a ramified cyclic extension of degree p and G = 𝔊(K/L). Let G = G_{t} and G_{t+1} = {1}. Suppose u ≤ t, Λ is in K, and

ΛO_{K} = 𝔓_{K}^{−u}.

We take {x_{1}, …, x_{n}} to be Λ and its conjugates under G. In this case we write

E^{i}(x) = E_{K/L}^{i}(Λ)

and

S^{i}(x) = S_{K/L}^{i}(Λ).

If 1 ≤ i ≤ p − 1 and γ_{i} is any integer less than or equal to

\frac{−iu + (p − 1) (t + 1)}{p}

we have

E_{K/L}^{i}(Λ) ≡ 0 (mod P_{L}^{γ_{i}}).

We may take

γ_{i} ≥ −\frac{iu}{p} + \frac{(p − 1)t}{p}.

If iu + t is not divisible by p this inequality may be supposed strict.

Suppose α_{1}, …, α_{p} are non-negative integers,

∑_{i=1}^{p} α_{i} = k

and

∑_{i=1}^{p} iα_{i} = ℓ.

If

γ = {∑_{i=1}^{p−1} γ_{i}α_{i}}−uα_{p}.

Then

∏_{i=1}^{p} {E^{i}(Λ)}^{α_{i}} ≡ O (ϖ_{K}^{γ}).  (11.3)

We have

γ ≥ −\frac{ℓ u}{p} + \frac{(p − 1)}{p}kt − \frac{(p − 1)}{p}α_{p}t.

The inequality is strict if α_{i} is non-zero for some i such that iu + t is not divisible by p.

We record now some inequalities that γ satisfies in various special cases. They will be needed later. We observe first of all that, if 1 ≤ i < p, γ_{i} is non-negative and is positive unless p divides iu + t.

(i) If ℓ = p and k = 2 then

1 + u + γ ≥ 1 + t.

In this case α_{p} = 0 and the left side is at least

1 + \frac{2(p − 1)}{p} t ≥ 1 + t.

If p is odd the inequality is strict. (ii) If ℓ = p and k = 2 then

γ ≥ 0.

Moreover the inequality is strict if p is odd. This statement is of course weaker than that of (i).

(iii) If ℓ = p, k ≥ 3, and p is odd, then

γ ≥ u + \frac{t − u}{p}

α_{p} is again 0. The left side is at least

−u + \frac{3(p − 1)}{p} t = u + \frac{t − u}{p} + \frac{1}{p} \{(3p − 4)t − (2p − 1)u\}.

The final term is non-negative. The inequality is strict if u ≠ t. If u = t and p does not divide u it is again strict for then α_{i} ≠ 0 for some i < p − 1 and for such an i the number iu + t is not divisible by p.

(iv) If k ≤ p then

(p − 1)u + γ ≥ u + \frac{t − u}{p}

except when α_{p} = k or α_{p} = p − 1. We have to show that

(p − 2)u + γ + \frac{u − t}{p} ≥ 0.

The left side is at least

\{p − 2 − \frac{ℓ}{p} + \frac{1}{p}\}u + \{ \frac{p − 1}{p} (∑_{i=1}^{p−1} α_{i}) − \frac{1}{p}\}t.

If α_{p} ≠ k the coefficient of t is positive and we need only show that it is at least as great as the negative of the coefficient of u or in other words that

(p − 1) (∑_{i=1}^{p−1} α_{i}) + (p − 2)p ≥ ℓ.

This follows from the assumption that α_{p} ≤ p − 2.

(v) If k ≤ p − 2 and α_{p} = k then

(p − 1)u + γ ≥ u.

In this case

γ ≥ −ku.

There are circumstances in which the estimates for γ_{i} and therefore those for γ can be substantially improved. We will discuss them shortly.

Suppose now that K/F is a totally ramified Galois extension and G = 𝔊(K/F) is the direct product of two cyclic groups of order p. By Lemma 11.1 the sequence of ramification groups is of the form

G = G_{−1} = G_{0} = G_{1} = … = G_{u} ≠ G_{u+1} = … = G_{t} ≠ G_{t+1} = {1}

with (u, p) = 1 and u ≡ t (mod p) or of the form

G = G_{−1} = G_{0} = G_{1} = … = G_{t} ≠ G_{t+1} = {1}

with (t, p) = 1. In the second case we take u = t. In the first case let L_{1} be the fixed field of G_{t} and in the second let L_{1} be any subfield of K of degree p over F. Let L_{2} be any subfield of K different from L_{1} which is also of degree p over F. Let G^{i} = 𝔊(K/L_{i}) and let

G^{i} = G_{s_{i}}^{i} ≠ G_{s_{i}+1}^{i} = {1}.

Then s_{1} = t and s_{2} = u. According to Proposition IV.4 of Serre’s book

δ_{K/F} = (p^{2} − p) (u + 1) + (p − 1) (t + 1)

and

δ_{K/L_{1}} = (p − 1) (t + 1)

and

δ_{K/L_{2}} = (p − 1) (u + 1).

Thus

δ_{L_{1}/F} = \frac{1}{p} (δ_{K/F} − δ_{K/L_{1}}) = (p − 1) (u + 1)

and

δ_{L_{2}/F} = \frac{1}{p} (δ_{K/F} − δ_{K/L_{2}}) = \frac{(p − 1)}{p} ((p − 1) (u + 1) + t + 1).

If \overline{G}^{i} = 𝔊(L_{i}/F) and

\overline{G}^{i} = \overline{G}_{t_{i}}^{i} ≠ \overline{G}_{t_{i}+1}^{i} = \{1\}

then t_{1} = u and

t_{2} = u + \frac{t − u}{p}.

Lemma 11.7

Suppose Δ belongs to K, v_{K}(Δ) = −u, and

Δ^{p} − Δ = a

belongs to L_{2}. If Y belongs to L_{2} then

v_{L_{1}} (S_{K/L_{1}}(Y Δ^{i})) ≥ (p − 1)t_{2} − it_{1} + v_{L_{2}}(Y)

and

v_{L_{1}}(E_{K/L_{1}}^{i}(Y Δ)) ≥ (p − 1)t_{2} + i(v_{L_{2}}(Y) − t_{1})

for 1 ≤ i ≤ p − 1.

We show first that if θ belongs to L_{1} and

θ = ∑_{i=0}^{p−1} Y_{i}Δ^{i}

with Y_{i} in L_{2} then

v_{L_{2}}(Y_{i}) ≥ it_{1} + v_{L_{1}}(θ)

Chapter 11 95 for 0 ≤ i ≤ p − 1. Since t_{1} = u and

v_{K}(θ) = min_{0≤i≤p−1} {v_{K}(Y_{i}) − iu}

the inequality is clear for i = 0. To prove it in general, we use induction one i. Suppose 0 < j ≤ p−1 and the inequality is valid for i < j.

Let

pO_{F} = 𝔓_{F}^{e} .

Applying the exercise at the beginning of the paragraph to the extension L_{2}/F we see that

pe ≥ (p − 1)t_{2} = (p − 1) (u + \frac{t − u}{p}) .

If ξ is any (p − 1)th root of unity then, by Lemmas 11.3 and 11.4, there is a σ in \overline{G}^{2} such that

Δ^{σ} = Δ + ξ + O(ϖ_{K}^{p^{2}e−(p−1)u}).

We may write

θ^{σ} − θ = ∑_{i=1}^{p−1} Y_{i}(Δ^{iσ} − Δ^{i})

as a linear combination

∑_{i=0}^{p−1} X_{i}Δ^{i}

with coefficients from L_{2}. Since

v_{L_{2}}(θ^{σ} − θ) ≥ v_{L_{1}}(θ) + t_{1}

we may apply the induction assumption to see that

v_{L_{2}}(X_{j−1}) ≥ (j − 1))t_{1} + v_{L_{1}}(θ^{σ} − θ) ≥ jt_{1} + v_{L_{1}}(θ).

On the other hand

Δ^{iσ} − Δ^{i} = (Δ + ξ)^{i} − Δ^{i} + O(ϖ_{K}^{p^{2}e−(p−1)u−(i−1)u})

so that θ^{σ} − θ is equal to

∑_{i=1}^{p−1} Y_{i} ∑_{k=0}^{i−1} \binom{i}{k} Δ^{k}ξ^{i−k} + η

with

η = O(θϖ_{K}^{p^{2}e−(p−2)u}).

Thus if

η = ∑_{i=0}^{p−1} Z_{i}Δ^{i}

with the Z_{i} in L_{2} we have

v_{K}(Z_{j−1}) ≥ (j − 1)u + v_{K}(Θ) + p^{2}e − (p − 2)u.

But

p^{2}e − (p − 2)u + (j − 1)u ≥ p(p − 1)u − (p − 2)u + (j − 1)u

which equals

((p − 1)^{2} + j)u ≥ pju.

Since

X_{j−1} = (∑_{i=j}^{p−1} Y_{i} \binom{i}{j} ξ^{i−j}) + Z_{j−1}

we have

v_{L_{2}} (∑_{i=j}^{p−1} Y_{i} \binom{i}{j} ξ^{i−j}) ≥ ju + v_{L_{1}}(θ)

for all ξ. We obtain the required estimate for v_{L_{2}}(Y_{j}) by summing over ξ.

We now show that

v_{L_{1}}(S_{K/L_{1}}(Y Δ^{i})) ≥ (p − 1)t_{2} − it_{1} + v_{L_{2}}(Y)

for Y in L_{2} and 1 ≤ i ≤ p − 1. All we need do is show that for any θ in the inverse different of L_{1}/F

S_{L_{1}/F} (θϖ_{L_{1}}^{−v_{L_{2}} (Y)+it_{1}−(p−1)t_{2}} S_{K/L_{1}}(Y Δ^{i})) ∈ O_{F}

or that if θ is in L_{1} and

v_{L_{1}}(θ) ≥ −(p − 1) (t_{1} + 1) + it_{1} − (p − 1)t_{2} − v_{L_{2}}(Y)

then

S_{L_{1}/F} (θ S_{K/L_{1}}(Y Δ^{i})) = S_{K/L}(θ Y Δ^{i})  (11.4)

is in O_{F}.

Let

θ = ∑_{j=0}^{p−1} Y_{j}Δ^{j}

with Y_{j} in L_{2} for 0 ≤ j ≤ p − 1. Then

θ Y Δ^{i} = ∑_{j=0}^{p−1−i} Y Y_{j}Δ^{j+i} + (a + Δ) ∑_{j=p−i}^{p−1} Y Y_{j}Δ^{j+i−p}.

Since

Δ^{p} − Δ = a

we have

E_{K/L_{2}}^{i}(Δ) = 0

for 1 ≤ i < p − 1 and

E_{K/L_{2}}^{p−1}(Δ) = (−1)^{p}.

The relations (11.2) imply that

S_{K/L_{2}}(Δ^{i}) = 0

for 1 ≤ i < p − 1 and that

S_{K/L_{2}}(Δ^{p−1}) = p − 1.

Thus (11.4) is equal to

(p − 1) S_{L_{2}/F} (Y Y_{p−1−i}) + S_{L_{2}/F} (paY Y_{p−i})

if i < p − 1 and to the sum of this and

S_{L_{2}/F} (Y Y_{p−1})

if i = p − 1.

We know that

v_{L_{2}}(Y_{j}) ≥ jt_{1} + v_{L_{1}}(θ)

for each j. Thus

v_{L_{2}}(Y Y_{p−1−i}) ≥ (p − 1 − i)t_{1} − (p − 1) (t_{1} + 1) + it_{1} − (p − 1)t_{2}

which is at least −(p − 1) (t_{2} + 1). So is

v_{L_{2}}(paY Y_{p−1}) ≥ (p − 1)t_{2} − t_{1} − (p − 1) (t_{1} + 1) + it_{1} + (p − i)t_{1} − (p − 1)t_{2}.

If i = p − 1

v_{L_{2}}(Y Y_{p−1}) ≥ (p − 1)t_{1} − (p − 1) (t_{1} + 1) + (p − 1)t_{1} − (p − 1)t_{2}

is also at least −(p − 1) (t_{2} + 1). All we need do now is observe that

S_{L_{2}/F} (𝔓_{L_{2}}^{−(p−1) (t_{2}+1)}) ⊆ O_{F} .

To complete the proof of the lemma we have to show that

v_{L_{1}}(E_{K/L_{1}}^{i}(Y Δ)) ≥ (p − 1)t_{2} + i(v_{L_{2}}(Y) − t_{1})

for 1 ≤ i ≤ p−1. This has been done for i = 1; so we proceed by induction. Applying the relations (11.2) we see that

(−1)^{i+1} iE_{K/L_{1}}^{i}(Y Δ) = ∑_{j=0}^{i−1}(−1)^{j} S_{K/L_{1}}^{i−j} (Y Δ) E_{K/L_{1}}^{j}(Y Δ).

According to the induction assumption and the first part of the lemma, with Y replaced by Y^{i−j}, a typical term in the sum on the right is O(ϖ_{L_{1}}^{v}) with

v = (p − 1)t_{2} − (i − j)t_{1} + (i − j)v_{L_{2}}(Y) + (p − 1)t_{2} + j(v_{L_{2}}(Y) − t_{1})

if j > 0 and

v = (p − 1)t_{2} − it_{1} + iv_{L_{2}}(Y)

if j = 0. The lemma follows.

We apply the second estimate with Y = 1 to improve, when Λ = Δ, L = L_{1}, and certain auxiliary conditions are satisfied, our estimates on the number γ appearing in (11.3).

(vi) Suppose p is odd and

ℓ = (p − 1)ν + j + 1.

If k ≥ ν + 2 and α_{p} ≤ k − 2 then

jt_{1} + γ ≥ pt_{2}.

If k ≥ ν + 2 and α_{p} ≤ k − 1 then

jt_{1} + γ ≥ (p − 1)t_{2} + t_{1}

and if k ≥ ν + 1 and α_{p} ≤ k − 1

jt_{1} + γ ≥ (p − 1)t_{2} − t_{1}.

In the present circumstances

γ_{i} ≥ (p − 1)t_{2} − it_{1}

for 1 ≤ i ≤ p − 1. Thus

jt_{1} + γ ≥ jt_{1} + ∑_{i=1}^{p−1} α_{i}((p − 1)t_{2} − it_{1}) − α_{p}t_{1}

which equals

jt_{1} + (p − 1)kt_{2} − ℓt_{1} − (p − 1)α_{p}(t_{2} − t_{1})

or

(p − 1)kt_{2} − (p − 1)νt_{1} − t_{1} − (p − 1)α_{p}(t_{2} − t_{1}).

If α_{p} ≤ k − 2, this is at least

2(p − 1)t_{2} + (p − 1) (k − 2 − ν)t_{1} − t_{1}

which in turn is at least pt_{2} if p ≥ 3 and k ≥ ν + 2. If α_{p} ≤ k − 1

γ ≥ (p − 1)t_{2} + (p − 1) (k − 1 − ν)t_{1} − t_{1}.

The required inequalities follow. We shall use all these estimates for γ in the next sequence of lemmas.

Lemma 11.8 If Δ is as in Lemma 11.7 and p is odd then

S_{L_{1}/F} N_{K/L_{1}}Δ ≡ S_{L_{2}/F} N_{K/L_{2}}Δ (mod 𝔓_{F}^{1+t_{2}}).

The assertion of the lemma may be reformulated as

S_{K/L_{2}}N_{K/L_{1}}Δ ≡ S_{K/L_{1}}N_{K/L_{2}}Δ (mod 𝔓_{L_{1}}^{1+pt_{2}}).

Notice that

pt_{2} = t + (p − 1)u.

Earlier we applied Newton’s identity to express S_{K/L_{1}}^{p}(Δ) in terms of the elementary symmetric functions of Δ and its conjugates. Since

pe ≥ (p − 1)t_{2}

we can apply the estimates (iii) for γ to see that

S_{K/L_{1}}^{p}(Δ)

is congruent to

pN_{K/L_{1}}Δ + \frac{p}{2} ∑_{j=1}^{p−1} E_{K/L_{1}}^{j}(Δ) E_{K/L_{1}}^{p−j}(Δ) + \{S_{K/L_{1}}(Δ)\}^{p}.  (11.5)

Since

Δ^{p} − Δ = a

we have

S_{K/L_{1}}(Δ) = S_{K/L_{1}}^{p}(Δ) − S_{L_{2}/F} (a).

According to Lemma 11.7 the left side belongs to 𝔓_{L_{1}}^{(p−1)t_{2}−t_{1}}. In particular it belongs to 𝔓_{L_{1}}. We need to know that it belongs to 𝔓_{L_{1}}^{1+t_{2}}. This is clear if p > 3 or t_{2} > t_{1}. To prove it in general we first observe that all terms but the last in (11.5) are congruent to 0 modulo 𝔓_{L_{1}}^{1+t_{2}}. The middle terms are taken care of by the estimates (ii) for γ. To take care of the first we have to show that

pe − u ≥ 1 + t_{2}.

We know that pe ≥ (p − 1)t_{2} and that if t = u the inequality is strict. We need only show that

(p − 1)t_{2} − u ≥ t_{2}

with a strict inequality if t ≠ u. This is clear since t_{2} ≥ u and t_{2} > u if t ≠ u. Thus

S_{K/L_{1}}Δ ≡ (S_{K/L_{1}}Δ)^{p} − S_{L_{2}/F} (a) (mod 𝔓_{L_{1}}^{1+t_{2}}).

We now need only show that

S_{L_{2}/F} (a) ≡ 0 (mod 𝔓_{L_{1}}^{1+t_{2}}).

The left side belongs to 𝔓_{L_{1}}^{pb} if b is any integer less than or equal to

\frac{−u + (p − 1)(t_{2} + 1)}{p}.

We may take

b ≥ \frac{−u + (p − 1)t_{2}}{p}

which is greater than or equal to \frac{1+t_{2}}{p} except when p = 3 and t = u. In this case, which is the one to worry about, t_{2} = u is prime to p and

\frac{−u + (p − 1)(t_{2} + 1)}{p} = \frac{u + 2}{3}

Chapter 11 100 has integral part at least \frac{u+1}{3}.

We apply (11.5) again to see that

S_{K/L_{1}}N_{K/L_{2}}Δ = S_{K/L_{1}}a = S_{K/L_{1}}^{p}(Δ) − S_{K/L_{1}}(Δ)

is congruent to

−S_{K/L_{1}}Δ + p N_{K/L_{1}}Δ + \frac{p}{2} ∑_{j=2}^{p−2} E_{K/L_{1}}^{j}(Δ) E_{K/L_{1}}^{p−j}(Δ).

We have still to consider

S_{K/L_{2}} N_{K/L_{1}} Δ.  (11.6)

There are some general remarks to be made first. Suppose Λ belongs to K and

v_{K}(Λ) = −u.

If x and z also belong to K and

xΛ^{j} ∈ O_{K}

and

z ∈ 𝔓_{K}^{1+t+(p^{2}−1)u}

then

N_{K/L_{1}}(x(Λ + z)^{j+1}) ≡ N_{K/L_{1}}(xΛ^{j+1}) (mod 𝔓_{L_{1}}^{1+pt_{2}}).

It is enough to show that

N_{K/L_{1}} (1 + \frac{z}{Λ})^{j+1} ≡ 1 (mod 𝔓_{L_{1}}^{1+t+pu}).

This follows from Lemma V.5 of Serre’s book and the relations

1 + t + p^{2}u ≥ 1 + t + pu

and

\frac{1 + t + p^{2}u + (p − 1) (t + 1)}{p} = 1 + t + pu.

According to Lemmas 11.3 and 11.4 there is for each σ ≠ 1 in G^{2} a (p − 1)th root of unity ξ = ξ(σ) such that

Δ^{σ} = (Δ + ξ)^{p} − a + O (ϖ_{K}^{2(p^{2}e−(p−1)u)}).

We have

p^{2}e − (p − 1)u ≥ (p − 1) {(p − 1)u + t} − (p − 1)u

which equals

(p − 1)t + (p − 2) (p − 1)u ≥ t + p(p − 2)u.

If t = u the first of these inequalities is strict and if t > u the last is. Thus p^{2}e − (p − 1)u ≥ 1 + t + p(p − 2)u Chapter 11 101 and

2(p^{2}e − (p − 1)u) ≥ {1 + t + (p^{2} − 1)u} + {1 + t + ((p − 1)^{2} − 2p)u}.

The second term is positive unless p = 3.

The expression

(Δ + ξ)^{p} − a

is equal to

Δ + ξ + ∑_{i=1}^{p−1} \binom{p}{i} ξ^{i}Δ^{p−i}.

But

\binom{p}{i} = \frac{p(p − 1) … (p − i + 1)}{i!} ≡ (−1)^{i+1} \frac{p}{i} (mod p^{2})

and

2p^{2}e − (p − 1)u ≥ 2(p^{2}e − (p − 1)u)

which is, as we have just seen, at least 1 + t + (p^{2} − 1)u. Thus if p > 3

Δ^{σ} ≡ (Δ + ξ) (1 − Z(ξ)) (mod 𝔓_{K}^{1+t+(p^{2}−1)u})

if

Z(ξ) = \frac{pΔ^{p−1}}{1 + ξ/Δ} ∑_{i=1}^{p−1} \frac{(−1)^{i}}{i} (\frac{ξ}{Δ})^{i} .

Expanding the denominator we obtain

Z(ξ) = pΔ^{p−1} ∑_{i=1}^{∞} a_{i} (\frac{ξ}{Δ})^{i} .

If i ≥ p − 1

a_{i} = (−1)^{i} ∑_{j=1}^{p−1} \frac{1}{j} ≡ 0 (mod p).

Clearly

Z(ξ) = O(pΔ^{p−2}) = O(ϖ_{K}^{p^{2}e−(p−2)u}).
If p = 3
3(p^{2}e − (p − 1)u) ≥ {1 + t + (p^{2} − 1)u} + {2(1 + t) + (2p^{2} − 6p + 1)u}.

The second term is at least 3u and in particular, is positive. Lemmas 11.3 and 11.4 show that

Δ^{σ} ≡ ((Δ + ξ)^{3} − a)^{3} − a (mod 𝔓_{K}^{1+t+(p^{2}−1)u}).

The right side equals

(Δ + ξ + 3ξΔ^{2} + 3ξ^{2}Δ)^{3} − a.

Expanding the cube and ignoring all terms in 𝔓_{K}^{1+t+(p^{2}−1)u} we obtain

Δ + ξ + 3Δ^{3} \{\frac{ξ}{Δ} + \frac{ξ^{2}}{Δ^{2}} \} + 9Δ^{4}ξ

Chapter 11 102 which we write as

(Δ + ξ) (1 − Z(ξ))

with

Z(ξ) = 3Δ^{2} ∑_{i=1}^{∞} a_{i} (\frac{ξ}{Δ})^{i} + 9Δ^{4} ∑_{i=1}^{∞} (\frac{−ξ}{Δ} )^{i} .

Since

2(p^{2}e − (p − 2)u) ≥ 2(p^{2}e − (p − 1)u) + 2u ≥ 1 + t + p^{2}u

and

p^{2}e − (p − 2)u ≥ 1 + t + (p − 1)^{2}u ≥ 1 + t + pu

for all odd p, lemma V.5 of Serre’s book shows that

N_{K/L_{1}}(x(Δ + ξ)^{j+1}(1 − Z(ξ))^{j+1}) ≡ {N_{K/L_{1}}(x(Δ + ξ)^{j+1})} {1 − S_{K/L_{1}}Z(ξ)}^{j+1} modulo 𝔓_{L_{1}}^{1+t+(p−1)u} if xΔ^{j} lies in O_{K}.

The expression (11.6) is equal to

N_{K/L_{1}}Δ + ∑_{{σ∈G^{2} \atop σ≠1}} N_{K/L_{1}}Δ^{σ}.

The preceding remarks show that, if p > 3, this is congruent to

N_{K/L_{1}}Δ + ∑_{ξ} N_{K/L_{1}}(Δ + ξ) {1 − S_{K/L_{1}}Z(ξ)}

modulo 𝔓_{L_{1}}^{1+t+(p−1)u}. Since

\frac{2p^{2}e + u + (p − 1)t}{p} ≥ 2(p − 1) (u + \frac{t − u}{p}) tu > t + pu

we have

N_{K/L_{1}}(Δ + ξ) S_{K/L_{1}}(a_{i}p Δ^{p−1−i}ξ^{i}) ∈ 𝔓_{L_{1}}^{1+pt_{2}}

if i ≥ p and we may replace S_{K/L_{1}}Z(ξ) by

∑_{i=1}^{p−1} pξ^{i} S_{K/L_{1}}(a_{i}Δ^{p−1−i})

if p > 3. Of course

N_{K/L_{1}}(Δ + ξ) = ∑_{i=0}^{p} ξ^{1−i} E_{K/L_{1}}^{i}(Δ).

Putting these observations together we see that, if p > 3, (11.6) is congruent modulo 𝔓_{L_{1}}^{1+pt_{2}} to the sum of

N_{K/L_{1}}Δ + (p − 1) {S_{K/L_{1}}Δ + N_{K/L_{1}}Δ}

and

−p(p − 1) ∑_{i=1}^{p−1} a_{i} S_{K/L_{1}}(Δ^{p−1−i}) E_{K/L_{1}}^{1+i}(Δ)

and

−p(p − 1) {pa_{p−1} S_{K/L_{1}}(Δ) + a_{p−2} S_{K/L_{1}}(Δ)}.

Since

pS_{K/L_{1}}(Δ) ∈ 𝔓_{L_{1}}^{1+pt_{2}}

the last expression may be ignored as may the term in the second corresponding to i = p−2. Since

a_{p−1} = ∑_{j=1}^{p−1} \frac{1}{j} ≡ 0 (mod p)

and

S_{K/L_{1}}(Δ^{0}) = p

and

3p^{2}e − t_{2} ≥ 1 + pt_{2}

the sum in the second expression need only be taken from 1 to p − 3. The relation (11.2) implies that

pS_{K/L_{1}}^{p−1−i}(Δ) E_{K/L_{1}}^{1+i}(Δ) ≡ (−1)^{i}p(p − 1 − i) E_{K/L_{1}}^{p−1−i}(Δ) E_{K/L_{1}}^{1+i}(Δ)

modulo 𝔓_{L_{1}}^{1+pt_{2}}. To complete the proof of Lemma 11.8, for p > 3, we need only show that

ia_{i−1} + (p − i) a_{p−i−1} ≡ (−1)^{i} (mod p)

for p − 2 ≥ p − i ≥ i ≥ 2. This amounts to showing that

i ∑_{j=1}^{i−1} \frac{1}{j} + (p − i) ∑_{j=1}^{p−i−1} \frac{1}{j} ≡ −1 (mod p).

We may replace the p − i in front of the second sum by −i. Making the obvious cancellations we obtain

−i ∑_{j=i}^{p−i−1} \frac{1}{j} ≡ −1 − i ∑_{j=i}^{p−i} \frac{1}{j}.

If \frac{1}{j} occurs in the sum on the right so does \frac{1}{p−j}.

The proof for p = 3 can proceed in exactly the same way provided we show that

9 ∑{N_{K/L_{1}}(Δ + ξ)} {S_{K/L_{1}} (ξ^{i}Δ^{4−i})}  (11.7)

lies in 𝔓_{L_{1}}^{1+3t_{2}} for i ≥ 1. Since

2p^{2}e − u ≥ 2(p − 1)t_{2} − u ≥ 3t_{2}

and one of the inequalities is strict

9N_{K/L_{1}}(Δ + ξ) ∈ 𝔓_{L_{1}}^{1+3t_{2}}.

The expression

ξ^{i} S_{K/L_{1}}(Δ^{4−i})

is clearly integral for i ≥ 4. By, for example, Lemma 11.7 it is also integral if i is 2 or 3. Thus i = 1 is the only case to cause a problem. If i = 1 we sum over ξ to see that (11.7) equals

18{E_{K/L_{1}}^{2}(Δ) S_{K/L_{1}}(Δ^{3}) + S_{K/L_{1}}(Δ^{3})}.

The terms appearing in the expression in brackets have been shown to lie in O_{L_{1}}.

There is one more lemma to be proved before we come to the basic fact of this paragraph. If x is in K we set

g(x) = S_{L_{1}/F} (N_{K/L_{1}}Δ S_{K/L_{1}}(x)) − S_{L_{2}/F} (N_{K/L_{2}}Δ S_{K/L_{2}}(x))

and

h(x) = S_{L_{1}/F} (N_{K/L_{1}}(xΔ)) − S_{L_{2}/F} (N_{K/L_{2}}(xΔ)).

In the following lemma p is supposed odd.

Lemma 11.9

(a) Suppose x is in L_{2}, 0 ≤ j ≤ p − 1, and xΔ^{j} lies in 𝔓_{K}^{1+t_{2}−t_{1}}. If j ≠ p − 2 then

g(xΔ^{j}) ≡ 0 (mod 𝔓_{F}^{1+t_{2}})

but if j = p − 2, there is an ω in L_{2} such that

ωx ≡ −x E_{K/L_{1}}^{p−1}(Δ) (mod 𝔓_{K}^{1+pt_{2}})

and

g(xΔ^{j}) ≡ −{S_{L_{2}/F} x − S_{L_{2}/F} (xω)} (mod 𝔓_{F}^{1+t_{2}}).

(b) Suppose x is in L_{2}, 0 ≤ j ≤ p − 1, and xΔ^{j} lies in 𝔓_{K}. If j ≠ p − 2

h(xΔ^{j}) ≡ 0 (mod 𝔓_{F}^{1+t_{2}})

but if j = p − 2

h(xΔ^{j}) ≡ (p − 1) (1 − {E_{K/L_{1}}^{p−1}(Δ)}^{p})N_{K/L_{1}}x

modulo 𝔓_{L_{1}}^{1+pt_{2}}.

The congruences modulo 𝔓_{F}^{1+t_{2}} are of course equivalent to congruences modulo 𝔓_{L_{1}}^{1+pt_{2}}. We start with part (a). If x belongs to O_{L_{2}} then

g(x) = S_{K/L_{1}}(xS_{K/L_{2}}(N_{K/L_{1}}Δ) − pxa).

Because of the previous lemma this is congruent to

S_{K/L_{1}}(xS_{L_{2}/F} a − pxa) = S_{L_{2}/F} xS_{L_{2}/F} a − pS_{L_{2}/F} (xa)

modulo 𝔓_{F}^{1+t_{2}}. We saw before that

S_{L_{2}/F} a ∈ 𝔓_{L_{1}}^{1+t_{2}}.

Chapter 11 105 The same argument shows that

S_{L_{2}/F} (xa) ∈ 𝔓_{L_{1}}^{1+t_{2}}.

p belongs to 𝔓_{L_{1}}^{(p−1)t_{2}}. Since the integral part of

\frac{(p − 1) (t_{2} + 1)}{p}

is at least

\frac{(p − 1)t_{2}}{p} ,

so does S_{L_{2}/F} x. This takes care of the case j = 0.

If 1 ≤ j = p − 1 then g(xΔ^{j}) is equal to

S_{L_{2}/F} (xS_{K/L_{2}}(Δ^{j} N_{K/L_{1}}Δ)) − (p − 1)δ S_{L_{2}/F} (xa)

where δ = 0 if j ≠ p − 1 and δ = 1 if j = p − 1. Consider

Z_{j} = x S_{K/L_{2}}(Δ^{j} N_{K/L_{1}}Δ).

It lies in L_{2} and is equal to

xΔ^{j} N_{K/L_{1}}Δ + ∑_{{σ∈G^{2} \atop σ≠1}} xΔ^{σ j} N_{K/L_{1}}Δ^{σ}.

We observe first of all that if Λ is in K, v_{K}(Λ) = −u, x is in L_{2}, xΛ^{j} lies in 𝔓_{K}^{1+t_{2}−t_{1}}, and z lies in 𝔓_{K}^{(p−1) (pt_{2}−t_{1})} then

x(Λ + z)^{j}N_{K/L_{1}}(Λ + z) ≡ xΛ^{j}N_{K/L_{1}}Λ (mod 𝔓_{K}^{1+pt_{2}})

provided p is greater than 3. To establish this congruence we show that

(1 + \frac{z}{Λ})^{j} N_{K/L_{1}} (1 + \frac{z}{Λ}) ≡ 1 (mod 𝔓_{K}^{(p−1)t_{2}+(p+1)t_{1}}).

To show this, one has only to observe that \frac{z}{Λ} and all its conjugates lie in 𝔓_{K}^{(p−1)pt_{2}−(p−2)t_{1}} and that

(p − 1)pt_{2} − (p − 2)t_{1} ≥ (p − 1)t_{2} + ((p − 1)^{2} − (p − 2))t_{1}

which equals

(p − 1)t_{2} + ((p − 1) (p − 2) + 1)t_{1} ≥ (p − 1)t_{2} + (p + 1)t_{1}

if p > 3.

Suppose for now that p > 3. Since p^{2}e − (p − 1)u ≥ (p − 1) (pt_{2} − t_{1}).

Chapter 11 106 Lemma 11.4 implies that Z_{j} is congruent to

xΔ^{j} N_{K/L_{1}}Δ + ∑_{ξ} x(Δ + ξ)^{j} N_{K/L_{1}}(Δ + ξ)

modulo 𝔓_{K}^{1+pt_{2}}

N_{K/L_{1}}(Δ + ξ) = ξ\{1 + \frac{1}{ξ} N_{K/L_{1}}Δ + ∑_{i=1}^{p−1} ξ^{−i}E_{K/L_{1}}^{i}(Δ)\}.

According to Lemma 11.7 this is congruent to

ξ + N_{K/L_{1}}Δ + ξ E_{K/L_{1}}^{p−1}(Δ)

modulo 𝔓_{K}^{pt_{2}}. Thus if xΔ^{j} belongs to 𝔓_{K}^{1+t_{2}−t_{1}}

Z_{j} ≡ xΔ^{j} N_{K/L_{1}}Δ + ∑_{ξ} x(Δ + ξ)^{j} (ξ + N_{K/L_{1}}Δ + ξ E_{K/L_{1}}^{p−1}(Δ))

modulo 𝔓_{K}^{1+pt_{2}}. We expand (Δ + ξ)^{j} and sum over ξ to obtain

pxΔ^{j} N_{K/L_{1}}Δ  (11.8)

if j < p − 2. If j = p − 2 we obtain

pxΔ^{j} N_{K/L_{1}}Δ + (p − 1)x (1 + E_{K/L_{1}}^{p−1}(Δ))

and if j = p − 1 we obtain

pxΔ^{j} N_{K/L_{1}}Δ + (p − 1)x {N_{K/L_{1}}Δ + (p − 1)Δ + (p − 1)Δ E_{K/L_{1}}^{p−1}(Δ)}. The expression (11.8) lies in

𝔓_{K}^{1+p^{2}e−pt_{1}}

provided xΔ^{j} lies in P_{K}.

p^{2}e − pt_{1} ≥ p(p − 1)t_{2} − pt_{1} ≥ pt_{2}.

Since

L_{2} ∩ 𝔓_{K}^{1+pt_{2}} = 𝔓_{L_{1}}^{1+t_{2}}

and

S_{L_{2}/F} (𝔓_{L_{2}}^{1+t_{2}}) ⊆ 𝔓_{F}^{1+t_{2}},

we have

g(xΔ^{j}) ≡ 0 (mod 𝔓_{F}^{1+t_{2}})

if 1 ≤ j < p − 2 and xΔ^{j} lies in 𝔓_{K}^{1+t_{2}−t_{1}}.

Since

E_{K/L_{1}}^{p−1}(Δ)

Chapter 11 107 lies in O_{L_{1}},

Z_{j} = (ω − 1)x

with

ω x = Z_{j} + x ≡ −x E_{K/L_{1}}^{p−1}(Δ) (mod 𝔓_{K}^{1+pt_{2}})

if j = p − 2. We may take ω in L_{2} and then

g(xΔ^{j}) ≡ −{S_{L_{2}/F} x − S_{L_{2}/F} (xω)} (mod 𝔓_{F}^{1+t_{2}}).

If j = p − 1 then
g(xΔ^{j}) = S_{L_{2}/F} (Z_{j} − (p − 1)xa)

and

Z_{j} − (p − 1)xa

is congruent to

(p − 1)x {N_{K/L_{1}}Δ + pΔ + (p − 1)Δ E_{K/L_{1}}^{p−1}(Δ) − Δ^{p}}

modulo 𝔓_{K}^{1+pt_{2}}. The product

{(p − 1)x} {pΔ}

lies in 𝔓_{K}^{1+pt_{2}} and

(p − 1) E_{K/L_{1}}^{p−1}(Δ) ≡ −E_{K/L_{1}}^{p−1}(Δ) (mod 𝔓_{K}^{1+pt_{2}}).

It is easily seen that

Δ^{p} + Δ E_{K/L_{1}}^{p−1}(Δ) − N_{K/L_{1}}Δ

is equal to

− ∑_{i=1}^{p−2}(−1)^{i} Δ^{p−i} E_{K/L_{1}}^{i}(Δ).

Recalling that xΔ^{p−1} is supposed to lie in 𝔓_{K} we appeal to Lemma 11.7 to see that the product of this expression with x lies in 𝔓_{K}^{1+pt_{2}}. Thus

g(xΔ^{j}) ≡ 0 (mod 𝔓_{F}^{1+t_{2}}).

If p = 3 and ξ = ξ(σ) then

Δ^{σ} = Δ + ξ + 3ξΔ^{2} + 3ξ^{2}Δ + z

with

z = O(ϖ_{K}^{2(p^{2}e−(p−1)u)}).

If

Δ_{σ} = Δ + ξ + 3ξΔ^{2}

then

Δ^{σ} = Δ_{σ} + 3ξ^{2}Δ + z.

If we can show that

3ξ^{2}Δ + z = O(ϖ_{K}^{(p−1)t_{2}+pt_{1}})

Chapter 11 108 it will follow that

xΔ^{σ j}N_{K/L_{1}}Δ^{σ} ≡ xΔ_{σ}^{j}N_{K/L_{1}}Δ_{σ} (mod 𝔓_{K}^{1+pt_{2}})  (11.9)

if xΔ^{j} lies in 𝔓_{K}^{1+t_{2}−t_{1}}

3ξ^{2}Δ = O(ϖ_{K}^{p^{2}e−t_{1}})

and

p^{2}e − t_{1} ≥ p(p − 1)t_{2} − t_{1} ≥ (p − 1)t_{2} + pt_{1}

because (p − 1)^{2} ≥ p. Moreover

2(p^{2}e − (p − 1)u) ≥ 2(p(p − 1)t_{2} − (p − 1)t_{1})

which is at least

(p − 1)t_{2} + ((2p − 1) (p − 1) − 2(p − 1))t_{1}

and

(2p − 1) (p − 1) − 2(p − 1) = (2p − 3) (p − 1) ≥ p.
We want to replace N_{K/L_{1}}Δ_{σ} by
N_{K/L_{1}}(Δ + ξ)

in the right side of (11.9). To do this we have to show that

N_{K/L_{1}} (\frac{Δ_{σ}}{Δ + ξ} ) ≡ 1 (mod P_{K}^{(p−1)t_{2}+(p+1)t_{1}}).

Since

\frac{Δ_{σ}}{Δ + ξ} = 1 + O(ϖ_{K}^{p^{2}e−t_{1}})

and

p^{2}e − t_{1} ≥ p(p − 1)t_{2} − t_{1},

we have only to verify that

p{p(p − 1)t_{2} − t_{1}} ≥ (p − 1)t_{2} + (p + 1)t_{1}  (11.10)

and that the integral part of

\frac{p(p − 1)t_{2} − t_{1} + (p − 1) (t + 1)}{p}  (11.11)

is at least

\frac{(p − 1)t_{2} + (p + 1)t_{1}}{p} .

The inequality (11.10) is clear. Since t ≥ t_{1} the integral part of (11.11) is at least

\frac{p(p − 1)t_{2} + (p − 2)t_{1}}{p} ≥ \frac{(p − 1)t_{2} + ((p − 1)^{2} + (p − 2))t_{1}}{p}

and

(p − 1)^{2} + (p − 2) ≥ p + 1.
Just as when p > 3 we may replace N_{K/L_{1}}(Δ + ξ) in (11.9) by
ξ + N_{K/L_{1}}Δ + ξ E_{K/L_{1}}^{p−1}(Δ).

Thus Z_{j} is congruent to

xΔ^{j}N_{K/L_{1}}Δ + ∑_{ξ} x(Δ + ξ + 3ξΔ^{2})^{j} (ξ + N_{K/L_{1}}Δ + ξ E_{K/L_{1}}^{2}(Δ))

modulo 𝔓_{K}^{1+pt_{2}} if p = 3, j is 1 or 2, and xΔ^{j} belongs to 𝔓_{K}^{1+t_{2}−t_{1}}. If j = 1 this expression is equal to

pxΔ^{j} N_{K/L_{1}}Δ + 2x(1 + 3Δ^{2}) (1 + E_{K/L_{1}}^{2}(Δ))  (11.12)

and if j = 2 it is equal to

pxΔ^{j} N_{K/L_{1}}Δ + 2x{2(1 + 3Δ^{2})Δ(1 + E_{K/L_{1}}^{2}(Δ)) + (1 + 3Δ^{2})^{2}N_{K/L_{1}}Δ}.  (11.13)

The term

pxΔ^{j} N_{K/L_{1}}Δ

can be ignored as before because it lies in 𝔓_{K}^{1+pt_{2}}. Also

3xΔ^{j+1} = O(ϖ_{K}^{1+p^{2}e+t_{2}−2t_{1}})

because xΔ^{j} lies in 𝔓_{K}^{1+t_{2}−t_{1}} and

p^{2}e + t_{2} − 2t_{1} ≥ p(p − 1)t_{2} − t_{1} ≥ pt_{2}.

We may also replace the factor 2 in (11.12) by 1. Thus (11.12) is congruent to

−x(1 + E_{K/L_{1}}^{2}(Δ))

modulo 𝔓_{K}^{1+pt_{2}}. At this point we may argue as we did for p > 3. To simplify (11.13), we observe that

9Δ^{4} N_{K/L_{1}}Δ = O(ϖ_{K}^{2p^{2}e−7t_{1}})

and that

2p^{2}e − 7t_{1} ≥ 12t_{2} − 7t_{1} ≥ 3t_{2}.

Moreover

3xΔ^{2} N_{K/L_{1}}Δ = O(ϖ_{K}^{1+p^{2}e−3t_{1}})

if xΔ^{2} belongs to 𝔓_{K} and

p^{2}e − 3t_{1} ≥ 6t_{2} − 3t_{1} ≥ 3t_{2}.

Thus (11.13) is congruent to

2x{2Δ(1 + E_{K/L_{1}}^{2}(Δ)) + N_{K/L_{1}}Δ}

Chapter 11 110 modulo 𝔓_{K}^{1+3t_{2}}. We may again argue as we did for p > 3.

We turn to the second part of the lemma. We observe first that if x belongs to L_{2}, y belongs to K, and

xy ∈ 𝔓_{K}

then

h(xy) ≡ h(y) N_{K/L_{1}}(x) (mod 𝔓_{F}^{1+t_{2}}).

The left side is

S_{L_{1}/F} (N_{K/L_{1}}xN_{K/L_{1}}yΔ) − S_{L_{2}/F} (x^{p} N_{K/L_{2}}yΔ).

Since N_{K/L_{2}}x = N_{L_{2}/F} x lies in F this equals

{N_{K/L_{1}}x}h(y) + S_{L_{2}/F} {N_{K/L_{2}}(yΔ) (N_{L_{2}/F} x − x^{p})}.

The second term is the trace from L_{2} to F of

\{N_{K/L_{2}}(xyΔ)\} \{\frac{N_{L_{2}/F} x − x^{p}}{N_{K/L_{2}}x} \}

if, as we may as well assume, x ≠ 0. All we need do is show that this expression lies in 𝔓_{L_{2}}^{1+t_{2}} for then its trace will lie in 𝔓_{F}^{1+t_{2}}. The first factor lies in 𝔓_{L_{2}}^{1−t_{2}}. The second factor is equal to

\{∏_{σ∈\overline{G}^{2}} x^{σ−1}\} − 1.

Since p ≥ 3 it will be sufficient to show that the image of the homomorphism

x ⟶ φ(x) = ∏_{σ∈\overline{G}^{2}} x^{σ−1}

of C_{L_{2}} into U_{L_{2}} is contained in U_{L_{2}}^{(p−1)t_{2}}. Let ρ be a generator of \overline{G}^{2} and let P (X) be the polynomial

∑_{i=1}^{p−1}(X^{i} − 1) = ∑_{i=0}^{p−1} X^{i} − p

then

φ(x) = x^{P (ρ)}.

Let

Q(X) = (X − 1)^{p−1}.

If 1 ≤ i ≤ p − 1 the ith coefficient of Q(X) is

(−1)^{p−1−i} \frac{(p − 1) … (p − i)}{i!} ≡ 1 (mod p).

Since both P (X) and Q(X) are divisible by X − 1 P (X) = Q(X) + p(X − 1)R(X)

Chapter 11 111 where R(X) is a polynomial with integral coefficients. For all z in C_{L_{2}},

z^{ρ−1} = 1 + w

with w = O(ϖ_{L_{2}}^{t_{2}}). Then

z^{p(ρ−1)} = (1 + w)^{p} ≡ 1 + w^{p} ≡ 1 (mod 𝔓_{L_{2}}^{pt_{2}})

and

z^{p(ρ−1)R(ρ)} ∈ U_{L_{2}}^{pt_{2}}.

If a ≥ 1 and

w ∈ 𝔓_{L_{2}}^{a}

then

(1 + w)^{p−1} = \frac{1 + w^{p}}{1 + w} = 1 + \frac{w^{p} − w}{1 + w} ≡ 1 (mod 𝔓_{L_{2}}^{a+t_{2}}).

One then shows easily by induction that, for all z in C_{L_{2}} and all n ≥ 1,

z^{(ρ−1)^{n}} ∈ U_{L_{2}}^{nt_{2}}.

If x lies in 𝔓_{K} we may take y = 1. Applying Lemma 11.8 we see that

h(x) ≡ N_{L_{2}/F} xh(1) ≡ 0 (mod 𝔓_{F}^{1+t_{2}}).

If 1 ≤ j ≤ p − 1, x lies in L_{2}, and xΔ^{j} lies in 𝔓_{K},

h(xΔ^{j}) ≡ P_{j} − Q_{j} (mod 𝔓_{F}^{1+t_{2}})

with

P_{j} = N_{L_{2}/F} xS_{L_{1}/F} (N_{K/L_{1}}Δ^{j+1})

and

Q_{j} = N_{L_{2}/F} x S_{L_{2}/F} (N_{K/L_{2}}Δ^{j+1}).

The expression P_{j} is congruent to

N_{L_{2}/F} x{N_{K/L_{1}}Δ^{j+1} + ∑_{ξ} N_{K/L_{1}}(Δ + ξ)^{j+1} {1 − S_{K/L_{1}}Z(ξ)}^{j+1}}  (11.14)

modulo 𝔓_{L_{1}}^{1+pt_{2}}. Since we are working modulo 𝔓_{L_{1}}^{1+pt_{2}} we need only consider

(1 − S_{K/L_{1}} Z(ξ))^{j+1}  (11.15)

modulo 𝔓_{L_{1}}^{pt_{2}+t_{1}}. Suppose first that p > 3. Then

Z(ξ) = O(ϖ_{K}^{p^{2}e−(p−2)u})

and

p^{2}e − (p − 2)u ≥ p(p − 1)t_{2} − (p − 2)u ≥ p(p − 2)t_{2}.

Chapter 11 112 Moreover the integral part of

\frac{p(p − 2)t_{2} + (p − 1) (t + 1)}{p}

is at least

(p − 2)t_{2} − \frac{(p − 1)}{p}t

and twice this is at least pt_{2} + t_{1}. We replace (11.15) by

1 − (j + 1) S_{K/L_{1}} Z(ξ).

Since

Z(ξ) ≡ pΔ^{p−1} ∑_{i=1}^{p−2} a_{i} (\frac{ξ}{Δ})^{i} (mod p^{2})  (11.16)

and

p^{2} = O(ϖ_{K}^{2p^{2}e})

while

2p^{2}e ≥ 2p(p − 1)t_{2} ≥ p(pt_{2} + t_{1}),

we may replace Z(ξ) by the right side of (11.16). By Lemma 11.7

pS_{K/L_{1}}(Δ^{p−1−i}) = O(ϖ_{L_{1}}^{pe+it_{2}})

if 1 ≤ i ≤ p − 2 and

pe + it_{2} ≥ (p − 1)t_{2} + it_{2} ≥ pt_{2} + t_{1}

if i ≥ 2. We replace Z(ξ) by

pa_{1}ξΔ^{p−2}.

We may write (11.14) as

N_{L_{2}/F} xN_{K/L_{1}}Δ^{j+1} \{1 + ∑_{ξ} N_{K/L_{1}} (1 + \frac{ξ}{Δ})^{j+1} \{1 − S_{K/L_{1}}(pa_{1}ξΔ^{p−2})\}\}.

When we expand

N_{K/L_{1}} (1 + \frac{ξ}{Δ})^{j+1}

and sum over ξ we will obtain

N_{L_{2}/F} xN_{K/L_{1}}Δ^{j+1}\{1 + ∑_{ξ} N_{K/L_{1}} (1 + \frac{ξ}{Δ})^{j+1}\}

which we write as

N_{L_{2}/F} x{N_{K/L_{1}}Δ^{j+1} = ∑_{ξ} N_{K/L_{1}}(Δ + ξ)^{j+1}}

plus a sum of terms of the form

αp N_{L_{2}/F} xN_{K/L_{1}}Δ^{j+1} E_{K/L_{1}}^{i} (\frac{1}{Δ}) S_{K/L_{1}}(Δ^{p−2})

Chapter 11 113 where α is rational and lies in O_{F} and i is at least 1. Since

E_{K/L_{1}}^{i} (\frac{1}{Δ}) = O(ϖ_{L_{1}}^{t_{1}})

for i ≥ 1 and

pS_{K/L_{1}}(Δ^{p−2}) = O(ϖ_{L_{1}}^{pt_{2}})

these supplementary terms may be ignored.

Now take p = 3.^{∗}

∗(1998) This is where and how the manuscript of Chapter 11 breaks off.

Chapter 12.

The Second Main Lemma

Suppose K is a normal extension of the local field F and G = 𝔊(K/F) is the direct product of two cyclic groups of prime order ℓ. Let X_{K} be a quasi-character of C_{K}. If σ belongs to G define X_{K}^{σ} by the relation

X_{K}^{σ}(α) = X_{K}(α^{σ^{−1}}).

Suppose that X_{K}^{σ} = X_{K} for all σ in G but that for no quasi-character X_{F} of C_{F} does X_{K} = X_{K/F}. If F ⊆ L ⊆ K and [K : L] = ℓ then X_{K} can be extended to a quasi-character of W_{K/L} because W_{K/L}/C_{K} is isomorphic to 𝔊(K/L) which is cyclic. If this quasi-character is X_{L} then X_{K} = X_{K/L}.

Lemma 12.1

Suppose L_{1} and L_{2} are two fields lying between F and K and [K : L_{1}] = [K : L_{2}] = ℓ. Suppose X_{L_{1}} is a quasi-character of C_{L_{1}}, X_{L_{2}} is a quasi-character of C_{L_{2}}, and

X_{K} = X_{K/L_{1}} = X_{K/L_{2}}.

Then

Δ(X_{L_{1}}, ψ_{L_{1}/F}) ∏_{μ_{F} ∈S(L_{1}/F)} Δ(μ_{F} , ψ_{F})

is equal to

Δ(X_{L_{2}}, ψ_{L_{2}/F}) ∏_{μ_{F} ∈S(L_{2}/F)} Δ(μ_{F} , ψ_{F}).

Because of the assumption on 𝔊(K/F) the field F must be non-archimedean. To prove the lemma in general it is enough to prove it for a given L_{1} and all L_{2}. There are three possibilities to consider.

(i) The sequence of groups of ramification takes the form

G = G_{−1} ≠ G_{0} = … = G_{t} ≠ G_{t+1} = … = {1}.

(ii) The sequence of groups of ramification takes the form

G = G_{−1} = G_{0} = G_{1} = … = G_{u} ≠ G_{u+1} = … = G_{t} ≠ G_{t+1} = … = {1}.

(iii) The sequence of groups of ramification takes the form

G = G_{−1} = G_{0} = G_{1} = … = G_{t} ≠ G_{t+1} = … = {1}.

In the first two cases we take G^{1} = 𝔊(K/L_{1}) to be G_{t}. In the third case the choice of L_{1} is immaterial.

If the relation X_{L_{i}}^{σ} = X_{L_{i}} obtains for one σ different from 1 in \overline{G}^{i} = 𝔊(L_{i}/F) it obtains for all such σ and X_{L_{i}} is of the form X_{L_{i}/F} for some quasi-character X_{F} of C_{F}. Then X_{K} = X_{K/F} which is contrary to assumption. Thus the characters X_{L_{i}}^{σ−1} with σ in \overline{G}^{i} are distinct. They are clearly trivial on N_{K/L_{i}} C_{K} so

\{X_{L_{i}}^{σ−1} | σ ∈ \overline{G}^{i}\} = S(K/L_{i}) = \{μ_{L_{i}/F} | μ_{F} ∈ S(L_{j}/F)\}.

Here j is 2 or 1 according as i is 1 or 2.

Let t_{i} ≥ −1 be that integer for which

\overline{G}^{i} = \overline{G}_{t_{i}}^{i}

while

\overline{G}_{t_{i}+1}^{i} = \{1\}.

Then

δ_{i} = δ(L_{i}/F) = (t_{i} + 1) (ℓ − 1).

In the first case L_{1}/F is unramified and L_{2}/F is ramified. We choose ϖ_{L_{2}} arbitrarily and take ϖ_{K} = ϖ_{L_{2}}. Also we set

ϖ_{L_{1}} = ϖ_{F} = N_{L_{2}/F} ϖ_{L_{2}}.

In the second and third cases K/L_{1} and K/L_{2} are ramified and K/F is totally ramified. We choose ϖ_{k} first and set

ϖ_{L_{i}} = N_{K/L_{i}}ϖ_{K}

and

ϖ_{F} = N_{K/F} ϖ_{K}.
Let m_{i} = m(X_{L_{i}}). The m(X_{L_{i}}^{σ}) = m_{i} and
m(X_{L_{i}}^{σ−1}) ≤ m_{i}.

Thus m(ν) ≤ m_{i} if ν belongs to S(K/L_{i}). If G^{i} = 𝔊(K/L_{i}) and if

G_{u_{i}}^{i} = G^{i}

while

G_{u_{i}+1}^{i} = {1}

then m(ν) = u_{i} + 1 if ν is non-trivial. Thus u_{i} + 1 ≤ m_{i}. Since νX_{L_{i}} is of the form X_{L_{i}}^{σ} for all ν in S(K/L_{i}),

m(νX_{L_{i}}) = m(X_{L_{i}}).

Lemma 8.8 and 8.12 imply that*

m(X_{K}) = ψ_{K/L_{i}} (m_{i} − 1) + 1.

This m(X_{K}) = m_{i} if K/L_{i} is unramified and

m(X_{K}) = ℓm_{i} − δ(K/L_{i})

* We are encountering once again the conflicting uses of the symbole ψ.

Chapter 12 116 if K/L_{i} is ramified. If n = n(ψ_{F}) then n_{i} = n(ψ_{L_{i}/F}) is n if L_{i}/F is unramified and is ℓn + δ_{i} if L_{i}/F is ramified.

In the first case

δ(K/L_{2}) = δ(L_{1}/F) = 0.

The relations

δ(K/F) = δ(K/L_{1}) + ℓδ(L_{1}/F) = δ(K/L_{1}) = (t + 1) (ℓ − 1)

and

δ(K/F) = δ(K/L_{2}) + δ(L_{2}/F) = δ_{2} = (t_{2} + 1) (ℓ − 1)

imply that t_{2} = t. Also

m(X_{K}) = m_{2} = ℓm_{1} − δ(K/L_{1}) = ℓm_{1} − δ_{2}

so that

m_{2} + n_{2} = ℓ(m_{1} + n_{1}).

Moreover

X_{L_{1}}(ϖ_{L_{1}}^{m_{1}+n_{1}}) = X_{K} (ϖ_{L_{2}}^{m_{1}+n_{1}})

is equal to

X_{L_{2}}(ϖ_{L_{2}}^{m_{2}+n_{2}}) = X_{L_{2}}(ϖ_{F}^{m_{1}+n_{1}}) \{X_{L_{2}} (∏_{σ∈\overline{G}^{2}} ϖ_{L_{2}}^{1−σ})\}^{m_{1}+n_{1}}

and

∏_{σ∈\overline{G}_{2}} X_{L_{2}}(ϖ_{L_{2}}^{1−σ}) = ∏_{μ_{F} ∈S(L_{1}/F)} μ_{L_{2}/F} (ϖ_{L_{2}})

is equal to

∏_{μ_{F} ∈S(L_{1}/F)} μ_{F} (ϖ_{F}) = (−1)^{ℓ−1}.

If

S′_{i} = S(L_{i}/F) − {1}

then

∏_{S′_{1}} μ_{F} (ϖ_{F}^{t_{1}+1+n}) = (−1)^{n(ℓ−1)}

and

∏_{S′_{2}} μ_{F} (ϖ_{F}^{t_{2}+1+n}) = 1.

Thus we have to show that

(−1)^{m_{1}(ℓ−1)} Δ_{1}(X_{L_{1}}, ψ_{L_{1}/F} ϖ_{F}^{m_{1}+n_{1}})

is equal to

Δ_{1}(X_{L_{2}}, ψ_{L_{2}/F} , ϖ_{F}^{m_{1}+n_{1}}) ∏_{S′_{2}} Δ_{1}(μ_{F} , ψ_{F} , ϖ_{F}^{t+1+n}).

In the second and third cases the relations m(X_{K}) = ℓm_{1} − δ(K/L_{1}) = ℓm_{2} − δ(K/L_{2})

Chapter 12 117 and

δ(K/F) = δ(K/L_{1}) + ℓδ_{1} = δ(K/L_{2}) + ℓδ_{2}

imply that m_{1} + δ_{1} = m_{2} + δ_{2} and hence that m_{1} + n_{1} = m_{2} + n_{2}. Thus

X_{L_{1}}(ϖ_{L_{1}}^{m_{1}+n_{2}}) = X_{K} (ϖ_{K}^{m_{1}+n_{1}}) = X_{L_{2}}(ϖ_{L_{2}}^{m_{2}+n_{2}}).

Since

∏_{S′_{i}} μ_{F} (ϖ_{F}^{t_{i}+1+n}) = 1

we have to show that

Δ_{1}(X_{L_{1}}, ψ_{L_{1}/F} , ϖ_{L_{1}}^{m_{1}+n_{1}}) ∏_{S′_{1}} Δ_{1}(μ_{F} , ψ_{F} , ϖ_{F}^{t_{1}+1+n})

is equal to

Δ_{1}(X_{L_{2}}, ψ_{L_{2}/F} , ϖ_{L_{2}}^{m_{2}+n_{2}}) ∏_{S′_{2}} Δ_{1}(μ_{F} , ψ_{F} , ϖ_{F}^{t_{2}+1+n}).

Suppose X′_{F} is a quasi-character of C_{F}. According to Lemma 10.1

Δ(X′_{L_{1}/F} , ψ_{L_{1}/F}) ∏_{μ_{F} ∈S(L_{1}/F)} Δ(μ_{F} , ψ_{F})

is equal to

∏_{μ_{F} ∈S(L_{1}/F)} Δ(μ_{F} X′_{F} , ψ_{F})  (12.1)

and

Δ(X′_{L_{2}/F} , ψ_{L_{2}/F}) ∏_{μ_{F} ∈S(L_{2}/F)} Δ(μ_{F} , ψ_{F})

is equal to

∏_{μ_{F} ∈S(L_{2}/F)} Δ(μ_{F} X′_{F} , ψ_{F}).  (12.2)

Suppose m′ = m(X′_{F}) = 2d′ + ε′ and d′ is greater than or equal to both 1 + t_{1} and 1 + t_{2}. Choose γ in F such that

γ O_{F} = P_{F}^{m′+n}

and then choose β = β(X_{F}). By Lemma 9.4 the expression (12.1) is equal to

\{Δ(X′_{F} , ψ_{F})\}^{ℓ} \{∏_{μ_{F} ∈S(L_{1}/F)} μ_{F} (\frac{γ}{β} )\}

and (12.2) is equal to

\{Δ(X′_{F} , ψ_{F})\}^{ℓ} \{∏_{μ_{F} ∈S(L_{2}/F)} μ_{F} (\frac{γ}{β} )\}.

Consequently

Δ(X′_{L_{1}/F} , ψ_{L_{1}/F}) \{∏_{μ_{F} ∈S(L_{1}/F)} μ_{F} (\frac{β}{γ} ) Δ(μ_{F} , ψ_{F})\}

Chapter 12 118 is equal to

Δ(X′_{L_{2}/F} , ψ_{L_{2}/F}) \{∏_{μ_{F} ∈S(L_{2}/F)} μ_{F} (\frac{β}{γ} ) Δ(μ_{F} , ψ_{F})\}.

Suppose that both m_{1} = m(X_{L_{1}}) and m_{2} = m(X_{L_{2}}) are at least 2 and let m_{i} = 2d_{i} + ε_{i}. Suppose that

m(X_{L_{i}}^{−1}X′_{L_{i}/F}) ≤ d_{i}

for i equal to 1 and 2. Then

m_{i} = m(X′_{L_{i}/F}) = ψ_{L_{i}/F} (m′ − 1) + 1.

If

X′_{L_{i}/F} (1 + x) = ψ_{L_{i}/F} (\frac{β_{i}x}{γ})

for x in 𝔓_{L_{i}}^{d_{i}+ε_{i}} then, by Lemma 9.4 again,

Δ(X_{L_{i}}, ψ_{L_{i}/F}) = X_{L_{i}}^{−1}X′_{L_{i}/F} (\frac{β_{i}}{γ} ) Δ (X′_{L_{i}/F} , ψ_{L_{i}/F}).

Thus to prove Lemma 12.1 in the present circumstances we have only to verify that

X_{L_{1}}^{−1}X′_{L_{1}/F} (\frac{β_{1}}{γ} ) ∏_{μ_{F} ∈S(L_{1}/F)} μ_{F} (\frac{γ}{β} )

is equal to

X_{L_{2}}^{−1}X′_{L_{2}/F} (\frac{β_{2}}{γ} ) ∏_{μ_{F} ∈S(L_{2}/F)} μ_{F} (\frac{γ}{β} ) .

Suppose first that ℓ is odd. Then

∏_{μ_{F} ∈S(L_{i}/F)} μ_{F} (\frac{γ}{β} ) = 1

and we need only verify that

X_{L_{1}}^{−1} (\frac{γ}{β_{1}} ) X′_{L_{2}/F} (\frac{γ}{β_{2}} ) = X_{L_{2}}^{−1} (\frac{γ}{β_{2}} ) X′_{L_{2}/F} (\frac{γ}{β_{2}} ) .

According to Lemmas 8.3 and 8.4 we may take β_{1} = β_{2} = β.

Certainly

X′_{L_{1}/F} (\frac{γ}{β} ) = X′_{F} (\frac{γ^{ℓ}}{β^{ℓ}} ) = X′_{L_{2}/F} (\frac{γ}{β} ) .

Since C_{F} is the product of N_{L_{1}/F} C_{L_{1}} and N_{L_{2}/F} C_{L_{2}} we may write \frac{γ}{β} as a product

\frac{γ}{β} = N_{L_{1}/F} δ_{1} N_{L_{2}/F} δ_{2}.

Consider

X_{L_{i}}(N_{L_{j}/F} δ_{j}) = X_{K}(δ_{j})

where j is 1 or 2 according as i is 2 or 1. The right side equals

X_{L_{j}}(δ_{j}^{ℓ}) = X_{L_{j}}(N_{L_{j}/F} δ_{j}) ∏_{σ∈𝔊(L_{j}/F)} X_{L_{j}}(δ_{j}^{1−σ}).

The product is equal to

∏_{μ_{F} ∈S(L_{i}/F)} μ_{L_{j}/F} (δ_{j})

which is 1 because ℓ is odd.

Before discussing the case ℓ = 2 we consider the circumstances under which, for a given X_{L_{1}} and X_{L_{2}}, a quasi-character X′_{F} with the properties described above exists. Lemma 12.2

(a) If L_{i}/F is unramified, x belongs to U_{L_{i}}^{u_{i}+1}, and

N_{L_{i}/F} (x) = 1

then

X_{L_{i}}(x) = 1.

(b) If L_{i}/F is ramified, K/L_{i} is unramified, x belongs to U_{L_{i}}^{t_{i}+1}, and

N_{L_{i}/F} (x) = 1

then

X_{L_{i}}(x) = 1.

(c) If L_{i}/F and K/L_{i} are ramified, x belongs to U_{L_{i}}^{u_{i}+t_{i}+1}, and

N_{L_{i}/F} (x) = 1

then

X_{L_{i}}(x) = 1.

Choose some non-trivial σ_{i} in \overline{G}^{i} = 𝔊(L_{i}/F). Then

μ_{L_{i}} = X_{L_{i}}^{σ_{i}^{−1}−1}

is a non-trivial character in S(K/L_{i}) and

m(μ_{L_{i}}) = u_{i} + 1.
Since L_{i}/F is cyclic there is a y in C_{L_{i}} such that
x = y^{σ_{i}−1}.

We shall show that y can be taken in U_{L_{i}}^{u_{i}+1}. Then

X_{L_{i}}(x) = μ_{L_{i}}(y) = 1.

Suppose L_{i}/F is unramified. If we cannot choose y in U_{L_{i}}^{u_{i}+1} there is a largest integer a ≥ −1 such that we can choose y in U_{L_{i}}^{a} where a is of course less than u_{i} + 1. Choose such a y. Then a is not -1 because we can always divide y by a power of ϖ_{F}. If a were 0 then y could not be congruent to an element of U_{F} modulo 𝔓_{F}. Then y^{σ_{i}−1} would not be in U_{F}^{1} . Since u_{i} + 1 > 0 in the present situation this is impossible. Let

y = 1 + εϖ_{F}^{a} .

Then ε cannot be congruent to an element of O_{F} modulo 𝔓_{F}. Thus

ε^{σ_{i}} − ε ≢ 0 (mod 𝔓_{L_{i}})

and

y^{σ_{i}−1} ≡ 1 + (ε^{σ_{i}} − ε) ϖ_{F}^{a} (mod 𝔓_{L_{i}}^{a+1})

is not in U_{L_{i}}^{a+1}. This is a contradiction.

Now suppose L_{i}/F is ramified and K/L_{i} is unramified. Then t_{i} + 1 ≥ 1 and u_{i} + 1 = 0. We need only show that y can be taken to be a unit. Write

y = εϖ_{L_{i}}^{b}

where ε is a unit. If b is congruent to 0 modulo ℓ we can divide y by some power of ϖ_{F} to obtain an element of U_{F} = U_{F}^{0} . To see that b must be congruent to 0 modulo ℓ we suppose the contrary. Then

y^{σ_{i}−1} = ε^{σ_{i}−1} (ϖ_{L_{i}}^{σ_{i}−1})^{b} ≡ (ϖ_{L_{i}}^{σ_{i}−1})^{b} (mod 𝔓_{L_{i}}^{t_{i}+1}).

If t_{i} = 0 the residue of ϖ_{L_{i}}^{σ_{i}−1} modulo 𝔓_{L_{i}} is a non-trivial ℓth root of unity and

(ϖ_{L_{i}}^{σ_{i}−1})^{b} ≢ 1 (mod 𝔓_{L_{i}}).

If t_{i} > 0 then

ϖ_{L_{i}}^{σ_{i}−1} = 1 + α ϖ_{L_{i}}^{t_{i}}

where α is a unit. Thus

(ϖ_{L_{i}}^{σ_{i}−1})^{b} ≡ 1 + α b ϖ_{L_{i}}^{t_{i}} (mod P_{L_{i}}^{t_{i}+1}).

The right side is not congruent to 0 modulo 𝔓_{L_{i}}^{t_{i}+1}.

Now suppose L_{i}/F and K/L_{i} are both ramified. Then ℓ = p and both u_{i} and t_{i} are at least 1. Again suppose that y cannot be chosen in U_{L_{i}}^{u_{i}+1} and let a be the largest integer such that y can be chosen in U^{a}. The argument just used shows that a ≥ 0. Since L_{i}/F is ramified

U_{L_{i}}^{k_{p}} = U_{F}^{k} U_{L_{i}}^{k_{p}+1}.

Therefore a is not divisible by p and in particular is at least 1. Let

y = 1 + εϖ_{L_{i}}^{a}

where β is a unit. Then

y^{σ_{i}−1} = (1 + ε^{σ_{i}}ϖ_{L_{i}}^{aσ_{i}}) (1 + εϖ_{L_{i}}^{a})^{−1}.

Let

ε^{σ_{i}} = ε + η ϖ_{L_{i}}^{t_{i}+1}

and

ϖ_{L_{i}}^{σ_{i}−1} = 1 + α ϖ_{L_{i}}^{t_{i}}

where α is a unit. Then y^{σ_{i}−1} is equal to

{1 + (ε + η ϖ_{L_{i}}^{t_{i}+1}) (1 + α ϖ_{L_{i}}^{t_{i}})^{a}ϖ_{L_{i}}^{a}} {1 + ε ϖ_{L_{i}}^{a}}^{−1}

which is congruent to

1 + aαε ϖ_{L_{i}}^{t_{i}+a}

modulo 𝔓_{L_{i}}^{t_{i}+a+1}. Therefore a ≥ u_{i} + 1. This is a contradiction. Lemma 12.3

If L_{1}/F is unramified we can choose X′_{F} such that

m(X_{L_{1}}^{−1}X′_{L_{1}/F}) = t + 1

and

m(X_{L_{2}}^{−1}X′_{L_{2}/F}) = t + 1.

If m(X_{L_{1}}) > t + 1 then m(X′_{F}) will equal m(X_{L_{1}}).

By the previous lemma we can define a quasi-character X′_{F} of

N_{L_{1}/F} U_{L_{1}}^{u_{1}+1}

by setting

X′_{F} (N_{L_{1}/F} x) = X_{L_{1}}(x).

We extend X′_{F} to a quasi-character, which we again denote by X′_{F}, of C_{F}. Then m(X_{L_{1}}^{−1}X′_{L_{1}/F}) ≤ u_{1} + 1.

However X_{K}^{−1}X′_{K/F} , X_{L_{1}}^{−1}X′_{L_{1}/F}, and X_{L_{2}}^{−1}X′_{L_{2}/F} satisfy the conditions of Lemma 12.1. Therefore

m(X_{L_{1}}^{−1}X′_{L_{1}/F}) ≥ u_{1} + 1.

Since L_{1}/F is unramified u_{1} and t_{2} are both equal to t. Thus

m(X_{L_{1}}^{−1}X′_{L_{1}/F}) = t + 1

and

m(X_{L_{2}}^{−1}X′_{L_{2}/F}) = ℓ(u_{1} + 1) − δ_{2} = ℓ(u_{1} + 1) − (ℓ − 1) (t_{2} + 1) = t + 1.

The last assertion of the lemma is clear. Lemma 12.4 Chapter 12 122

If K/F is totally ramified then

m(X_{L_{i}}) ≥ t_{i} + u_{i} + 1.

There exists X′_{F} such that

m(X_{L_{i}}^{−1}X′_{L_{i}/F}) = t_{i} + u_{i} + 1

for i equal to 1 and 2.

In the present circumstances t_{i} and u_{i} are both at least 1. Choose a non-trivial σ_{i} in \overline{G}^{i} and let

μ_{L_{i}} = X_{L_{i}}^{σ_{i}^{−1}−1}

as before. Choose y in U_{L_{i}}^{u_{i}} so that

μ_{L_{i}}(y) ≠ 1.

Then

X_{L_{i}}(y^{σ_{i}−1}) ≠ 1.

However if

y = 1 + ε ϖ_{L_{i}}^{u_{i}}

where ε is a unit then

y^{σ_{i}−1} ≡ 1 + u_{i}α ε ϖ_{L_{i}}^{t_{i}+u_{i}} (mod 𝔓_{L_{i}}^{t_{i}+u_{i}+1})

if

ϖ_{L_{i}}^{σ_{i}−1} = 1 + α ϖ_{L_{i}}^{t_{i}}.

In particular

y^{σ_{i}−1} ∈ U_{L_{i}}^{t_{i}+u_{i}}

so that

m(X_{L_{i}}) ≥ t_{i} + u_{i} + 1.

Just as in the previous lemma we can find a quasi-character X′_{F} of C_{F} such that

m(X_{L_{1}}^{−1}X′_{L_{1}/F}) = t_{1} + u_{1} + 1.

We have seen that m_{1} + δ_{1} = m_{2} + δ_{2}. The same argument shows that

m(X_{L_{1}}^{−1}X′_{L_{1}/F}) + δ_{1} = m(X_{L_{2}}^{−1}X′_{L_{2}/F}) + δ_{2}.

To complete the proof of the lemma we show that

t_{1} + u_{1} + δ_{1} = t_{2} + u_{2} + δ_{2}.

Since

δ_{i} = (ℓ − 1) (t_{i} + 1)

we have only to show that

u_{1} + 1 + ℓ(t_{1} + 1) = u_{2} + 1 + ℓ(t_{2} + 1).

Multiplying the left or the right side by ℓ−1 we obtain δ(K/F). The equality follows immediately.

Lemmas 12.3 and 12.4 together with the remarks which provoked them allow us to prove Lemma 12.1 in many, but by no means all, cases. We shall not however apply these lemmas immediately. We shall rather begin the systematic exposition of the proof of Lemma 12.1 taking up the cases to which these lemmas apply in their turn.

Suppose first that L_{1} is unramified over F. As before m_{i} = m(X_{L_{i}}). Then

m_{2} = m_{1} + (ℓ − 1) (m_{1} − t − 1) ≥ m_{1}

because u_{1} = t. Since the number m_{1} is at least t + 1 and t ≥ 0 it is at least 1. If m_{1} = 1 then t = 0 and m_{2} = 1. Once we have treated this case, as we shall immediately, we may suppose that m_{2} ≥ m_{1} > 1.

If m_{2} = 1 let
λ = O_{L_{2}}/𝔓_{L_{2}} = O_{F} /𝔓_{F}

and let

κ = O_{K}/𝔓_{K} = O_{L_{1}}/𝔓_{L_{1}}.

κ is an extension of λ. The restriction of X_{L_{1}} to U_{L_{1}} defines a character X_{λ} of λ^{∗} and the restriction of X_{L_{2}} to U_{K} defines a character X_{κ} of κ^{∗}. The restriction of X_{K} to U_{K} defines a character of κ^{∗} which is equal to X_{κ/λ} and to X_{κ}^{ℓ} so that

X_{κ}^{ℓ} = X_{κ/λ}.

As σ varies over \overline{G}^{2}, ϖ_{L_{2}}^{σ−1}, taken modulo P_{L_{2}}, varies over the ℓth roots of unity in λ and if

X_{L_{2}}^{σ^{−1}−1} = ν

then

X_{λ}(ϖ_{L_{2}}^{σ−1}) = ν(ϖ_{L_{2}}).

The right side is not 1 if σ ≠ 1 because ν is then non-trivial. Thus the restriction of X_{λ} to the ℓth roots of unity is not trivial. To every μ_{F} in S(L_{2}/F) is associated a character μ_{λ} of λ^{∗} which is of order 1 if μ_{F} = 1 and of order ℓ otherwise. If ψ_{λ} is the additive character of λ defined by

ψ_{λ}(x) = ψ_{F} (\frac{x}{ϖ_{F}^{1+n}} )

then

Δ_{1}(μ_{F} , ψ_{F} , ϖ_{F}^{1+n}) = A [−τ(μ_{λ}, ψ_{λ})]

if μ_{F} is not trivial. Moreover

ψ_{L_{2}/F} (\frac{x}{ϖ_{F}^{1+n}} ) = ψ_{λ}(ℓ x)

and

Δ_{1}(X_{L_{2}}, ψ_{L_{2}/F} , ϖ_{F}^{1+n}) = A [−X_{λ}(ℓ)τ(X_{λ}, ψ_{λ})].

Finally

Δ_{1}(X_{L_{1}}, ψ_{L_{1}/F} , ϖ_{F}^{1+n}) = A [−τ(X_{κ}, ψ_{κ/λ})].

Thus the required identity is a consequence of the relation

τ(X_{κ}, ψ_{κ/λ}) = X_{λ}(ℓ)τ(X_{λ}, ψ_{λ}) ∏_{μ_{λ}≠1} τ(μ_{λ}, ψ_{λ})

which we proved as Lemma 7.9.

Retaining the assumption that L_{1}/F is unramified we now suppose that m_{1} > 1. There are two possibilities. (a)

m_{1} ≥ 2(t + 1)

(b)

t + 1 ≤ m_{1} < 2(t + 1).

The second possibility occurs only when t > 0.

If m_{j} ≥ 2(t + 1) choose X′_{F} so that

m(X_{L_{i}}^{−1}X′_{L_{i}/F}) = t + 1

for i = 1 and 2. It is clear that

m(X_{L_{1}}^{−1}X′_{L_{1}/F}) ≤ d_{1}

if m_{i} = 2di + ε_{i}. Since m_{2} ≥ m_{1} we also have

m(X_{L_{2}}^{−1}X′_{L_{2}/F}) ≤ d_{2}.

Moreover

m′ = m(X′_{F}) = m_{1}

so that d′ is greater than or equal to both 1 + t_{2} = 1 + t and 1 + t_{1} = 0. Lemma 12.1 for L_{1/F} unramified and m_{1} ≥ 2(t + 1), follows immediately if ℓ is odd. Suppose ℓ = 2.

If t = 0 we can invoke Lemmas 8.3 and 8.7 to see that if β = β(X′_{F}) we may choose β_{1} = β(χ_{L_{1}/F}) and β_{2} = β(X′_{L_{2}/F}) equal to β. If μ_{F}^{(1)} is the non-trivial element of S(L_{1}/F) and μ_{F}^{(2)} is the non-trivial element of S(L_{2}/F) we have only to show that

X_{L_{1}} (\frac{γ}{β} ) X′_{L_{1}/F} (\frac{β}{γ} ) μ_{F}^{(1)} (\frac{γ}{β} )

is equal to

X_{L_{2}} (\frac{γ}{β} ) X′_{L_{2}/F} (\frac{β}{γ} ) μ_{F}^{(2)} (\frac{γ}{β} ) .

Certainly

X′_{L_{1}/F} (\frac{β}{γ} ) = X′_{L_{2}/F} (\frac{β}{γ} )

and we need only show that if δ is in C_{F} then

X_{L_{1}} (δ) μ_{F}^{(1)} (δ) = X_{L_{2}}(δ) μ_{F}^{(2)} (δ).

We may write

δ = N_{L_{1}/F} δ_{1} N_{L_{2}/F} δ_{2}.

Then

μ_{F}^{(i)}(N_{L_{1}/F} δ_{i}) = 1

and, if j is 1 or 2 according as i is 2 or 1,

X_{L_{i}}(N_{L_{j}/F} δ_{j}) = X_{K}(δ_{j}) = X_{L_{j}}(δ_{j}^{2})

which equals

X_{L_{j}}(N_{L_{j}/F} δ_{j})μ_{L_{j}/F}^{(i)} (δ_{j}) = X_{L_{j}}(N_{L_{j}/F} δ_{j})μ_{F}^{(i)}(N_{L_{j}/F} δ_{j}).

The required equality follows immediately.

If t is positive we may still choose β_{1} = β. If m_{1} − t − 1 = v then, by Lemma 8.6, we may choose β_{2} in the form

β_{2} = β + η

with η in 𝔓_{L_{2}}^{v}. Since v ≥ t + 1

X_{L_{2}}^{−1}(β_{2})X′_{L_{2}/F} (β_{2}) = X_{L_{2}}^{−1}(β)X′_{L_{2}/F} (β).

This observation made, we can proceed as before.

Some preparation is necessary before we discuss the second possibility. Suppose that t is positive so that ℓ is equal to the residual characteristic p. The finite field λ_{1} = O_{L_{1}}/P_{L_{1}} is an extension of degree p of ϕ = O_{F} /P_{F}.

The map

x ⟶ x^{p} − x

is an additive endomorphism of ϕ with the prime field as kernel. Choose a y in ϕ which is not in the image of this map and consider the equation

x^{p} − x = y.

If x, in some extension field of ϕ, satisfies this equation and ϕ has q elements then x^{q} ≠ x. However

(x^{q} − x)^{p} − (x^{q} − x) = (x^{p} − x)^{q} − (x^{p} − x) = y^{q} − y = 0.

So

x^{q} − x = z

where z is a non-zero element of the prime field. Then

x^{q^{2}} = (x + z)^{q} = x^{q} + z = x + 2z

and in general

x^{q^{n}} = x + nz.

Thus the lowest power n of q such that x^{q^{n}} = x is n = p and x determines an extension of degree p. Consequently x may be chosen to lie in λ_{1} and then λ_{1} = ϕ(x).

Let E^{r}(x) be the rth elementary symmetric function of x and its conjugates. Since

x^{p} − x + (−1)^{p}N_{λ_{1}/ϕ}x = 0  (12.3)

we have

E^{r}(x) = 0  (12.4)

if 1 ≤ r < p − i,

E^{p−1}(x) = (−1)^{p}  (12.5)

and, of course,

E^{p}(x) = N_{λ_{1}/ϕ}x.  (12.6)

If λ is a non-zero element of the prime field we can replace y by λy. Then x is to be replaced by λx. Also we can replace x by x + λ without changing y.

Let R(L_{1}) be the set of (q^{p} − 1)th roots of unity in L_{1}. Choose a γ in R(L_{1}) whose image in λ_{1} is x. If we are dealing with fields of power series γ will also satisfy the equations (12.3), (12.4), (12.5) and (12.6). Let us see how these equations are to be modified for fields of characteristic zero. F and L_{1} are then extensions of the p-adic field ℚ_{p}. Let F^{0} and L_{1}^{0} be the maximal unramified extensions of ℚ_{p} contained in F and L_{1} respectively. R(L_{1}) is a subset of L_{1}^{0} and p generates the ideal 𝔓_{F^{0}} and the ideal 𝔓_{L_{1}^{0}}. Thus

γ^{p} − γ + (−1)^{p} N_{L_{1}/F} γ ≡ 0 (mod p)

and

E^{r}(γ) ≡ 0 (mod p)

if 1 ≤ r < p − 1 while

E^{p−1}(γ) ≡ (−1)^{p} (mod p).

Let

S^{r}(γ) = ∑_{σ∈𝔊(L_{1}/F)} γ^{rσ}.

The following relations are special cases of Newton’s formulae.

S^{1}(γ) − E^{1}(γ) = 0
S^{2}(γ) − E^{1}(γ) S^{1}(γ) + 2E^{2}(γ) = 0
S^{3}(γ) − E^{1}(γ) S^{2}(γ) + E^{2}(γ) S^{1}(γ) − 3E^{3}(γ) = 0
⋮
S^{p−1}(γ) − E^{1}(γ) S^{p−2}(γ) + … + (−1)^{p−1}(p − 1)E^{p−1}(γ) = 0
S^{p}(γ) − E^{1}(γ) S^{p−1}(γ) + … + (−1)^{p}p E^{p}(γ) = 0.

We infer that

S^{r}(γ) ≡ 0 (mod p)

if 1 ≤ r < p − 1 and that

S^{p−1}(γ) ≡ (−1)^{p}(p − 1) E^{p−1}(γ) (mod p).

Combining the first of these congruences with Newton’s formulae we obtain

S^{r}(γ) ≡ r(−1)^{r+1} E^{r}(γ) (mod p^{2})

if 1 ≤ r ≤ p − 1. If p is odd

S^{p}(γ) − p E^{p}(γ) ≡ E^{1}(γ) S^{p−1}(γ) − E^{p−1}(γ) S^{1}(γ) (mod p^{2}).

The right side is equal to

E^{1}(γ) (S^{p−1}(γ) − E^{p−1}(γ)) ≡ 0 (mod p^{2}).

If p is even

S^{2}(γ) + 2E^{2}(γ) = {E^{1}(γ)}^{2}.

Since

E^{1}(γ) ≡ 1 (mod 2)

we have

S^{2}(γ) + 2E^{2}(γ) ≡ 1 (mod 4).

If σ ≠ 1 belongs to 𝔊(L_{1}/F) there is a (p − 1)th root of unity ζ such that

γ^{σ} − γ ≡ ζ (mod p).

By a suitable choice of y the root ζ can be made to equal, for a given σ, any chosen (p − 1)th root of unity.

The above relations are of course also valid when F is a field of power series.

Choose a non-trivial character μ_{F} in S(L_{2}/F) and choose α so that

μ_{F} (1 + x) = ψ_{F} (\frac{α x}{ϖ_{F}^{t+1+n}} )

Chapter 12 128 if x is in 𝔓_{F}^{s}. Here s is the least integer greater than or equal to \frac{t+1}{2}. If ζ is a (p − 1)th root of unity we define μ_{F}^{ζ} to be μ_{F}^{j} if j is the unique integer such that

ζ ≡ j (mod p).

As we observed in the proof of Lemma 8.5

μ_{F}^{ζ} (1 + x) = ψ_{F} (\frac{α ζx}{ϖ_{F}^{1+t+n}} )

if x is in 𝔓_{F}^{s}.

Let m_{i} = 2d_{i} + ε_{i} as usual. If β_{1} in L_{1} is such that

X_{L_{1}}(1 + x) = ψ_{L_{1}/F} (\frac{αβ_{1}x}{ϖ_{F}^{m_{1}+n_{1}}} )

for x in 𝔓_{L_{1}}^{d_{1}+ε_{1}} then, if σ ≠ 1 belongs to \overline{G}^{1} = 𝔊(L_{1}/F) and x belongs to 𝔓_{L_{1}}^{d_{1}+ε_{1}},

ψ_{L_{1}/F} (\frac{α(β_{1}^{σ} − β_{1})x}{ϖ_{F}^{m_{1}+n_{2}}}) = X_{L_{1}}^{σ−1}(1 + x)

is equal to

ψ_{L_{1}/F} (\frac{αζ_{σ}x}{ϖ_{F}^{1+t+n}} )

if ζ_{σ} is such that

X_{L_{1}}^{σ−1} = μ_{F}^{ζ_{σ}}.

Thus if

v = m_{1} − 1 − t

we have

β_{1}^{σ} − β_{1} ≡ ζ_{σ} ϖ_{F}^{v} (mod 𝔓_{L_{1}}^{d_{1}}).

It is clear that

ζ_{στ} ≡ ζ_{σ}^{τ} + ζ_{σ} (mod 𝔓_{L_{1}}).

Suppose

γ^{σ} − γ ≡ ξ_{σ} (mod p)

where ξ_{σ} is also a (p − 1)th root of unity. Then

ξ_{στ} ≡ ξ_{σ}^{τ} + ξ_{σ} (mod 𝔓_{L_{1}}).

We observed that we could arrange that

ξ_{σ} = ζ_{σ}

for one non-trivial σ. Once we do this the equality will hold for all σ. Then γ^{pσ} − γ^{p} ≡ ξ_{σ} (mod p) and

(β_{1} − γ^{p}ϖ_{F}^{v})^{σ} ≡ β_{1} − γ^{p}ϖ_{F}^{v} (mod 𝔓_{L_{1}}^{d_{1}})

for all σ because, as we observed in the proof of Lemma 8.5, p belongs to P_{L_{1}}^{r} if r + s = t + 1 and

2(r + v) ≥ t + 2v = 2m_{1} − 2 − 2t + t

which is at least

(m_{1} − 1) + (m_{1} − 1 − t) ≥ m_{1} − 1

so that r + v ≥ d_{1}. Since L_{1}/F is unramified there is therefore a β in F such that

β_{1} − γ^{p}ϖ_{F}^{v} ≡ β (mod 𝔓_{L_{1}}^{d_{1}}).

We may suppose that

β_{1} = β + γ^{p}ϖ_{F}^{v} .

β is a unit unless v = 0. If v = 0 then, by replacing γ by a root of unity congruent to γ + 1 modulo 𝔓_{L_{1}} if necessary, we can still arrange that β is a unit. β is congruent to a norm N_{L_{2}/F} β′ modulo P_{F}^{t}. Since d_{1} ≤ t we^{∗}

∗(1998) At the moment this is all that could be found of Chapter 12.

Chapter Thirteen.

The Third Main Lemma

Suppose K/F is Galois and G = 𝔊(K/F). Suppose G = HC when H ≠ {1}, H ∩ C = {1}, and C is a non-trivial abelian normal subgroup of G which is contained in every non-trivial normal subgroup of G.

Lemma 13.1

Let E be the fixed field of H and let X_{F} be a quasi-character of C_{F}. If m = m(X_{F}) then

m(X_{E/F}) = ψ_{E/F} (m − 1) + 1.

Set

m′ = m(X_{E/F}) − 1.

Observe that m′ − 1 is the greatest lower bound of all real numbers v > −1 such that X_{E/F} is trivial on U_{E}^{v} and that m − 1 is the greatest lower bound of all real numbers u such that X_{F} is trivial on U_{F}^{u}. Since

N_{E/F} (U_{E}^{ψ_{E/F} (u)}) ⊆ U_{F}^{u}

we see immediately that

m′ − 1 ≤ ψ_{E/F} (m − 1).

To prove the lemma we need only show that

N_{E/F} (U_{E}^{ψ_{E/F} (m−1)}) ⊇ U_{F}^{m−1}.

We show this with m − 1 replaced by any u ≥ −1.

By Lemma 6.15, τ_{K/F} maps W_{K/F}^{u} onto U_{F}^{u}. The projection of W_{K/F}^{u} on G is a normal subgroup of G. Thus it is either {1} or a subgroup containing C. If it is {1} then

W_{K/F}^{u} = W_{K/F}^{u} ∩ C_{K} = U_{K}^{ψ_{K/F} (u)}

and

U_{F}^{u} = N_{K/F} (U_{K}^{ψ_{K/F} (u)}) = N_{E/F} (N_{K/E}U_{K}^{ψ_{K/F} (u)})

which, by Lemma 6.6, is contained in

N_{E/F} (U_{E}^{ψ_{E/F} (u)}).

Suppose the projection is not {1}. If L is the fixed field of C the group W_{K/F}^{u} contains

{wvw^{−1}v^{−1} | w ∈ W_{K/F} , v ∈ W_{K/F}^{u} ∩ W_{K/L}}.  (13.1)

Since C is generated by

{σρσ^{−1}ρ^{−1} | σ ∈ G, ρ ∈ C}

the group generated by the set (13.1) contains a set of representatives for the cosets of C_{K} in W_{K/L}. This group clearly lies in the kernel of τ_{K/F}. Thus every element of W_{K/F}^{u} is congruent modulo the kernel of τ_{K/F} in W_{K/F}^{u} to an element of

W_{K/F}^{u} ∩ W_{K/E} = W_{K/E}^{ψ_{E/F} (u)}

and

U_{F}^{u} = τ_{K/F} (W_{K/F}^{u}) = τ_{K/F} (W_{K/E}^{ψ_{E/F} (u)})

which is

N_{E/F} (τ_{K/E}(W_{K/E}^{ψ_{E/F} (u)}))

and this set is contained in

N_{E/F} (U_{E}^{ψ_{E/F} (u)}).

Suppose F_{1} is non-archimedean, K_{1}/F_{1} is Galois and F_{1} ⊆ E_{1} ⊆ K_{1}. Let μ_{1} be a character of 𝔊(K_{1}/E_{1}). We may also regard μ_{1} as a character of C_{E_{1}}. Let σ be an element of 𝔊(K_{1}/F_{1}) and define the character of μ_{1}^{σ} of 𝔊(K_{1}/E_{1}^{σ}) by

μ_{1}^{σ}(ρ) = μ_{1}(σρσ^{−1})

for ρ ∈ 𝔊(K_{1}/E_{1}^{σ}) or, what amounts to the same

μ_{1}^{σ}(α) = μ_{1}(α^{σ^{−1}})

for α ∈ C_{E_{1}^{σ}}. Since

ψ_{E_{1}^{σ}/F_{1}}(α) = ψ_{E_{1}/F_{1}}(α^{σ^{−1}})

the next lemma is a congruence of the definitions.

Lemma 13.2

Δ(μ_{1}^{σ}, ψ_{E_{1}^{σ}/F_{1}}) = Δ(μ_{1}, ψ_{E_{1}/F_{1}}).

We return to the extension K/L and the group G. Let T be a set of representatives for the orbits under G of the non-trivial characters in S(K/L). If μ ∈ T let G_{μ} be the isotropy group of μ and let F_{μ} be the fixed field of G_{μ}. Let H_{μ} = H ∩ G_{μ}. Since C is contained in G_{μ} we have G_{μ} = H_{μ} · C. Then μ may also be regarded as a character of C. Let μ′ be the character of G_{μ} defined by

μ′(hc) = μ(c)

if h ∈ H_{μ} and c ∈ C. Eventually we must show that

Δ(X_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})  (13.2)

is equal to

Δ(X_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′X_{F_{μ}/F} , ψ_{F_{μ}/F})  (13.3)

if X_{F} is a quasi-character of C_{F}. At the moment we content ourselves with a special case. The next lemma will be referred to as the Third Main Lemma.

Lemma 13.3

If K/F is tamely ramified the expressions (13.2) and (13.3) are equal.

As we observed in Lemma 6.4 the extension L/F will be unramified and ℓ = [C : 1] will be a prime. Choose a generator ϖ_{F} of 𝔓_{F}. Since F_{μ}/F is unramified we may choose ϖ_{F_{μ}} = ϖ_{F}. Choose ϖ_{E} so that N_{E/F} ϖ_{E} = ϖ_{F}. Certainly

δ_{L/F} = δ_{K/E} = 0

while

δ_{K/L} = ℓ − 1.

Since

δ_{K/F} = δ_{K/L} + ℓ δ_{L/F} = δ_{K/E} + δ_{E/F}

we conclude that

δ_{E/F} = ℓ − 1.

Clearly

∑_{μ}[F_{μ} : F] = ∑_{μ}[H : H_{μ}] = ∑_{μ}[G : G_{μ}]

is just the number of non-trivial characters in S(K/L), that is ℓ − 1. Moreover m(μ′) = 1. Let E_{μ} be the fixed field of H_{μ}. Then

N_{E_{μ}/F_{μ}}(ϖ_{E}) = N_{E/F} (ϖ_{E}) = ϖ_{F} .

Thus, as an element of C_{F_{μ}}, ϖ_{F} lies in the image of W_{K/E_{μ}} under τ_{K/F_{μ}} and hence μ′(ϖ_{F}) = 1. Also

n(ψ_{E/F}) = ℓn(ψ_{F}) + δ_{E/F} = ℓn + (ℓ − 1)

while

n(ψ_{F_{μ}/F}) = n.
If m = m(X_{F}) = 0 then
m(X_{E/F}) = m(X_{F_{μ}/F}) = 0

and

X_{E/F} (ϖ_{E}^{ℓn+ℓ−1}) = X_{F} (ϖ_{F}^{ℓn+ℓ−1}) = X_{F} (ϖ_{F}^{n}) ∏_{μ} X_{F_{μ}/F} (ϖ_{F}^{1+n})

so that the lemma amounts to the equality

∏_{μ} Δ_{1}(μ, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}) = ∏_{μ} Δ_{1}(μ, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}).

If m > 0 then, by Lemma 6.4,

m(X_{E/F}) = ℓm − (ℓ − 1)

and

m(X_{E/F}) + n(ψ_{E/F}) = ℓ(m + n).

Since K/E is unramified

m(X_{K/F}) = m(X_{E/F}) = ℓm − (ℓ − 1).

However

X_{K/F} = (μ′X_{F_{μ}/F})_{K/F_{μ}}

so that

ψ_{K/F_{μ}}(m(μ′X_{F_{μ}/F}) − 1) ≥ m(X_{K/F}) − 1 = ℓ(m − 1)

or

m(μ′X_{F_{μ}/F}) − 1 ≥ φ_{K/F_{μ}}(ℓ(m − 1)) = m − 1.

Consequently

m(μ′X_{F_{μ}/F}) ≥ m.

Since it is clearly less than or equal to m it is equal to m. Because

X_{E/F} (ϖ_{F}^{m+n}) = X_{F} (ϖ_{F}^{ℓ(m+n)}) = X_{F} (ϖ_{F}^{m+n}) ∏_{μ} X_{F_{μ}/F} (ϖ_{F}^{m+n})

we have to show that

Δ_{1}(X_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∏ Δ_{1}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{m+n})

is equal to

Δ_{1}(X_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏ Δ_{1}(μ′X_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n}).
Let ϕ be the field O_{F} /P_{F}, let λ = O_{L}/P_{L}, let q be the number of elements in ϕ, and let
f = [λ : ϕ] = [L : F].

Let θ be the homomorphism of C into λ^{∗} introduced in Chapter IV of Serre’s book. Thus

θ(c) = ϖ_{E}^{c−1} (mod 𝔓_{F})

so that if h ∈ H

θ(h^{−1}ch) ≡ (ϖ_{E}^{h^{−1}c−h^{−1}}) ≡ (ϖ_{E}^{c−1})^{h} ≡ θ(c)^{h}.

Let h_{0} be that element of H such that

α^{h_{0}} = α^{q}

if α ∈ λ and let c_{0} be a generator of C. Then θ(c_{0}) has order ℓ and, since the centralizer of C in H is {1},

θ(h_{0}^{−r}ch_{0}^{r}) = θ(c_{0})^{q^{r}}

is θ(c_{0}) if and only if f divides r. On the other hand, it is θ(c_{0}) if and only if ℓ divides q^{r} − 1. Thus the order of q modulo ℓ is f. We also observe that both C and its dual group are cyclic of prime order so that any element of H which fixed an element of T would act trivially on the dual group and therefore on C itself. It follows that F_{μ} = L for all μ in T.

Suppose first that m = 1. Let ψ_{ϕ} be the character

ψ_{ϕ}(x) = ψ_{F} (\frac{x}{ϖ_{F}^{1+n}} )

on ϕ. Since O_{E}/𝔓_{E} is naturally isomorphic to O_{F} /𝔓_{F} and the map x ⟶ N_{E/F} x gives the map x ⟶ x^{ℓ} of ϕ into itself while the map x ⟶ S_{E/F} x induces the map x ⟶ ℓx the required identity reduces to the equality of

X_{ϕ}(ℓ^{ℓ})τ(X_{ϕ}^{ℓ}, ψ_{ϕ}) ∏_{μ∈T} τ(μ_{λ}, ψ_{λ/ϕ})

and

τ(X_{ϕ}, ψ_{ϕ}) ∏_{μ∈T} τ(μ_{λ}X_{λ/ϕ}, ψ_{λ/ϕ}).

This equality has been proved in Lemma 7.8.

Now let m be greater than 1. Since F_{μ} = 1 for all μ we are trying to show that

Δ_{1}(X_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∏ Δ_{1}(μ, ψ_{L/F} , ϖ_{F}^{m+n})

is equal to

Δ_{1}(X_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏ Δ_{1}(μX_{L/F} , ψ_{L/F} , ϖ_{F}^{m+n}).

Since the action of H on C is not trivial ℓ cannot be 2. If μ lies in T and μ^{−1} lies in the orbit of ν then

Δ_{1}(ν, ψ_{L/F} , ϖ_{F}^{m+n}) = Δ_{1}(μ^{−1}, ψ_{L/F} , ϖ_{F}^{m+n})

is μ(−1) times the complex conjugate of

Δ_{1}(μ, ψ_{L/F} , ϖ_{F}^{m+n}).

Since the order of μ is ℓ, μ(−1) = 1 and, if μ ≠ ν, the product of the two terms corresponding to μ and ν is 1. If

μ^{−1} = μ^{q^{r}}

with 0 ≤ r < f lies in the orbit of μ then ℓ divides q^{r} + 1. Thus ℓ divides q^{2r} − 1 and 2r = f. By Lemma 7.1

| τ(μ_{λ}, ψ_{λ/ϕ}) | = \sqrt{q}^{f} = q^{r}

and

τ(μ_{λ}, ψ_{λ/ϕ}) = −Δ_{1}(μ, ψ_{L/F} , ϖ_{F}^{1+n})q^{r}

if ψ_{λ/ϕ} has the same meaning as before and μ_{λ} is the character of λ^{∗} induced by μ. Since

δ = Δ_{1}(μ, ψ_{L/F} , ϖ_{F}^{1+n})

is its own complex conjugate, it is ±1. If α ∈ ϕ then

μ^{−1}(α) = μ(α^{q^{r}}) = μ(α).

Chapter 13 135 Since u(α) is an ℓth root of unity it is 1. Thus

τ(μ_{λ}, ψ_{λ/ϕ}) = τ(μ_{λ}).

However it follows from Lemma 7.1 that

τ(μ_{λ}) ≡ 1 (mod η)

where η is a number in 𝔨_{p(q^{f} −1)} which is not a unit and whose only prime divisors are divisors of ℓ. Thus

−δq^{r} = τ(μ_{λ}) ≡ 1 (mod ℓ)

and δ = 1. We are reduced to showing that

Δ_{1}(X_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n})

is equal to

Δ_{1}(X_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏ Δ_{1}(μX_{L/F} , ψ_{L/F} , ϖ_{F}^{m+n})

Let β = β(X_{F}). By repeated applications of Lemma 8.9 we see that we may take

β(X_{L/F}) = β(X_{K/F}) = β(μX_{L/F}) = β.

If β(X_{E/F}) is chosen we could also take

β(X_{K/F}) = β(X_{E/F}).

Thus if

m′ = m(X_{E/F}) = m(X_{K/F}) = 2d′ + ε′

we have

β ≡ β(X_{E/F}) (mod 𝔓_{K}^{d′}).

Since both sides of the congruence lie in E

β ≡ β(X_{E/F}) (mod 𝔓_{E}^{d′})

and we may take

β(X_{E/F}) = β.

Then

Δ_{2}(X_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) = ψ_{F} (\frac{ℓβ}{ϖ_{F}^{m+n}} ) X_{F}^{−1}(β^{ℓ})

while

Δ_{2}(X_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏_{μ∈T} Δ_{2}(μX_{F} , ψ_{L/F} , ϖ_{F}^{m+n})

is equal to

ψ_{F} (\frac{ℓβ}{ϖ_{F}^{m+n}} ) X_{F}^{−1}(β^{ℓ}).

Chapter 13 136

To complete the proof of the lemma we have to show that

Δ_{3}(X_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏_{μ∈T} Δ_{3}(μX_{F} , ψ_{L/F} , ϖ_{F}^{m+n})  (13.4)

is equal to

Δ_{3}(X_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n})

when one, and hence both, of m and m′ is odd.

As remarked in Lemma 9.4

Δ_{3}(μX_{L/F} , ψ_{L/F} , ϖ_{F}^{m+n}) = Δ_{3}(X_{L/F} , ψ_{L/F} , ϖ_{F}^{m+n}).

According to Lemma 9.6 the right side is equal to

εΔ_{3}(X_{F} , ψ_{F} , ϖ_{F}^{m+n})^{[L:F]}

where ε is 1 if f = [L : F] is odd and -1 if it is even. Thus (13.4) is equal to

ε^{\frac{ℓ−1}{f}} \{Δ_{3}(X_{F} , ψ_{F} , ϖ_{F}^{m+n})\}.

As before

ϕ = O_{F} /𝔓_{F} = O_{E}/𝔓_{E}.

Let φ_{ϕ} be the function on ϕ defined by

φ_{ϕ}(x) = ψ_{F} (\frac{βx}{ϖ_{F}^{d+1+n}} ) X_{F}^{−1}(1 + ϖ_{F}^{d} x)

if m = 2d + 1. Then m′ = 2ℓd + 1 so that d′ = ℓd. Let φ′_{ϕ} be the function on ϕ defined by

φ′_{ϕ}(x) = ψ_{E/F} (\frac{βx}{ϖ_{F}^{d+1+n}} ) X_{E/F}^{−1} (1 + ϖ_{F}^{d} x).

Because of Lemma 9.3, to complete the proof of the lemma we have only to show that

ε^{\frac{ℓ−1}{f}} A[σ(φ_{ϕ})^{ℓ}] = A[σ(φ′_{ϕ})].

Since d′ ≥ m and

\frac{3d′ + ℓ − 1}{ℓ} ≥ m

we have

N_{K/L}(1 + ϖ_{F}^{d} x) ≡ 1 + ϖ_{F}^{d} S_{K/L}x + ϖ_{F}^{2d}E_{K/L}^{2}(x) (mod 𝔓_{L}^{m})

if E_{K/L}^{2}(x) is the second elementary symmetric function of x and its conjugates over L. Thus N_{E/F} (1 + ϖ_{F}^{d} x) ≡ 1 + ϖ_{F}^{d} S_{E/F} x + ϖ_{F}^{2d}E_{E/F}^{2} (x) (mod 𝔓_{F}^{m}).

This in turn is congruent to

(1 + ϖ_{F}^{d} S_{E/F} x) (1 + ϖ_{F}^{2d}E_{K/F}^{2} (x)).

Thus

φ′_{ϕ}(x) = φ_{ϕ}(ℓx)ψ_{ϕ}(−E_{λ/ϕ}^{2}(x))

if

ψ_{ϕ}(x) = φ_{F} (\frac{βx}{ϖ_{F}^{1+n}} )

or

φ′_{ϕ}(x) = φ_{ϕ}(ℓx)ψ_{ϕ} (\frac{−ℓ(ℓ − 1)}{2} x^{2}) = \{φ_{ϕ}(x)\}^{ℓ}.

Suppose first that p is odd and let

φ_{ϕ}(x) = ψ_{ϕ} (\frac{x^{2} − 2αx}{2})

so that

φ′_{ϕ}(x) = ψ_{ϕ} (\frac{ℓx^{2} − 2ℓαx}{2}) = ψ_{ϕ} (\frac{ℓ(x − α)^{2}}{2} ) ψ_{ϕ} (\frac{−ℓα^{2}}{2} ) .

Referring to the observations in paragraph 9 we see that we must show that

ε^{\frac{ℓ−1}{f}} ν_{ϕ}(−1)^{\frac{ℓ−1}{2}} ψ_{ϕ} (\frac{−ℓα^{2}}{2} ) = ν_{ϕ}(ℓ)ψ_{ϕ} (\frac{−ℓα^{2}}{2} )

or

ε^{\frac{ℓ−1}{f}} = ν_{ϕ}(−1)^{\frac{ℓ−1}{2}} ν_{ϕ}(ℓ)

if ν_{ϕ} is the quadratic character of ϕ^{∗}. Let q be the number of elements in ϕ. If q is an even power of p the right side is 1 and if q is an odd power of p the right side is, by the law of quadratic reciprocity, ω(p) if ω is the quadratic character of the field with ℓ elements. Thus in all cases the right side is ω(q). If f is odd then q^{f} is a quadratic residue of ℓ if and only if q is. Since

q^{f} − 1 ≡ 0 (mod ℓ).

q is a quadratic residue and both sides of the equation are 1. If f is odd the left side is (−1)^{\frac{ℓ−1}{f}}. Since f is the order of q modulo ℓ this is ω(q).

Now suppose that p = 2. If

ψ_{ϕ}(−x^{2}) = ψ_{ϕ}(αx)

then, by the remarks in the proof of Lemma 9.7, we have to show that

ε^{\frac{ℓ−1}{f}} φ_{ϕ}(α)^{\frac{ℓ−1}{2}} = 1

if ℓ ≡ 1(mod 4) and that

ε^{\frac{ℓ−1}{f}} φ_{ϕ}(α)^{\frac{ℓ+1}{2}} = 1

Chapter 13 138 if ℓ ≡ 3 (mod 4). We also saw in paragraph 9 that

{φ_{ϕ}(α)}^{2} = ψ_{ϕ}(α^{2})

was +1 or -1 according as q is or is not an even power of p. By the second supplement to the law of quadratic reciprocity

φ_{ϕ}(α)^{\frac{ℓ−1}{2}} = ω(q)

if ℓ ≡ 1 (mod 4) and

φ_{ϕ}(α)^{\frac{ℓ+1}{2}} = ω(q)

if ℓ ≡ 3 (mod 4). We have just seen that

ε^{\frac{ℓ−1}{f}} = ω(q).

The lemma is proved.

Chapter Fourteen.

The Fourth Main Lemma.

In the previous paragraph we said that we would eventually have to show that

Δ(χ_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})  (14.1)

is equal to

Δ(χ_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}).  (14.2)

However we verified that the two expressions are equal only when K/F is tamely ramified. In this paragraph we shall show that they are equal if Theorem 2.1 is valid for all pairs K′/F′ in P(K/F) for which [K′ : F′] < [K : F].

Lemma 14.1

Suppose K/F is wildly ramified and Theorem 2.1 is valid for all pairs K′/F′ in P(K/F) for which [K′ : F′] < [K : F]. If χ_{F} is any quasi-character of C_{F} the expressions (14.1) and (14.2) are equal.

If a and b are two non-zero complex numbers and m is a positive integer we again write a ∼_{m} b if, for some non-negative integer r, \frac{a}{b} is an m^{r}th root of unity. Define the non-zero complex number ρ by demanding that

Δ(χ_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})

be equal to

ρΔ(χ_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}).

We have to show that ρ = 1. Lemma 14.1 will be an easy consequence of the following four lemmas.

Lemma 14.2

If m(χ_{F}) is 0 or 1 then ρ = 1 and in all cases p ∼_{2p} 1.

Lemma 14.3

If [G : G_{1}] is a power of 2 then ρ ∼_{p} 1.

Lemma 14.4

If the induction assumption is valid, if F ⊆ F′ ⊆ L, if F′/F is normal, and if [F′ : F] = ℓ is a prime then p ∼_{ℓ} 1.

Lemma 14.5

Suppose H = H_{1}H_{2} where H_{2} is a cyclic normal subgroup of H, [H_{2} : 1] is a power of a prime ℓ, and [H_{1} : 1] is prime to ℓ. If the induction assumption is valid ρ ∼_{ℓ} 1.

Grant these four lemmas for a moment and observe that if m and n are relatively prime then ρ ∼_{m} 1 and ρ ∼_{n} 1 imply that ρ = 1. If ℓ is a prime which divides [G : G_{0}] there is a field F′ containing F and contained in L so that F′/F is normal and [F′ : F] = ℓ. Thus Lemma 14.1 follows from Lemma 14.4 unless [G : G_{0}] is a prime power. Lemma 14.1 follows from Lemmas 14.2 and 14.4 unless [G : G_{0}] is a power of 2 or p. Suppose [G : G_{0}] is a power of 2 or p. Then ρ ∼_{[G:G_{0}]} 1 except perhaps when [G : G_{0}] = 1. If ℓ is a prime which does not divide [G : G_{0}] but does divide [G_{0} : G_{1}] let H_{2} be the ℓ-Sylow subgroup of G_{0}/G_{1}. H_{2} is a normal subgroup of G/G_{1} which we may identify with H and H/H_{2} has order prime to H_{2}. Thus, by a well-known theorem of Schur [7], H = H_{1}H_{2} where H_{1} ∩ H_{2} = {1} and H_{1} has order prime to H_{2}. It follows from Lemma 14.5 that ρ = 1 unless [G : G_{0}] = 1 or [G : G_{1}] is a power of 2 or p. If [G : G_{0}] = 1 and ℓ is a prime dividing [G_{0} : G_{1}] there is a field F′ with F ⊆ F′ ⊆ L such that F′/F is normal and [F′ : F] = ℓ. Thus if [G : G_{0}] = 1 it follows from Lemma 14.4 that ρ = 1 unless [G : G_{1}] is a power of 2. However if [G : G_{1}] is a power of 2 there certainly is an F′ in L with [F′ : F] = 2. It follows from Lemmas 14.3 and 14.4 that ρ = 1 in this case unless p = 2. If [G : G_{1}] is a power of p then G_{0} = G_{1} and G/G_{1} is abelian. By assumption the abelian p-group G/G_{1} acts on the p-group C = G_{1} faithfully and irreducibly. This is impossible.

We prove Lemma 14.2 first. Let t ≥ 1 be such that C = G_{t} while G_{t+1} = {1}. Let θ_{t} be the homomorphism of G_{t} into 𝔓_{K}^{t}/𝔓_{K}^{t+1} and θ_{0} the homomorphism of G_{0}/G_{1} into U_{K}^{0}/U_{K}^{1} introduced in Serre’s book. If σ ∈ G_{0} and γ ∈ G_{t} then

θ_{t}(σγσ^{−1}γ^{−1}) = (θ_{0}^{t}(σ) − 1)θ_{t}(γ).

If σ is not in G_{1} then θ_{0}^{t}(σ) is not 1 and γ → σγσ^{−1}γ^{−1} is a one-to-one map of C onto itself. Thus, if σ ∈ G_{0},

μ(σγσ^{−1}) = μ(γ)

implies μ = 1 or σ ∈ G_{1}. Consequently if μ ≠ 1, G_{μ} ∩ G_{0} = G_{1} and L/F_{μ} is unramified. Since μ = μ′_{L/F_{μ}},

m(μ′) = m(μ) = t + 1.

Observe also that t must be relatively prime to [G_{0} : G_{1}]. In particular if t is even [G_{0} : G_{1}] is odd.

The relations

δ_{K/L} = ([G_{t} : 1] − 1) (t + 1)
δ_{L/F} = [G_{0} : G_{1}] − 1
δ_{K/E} = [G_{0} : G_{1}] − 1

and

δ_{K/F} = δ_{K/L} + [G_{1} : 1] δ_{L/F} = δ_{K/E} + [G_{0} : G_{1}]δ_{E/F}

Chapter 14 141 obtained from Proposition 4 of Chapter IV of Serre’s book, imply that

δ_{E/F} = ([G_{1}; 1] − 1) (\frac{t}{[G_{0} : G_{1}]} + 1) .

If n = n(ψ_{F}) then

n(ψ_{F_{μ}/F}) = [G_{0} : G_{1}]n + [G_{0} : G_{1}] − 1

and

n′ = n(ψ_{E/F}) = [G_{1} : 1]n + ([G_{1} : 1] − 1) (\frac{t}{[G_{0} : G_{1}]} + 1) .

Choose a generator ϖ_{K} of 𝔓_{K} and a generator ϖ_{E} of 𝔓_{E}. Then set ϖ_{L} = N_{K/L}ϖ_{K} and ϖ_{F} = N_{E/F} ϖ_{E}. There is a unit δ in K such that

\frac{ϖ_{F}^{1+n}}{ϖ_{E}^{1+n′}} = δ \frac{ϖ_{K}^{t}}{ϖ_{L}^{t}} .

Taking the norm from K to L of both sides we see that if q = [G_{1} : 1] and k = [G_{0} : G_{1}] then

\frac{ϖ_{L}^{t(q−1)}}{ϖ_{F}^{t \frac{(q−1)}{k}}} = N_{K/L}δ.

Let m = m(χ_{F}). If m − 1 is equal to \frac{t}{[G_{0}:G_{1}]} then [G_{0} : G_{1}] divides t and [G_{0} : G_{1}] is 1. Suppose that

m < \frac{t}{[G_{0} : G_{1}]} + 1.

Then

m(χ_{F_{μ}/F}) < ψ_{F_{μ}/F} (\frac{t}{[G_{0} : G_{1}]}) + 1.

However ψ_{F_{μ}/F} = φ_{L/F_{μ}} ∘ ψ_{L/F} = ψ_{L/F} so that

ψ_{F_{μ}/F} (u) = [G_{0} : G_{1}]u

if u ≥ 0. Thus m(χ_{F_{μ}/F}) < t + 1 and m(μ′χ_{F_{μ}/F}) = t + 1. Moreover, by Lemmas 13.1 and 6.4, m′ = m(χ_{E/F}) = m. Choose a generator ϖ_{F_{μ}} of 𝔓_{F_{μ}}. Then

N_{F_{μ}/F} (ϖ_{F_{μ}}^{t}) = γ_{μ} ϖ_{F}^{t \frac{[Fμ:F]}{k}}

where γ_{μ} is a unit. The order of ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t} in F_{μ} is 1 + t + n(ψ_{F_{μ}/F}). Observing that

∑_{μ}[F_{μ} : F] = q − 1

we see that

Δ_{1}(χ_{E/F} , ψ_{E/F′}ϖ_{E}^{m′+n′}) ∏_{μ∈T} Δ_{1}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t})

Chapter 14 142 is equal to

p{∏_{μ} χ_{F} (γ_{μ})} {Δ_{1}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}} {∏_{μ} Δ_{1}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t})}.

It is now clear that ρ = 1 if m = 0.

If

m ≥ \frac{t}{[G_{0} : G_{1}]} + 1

so that in particular m ≥ 2, then

m′ = m(χ_{E/F}) = [G_{1} : 1]m − ([G_{1} : 1] − 1) (\frac{t}{[G_{0} : G_{1}]} + 1)

is also greater than or equal to 2 and

m′ + n′ = [G_{1} : 1] (m + n).

Since m′ ≥ 2 and K/E is tamely ramified

m(χ_{K/F}) = ψ_{K/E}(m(χ_{E/F}) − 1) + 1 = ψ_{K/F} (m − 1) + 1.

Since

m(χ_{F_{μ}/F}) ≤ ψ_{F_{μ}/F} (m − 1) + 1

and

ψ_{F_{μ}/F} (m − 1) + 1 ≥ ψ_{F_{μ}/F} (\frac{t}{k} ) + 1 = t + 1

we have

m(μ′χ_{F_{μ}/F}) ≤ ψ_{F_{μ}}(m − 1) + 1.

However

χ_{K/F} = (μ′χ_{F_{μ}/F})_{K/F_{μ}}

so that

ψ_{K/F_{μ}}(ψ_{F_{μ}/F} (m − 1)) + 1 = ψ_{K/F} (m − 1) + 1 = m(χ_{K/F})

is at most

ψ_{K/F} (m(μ′χ_{F_{μ}/F}) − 1) + 1.

Thus

m(μ′χ_{F_{μ}/F}) = ψ_{F_{μ}/F} (m − 1) + 1.

Consequently

m(μ′χ_{F_{μ}/F}) + n(ψ_{F_{μ}/F}) = [G_{0} : G_{1}] (m + n).

Since the range of each μ′ lies in the group of qth roots of unity Δ_{1}(χ_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∏_{μ} Δ_{1}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{n+1}ϖ_{F_{μ}}^{t}) is equal to

σΔ_{1}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏_{μ} Δ_{1}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n})

with σ ∼_{p} ρ.

The next step in the proof of the lemma is to establish a simple identity. As usual let r be the integral part of \frac{t+1}{2} and let r + s = t + 1. Choose β(μ′) so that

ψ_{F_{μ}/F} (\frac{β(μ′)x}{ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}} ) = μ′(1 + x)

for x in 𝔓_{F_{μ}}^{s}. There is a unit α_{μ} in L such that α_{μ}ϖ_{F_{μ}}^{t} = ϖ_{L}^{t}. Then

ψ_{L/F} (\frac{α_{μ}β(μ′)x}{ϖ_{F}^{1+n}ϖ_{L}^{t}}) = μ(1 + x)

for x in 𝔓_{L}^{s}. We take β(μ) = α_{μ}β(μ′). If σ ∈ G a possible choice for β(μ^{σ}) is

β(μ)^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} .

Let ϕ = O_{F} /𝔓_{F} = O_{E}/𝔓_{E} and let ψ_{ϕ} be the additive character of ϕ defined by

ψ_{ϕ}(x) = ψ_{F} (\frac{x}{ϖ_{F}^{1+n}} ) .

There is a unique α in ϕ such that

ψ_{ϕ}(αx) = ψ_{ϕ}(x^{q}).

Finally let

ω_{1} = S_{E/F} (\frac{ϖ_{F}^{1+n}}{ϖ_{E}^{1+n′}} ) .

I want to show that

∏_{μ} γ_{μ} = \frac{ω_{1}^{q}}{α^{q}} ∏_{μ} N_{F_{μ}/F} β(μ′)  (14.3)

in ϕ.

Let λ = O_{L}/P_{L} = O_{K}/P_{K}. If

ω = S_{K/L} (\frac{ϖ_{K}^{t}}{ϖ_{L}^{t}} )

then ω_{1} = δω in λ. We need the following lemma. Lemma 14.6

Suppose K′/F′ is an abelian extension and G′ = 𝔊(K′/F′). Suppose there is a t ≥ 1 such that G′ = G′_{t} and G′_{t+1} = {1}. Let ϖ_{K′} be a generator of 𝔓′_{K}, let ϖ_{F′} = N_{K′/F′}(ϖ_{K′}), and let

ω = S_{K′/F′} (\frac{ϖ_{K′}^{t}}{ϖ_{F′}^{t}} ) .

Also let q = [K′ : F′]. There are numbers a, …, f in O_{F′} such that for all x in O_{F′}

N_{K′/F′}(1 + xϖ_{K′}^{t})

is congruent to

1 + (x^{q} + ax^{\frac{q}{p}} + … + fx^{p} + ωx)ϖ_{F′}^{t}

modulo 𝔓_{F′}^{t+1}.

Suppose F′ ⊆ L′ ⊆ K′ and the lemma is true for K′/L′ and L′/F′. The lemma for K′/F′ follows from the relations

[K′ : F′] = [K′ : L′] [L′ : F′]

and

N_{K′/F′}(1 + xϖ_{K′}^{t}) = N_{L′/F′}(N_{K′/L′}(1 + xϖ_{K′}^{t}))

and

S_{K′/F′} (\frac{ϖ_{K′}^{t}}{ϖ_{F′}^{t}} ) ≡ S_{L′/F′} (\frac{ϖ_{L′}^{t}}{ϖ_{F′}^{t}} ) S_{K′/L′} (\frac{ϖ_{K′}^{t}}{ϖ_{L′}^{t}} ) .

The lemma for extensions of prime order is proved in Serre’s book.

Suppose then

N_{K/L}(1 + xϖ_{K}^{t}) ≡ 1 + (x^{q} + … + ϖx)ϖ_{L}^{t} (mod 𝔓_{L}^{t+1})

for x in O_{L}. Since

ψ_{λ/ϕ}(αx) = ψ_{ϕ}(αS_{λ/ϕ}(x)) = ψ_{ϕ}((S_{λ/ϕ}(x))^{q})

which in turn equals

ψ_{ϕ}(S_{λ/ϕ}x^{q}) = ψ_{λ/ϕ}(x^{q})

we conclude that

ψ_{λ/ϕ}(y(x^{q} + … + ωx)) = ψ_{λ/ϕ}((αy^{\frac{1}{q}} + … + ωy)x).

Also

(αy^{\frac{1}{q}} + … + ωy)^{q} = α^{q}y + … + ω^{q}y^{q}

is a polynomial Q(y) in y.

For each ν in S(K/L), we choose β_{1}(ν) so that

ψ_{L/F} (\frac{β_{1}(ν)x}{ϖ_{F}^{1+n}ϖ_{L}^{t}} ) = ν(1 + x)

for x in P_{L}^{s}. Since k^{p} = k in λ

ψ_{λ/ϕ}(kβ_{1}(ν)(x^{q} + … + ωx) = ψ_{λ/ϕ}(β_{1}(ν) [(kx)^{q} + … + ωkx])

if x is in O_{F}. The left side is also equal to

ψ_{L/F} (\frac{β_{1}(ν) P_{K/L}(xϖ_{K}^{t})}{ϖ_{F}^{1+n} ϖ_{L}^{t}}) = 1.

Chapter 14 145 Thus Q(kβ_{1}(ν)) = 0. Since β_{1}(ν_{1}) = β_{2}(ν_{2}) (mod 𝔓_{L}) implies ν_{1} = ν_{2} we have found all the roots of Q(y) = 0. Thus

\frac{α^{q}}{ω^{q}} ≡ ∏_{ν≠1} kβ_{1}(ν) ≡ ∏_{ν≠1} β_{1}(ν) (mod 𝔓_{L}).

Let M_{μ} be a set of representatives for the cosets of G_{μ} in G. Then

∏_{μ∈T} N_{F_{μ}/F} β(μ′) = ∏_{μ∈T} ∏_{σ∈M_{μ}} β(μ′)^{σ}

is congruent to

\{∏_{ν≠1} β_{1}(ν)\} \{∏_{μ∈T} ∏_{σ∈M_{μ}} α_{μ}^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} \}^{−1}

modulo 𝔓_{L}. To verify the identity (14.3) we have to show that

\{∏_{μ} ∏_{σ∈M_{μ}} α_{μ}^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} \} \{∏_{μ} γ_{μ}\} ≡ δ^{q} (mod 𝔓_{K}).

Since

γ_{μ} = \{∏_{σ∈M_{μ}} \frac{ϖ_{L}^{σ t}}{α_{μ}^{σ}} \}ϖ_{F}^{−t \frac{[Fμ:F]}{k}}

the congruence reduces to

\frac{ϖ_{L}^{t(q−1)}}{ϖ_{F}^{t \frac{(q−1)}{k}}} ≡ δ^{q} (mod 𝔓_{K})

which is valid because the left side is N_{K/L}δ and

N_{K/L}δ ≡ δ^{q} (mod 𝔓_{K}).

If m = 1 then m(χ_{F_{μ}/F}) ≤ 1 and we can take β(μ′χ_{F_{μ}/F}) = β(μ′). Then
Δ_{2}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t})

is equal to

χ_{F} (N_{F_{μ}/F} (β(μ′))) Δ_{2}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}).

Lemma 9.4 implies that

Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) = Δ_{3}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}).

If x belongs to O_{E} then

ψ_{E/F} (\frac{x}{ϖ_{E}^{1+n′}} ) = ψ_{ϕ}(ω_{1}x).

If χ_{ϕ} is the character of ϕ^{∗} determined by χ_{F} then Δ_{1}(χ_{F} , ψ_{F} , ϖ_{F}^{1+n}) = A[−τ(χ_{ϕ}, ψ_{ϕ})] Chapter 14 146 and

Δ_{1}(χ_{E/F} , ψ_{E/F} , ϖ_{E}^{1+n′}) = χ_{ϕ}(ω_{1}^{q}) A[−τ(χ_{ϕ}^{q}, ψ_{ϕ})].

The right side of this expression is equal to

χ_{ϕ}(ω_{1}^{q}α^{−q}) A[−τ(χ_{ϕ}, ψ_{ϕ})].

The identity (14.3) now shows that ρ = 1 when m = 1.

Suppose that

1 < m < \frac{t}{[G_{0} : G_{1}]} + 1.

Let β be a given choice of β(χ_{F}). Then

β(χ_{E/F}) ≡ P_{E/F}^{∗} (β, ϖ_{E}^{m′+n′}, ϖ_{F}^{m+n}) (mod 𝔓_{E}^{d′})

if m′ = 2d′ + ε′. On the other hand

ψ_{L/F} (m − 1) + 1 + n(ψ_{L/F}) = [G_{0} : G_{1}](m − 1) + 1 + [G : G_{0}]n + [G_{0} : G_{1}] − 1

which equals [G_{0} : G_{1}](m + n) and Lemmas 8.3, 8.4 and 8.7 imply that

P_{L/F}^{∗} (β, ϖ_{F}^{m+n}, ϖ_{F}^{m+n}) ≡ β (mod 𝔓_{L}^{d_{1}})

if

ψ_{L/F} (m − 1) + 1 = 2d_{1} + ε_{1}.

If

ψ_{K/F} (m − 1) + 1 = 2d′_{1} + ε′_{1}

then

P_{K/E}^{∗}(β(χ_{E/F} , ϖ_{E}^{m′+n′}, ϖ_{E}^{m′+n′}) ≡ β(χ_{E/F}) (mod 𝔓_{K}^{d′_{1}}).

Thus

β(χ_{E/F}) ≡ P_{K/F}^{∗} (β, ϖ_{E}^{m′+n′}, ϖ_{F}^{m+n}) (mod 𝔓_{K}^{d′_{1}}).

Let

v = t + 1 − (ψ_{L/F} (m − 1) + 1) = t − [G_{0} : G_{1}] (m − 1).

If

γ = \frac{ϖ_{L}^{t−v}}{ϖ_{F}^{m−1}}

and

γ′ = \frac{ϖ_{E}^{m′+n′}ϖ_{K}^{v}}{ϖ_{F}^{1+n}ϖ_{L}^{t}}

then

P_{K/F}^{∗} (β, ϖ_{E}^{m′+n′}, ϖ_{F}^{m+n}) ≡ P_{K/L}^{∗}(β, ϖ_{E}^{m′+n′}, ϖ_{F}^{m+n}) (mod 𝔓_{K}^{d′_{1}})

is congruent to

γ′P_{K/L}^{∗}(γβ, ϖ_{F}^{1+n}ϖ_{L}^{t}ϖ_{K}^{−v}, ϖ_{F}^{1+n}ϖ_{L}^{t−v})

modulo 𝔓_{K}^{d′_{1}}.

It is clear that

Δ_{2}(χ_{E/F} , ψ_{E/F} , ϖ_{E}^{m′+n′}) ∼_{p} χ_{F}^{−1}(N_{E/F} (β(χ_{E/F})))

and that

Δ_{2}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{n+1}ϖ_{F_{μ}}^{t}) ∼_{p} 1.

If we choose

β(μ′χ_{F_{μ}/F}) = β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}}

then

Δ_{2}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{n+1}ϖ_{F_{μ}}^{t}) ∼_{p} χ_{F}^{−1} (N_{F_{μ}/F} (β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}} )) .

Moreover

Δ_{2}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∼_{p} χ_{F}^{−1}(β).

Let

Δ_{2}(χ_{E/F} , ψ_{E/F} , ϖ_{E}^{m′+n′}) ∏_{μ} Δ_{2}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{n+1}ϖ_{F_{μ}}^{t})

equal

τΔ_{2}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) {∏_{μ} Δ_{2}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{n+1}ϖ_{F_{μ}}^{t})} {∏_{μ} χ_{F} (γ_{μ})}.

Since

χ_{F} (u) ∼_{p} 1

if u ∈ U_{F}^{1} , all we need do to show that τ ∼_{p} 1 is prove that

β ∏_{μ} N_{F_{μ}/F} (β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}} )

is congruent to

N_{E/F} (β(χ_{E/F})) ∏_{μ} γ_{μ}

modulo 𝔓_{F}.

As before we choose β(μ) = α_{μ}(β(μ′). If ν = μ^{σ} a possible choice for β(ν) is

α_{μ}^{σ}β(μ′)^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} .

We can also choose

β(μχ_{L/F}) = α_{μ}β(μ′) + β \frac{ϖ_{L}^{t}}{ϖ_{F}^{m−1}} = α_{μ} (β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}} ) .

Chapter 14 148 Then a possible choice for B(μ^{σ}χ_{L/F}) is

α_{μ}^{σ}\{β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}} \}^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} = α_{μ}^{σ}(β(μ′)^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} + β \frac{ϖ_{L}^{t}}{ϖ_{F}^{m−1}} .

We apply Lemma 8.10 with F replace by L,

δ = ϖ_{F}^{1+n}ϖ_{L}^{t}

and ε_{1} = ϖ_{K}^{v}. It implies that

N_{K/L} (\frac{β(χ_{E/F})}{γ′})

is congruent to

γ_{β} ∏_{μ∈T} ∏_{σ∈M_{μ}} \{α_{μ}^{σ} (β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}} )^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} \}

modulo 𝔓_{L}. The last expression is equal to

γ_{β}\{∏_{μ∈T} N_{F_{μ}/F} (β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{t}} )\} \{∏_{μ∈T} ∏_{σ∈M_{μ}} α_{μ}^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} \}

and we have to show that

γN_{K/L}(γ′) \{∏_{μ} γ_{μ}\} \{ ∏_{μ∈T} ∏_{σ∈M_{μ}} α_{μ}^{σ} \frac{ϖ_{L}^{t}}{ϖ_{L}^{σ t}} \}

is congruent to 1 modulo 𝔓_{L}. First of all

N_{K/L}(γ′) = ϖ_{F}^{m′+n′−q(1+n)}ϖ_{L}^{−qt+v} = γ^{−1} \frac{ϖF^{\frac{(q−1)t}{k}}}{ϖ_{L}^{(q−1)t}} .

Since

γ_{μ} = \{∏_{σ∈M_{μ}} \frac{α_{μ}^{σ}}{ϖ_{L}^{σ t}} \}^{−1} ϖ_{F}^{−t \frac{[Fμ:F]}{k}}

the required relation follows.

Define η by setting

ηΔ_{3} (χ_{E/F} , ψ_{E/F} , ϖ_{E}^{m′+n′}) ∏_{μ∈T} Δ_{3} (μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t})

equal to

Δ_{3}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏_{μ∈T} Δ_{3} (μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) .

We now know that η ∼_{p} ρ. We shall show that η ∼_{p} 1. This will prove not only the assertion of Lemma 14.2 but also that of Lemma 14.3, provided of course that

m − 1 < \frac{t}{[G_{0} : G_{1}]}.

Lemmas 9.2 and 9.3 imply directly that η ∼_{p} 1 if p is 2.

Suppose p is odd. Lemma 9.4 implies that

Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) ∼_{p} Δ_{3}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}).

Since m′ = m all we need do is show that

Δ_{3}(χ_{E/F} , ψ_{E/F} , ϖ_{E}^{m′+n′}) ∼_{p} Δ_{3}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n})

when m is odd. Let ϕ = O_{F} /𝔓_{F} = O_{E}/𝔓_{E}. Let β′ = β(χ_{E/F}) and let β = β(χ_{F}). If ψ′_{ϕ} is the character of ϕ defined by

ψ′_{ϕ}(x) = ψ_{F} (\frac{β x}{ϖ_{F}^{n+1}} )

and ψ′′_{ϕ} is the character of ϕ defined by

ψ′′_{ϕ}(x) = ψ_{E/F} (\frac{β′x}{ϖ_{E}^{n′+1}} )

and if ψ′′_{ϕ}(x) = ψ′_{ϕ}(δx) then, by Lemmas 9.2 and 9.3, all we have to do is show that δ is a square in ϕ. If

ω_{1} = S_{E/F} (\frac{ϖ_{F}^{n+1}}{ϖ_{E}^{n′+1}} )

then δ = ω_{1} \frac{β′}{β} in ϕ. To show that δ is a square we show that δ^{q} is a square.

δ^{q} ≡ N_{E/F} α ≡ β^{1−q} \frac{N_{E/F} (β′)}{β}ω_{1}^{q}.

We saw that

\frac{N_{E/F} (β′)}{β} = \{∏_{μ} γ_{μ}\}^{−1} \{∏_{μ} N_{F_{μ}/F} (β(μ′) + β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}} )\}

in ϕ. But

β \frac{ϖ_{F_{μ}}^{t}}{ϖ_{F}^{m−1}} ≡ 0 (mod 𝔓_{L})

because t > [G_{0} : G_{1}](m − 1). We also saw that

\{∏_{μ} γ_{μ}\}^{−1} \{∏_{μ} N_{F_{μ}/F} β(μ′)\} = \frac{α^{q}}{ϖ_{1}^{q}}

in ϕ. Since β^{1−q} is clearly a square, we need only check that α is a square. The character

x ⟶ ψ_{ϕ} (x^{q} − αx)

is identically 1, so that the kernel of the map

x ⟶ x^{q} − αx

Chapter 14 150 is non-trivial. Thus α = x^{q−1} for some x in ϕ.

Now suppose that

m − 1 ≥ \frac{t}{[G_{0} : G_{1}]}.

We have to show that the complex number σ defined at the beginning of the proof satisfies σ ∼_{2p} 1. To prove Lemma 14.3 we will have to show that σ ∼_{p} 1 if [G : G_{1}] is a power of 2.

Given β = β(χ_{F}) we may choose β(χ_{L/F}) = β(χ_{F_{μ}/F}) = β. Moreover

β(χ_{E/F}) ≡ P_{E/F}^{∗} (β, ϖ_{F}^{m+n}, ϖ_{F}^{m+n}) (mod 𝔓_{E}^{d′})

if m′ = m(χ_{E/F}) = 2d′ + ε′. By Lemmas 8.3, 8.4 and 8.7

P_{K/F}^{∗} (β, ϖ_{F}^{m+n}, ϖ_{F}^{m+n}) ≡ P_{K/E}^{∗}(β(χ_{E/F}), ϖ_{F}^{m+n}, ϖ_{F}^{m+n}) ≡ β(χ_{E/F})

modulo P_{K}^{d′_{1}} if

ψ_{K/F} (m − 1) + 1 = 2d′_{1} + ε′_{1}.

Thus

β(χ_{E/F}) ≡ P_{K/L}^{∗}(β, ϖ_{F}^{m+n}, ϖ_{F}^{m+n}) (mod 𝔓_{K}^{d′_{1}}).

If

ψ_{F_{μ}/F} (m − 1) + 1 = 2d_{μ} + ε_{μ}

and

ψ_{F_{μ}/F} (\frac{α(μ′)x}{ϖ_{F}^{m+n}}) = μ′(1 + x)

for x in 𝔓_{F_{μ}}^{d_{μ}+ε_{μ}} we may take

β(μ′χ_{F_{μ}/F}) = β + α(μ′).

If

ψ_{L/F} (m − 1) + 1 = 2d_{1} + ε_{1}

then

μ(1 + x) = ψ_{L/F} (\frac{α(μ′)x}{ϖ_{F}^{m+n}})

for x in 𝔓_{L}^{d_{1}+ε_{1}}. If ν = μ^{σ} then

ν(1 + x) = ψ_{L/F} (\frac{α(μ′)^{σ}x}{ϖ_{F}^{m+n}})

for x in 𝔓_{L}^{d_{1}+ε_{1}}. Lemma 8.2 implies that

N_{E/F} (β(χ_{E/F})) = N_{K/L}(β(χ_{E/F}))

is congruent to

β ∏_{μ∈T} ∏_{σ∈M_{μ}}(β + α(μ′)^{σ})

modulo 𝔓_{L}. The last expression is equal to

β ∏_{μ∈T} N_{F_{μ}/F} (β + α(μ′)).

Moreover

Δ_{2}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∼_{p} χ_{F}^{−1}(β)
Δ_{2}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) ∼_{p} 1
Δ_{2}(χ_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∼_{p} χ_{F}^{−1}(N_{E/F} (β(χ_{E/F})))
Δ_{2}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n}) ∼_{p} χ_{F}^{−1}(N_{F_{μ}/F} (β + α(μ′))).

Define τ by demanding that

Δ_{3}(χ_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∏_{μ} Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t})

be equal to

τ Δ_{3}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏_{μ} Δ_{3}(μ′, χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n}).

Since χ_{F} (u) ∼_{p} 1 if u ∈ U_{F}^{1} the preceding discussion shows that σ ∼_{p} τ. Lemmas 9.2 and 9.3 show that τ ∼_{2p} 1. Lemma 14.2 is now completely proved. To prove Lemma 14.3 we have to show that τ ∼_{p} 1 if [G : G_{1}] is a power of 2. We may suppose that p is odd.

There are a number of possibilities.

(i.a) t is even and m is odd. [G_{0} : G_{1}] must be odd and hence 1, for we are now assuming that [G : G_{1}] is a power of 2. Since

m(χ_{F_{μ}/F}) = [G_{0} : G_{1}] (m − 1) + 1

and

m(χ_{E/F}) = [G_{1} : 1] (m − 1) − \frac{([G_{1} : 1] − 1)t}{[G_{0} : G_{1}]} + 1

both m(χ_{F_{μ}/F}) and m(χ_{E/F}) are odd. (i.b) t is even and m is even. Again [G_{0} : G_{1}] is 1. This time both m(χ_{F_{μ}/F}) and m(χ_{E/F}) are

even. (ii.a) t is odd and m is odd. Then m(χ_{F_{μ}/F}) is odd. If

\frac{[G_{1} : 1] − 1}{[G_{0} : G_{1}]}

is even m(χ_{E/F}) is odd. Otherwise it is even. (ii.b) t is odd and m is even. If [G_{0} : G_{1}] is odd, that is 1, then m(χ_{F_{μ}/F}) is odd and m(χ_{E/F}) is

even. If [G_{0} : G_{1}] is even then m(χ_{F_{μ}/F}) is odd and m(χ_{E/F}) is even or odd according as

\frac{[G_{1} : 1] − 1}{[G_{0} : G_{1}]}

is even or odd.

If t is odd then clearly

∏_{μ} Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) ∼_{p} 1.

We are going to show that this is also true if t is even. Then L/F, and hence F_{μ}/F, is unramified. Let ϕ_{μ} = O_{F_{μ}}/𝔓_{F_{μ}}. If

ψ_{ϕ_{μ}}(x) = ψ_{F_{μ}/F} (\frac{β(μ′)x}{ϖ_{F}^{1+n}})

and if φ_{ϕ_{μ}} is a nowhere vanishing function on ϕ_{μ} satisfying

φ_{ϕ_{μ}}(x + y) = φ_{ϕ_{μ}}(x) φ_{ϕ_{μ}}(y)ψ_{ϕ_{μ}}(xy)

then

Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) ∼_{p} A[−σ(φ_{ϕ_{μ}})].

If α belongs to ϕ_{μ}^{∗} let ν_{ϕ_{μ}}(α) equal +1 or -1 according as α is or is not a square in ϕ_{μ}. If ϕ = O_{F} /𝔓_{F} then

ν_{ϕ_{μ}}(α) = ν_{ϕ} (N_{ϕ_{μ}/ϕ}(α)).

If

ψ_{ϕ}(x) = ψ_{F} (\frac{x}{ϖ_{F}^{1+n}} )

then, according to paragraph 9,

Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) ∼_{p} ν_{ϕ}(N_{F_{μ}/F} (β(μ′))) {A[−σ(φ_{ϕ})]}^{[F_{μ}:F]}

if φ_{ϕ} is any nowhere vanishing function on ϕ satisfying

φ_{ϕ}(x + y) = φ_{ϕ}(x) φ_{ϕ}(x) ψ_{ϕ}(xy).

Thus if a is the number of μ in T

∏_{μ} Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t})

is equal to

η(−1)^{a} ν_{ϕ} (∏_{μ} N_{F_{μ}/F} (β(μ′))) A[σ(φ_{ϕ})^{q−1}]

where η ∼_{p} 1 and q = [G_{1} : 1].

We saw in paragraph 9 that

A[σ(φ_{ϕ})^{2}] ∼_{p} ν_{ϕ}(−1) A[|σ(φ_{ϕ})|^{2}] = ν_{ϕ}(−1).

Since t is even G_{0} = G_{1} and G/G_{1} = G/G_{0} is abelian. If σ ∈ G

{μ ∈ S(K/L) | μ = μ^{σ}}

is a subgroup of S(K/L) invariant under G. It is necessarily either S(K/L) or {1}. If σ is not in G_{1} it is not S(K/L). Thus G_{μ}, the isotropy group of μ, is G_{1} for all μ in T and F_{μ} = L. Moreover

∏_{μ} N_{F_{μ}/F} (β(μ′)) = ∏_{μ} ∏_{σ∈G/G_{1}} β(μ′)^{σ}.

We may regard C = G_{1} as a vector space over the field with p elements. If σ ∈ G/G_{1} and the order of σ divides p − 1 then all the eigenvalues of the linear transformation c ⟶ σcσ^{−1} lie in the prime field. Since the linear transformation also has order dividing p − 1 it is diagonalizable. Since G/G_{1} is abelian and acts irreducibly on C the linear transformation is a multiple of the identity. In particular if σ_{0} is the unique element of order 2 then σ_{0}cσ_{0}^{−1} = c^{−1} for all c. As a consequence μ^{σ_{0}} = μ^{−1} and

β(μ′)^{σ_{0}} ≡ −β(μ′) (mod 𝔓_{L})

if we choose, as we may since F_{μ}/F is unramified, ϖ_{F_{μ}} = ϖ_{F}. If D is the group {1, σ_{0}} and M is a set of representatives for the cosets of D in G/G_{1} then

∏_{μ∈T} ∏_{σ∈G/G_{1}} β(μ′)^{σ} = γγ^{σ_{0}}

if

γ = ∏_{μ∈T} ∏_{σ∈M} β(μ′)^{σ}.

Clearly

γγ^{σ_{0}} = (1−)^{\frac{q−1}{2}} γ^{2} (mod 𝔓_{L}).

If χ is the non-trivial character of D and

v : G/G_{0} ⟶ D

is the transfer then

γ^{σ} = χ(v(σ))^{a}γ

for all σ in G/G_{0}. ν_{ϕ}(γ^{2}) = 1 if and only if χ(v(σ))^{a} is 1 for all σ. If σ is a generator of G/G_{0} then

v(σ) = σ^{\frac{[G:G_{0}]}{2}} = σ_{0}

so that ν_{ϕ}(γ^{2}) = (−1)^{a}. Putting all these facts together we see that

∏_{μ} Δ_{3}(μ′, ψ_{F_{μ}/F} , ϖ_{F}^{1+n}ϖ_{F_{μ}}^{t}) ∼_{p} 1.

Observe that if we had taken ϖ_{F_{μ}} to be δ_{μ}ϖ_{F} then N_{F_{μ}/F} β(μ′) would have to be multiplied by

{N_{F_{μ}/F} δ_{μ}}^{t}

which is a square modulo 𝔓_{F} because t is even. Thus the result is valid for all choices of ϖ_{F_{μ}}.

Eventually we will have to discuss the various possibilities separately. There are however a number of comments we should make first. If m is odd and m(χ_{F_{μ}/F}) is odd then

Δ_{3}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n}) ∼_{p} ν_{ϕ_{μ}}(β + α(μ′)) A[−σ(φ_{ϕ_{μ}})]

Chapter 14 154 if

ψ_{ϕ_{μ}}(x) = ψ_{F_{μ}/F} (\frac{x}{ϖ_{F}^{1+n}} ) .

Observe that, because m is odd, we may take the number δ in Lemma 9.3 to be ϖF^{\frac{m−1}{2}} . Of course φ_{ϕ_{μ}} is any function on ϕ_{μ} which vanishes nowhere and satisfies

φ_{ϕ_{μ}}(x + y) = φ_{ϕ_{μ}}(x) φ_{ϕ_{μ}}(y)ψ_{ϕ_{μ}}(xy).

Applying Lemma 9.1 we see that

A[−σ(φ_{ϕ_{μ}})] ∼_{p} −ν_{ϕ} (k^{\frac{[Fμ:F]}{k}} ) A[σ(φ_{ϕ})^{\frac{[Fμ:F]}{k}} ]

if k = [G_{0} : G_{1}], if

ψ_{ϕ}(x) = ψ_{F} (\frac{x}{ϖ_{F}^{1+n}} )

and if φ_{ϕ} bears the usual relation to ψ_{ϕ}. We use, of course, the relation

kS_{ϕ_{μ}/ϕ}(x) = S_{F_{μ}/F} (x).

Observe also that

ν_{ϕ_{μ}}(β + α(μ′)) = ν_{ϕ}(N_{ϕ_{μ}/ϕ}(β + α(μ′))).

If m is odd

Δ_{3}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∼_{p} −ν_{ϕ}(β) A[σ(φ_{ϕ})].

If both m and m′ = m(χ_{E/F}) are odd and if β′ = β(χ_{E/F}) then

Δ_{3}(χ_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∼_{p} −ν_{ϕ}(β′) A[σ(φ′_{ϕ})]

if φ′_{ϕ} bears the usual relation to the character

ψ′_{ϕ}(x) = ψ_{E/F} (\frac{x}{ϖ_{F}^{1+n}ϖE^{\frac{(q−1)t}{k}}} ) .

There is a unit ε in O_{K} such that ϖ_{E} = εϖ_{K}^{q}. If σ ∈ C then

ϖ_{E}^{σ−1} = ε^{σ−1}ϖ_{K}^{(σ−1)q} ≡ 1 (mod 𝔓_{K})

because t ≥ 1. Thus the multiplicative congruence

ϖ_{E}^{q} ≡ N_{E/F} ϖ_{E} = ϖ_{F} (mod^{∗} 𝔓_{E})

is satisfied and

\frac{1}{ϖE^{\frac{(q−1)t}{k}}} ≡ \frac{ϖ_{F}^{1+n}}{ϖ_{E}^{1+n′}} (mod^{∗} 𝔓_{E}).

If

ω_{1} = S (\frac{ϖ_{F}^{1+n}}{ϖ_{E}^{1+n′}} )

as before, then

ψ′_{ϕ}(x) = ψ_{F} (\frac{ω_{1}x}{ϖ_{F}^{1+n}} ) .

Since

ν_{ϕ}(β′) = ν_{ϕ}(β′)^{q} = ν_{ϕ}(N_{E/F} β′)

we have

Δ_{3}(χ_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∼_{p} −ν_{ϕ}(N_{E/F} β′) ν_{ϕ}(ω_{1}) A[σ(φ_{ϕ})].

Define η by demanding that

Δ_{3}(χ_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n})

be equal to

ηΔ_{3}(χ_{F} , ψ_{F} , ϖ_{F}^{m+n}) ∏_{μ} Δ_{3} (μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n}).

We have to show that η ∼_{p} 1. If both t and m are even this is clear. If t is even and m is odd we are to show that

ν_{ϕ}(N_{E/F} β′)ν_{ϕ}(ω_{1}) ∼_{p} (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}} ν_{ϕ}(k)^{\frac{q−1}{k}} ν_{ϕ}(β) ∏_{μ} ν_{ϕ}(N_{ϕ_{μ}/ϕ}(β + α(μ′)))

if a is the number of elements in T. Since t is even k is 1. As before

β ∏_{μ} N_{ϕ_{μ}/ϕ}(β + α(μ′)) = β ∏_{μ} ∏_{σ∈G/G_{1}}(β + α(μ′)^{σ})

is congruent to N_{E/F} β′ modulo 𝔓_{K}. All we need do is show that

ν_{ϕ}(ω_{1}) = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2}} .

Since t is even each γ_{μ} is a square in ϕ. Applying the identity (14.3) we see that

ν_{ϕ}(ω_{1}) = ν_{ϕ}(ω_{1}^{q}) = ν_{ϕ}(α)ν_{ϕ}^{−1} (∏_{μ} N_{F_{μ}/F} β(μ′)) .

We have seen that α is a square in ϕ so that ν_{ϕ}(α) = 1. We also saw that

ν_{ϕ} (∏_{μ} N_{F_{μ}/F} β(μ′)) = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2}}

when t is even. The required relation follows.

We suppose henceforth that t is odd. The discussion will be fairly complicated. Suppose first that m is also odd. Then

[G_{0} : G_{1}](m − 1) ≠ t

Chapter 14 156 and

m − 1 > \frac{t}{[G_{0} : G_{1}]}

so that

β + α(μ′) ≡ β (mod 𝔓_{L})

and

∏_{μ} Δ_{3}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n}) ∼_{p} (−1)^{a} ν_{ϕ} (k^{\frac{q−1}{k}} ) ν_{ϕ} (β^{\frac{q−1}{k}} ) A[σ(φ_{ϕ})]^{\frac{q−1}{k}} .

Thus if \frac{q−1}{k} is odd we have to show that

(−1)^{a+1} ν_{ϕ}(k) ν_{ϕ}(−1)^{\frac{q−1}{2k} + \frac{1}{2}} ∼_{p} 1  (14.4)

and if \frac{q−1}{k} is even we have to show that

ν_{ϕ}(ω_{1}) ∼_{p} (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}} .  (14.5)

Now suppose m is even. If [G_{0} : G_{1}] is 1 there is nothing to prove. If k = [G_{0} : G_{1}] is even then

m − 1 > \frac{t}{[G_{0} : G_{1}]}

and

β + α(μ′) ≡ β (mod 𝔓_{L}).

If

ψ′_{ϕ_{μ}}(x) = ψ_{F_{μ}/F} (β \frac{ϖ_{F_{μ}}^{k(m−1)}x}{ϖ_{F}^{m+n}} )

and φ′_{ϕ_{μ}} is a function on ϕ_{μ} which vanishes nowhere and satisfies

φ′_{ϕ_{μ}}(x + y) = φ′_{ϕ_{μ}}(x) φ′_{ϕ_{μ}}(y) ψ′_{ϕ_{μ}}(xy)

then

Δ_{3}(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F} , ϖ_{F}^{m+n}) ∼_{p} A[−σ(φ′_{ϕ_{μ}})].

If

ε_{μ}ϖ_{F}^{m−1} = ϖ_{F_{μ}}^{k(m−1)}

then ε_{μ} is a unit and

ψ_{ϕ_{μ}}(x) = ψ_{ϕ_{μ}/ϕ}(kβε_{μ}x)

if, as before,

ψ_{ϕ}(x) = ψ_{F} (\frac{x}{ϖ_{F}^{1+n}} ) .

By Lemma 9.1, A[−σ(φ_{ϕ_{μ}})] is equal to −ν_{ϕ} (k^{\frac{[Fμ:F]}{k}} ) ν_{ϕ} (β^{\frac{[Fμ:F]}{k}} ) ν_{ϕ}(N_{ϕ_{μ}/ϕ}ε_{μ}) A[σ(φ_{ϕ})]^{\frac{[Fμ:F]}{k}} .

If \frac{q−1}{k} is even we have to show that

(−1)^{a} ν_{ϕ} (∏_{μ} N_{ϕ_{μ}/ϕ}ε_{μ}) ν_{ϕ}(−1)^{\frac{q−1}{2k}} ∼_{p} 1  (14.6)

If \frac{q−1}{k} is odd then m′ = m(χ_{E/F}) is odd. If

ψ′′_{ϕ}(x) = ψ_{E/F} (β′ \frac{ϖ_{E}^{m′−1}x}{ϖ_{F}^{m+n}})

and φ′′_{ϕ} bears the usual relation to ψ′′_{ϕ} then

Δ_{3}(χ_{E/F} , ψ_{E/F} , ϖ_{F}^{m+n}) ∼_{p} A[−σ(φ′′_{ϕ})].

Now ν_{ϕ}(β′) = ν_{ϕ}(β′)^{q} and

(β′)^{q} ≡ N_{E/F} β′

which in turn is congruent to

β ∏_{μ∈T} ∏_{σ∈M_{μ}}(β + α(μ′)^{σ}) ≡ β^{q}

modulo 𝔓_{K}. Let

ε_{1}ϖ_{F}^{m+n} = ϖ_{E}^{q(m+n)}

and, as before,

ω_{1} = S_{E/F} (\frac{ϖ_{F}^{1+n}}{ϖ_{E}^{1+n′}} )

then

A[−σ(φ′′_{ϕ})] ∼_{p} ν_{ϕ}(ω_{1}) ν_{ϕ}(ε_{1}) ν_{ϕ}(β) A[−σ(φ_{ϕ})].

We saw that

ϖ_{E}^{q} ≡ ϖ_{F} (mod^{∗} 𝔓_{E})

so that

ε_{1} ≡ 1 (mod 𝔓_{E}).

Thus we have to show that

ν_{ϕ}(ω_{1}) ∼_{p} (−1)^{a+1} ν_{ϕ}(k) ν_{ϕ}(−1)^{\frac{q−1}{2k} − \frac{1}{2}} ν_{ϕ} (∏_{μ} N_{ϕ_{μ}/ϕ}ε_{μ}) .  (14.7)

The four identities (14.4), (14.5), (14.6), and (14.7) look rather innocuous. However to prove them is not an entirely trivial matter. We first consider the case that G/G_{1} is abelian. If σ_{0} ∈ G/G_{1} is of order 2, the argument used before shows that σ_{0}cσ_{0}^{−1} = c^{−1} for all c in C. Since the representation of G/G_{1} on C is faithful G/G_{1} has only one element of order 2 and is therefore cyclic. In this case F_{μ} = L for all μ and a = \frac{q−1}{[G:G_{1}]}. We may choose ϖ_{F_{μ}} = ϖ_{L}. If

N_{L/F} ϖ_{L} = γϖ_{F}^{[G:G_{0}]}

then γ_{μ} = γ^{t} and

∏_{μ} γ_{μ} = γ^{at}.

If [G_{0} : G_{1}] = 1 we may choose ϖ_{L} = ϖ_{F} so that γ = 1. The argument used before shows that

ν_{ϕ} (∏_{μ∈T} ∏_{σ∈G/G_{1}} β(μ′)^{σ}) = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2}} .

The identity (14.3) shows that

ν_{ϕ}(ω_{1}) = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}} .

The identity (14.5) which is the only one of concern here becomes

ν_{ϕ}(−1)^{\frac{q−1}{2}} = ν_{ϕ}(−1)^{\frac{q−1}{2k}}

which is clear because k = [G_{0} : G_{1}] = 1.

Now take [G : G_{0}] = 1. We may choose ϖ_{E} = N_{K/E}ϖ_{K} so that ϖ_{F} = N_{L/F} ϖ_{L} and γ is again 1. It is perhaps worth pointing out these special choices are not inconsistent with any choices yet made in this paragraph. This is necessary because the arguments appearing in the functions Δ_{2} must be the same as those appearing in the functions Δ_{3}. We previously defined

δ = \frac{ϖ_{F}^{1+n}}{ϖ_{E}^{1+n′}} · \frac{ϖ_{L}^{t}}{ϖ_{K}^{t}}

and showed that

N_{K/L}δ = \frac{ϖ_{L}^{t(q−1)}}{ϖF^{\frac{t(q−1)}{k}}} .

Observe that

\frac{ϖ_{L}^{k}}{ϖ_{F}} = ∏_{σ∈G/G_{1}} ϖ_{L}^{1−σ} ≡ ∏ θ_{0}(σ)^{−1} ≡ −1 (mod 𝔓_{L})

because

{θ_{0}(σ) | σ ∈ G/G_{1}}

is just the set of kth roots of unity in ϕ and k is a power of 2. It is not 1 because

[G_{0} : G_{1}] = [G : G_{1}] > 1.

Since

δ^{q} ≡ N_{K/L}δ (mod 𝔓_{K})

we have

ν_{ϕ}(δ) = ν_{ϕ}(−1)^{\frac{q−1}{k}} .

If as before

ψ_{L/F} (\frac{β_{1}(ν)x}{ϖ_{F}^{1+n}ϖ_{L}^{t}} ) = ν(1 + x)

for x in 𝔓_{L}^{s} and ν in S(K/L) then, as we saw when proving the identity (14.3),

ω_{1}^{q} ≡ δ^{q}α^{q} {∏_{ν≠1} β_{1}(ν)} (mod 𝔓_{L}).

Thus

ν_{ϕ}(ω_{1}) = ν_{ϕ}(−1)^{\frac{q−1}{k}} ∏_{ν≠1} ν_{ϕ}(β_{1}(ν)).

We can choose \frac{q−1}{p−1} elements ν_{i} in S(K/L) so that every non-trivial element of S(K/L) is of the form ν_{i}^{j}, 0 < j < p. Then

∏_{ν≠1} ν_{ϕ}(β_{1}(ν)) = ν_{ϕ} (∏i=1^{\frac{q−1}{p−1}} ∏_{j=1}^{p−1} jβ(ν_{i})) = ν_{ϕ}(−1)^{\frac{q−1}{p−1}}

because

ν_{ϕ}(β(ν_{i}))^{p−1} = 1.

When m is even

ν_{ϕ}(ε_{μ}) = ν_{ϕ} (\frac{ϖ_{L}^{k}}{ϖ_{F}} ) = ν_{ϕ}(−1).

Since a = \frac{q−1}{k} the identities (14.4), (14.5), (14.6) and (14.7) become
ν_{ϕ}(k) ν_{ϕ}(−1)^{\frac{q−1}{2k} + \frac{1}{2}} = 1  (14.4′)
ν_{ϕ}(−1)^{\frac{q−1}{p−1}} = ν_{ϕ}(−1)^{\frac{q−1}{2k}}  (14.5′)
ν_{ϕ}(−1)^{\frac{q−1}{2k}} = 1  (14.6′)
ν_{ϕ}(−1)^{\frac{q−1}{k}} ν_{ϕ}(−1)^{\frac{q−1}{p−1}} = ν_{ϕ}(k) ν_{ϕ}(−1)^{\frac{q−1}{2k} + \frac{1}{2}} ν_{ϕ}(−1)^{\frac{q−1}{k}} .  (14.7′)

If p ≡ 1 (mod 4) the identities (14.5′) and (14.6′) are clearly valid. Moreover for (14.5′) and (14.6′) the number \frac{q−1}{k} is even. Since k is a positive power of 2, q is an even power of p if p ≡ 3 (mod 4). If q = p^{2f} then

\frac{q − 1}{p − 1} = 1 + p + … + p^{2f−1} ≡ 0 (mod 4)

and the left side of (14.5′) is 1. If \frac{q−1}{2k} is even (14.5′) and (14.6′) are now clear. If it is odd, 4 divides k because 8 divides q − 1. But {θ_{0}(σ) | σ ∈ G_{0}/G_{1}} is the set of kth roots of unity in O_{L}/𝔓_{L} = ϕ so -1 is a square in ϕ, ν_{ϕ}(−1) = 1, and the relations are valid in this case too. The relations (14.4′) and (14.7′) are obvious if the degree of ϕ over the prime field ϕ_{0} is even. Since ϕ^{×} contains the kth roots of unity and k is a power of 2 the degree can be odd only if k divides p − 1. Since

\frac{q − 1}{k} = \frac{q − 1}{p − 1} · \frac{p − 1}{k}

and \frac{q−1}{k} is now odd \frac{p−1}{k} must also be odd and by quadratic reciprocity

ν_{ϕ}(k) = ν_{ϕ_{0}}(k) = ν_{ϕ_{0}}(−1) ν_{ϕ_{0}} (\frac{p − 1}{k}) = ν_{ϕ_{0}}(−1) ν_{ϕ_{0}}(−1)^{\frac{p−1}{2k} − \frac{1}{2}}

because

p ≡ 1 (mod \frac{p − 1}{k}) .

If p ≡ 1 (mod 4) the two relations are now clear. If p ≡ 3 (mod 4) and q = p^{f}

\frac{q − 1}{p − 1} = 1 + p + … + p^{f−1}

must be odd. It is therefore congruent to 1 modulo 4. (14.4′) becomes

ν_{ϕ_{0}}(−1) ν_{ϕ_{0}}(−1)^{\frac{p−1}{k}} = 1

and (14.7′) becomes

ν_{ϕ_{0}}(−1) = ν_{ϕ_{0}}(−1).

There is no question that both these relations are valid.

We have still to treat the case that G/G_{1} is abelian while neither [G : G_{0}] nor [G_{0} : G_{1}] is 1. Then

N_{F_{μ}/F} β(μ′) ≡ {∏_{σ∈G/G_{0}} β(μ′)}^{k} (mod 𝔓_{F_{μ}})

is a square in ϕ and the identity (14.3) implies that

ν_{ϕ}(ω_{1}) = ν_{ϕ}(γ^{a})

C_{F} /N_{L/F} C_{L} is cyclic of order [G : G_{1}]. It has a generator which contains an element of the form γ_{1}ϖ_{F}. Moreover the coset of

(γ_{1}ϖ_{F})^{[G:G_{0}]} N_{L/F} ϖ_{L}^{−1} = γ_{1}^{[G:G_{0}]}γ^{−1}

is a generator of U_{F} /U_{F} ∩ N_{L/F} C_{L}. The order of this group is a power of 2 and p is odd so every element of U_{F} ∩ N_{L/F} C_{L} is a square. Consequently γ cannot be a square and ν_{ϕ}(γ) = −1. If m is even and F′ is the fixed field of G_{0} then

ε_{μ} = (\frac{ϖ_{L}^{k}}{ϖ_{F}} )^{m−1} = (\frac{N_{L/F′}ϖ_{L}}{ϖ_{F}} )^{m−1} (∏_{σ∈G_{0}/G_{1}} ϖ_{L}^{1−σ})^{m−1}

which is congruent to

− (\frac{N_{L/F′}ϖ_{L}}{ϖ_{F}} )^{m−1}

modulo 𝔓_{L}. Since [F′ : F] is even

N_{ϕ_{μ}/ϕ} ε_{μ} = N_{F′/F} ε_{μ} = (\frac{N_{L/F} ϖ_{L}}{ϖ_{F}^{[G:G_{0}]}} )^{m−1} = γ^{m−1}

and

ν_{ϕ}(N_{ϕ_{μ}/ϕ} ε_{μ}) = ν_{ϕ}(γ) = −1.

Because

a = \frac{q − 1}{[G : G_{1}]}

is integral, \frac{q−1}{k} is even, and we need only worry about the identities (14.5) and (14.6). They both reduce to

ν_{ϕ}(−1)^{\frac{q−1}{2k}} = 1.
To prove this we show that \frac{q−1}{2k} is even if ν_{ϕ}(−1) = −1. Since
k = [U_{F} : U_{F} ∩ N_{L/F} C_{L}]

and this index must divide the order of ϕ^{∗} the number ν_{ϕ}(−1) is -1 only if k = 2. Of course p will be congruent to 3 modulo 4. Since 4 divides q − 1, q is an even power of p and q ≡ 1 (mod 8). Thus

\frac{q − 1}{2k} = \frac{q − 1}{4}

is even.

Now suppose that G/G_{1} is not abelian. Let σ ⟶ x(σ) be a given isomorphism of G_{0}/G_{1} with ℤ/kℤ and let x ⟶ σ(x) be its inverse. Let τ ⟶ λ(τ) be that homomorphism of G/G_{0} into the units of ℤ/kℤ which satisfies

x(τστ^{−1}) = λ(τ) x(σ).

There is precisely one element of order 2 in G_{0}/G_{1}, namely σ (\frac{k}{2} ), and it lies in the center of G/G_{1}. Since G/G_{0} is cyclic, G/G_{1} is non-abelian only if k > 2. Choose a fixed σ_{0} in G which generates G/G_{0} and set

μ_{0} = λ(σ_{0})

and

y_{0} = x(σ_{0}^{[G:G_{0}]}).

We shall sometimes regard C as a vector space over the field with p elements. If σ belongs to G/G_{1} let π(σ) be the linear transformation

c ⟶ σcσ^{−1}.

The dual space will be identified with S(K/L) and π^{∗} will be the representation contragredient to π.

The relation

N_{F_{μ}/F} β(μ′) ≡ {∏_{σ∈G/G_{μ}G_{0}} β(μ′)^{σ}}^{k}

together with the identity (14.3) implies that

ν_{ϕ}(ω_{1}) = ν_{ϕ} (∏_{μ} γ_{μ}) .

Moreover if m is even and F′_{μ} is the fixed field of G_{μ}G_{0}

ε_{μ} = (\frac{ϖ_{F_{μ}}^{k}}{ϖ_{F}} )^{m−1} = \{\frac{N_{F_{μ}/F′_{μ}}ϖ_{F_{μ}}}{ϖ_{F}} \}^{m−1}\{∏_{σ∈G_{0}/G_{1}} ϖ_{F_{μ}}^{1−σ}\}^{m−1}

which is congruent to

\{−\frac{N_{F_{μ}/F′_{μ}}ϖ_{F_{μ}}}{ϖ_{F}} \}^{m−1}

modulo 𝔓_{F_{μ}}. Since

\{N_{ϕ_{μ}/ϕ} ε_{μ}\}^{t} = N_{F′_{μ}/F} \{−\frac{N_{F_{μ}/F′_{μ}}ϖ_{F_{μ}}}{ϖ_{F}} \}^{(m−1)t}

which equals

(−1)^{t[ϕ_{μ}:ϕ]} γ_{μ}^{m−1}

and t is odd

ν_{ϕ}(N_{ϕ_{μ}/ϕ} ε_{μ}) = ν_{ϕ}(−1)^{[ϕ_{μ}:ϕ]} ν_{ϕ}(γ_{μ}).

These relations will be used frequently and without comment.

I want to discuss the case [G : G_{0}] = 2 and μ_{0} ≡ −1 (mod 4) first. Since

(−μ_{0})^{2} = μ_{0}^{2} = λ(σ_{0}^{2}) ≡ 1 (mod k)

we must have

−μ_{0} ≡ 1 (mod k)

or, if k > 4,

−μ_{0} ≡ \frac{k}{2} + 1 (mod k).

Then

μ_{0} ≡ −1 (mod k)

or

μ_{0} ≡ \frac{k}{2} − 1 (mod k).

Since

μ_{0} − 1 ≡ 2 (mod 4)

the centralizer of σ_{0} in G_{0}/G_{1} consists of the identity and σ (\frac{k}{2} ). Thus x(σ_{0}^{2}) is 0 or \frac{k}{2} .

Suppose μ_{0} ≡ −1 (mod k) and x(σ_{0}^{2}) = \frac{k}{2} . If σ belongs to G_{0}/G_{1} then σ_{0}σσ_{0}^{−1} = σ^{−1} and (σ_{0}σ)^{2} = σ_{0}^{2}. Thus σ (\frac{k}{2} ) is the only element of order 2 in G/G_{1}. If σ belongs to G/G_{1} then σ has a non-trivial fixed point in S(K/L) if and only if π(σ) has 1 as an eigenvalue. If σ ≠ 1 there is an integer n such that σ^{n} has order 2. Then π(σ^{n}) also has 1 as an eigenvalue. Thus if any non-trivial element of G/G_{1} has a non-trivial fixed point there is an element τ of order 2 such that π(τ) has 1 as an eigenvalue. The usual argument shows that

π (σ (\frac{k}{2} )) = −I

so that, in the case under consideration, only the identity has fixed points. Then

a = \frac{q − 1}{[G : G_{1}]}.

In particular \frac{q−1}{k} is even. We choose ϖ_{F_{μ}} = ϖ_{L} and let

γϖ_{F}^{[G_{:}G_{0}]} = N_{L/F} ϖ_{L}.

Only identities (14.5) and (14.6) are to be considered. (14.5) reduces to

ν_{ϕ}(γ)^{at} = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}}

and (14.6) reduces to

(−1)^{a} ν_{ϕ}(−1)^{a[G:G_{0}]} ν_{ϕ}(γ)^{at} ν_{ϕ}(−1)^{\frac{q−1}{2k}} = 1.

Since [G : G_{0}] is even they are equivalent. Suppose ϕ has r elements. If x ∈ λ = O_{L}/𝔓_{L} then x^{σ_{0}} = x^{r^{f}} for some f. If σ belongs to G_{0}/G_{1} then

θ_{0}(σ)^{μ_{0}r^{f}} = θ_{0}(σ_{0}σσ_{0}^{−1})^{r^{f}} ≡ (\frac{ϖ_{L}^{σ_{0}σ}}{ϖ_{L}^{σ_{0}}} )^{r^{f} σ_{o}^{−1}} ≡ θ_{0}(σ).

Thus

μ_{0}^{r^{f}} ≡ 1 (mod k)

and

r ≡ −1 (mod 4)

so that ν_{ϕ}(−1) = −1. Since, in the present case,

a = \frac{q − 1}{2k}

the identities become

ν_{ϕ}(γ)^{at} = 1.

The map

τ_{L/F} : W_{L/F} ⟶ C_{F}

determines a map of G/G_{1} onto C_{F} /N_{L/F} C_{L}. The image of σ_{0} contains an element of the form γ_{1}ϖ_{F} where γ_{1} is a unit. The image of σ_{0}^{2} is 1 because the commutator subgroup contains

{σ((μ_{0} − 1)x)} = {σ(x) | x ≡ 0 (mod 2)}

and in particular contains σ_{0}^{2}. Since [G : G_{0}] = 2 the number γγ_{1}^{−2} lies in U_{F} ∩ N_{L/F} C_{L}. The index of the commutator subgroup of G/G_{1} in G/G_{1} is 4 so

[U_{F} : U_{F} ∩ N_{L/F} C_{L}] = 2.

Consequently γγ_{1}^{−2} and γ are both squares and ν_{ϕ}(γ) = 1.

Now suppose μ_{0} ≡ −1 (mod k) and x(σ_{0}^{2}) = 0. Every element of the form σ_{0}σ, σ ∈ G_{0}/G_{1}, has order 2. If π(σ_{0}σ) = −I then σ_{0}σ lies in the center of G/G_{1} which is impossible. Thus π(σ_{0}σ) has 1 as an eigenvalue. If τ ∈ G_{0}/G_{1} then

τ^{−1}σ_{0}στ = σ_{0}στ^{2}

so there are two conjugacy classes in the set σ_{0}G_{0}/G_{1}. One has σ_{0} as representative and the other has σ_{1} = σ_{0}σ(1).

Let V be a non-trivial subspace of S(K/L) invariant and irreducible under the action of G_{0}/G_{1}. Suppose first that V is also invariant under π^{∗}(σ_{0}) so that V = S(K/L). Choose v_{0} ≠ 0 so that π^{∗}(σ_{0})v_{0} = v_{0}. Let λ′ be the field obtained by adjoining the kth roots of unity to the prime field. Certainly λ′ ⊆ λ and, since

θ_{0}(σ)^{σ_{0}} = θ_{0}(σ_{0}^{−1}σσ_{0}),

λ′ is not contained in ϕ. Let ϕ′ = ϕ ∩ λ′. We may regard {1, σ_{0}} as 𝔊(λ′/ϕ′). The map φ which sends σ in G_{0}/G_{1} to (θ_{0}^{−1}(σ), 1) and σσ_{0} to (θ_{0}^{−1}(σ), σ_{0}) is an isomorphism of G/G_{1} with the semi-direct product of the kth roots of unity in λ′ and 𝔊(λ′/ϕ′). There is a unique map, again denoted by φ, of V onto λ′ such that φ(v_{0}) = 1 while

φ(π^{∗}(τ)v) = φ(τ)φ(v)

for τ in G/G_{1}. Of course the kth roots of unity act on λ′ by left multiplication. The Galois group acts by σ_{0}α = α^{σ_{0}^{−1}}. Putting the actions together we get an action of the semi-direct product. To study the action of G/G_{1} on V we study the equivalent action of the semi-direct product in λ′.

It is best to consider a more general situation. Suppose ϕ′ is a finite field with p^{f} elements, λ′ is an extension of ϕ′ with p^{ℓ} elements and Γ is the semi-direct product of the group of kth roots of unity, where k divides p^{ℓ} − 1, and 𝔊(λ′/ϕ′). Γ acts on λ′ as before. Let ℓ = nf. If 0 ≤ j_{1} < n, j = (j_{1}, n), and ρ is the automorphism x ⟶ x^{p^{f}} of λ′/ϕ′ then the number of elements of λ′ fixed by a member of Γ of the form (α, ρ^{j_{i}}) where α is a kth root of unity is the same as the number of elements fixed by some other member of the form (β, ρ^{−j}). Indeed if

b \frac{j_{1}}{j} ≡ −1 (mod \frac{n}{j})

and b is prime to the order of (α, ρ^{j_{1}}) we can take

(β, ρ^{−j}) = (α, ρ^{j_{1}})^{b}.

Let θ be a generator of the multiplicative group of λ′. The equation

β θ^{mρ^{j}} = θ^{m}

can be solved for β if and only if θ^{m(p^{jf} −1}) has order dividing k, that is, if and only if p^{ℓ} −1 divides km(p^{jf} − 1) or, if

u = \frac{p^{ℓ} − 1}{k}

if and only if u divides m(p^{jf} − 1). Let u(j) be the greatest common divisor of u and p^{jf} − 1. u divides m(p^{jf} − 1) if and only if \frac{u}{u(j)} divides m. The number of such m with 0 ≤ m < p^{ℓ} − 1 is

\frac{u(j)}{u} (p^{ℓ} − 1) = u(j)k.

Once m and j are chosen α is determined. The number of non-zero x in λ′ which are fixed by some (β, ρ^{−j}) where j divides n but by no (β, ρ^{−i}) where i properly divides j is

∑_{i|j} μ (\frac{j}{i} ) u(i)k

if μ(·) is the Möbius function. The number of orbits formed by such x is

\frac{1}{jk} ∑_{i|j} μ (\frac{j}{i} ) u(i)k

so that the total number of orbits of Γ in the multiplicative group of λ′ is

a = ∑_{i|n} ∑_{j|\frac{n}{i}} \frac{μ(j)}{ij} u(i)

which equals

∑_{i|n} ∏_{π|\frac{n}{i}} (1 − \frac{1}{π} ) \frac{u(i)}{i}.

The product is over primes. Lemma 14.7

If \frac{p^{ℓ}−1}{k} is odd then

(−1)^{a+1} ν_{ϕ′}(k) ν_{ϕ′}(−1)^{\frac{p^{ℓ}−1}{2k} + \frac{1}{2}} = 1.

The identity of the lemma is equivalent to

(−1)^{a+1} ν_{ϕ′}(u) ν_{ϕ′}(−1)^{\frac{u−1}{2}} = 1

because

ν_{ϕ′}(k) = ν_{ϕ′}(−1) ν_{ϕ′}(u).

By the law of quadratic reciprocity the left side of the identity is equal to

(−1)^{a+1} (p^{f}|u)

if (p^{f}|u) is Jacobi’s symbol. If u = 1 there is only one orbit so

(−1)^{a+1} = 1.

Of course (p^{f}|1) = 1 so the identity is clear in this case.

We prove it in general by induction on the number of prime factors of u. Let π_{0} be a prime factor of u and let u = π_{0}^{x}v with v prime to π_{0}. Let v(j) be the analogue of u(j). Then u(j) = π_{0}^{x(j)}v(j). Leb b be the analogue of a. Then

ν = a − b = ∑_{i|n} ∏_{π|\frac{n}{i}} (1 − \frac{1}{π} ) \frac{v(i)}{i} (π_{0}^{x(i)} − 1).

Observe that π_{0} and all v(i) are odd. To prove the lemma by induction we must show that

(−1)^{ν} (p^{f} | π_{0}^{x}) = 1.  (14.8)

Let

n = 2^{y}n_{1}

with n_{1} odd. There are two possibilities to be considered.

(i)

π_{0} ≡ 1 (mod 2^{y+1}).

Since the order of p^{f} modulo π_{0} divides n the quotient of π_{0} − 1 by this order is even and p^{f} is a quadratic residue of π_{0}. Also if i divides n

\frac{π_{0}^{x(i)} − 1}{i}

is divisible, in the 2-adic field, by 4 if 2 divides \frac{n}{i} and is always divisible by 2. Thus ν is even and (14.8) is valid.

(ii)

π_{0} = 1 + 2^{c}w

with c ≤ y and w odd. Let i ≠ n_{1} divide n_{1} and consider

∑_{j=0}^{y} ∏_{π|\frac{n}{2^{j} i}} (1 − \frac{1}{π} ) \frac{v(2^{j}i)}{2^{j}i} (π_{0}^{x(2^{j}i)} − 1).  (14.9)

If x(2^{y}i) = 0 the sum is zero. If x(2^{y}i) ≠ 0 let z be the smallest integer for which x(2^{z}i) ≠ 0. If j < z then x(2^{j}i) = 0. If j ≥ z

p^{2^{j}if} − 1 = (p^{2^{z}if} − 1) (∑_{d=0}^{2^{j−z}−1} p^{2^{z}cif}) .

The residue of the sum modulo π_{0} is 2^{j−z}. Thus

x(2^{j}i) = x(2^{z}i)

if j ≥ z and (14.9) is equal to

\frac{1}{i} \{∏_{π|\frac{n_{1}}{i}} (1 − \frac{1}{π} )\} \{\frac{v(2^{y}i)}{2^{y}} + ∑_{j=z}^{y−1} \frac{v(2^{j}i)}{2^{j+1}}\} (π_{0}^{x(2^{y}i)} − 1) .

We write

\frac{v(2^{y}i)}{2^{y}} + ∑_{j=z}^{y−1} \frac{v(2^{j}i)}{2^{j+1}}

as

\frac{v(2^{z}i)}{2^{z}} + ∑_{j=z+1}^{y} \frac{v(2^{j}i) − v(2^{j−1}i)}{2^{j}}.

If k is replaced by \frac{p^{ℓ}−1}{v} the number of elements of λ^{∗} fixed by some (α, ρ^{−2^{j}i}) but by no (α, ρ^{−2^{j−1}i}) is

\{v(2^{j}i) − v(2^{j−1}i)\} \frac{p^{ℓ} − 1}{v}.

The collection of such elements is invariant under the group obtained by replacing k by \frac{p^{ℓ}−1}{v} and ϕ′ by the field with p^{if} elements. The isotropy group of each such point has a generator of the form (α, ρ^{−2^{j}i}) and, therefore, has order \frac{n}{2^{j}i} and index \frac{2^{j}(p^{ℓ}−1)}{v}. Thus \frac{2^{j}(p^{ℓ}−1)}{v} divides

\{v(2^{j}i) − v(2^{j−1}i)\} \frac{p^{ℓ} − 1}{v}

so that 2^{j} divides

v(2^{j}i) − v(2^{j−1}i).

Since \frac{n_{1}}{i} is divisible by at least one prime, the expression (14.9) is congruent, in the 2-adic field, to

\frac{1}{i} \{∏_{π|\frac{n_{1}}{i}} (1 − \frac{1}{π} )\} \frac{v(2^{z}i)}{2^{z}} (π^{x(2^{y}i)} − 1)

modulo 4. Since z ≤ c and the product is not empty this is congruent, in the 2-adic field again, to 0 modulo 2. Thus ν is even or odd according as

∑_{j=0}^{y}\{∏_{π|2^{y−j}} (1 − \frac{1}{π} )\} \frac{v(2^{j}n_{1})}{2^{j}n_{1}} (π_{0}^{x(2^{j}n_{1})} − 1)

is or is not divisible by 2 in the 2-adic field. Consequently

ν ≡ ∑_{j=0}^{y} \{∏_{π | 2^{y−j}} (1 − \frac{1}{π} )\} \frac{v(2^{j}n_{1})}{2^{j}} (π_{0}^{x(2^{j}n_{1})} − 1) (mod 2).

Of course x(2^{y}n_{1}) = x ≠ 0. Let z again be the smallest integer for x(2^{z}n_{1}) ≠ 0. Then z ≤ c and

x(2^{j}n_{1}) = x(2^{z}n_{1})

if j ≥ z. The sum above is equal to

\{\frac{v(2^{z}n_{1})}{2^{z}} + ∑_{j=z+1}^{y} \frac{v(2^{j}n_{1}) − v(2^{j−1}n_{1})}{2^{j}}\} (π_{0}^{x} − 1).

As before this is congruent modulo 2 to

\frac{v(2^{z}n_{1})}{2^{z}} (π_{0}^{x} − 1).

If z < c this is even and the order of p modulo π_{0} divides \frac{π_{0}−1}{2} so that (p|π_{0}) = 1. If z = c then

\frac{π_{0}^{x} − 1}{2^{z}} = \frac{1}{2^{c}} ∑_{i=1}^{x} \binom{x}{i} (2^{c}w)^{i} ≡ x (mod 2)

so that ν ≡ x (mod 2). However the order of p^{f} modulo π_{0} is divisible by 2^{z} so that it does not divide \frac{π_{0}−1}{2} and

(p^{f} | π_{0}^{x}) = (−1)^{x}.

The relation (14.8) is now easily verified.

We return to the original problem. Since λ′ is a quadratic extension of ϕ′ and λ′ is not contained in ϕ the degree of ϕ over ϕ′ is odd. Since V and λ′ have the same number of elements q = p^{ℓ}. If \frac{q−1}{k} is odd the relation (14.4) follows immediately from the equality

(−1)^{a+1} ν_{ϕ}(k) ν_{ϕ}(−1)^{\frac{q−1}{2k} + \frac{1}{2}} = (−1)^{a+1} ν_{ϕ′}(k) ν_{ϕ′}(−1)^{\frac{q−1}{2k} + \frac{1}{2}}

and the preceding lemma.

The number of μ in T with isotropy group of order 2 is u(1) and the number of μ with trivial isotropy group is \frac{u(2)−u(1)}{2}. For points of the second type [ϕ_{μ} : ϕ] = 2 and for points of the first type [ϕ_{μ} : ϕ] = 1. Since, as we verified earlier,

ν_{ϕ}(ω_{1}) = ν_{ϕ} (∏_{μ} γ_{μ})

and

ν_{ϕ}(N_{ϕ_{μ}/ϕ} ε_{μ}) = ν_{ϕ}(−1)^{[ϕ_{μ}:ϕ]} ν_{ϕ}(γ_{μ})

the identity (14.7) reduces to

u(1) ≡ 1 (mod 2)

which is true because u(1) divides u = \frac{p^{ℓ}−1}{k} which, when (14.7) is under consideration, is odd by assumption.

The identity (14.5) may be formulated as

ν_{ϕ} (∏_{μ} γ_{μ}) = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}}

and (14.6) as

ν_{ϕ} (∏_{μ} γ_{μ}) ν_{ϕ}(−1)^{∑_{μ}[ϕ_{μ}:ϕ]} = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}} .

For these two identities \frac{q−1}{k} is even. Again

∑[ϕ_{μ} : ϕ] ≡ u(1) (mod 2).

But

u(2) = u = \frac{p^{ℓ} − 1}{k} = \frac{q − 1}{k}

and

2a = u(1) + u(2)

so u(1) is even. It will be enough to verify (14.5).

We may choose T so that if μ is in T then its isotropy group is trivial or contains one of σ_{0} or σ_{1}. If σ_{0} lies in the isotropy group of μ and ν in the orbit of μ corresponds to θ^{m} in λ′ then

α^{2}θ^{mρ} = θ^{m}

for some kth root of unity α. This is possible if and only if p^{ℓ} − 1 divides \frac{km}{2} (p^{f} − 1) or 2u divides

m(p^{f} − 1). This is the same as requiring that \frac{2u}{u(1)} divide \frac{m(p^{f} −1)}{u(1)}. The number u is even. We have already observed that if r is the number of elements in ϕ so that x^{σ_{0}} = x^{r} for x in ϕ then

μ_{0}r ≡ 1 (mod k)

and in particular

μ_{0}r ≡ 1 (mod 4).

Since [ϕ : ϕ′] is odd and μ_{0} ≡ −1 (mod 4) the highest power of 2 dividing p^{f} − 1 is 2. Thus \frac{2u}{u(1)}

and \frac{p^{f} −1}{u(1)} are relatively prime so that \frac{2u}{u(1)} divides \frac{m(p^{f} −1)}{u(1)} if and only if \frac{2u}{u(1)} divides m. There are

\frac{u(1)k}{2} such m with 0 ≤ m < p^{ℓ} −1. The corresponding characters ν fall into \frac{u(1)}{2} orbits. Thus there are \frac{u(1)}{2} elements in T whose isotropy group contains σ_{0} and \frac{u(1)}{2} whose isotropy group contains σ_{1}. Let L_{0} be the fixed field of σ_{0} and L_{1} the fixed field of σ_{1}. Let ϖ_{L_{0}} and ϖ_{L_{1}} generate 𝔓_{L_{0}} and 𝔓_{L_{1}} respectively and let

N_{L_{0}/F} ϖ_{L_{0}} = γ_{0} ϖ_{F}
N_{L_{1}/F} ϖ_{L_{1}} = γ_{1}ϖ_{F}
N_{L/F} ϖ_{L} = γϖ_{F}^{2} .

We have to show that

ν_{ϕ} (γ0^{\frac{u(1)}{2}} γ1^{\frac{u(1)}{2}} γ^{\frac{u(2)−u(1)}{2}} ) = (−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}} .

First we prove a lemma, special cases of which we have already seen.

Lemma 14.8

Suppose L/F is normal but non-abelian and [L : F] is a power of 2. Suppose H = 𝔊(L/F) and the first ramification group H_{1} is {1} but [H : H_{0}] > 1 and [H_{0} : H_{1}] > 1. Let ϖ_{L} generate the prime ideal of O_{L}, let ϖ_{F} generate the prime ideal of O_{F}, and let

N_{L/F} ϖ_{L} = γϖ_{F}^{[H:H_{0}]}.

Then γ is a square in U_{F}.

The hypotheses imply that the residue field has odd characteristic. Let A be the fixed field of H_{0} and L′ be the fixed field of the commutator subgroup of H. Then A ⊆ L′ and if

ϖ_{L′} = N_{L/L′}ϖ_{L}

then

N_{L′/F} ϖ_{L′} = γϖ_{F}^{[A:F]}.

Of course [A : F] = [H : H_{0}]. Since H is nilpotent but not abelian L′ cannot be a cyclic extension. If γ is not a square in U_{F} then γ^{−1} generates U_{F} /U_{F} ∩ N_{L′/F} C_{L}. Since

ϖ_{F}^{[A:F]} ≡ γ^{−1} (mod N_{L′/F} C_{L′}).

ϖ_{F} would then generate C_{F} /N_{L′/F} C_{L′} which is impossible.

Returning to the problem at hand, we observe that the quotient of G/G_{1} by the squares in G_{0}/G_{1} is a group of order 4 in which every square is 1. The fixed field F′ of this group is the composite of all quadratic extensions of F. F_{0} = F′ ∩ L_{0} and F_{1} = F′ ∩ L_{1} are the two different ramified quadratic extensions of F. Define

ϖ_{F_{0}} = N_{L_{0}/F_{0}}ϖ_{L_{0}}

and

ϖ_{F_{1}} = N_{L_{1}/F_{1}}ϖ_{L_{1}}.

Then

N_{F_{0}/F} ϖ_{F_{0}} = γ_{0}ϖ_{F}

and

N_{F_{1}/F} ϖ_{F_{1}} = γ_{1}ϖ_{F} .

We need to show that

ν_{ϕ}(γ_{0}γ_{1}) = ν_{ϕ} (\frac{γ_{0}}{γ_{1}} ) = −1.

If not, \frac{γ_{0}}{γ_{1}} is a square and thus in N_{F_{1}/F} C_{F_{1}}. Then γ_{0}ϖ_{F} belongs to

N_{F_{0}/F} C_{F_{0}} ∩ N_{F_{1}/F} C_{F_{1}} = N_{F′/F} C_{F′}.

This is impossible because F′ contains an unramified extension.

We observed before that since

μ_{0} = λ(σ_{0}) ≡ −1 (mod 4)

the number ν_{ϕ}(−1) is −1. The identity (14.5) reduces to

(−1)^{\frac{u(1)}{2}} = (−1)^{a} (−1)^{\frac{q−1}{2k}} .

Since

a = \frac{u(1)}{2} + \frac{u(2)}{2}

and

u(2) = \frac{q − 1}{k}

this relation is clearly valid.

We continue to suppose that μ_{0} ≡ −1 (mod k) and that σ_{0}^{2} = 1 but now we suppose that V is not invariant under π^{∗}(σ_{0}). Since π^{∗}(σ_{0})V ∩ V and π^{∗}(σ_{0})V + V are both invariant under G/G_{1} the first is 0 and the second is S(K/L) so that S(K/L) is the direct sum V ⊕ π^{∗}(σ_{0})V. Let V have p^{ℓ} elements so that q = p^{2ℓ}. If λ′ is again the field generated over the prime field by the kth roots of unity λ′ has p^{ℓ} elements. If ϕ′ = λ′ ∩ ϕ has p^{f} elements then p^{ℓ} = p^{2f} so that p^{ℓ} ≡ 1 (mod 8). Also k divides p^{ℓ} − 1 so that

\frac{q − 1}{k} = (\frac{p^{ℓ} − 1}{k}) (p^{ℓ} + 1)

is even.

If σ ∈ G_{0}/G_{1} the non-zero fixed points of σ_{0}σ are the elements of the form

v ⊕ π^{∗}(σ_{0}σ)v

with v ≠ 0. There are (p^{ℓ} − 1)k of them altogether and they fall into p^{ℓ} − 1 orbits. The remaining

(p^{2ℓ} − 1) − (p^{ℓ} − 1)k

elements fall into

\frac{1}{2k} \{(p^{2ℓ} − 1) − (p^{ℓ} − 1)k\}

orbits. Thus

a = \frac{p^{ℓ} − 1}{2} + \frac{p^{2ℓ} − 1}{2k}.

Since, for the same reasons as before, ν_{ϕ}(−1) = −1 the identity (14.5) becomes

ν_{ϕ} (∏_{μ} γ_{μ}) = (−1)^{\frac{p^{ℓ}−1}{2}}  (14.10)

while (14.6) becomes

ν_{ϕ} (∏_{μ} γ_{μ}) ν_{ϕ}(−1)^{Σ[ϕ_{μ}:ϕ]} = (−1)^{\frac{p^{ℓ}−1}{2}} .

Since

∑ [ϕ_{μ} : ϕ] ≡ p^{ℓ} − 1 ≡ 0 (mod 2)

only (14.5) need be proved. (14.4) and (14.7) are not to be considered because \frac{q−1}{k} is even.

We proceed as before. The points in T can be chosen so that their isotropy groups are either trivial or contain σ_{0} or σ_{1}. \frac{p^{ℓ}−1}{2} will have isotropy groups containing σ_{0} and \frac{p^{ℓ}−1}{2} will have isotropy groups containing σ_{1}. The argument used above shows that the left side of (14.10) is equal to

(−1)^{\frac{p^{ℓ}−1}{2}}

as desired.

Now suppose k ≥ 8 and

μ_{0} = λ(σ_{0}) ≡ \frac{k}{2} − 1 (mod k).

We are of course still supposing that [G : G_{0}] = 2. If σ belongs to G_{0}/G_{1} then

σσ_{0}σ^{−1} = σ_{0}σ^{\frac{k}{2} −2}

and

(σ_{0}σ)^{2} = σ_{0}^{2}σ^{\frac{k}{2}} .

Thus

x((σ_{0}σ)^{2}) = x(σ_{0}^{2}) + \frac{k}{2} x(σ).

Since x(σ_{0}^{2}) is 0 or \frac{k}{2} , we can make the sum on the right 0. Replacing σ_{0} by σ_{0}σ if necessary, we suppose that σ_{0}^{2} = 1. Then (σ_{0}σ)^{2} = 1 if and only if

\frac{k}{2} x(σ) ≡ 0 (mod k)

which is so if and only if σ_{0}σ is conjugate to σ.

Take V in S(K/L) as before. If V is invariant under π^{∗}(σ_{0}) and λ′ with p^{ℓ} elements and ϕ′ with p^{f} elements have the same meaning as before then

p^{f} = \frac{k}{2} − 1 + wk

for some integer w so that

q − 1 = p^{2f} − 1 = k · \frac{k}{4} − k + 2wk (\frac{k}{2} − 1) + (wk)^{2}

and

\frac{q − 1}{k} ≡ \frac{k}{4} − 1 (mod 2)

is odd. Thus the identities (14.5) and (14.6) are not to be considered. The identities (14.4) and (14.7) follow from Lemma 14.7 exactly as above.

Suppose then S(K/L) is the direct sum V ⊕ π^{∗}(σ_{0})V. If V has p^{ℓ} elements then q = p^{2ℓ} and

\frac{q − 1}{k} = \frac{p^{ℓ} − 1}{k} (p^{ℓ} + 1)

is even because k divides p^{ℓ} − 1. The non-zero elements of S(K/L) which are fixed points of some σ_{0}σ with σ a square in G_{0}/G_{1} are the elements

v ⊕ π^{∗}(σ_{0}σ)v

with v ≠ 0. There are (p^{ℓ} − 1) \frac{k}{2} such elements and they fall into \frac{p^{ℓ}−1}{2} orbits. The remaining

(p^{2ℓ} − 1) − (p^{ℓ} − 1) \frac{k}{2}

non-zero elements have trivial isotropy group and fall into

\frac{1}{2k} \{(p^{2ℓ} − 1) − (p^{ℓ} − 1) \frac{k}{2} \}

orbits. Thus

a = \frac{p^{2ℓ} − 1}{2k} + \frac{p^{ℓ} − 1}{4}.

Since, as before, ν_{ϕ}(−1) = −1 the identity (14.5) becomes

ν_{ϕ} (∏_{μ} γ_{μ}) = (−1)^{\frac{p^{ℓ}−1}{4}}  (14.11)

while (14.6) becomes

ν_{ϕ} (∏_{μ} γ_{μ}) (−1)^{Σ[ϕ_{μ}:ϕ]} = (−1)^{\frac{p^{ℓ}−1}{4}} .

Again

∑ [ϕ_{μ} : ϕ] ≡ \frac{p^{ℓ} − 1}{2} ≡ 0 (mod 2)

so that it is enough to prove (14.11). The identities (14.4) and (14.7) need not be considered.

If λ′ and ϕ′ are defined as before and ϕ′ has p^{f} elements then λ′ has p^{ℓ} = p^{2f} elements so that

p^{ℓ} ≡ 1 (mod 8)

and \frac{p^{ℓ}−1}{4} is even. We may suppose that each μ in T either has trivial isotropy group or is fixed by σ_{0}. Lemma 14.8 shows that those μ with trivial isotropy group contribute nothing to the left side of (14.11). If L_{0} is the fixed field of σ_{0} and

N_{L_{0}/F} ϖ_{L_{0}} = γ_{0}ϖ_{F}

the left side of K is

ν_{ϕ}(γ_{0})^{\frac{p^{ℓ}−1}{2}}

which is 1. The truth of the identity is now clear.

We return to the general case so that [G : G_{0}] may be greater than 2 and μ_{0} may be congruent to 1 modulo 4. Of course [G : G_{0}] is still even. Let

λ_{0} = λ (σ_{0}^{\frac{1}{2} [G:G_{0}]})

so that

λ_{0} = μ_{0}^{\frac{1}{2} [G:G_{0}]}.

If [G : G_{0}] > 2 then

λ_{0} ≡ 1 (mod 4).

If [G : G_{0}] = 2 then λ_{0} = μ_{0}. Since the case that [G : G_{0}] = 2 and μ_{0} ≡ −1 (mod 4) is completely settled we may suppose that λ_{0} ≡ 1 (mod 4). Set

τ_{0} = σ_{0}^{\frac{1}{2} [G:G_{0}]}.

Any element of G/G_{1} which does not lie in G_{0}/G_{1} and whose square is 1 is of the form σ(x)τ_{0}. If

σ_{0}^{[G:G_{0}]} = σ(y_{0})

then

(σ(x)τ_{0})^{2} = σ((λ_{0} + 1)x)τ_{0}^{2} = σ(y_{0} + (λ_{0} + 1)x).

Since G/G_{1} is not cyclic y_{0} is even. Since

λ_{0} + 1 ≡ 2 (mod 4)

there are exactly two solutions of the equation

y_{0} + (λ_{0} + 1)x ≡ 0 (mod k).

Let x_{0} be one of them. Then x_{0} + \frac{k}{2} is the other. We may suppose that k does not divide x_{0}. Set

ρ_{0} = σ(x_{0})τ_{0}.

We observed before that if σ ≠ 1 belongs to G/G_{1} and π^{∗}(σ) has a non-zero fixed point then some power of σ is of order 2 and has a non-zero fixed point. Since σ (\frac{k}{2} ) has no non-zero fixed point this power must be ρ_{0} or σ(\frac{k}{2} )ρ_{0}. Since σ(\frac{k}{2} ) lies in the center of G/G_{1}, σ must lie in the centralizer of ρ_{0}.

The group

\{1, σ(\frac{k}{2} ), ρ_{0}, σ(\frac{k}{2} )ρ_{0}\}

is of order 4 and every element in it is of order 2 so it cannot be contained in the center of G/G_{1}. However it is a normal subgroup and its centralizer H^{∗} has index 2 in G/G_{1}. G/G_{1} may be identified with H. Every element σ of H such that π^{∗}(σ) has a non-zero fixed point lies in H^{∗}. S(K/L) is the direct sum of V and W where

V = {v | π^{∗}(ρ_{0})v = v}
W = {w | π^{∗}(ρ_{0})w = −w}.

If σ in H does not belong to H^{∗} then π^{∗}(σ)V = W and π^{∗}(σ)W = V. The number of non-zero orbits of H in V ∪ W is the same as the number a′ of non-zero orbits of H^{∗} in V. If V has p^{ℓ} elements so that q = p^{2ℓ} the number of non-zero orbits of H in V ⊕ W − (V ∪ W) is

a′′ = \frac{(p^{ℓ} − 1)^{2}}{[G : G_{1}]} = \frac{p^{ℓ} − 1}{k} · \frac{p^{ℓ} − 1}{[G : G_{0}]}.

The action of H^{∗} on V must be irreducible although it is not faithful. However the action of H^{∗} ∩ H_{0} = H_{0}^{∗} is faithful.

Let F′ be the fixed field of H^{∗} in L or, what is the same, of H^{∗}C in K. Let C^{1} ⊆ C be the orthogonal complement of V and let H^{1} be the subgroup of H which acts trivially on V. H^{1}C^{1} is a normal subgroup of H^{∗}C and its fixed field K′ is normal over F′. If H′ = H^{∗}/H^{1} and C′ = C/C^{1} then G′ = 𝔊(K′/F′) = H′C′. Moreover H′ ∩ C′ = \{1\} and H′ ≠ \{1\} because σ (\frac{k}{2} ) does not lie in H^{1}. Since the action of H′ on C′ is faithful and irreducible C′ is contained in every non-trivial normal subgroup of G′. To complete the proof of the four identities (14.4), (14.5), (14.6), and (14.7) we use induction on [K : F].

Let k′ be the order of H′_{0} and let ϕ′ = O_{F′}/𝔓_{F′}. If K/F is replaced by K′/F′ the identity (14.4) becomes

(−1)^{a′+1} ν_{ϕ′}(k′) ν_{ϕ′}(−1)^{\frac{p^{ℓ}−1}{2k′} + \frac{1}{2}} = 1.  (14.4′′)

T is to be replaced by T′, a set of representatives for the non-zero orbits of H′ or H^{∗} in V, which may be identified with the character group of C′. We may suppose that T′ is a subset of T. Because H′_{0} ≠ {1} the identity (14.5) for the field K′/F′ may be written as

ν_{ϕ′} (∏_{μ∈T′} γ′_{μ}) = (−1)^{a′} ν_{ϕ′}(−1)^{\frac{p^{ℓ}−1}{2k′}} .  (14.5′′)

Of course

N_{F_{μ}/F′}(ϖ_{F_{μ}}^{t}) = γ′_{μ}ϖF′^{\frac{t[Fμ:F′]}{k′}} .

Recall that t is odd. By Proposition IV.3 of Serre’s book, t has the same significance for K′/F′ as it had for K/F. The identity (14.6) may be written as

(−1)^{a′} ν_{ϕ′} (∏_{μ∈T′} γ′_{μ}) ν_{ϕ′}(−1)^{Σ_{μ∈T′} [ϕ_{μ}:ϕ′]} ν_{ϕ′}(−1)^{\frac{p^{ℓ}−1}{2k′}} = 1.  (14.6′′)

and (14.7) as

(−1)^{a′+1} ν_{ϕ′}(k′) ν_{ϕ′}(−1)^{\frac{p^{ℓ}−1}{2k′} − \frac{1}{2}} ν_{ϕ′}(−1)^{Σ_{μ∈T′} [ϕ_{μ}:ϕ′]} = 1  (14.7′′)

Assuming (14.4′′), (14.5′′), (14.6′′), and (14.7′′) we are going to prove (14.4), (14.5), (14.6), and (14.7).

Since H′_{0} is isomorphic to H_{0}^{∗} either k′ = k or k′ = \frac{k}{2} . Suppose first that \frac{q−1}{k} is odd. Then k′ = \frac{k}{2} for if not

\frac{q − 1}{k} = (\frac{p^{ℓ} − 1}{k}) (p^{ℓ} + 1)

would be even. Thus H_{0} = G_{0}/G_{1} is not contained in H^{∗} and F′/F is ramified so that ϕ′ = ϕ. Since

\frac{q − 1}{k} = (\frac{p^{ℓ} − 1}{k′}) \frac{(p^{ℓ} + 1)}{2}

the number \frac{p^{ℓ}−1}{k′} is odd. To prove (14.4) we have to show that

(−1)^{a′′} ν_{ϕ}(2) ν_{ϕ}(−1)^{δ} = 1

if

δ = \frac{p^{2ℓ} − 1}{2k} − \frac{p^{ℓ} − 1}{k} = \frac{p^{ℓ} − 1}{k} \{\frac{p^{ℓ} + 1}{2} − 1\}.

Since G_{0}/G_{1} is not contained in H^{∗}, τ_{0} does not commute with G_{0}/G_{1} and the map λ of G/G_{0} into the units of ℤ /k ℤ is faithful. Thus

λ_{0} ≢ 1 (mod k).

But

λ_{0} ≡ 1 (mod 4)

so that k ≥ 8. In general if k ≥ 4, the group of units of ℤ /k ℤ is the product of {1, −1} and

{α | α ≡ 1 (mod 4)}.

If

α = 1 + 2^{b}x

with x odd and 4 ≤ 2^{b} ≤ k then

α^{2} = 1 + 2^{b+1}y

with y odd. One shows easily by induction that the order of α is 2^{−b}k so that

{α | α ≡ 1 (mod 4)}

is cyclic of order \frac{k}{4} . This implies in the particular case under consideration that [G : G_{0}] divides \frac{k}{4} . Write

a′′ = (\frac{p^{ℓ} − 1}{k′}) (\frac{p^{ℓ} − 1}{2[G : G_{0}]}) .

a′′ is odd if and only if

2[G : G_{0}] = k′.

We consider various cases separately. As before μ_{0} = λ(σ_{0}). If ϕ has p^{f} elements then

μ_{0}p^{f} ≡ 1 (mod 8)

(i)

μ_{0} ≡ 1 (mod 8).

Then

ν_{ϕ}(2) = ν_{ϕ}(−1) = 1

and the order of μ_{0} in the units of ℤ /k ℤ which is equal to [G : G_{0}] divides \frac{k}{8} . Thus a′′ is even. The identity (14.4) follows.

(ii)

μ_{0} ≡ 3 (mod 8).

Then

ν_{ϕ}(2) = ν_{ϕ}(−1) = −1.

Since μ_{0} ≡ 3 (mod 8) the numbers μ_{0} and λ_{0} are different. Thus λ_{0} is a square and hence congruent to 1 modulo 8. Then k > 8 and

p^{ℓ} ≡ 1 (mod 8).

Then

δ ≡ \frac{p^{ℓ} + 1}{4} − \frac{1}{2} ≡ 0 (mod 2).

Since μ_{0} ≠ λ_{0} the index [G : G_{0}] is not 2. Thus the order of μ_{0} is at least 4 and is therefore the order of −μ_{0}. Since −μ_{0} ≡ 5 (mod 8) its order is \frac{k}{4} and

[G : G_{0}] = \frac{k}{4} .

Consequently a′′ is odd. Again (14.4) is satisfied. (iii)

μ_{0} ≡ 5 (mod 8).

Then ν_{ϕ}(2) = −1 while ν_{ϕ}(−1) = 1. The order of μ_{0} which equals [G : G_{0}] is again \frac{k}{4} so that a′′ is odd and (14.4) is satisfied.

(iv)

μ_{0} ≡ 7 (mod 8).

Then ν_{ϕ}(2) = 1 while ν_{ϕ}(−1) = −1. Again k > 8 and

δ ≡ 0 (mod 2).

The order of μ_{0} is again at least 4 and therefore equal to the order of −μ_{0} and that divides \frac{k}{8} . Thus [G : G_{0}] divides \frac{k}{8} and a′′ is even. (14.4) follows once more.

Since ϕ′ = ϕ all we need to prove (14.7) once (14.4) and (14.7′′) are granted is show that

∑_{μ∈T −T′} [ϕ_{μ} : ϕ] ≡ 0 (mod 2).

This is clear because, for these μ, F_{μ} = L and ϕ_{μ} = O_{L}/𝔓_{L} is of even degree over ϕ.

Finally we have to assume that \frac{q−1}{k} is even and prove (14.5) and (14.6). First a lemma.

Lemma 14.9

If \frac{q−1}{k} is even,

λ (σ_{0}^{\frac{1}{2} [G:G_{0}]}) ≡ 1 (mod 4),

and G/G_{0} acts faithfully on G_{0}/G_{1}, then

(−1)^{a} ν_{ϕ}(−1)^{\frac{q−1}{2k}} = 1.

Since the action is faithful G_{0}/G_{1} is not contained in H^{∗} and k′ = \frac{k}{2} . As before λ_{0} ≡ 1 (mod 4) and λ_{0} ≢ 1 (mod k) together imply that k ≥ 8 and k′ ≥ 4. Since k′ divides p^{ℓ} − 1,

p^{ℓ} ≡ 1 (mod 4)

and \frac{p^{ℓ}+1}{2} is odd. Since

\frac{q − 1}{k} = (\frac{p^{ℓ} − 1}{k′}) (\frac{p^{ℓ} + 1}{2})

the number \frac{p^{ℓ}−1}{k′} is even.

If σ belongs to H^{∗} and σ acts trivially on H_{0}^{∗} then

λ(σ) ≡ 1 (mod \frac{k}{2} )

so that

λ(σ^{2}) ≡ 1(mod k)

and σ^{2} belongs to H_{0}. Thus σ belongs to ρ_{0}H_{0} ∪ H_{0}. Since ρ_{0} belongs to H^{1} the image of σ in H′ lies in H′_{0}. Thus G′/G′_{0} acts faithfully on G′_{0}/G′_{1}. If σ belongs to H^{1} then σ acts trivially on H_{0}^{∗} because the representation of H_{0}^{∗} on V is faithful. Thus H^{1} is contained in ρ_{0} H_{0} ∪ H_{0} and is therefore just {ρ_{0}, 1}. Thus

[G′ : G′_{0}] = [H′ : H′_{0}] = [H^{∗} : H_{0}^{∗}H^{1}] = \frac{1}{2} [G : G_{0}].

Suppose that

(−1)^{a} ν_{ϕ′}(−1)^{\frac{p^{ℓ}−1}{2k′}} = 1.  (14.12)

Since ϕ′ = ϕ and, because k′ ≥ 4 divides p^{ℓ} − 1,

\frac{q − 1}{2k} = (\frac{p^{ℓ} − 1}{2k′}) (\frac{p^{ℓ} + 1}{2}) ≡ (\frac{p^{ℓ} − 1}{2k′}) (mod 2),

all we need do to establish the lemma is to show that

a′′ ≡ 0 (mod 2).

As before [G : G_{0}] divides \frac{k}{4} . If

\frac{k}{4} = n[G : G_{0}]

then

a′′ = \frac{1}{k} \frac{(p^{ℓ} − 1)^{2}}{[G : G_{0}]} = n (\frac{p^{ℓ} − 1}{k′})^{2}

is certainly even because 2k′ divides p^{ℓ} − 1.

If [G : G_{0}] ≥ 4 let
λ′_{0} = λ (σ_{0}^{\frac{1}{4} [G:G_{0}]}) .

If

λ′_{0} ≡ 1 (mod 4)

we may suppose that (14.12) is true by induction. If [G : G_{0}] = 4 and

λ′_{0} ≡ 3 (mod 4)

or if [G : G_{0}] = 2 we must establish it directly.

Suppose first that [G : G_{0}] = 2. If ϕ has p^{f} elements then

λ_{0} ≡ μ_{0} ≡ p^{f} ≡ 1 (mod 4)

so that ν_{ϕ}(−1) = 1. It is clear that in this case

a′ = \frac{p^{ℓ} − 1}{k′}.

a′ is thus even and (14.12) is valid.

Now suppose [G : G_{0}] = 4 so that [G′ : G′_{0}] = 2. If σ′_{0} generates G′ modulo G′_{0} then λ′_{0} is the image of σ′_{0} in the group of units of ℤ /k′ ℤ. We have already studied the case that λ′_{0} ≡ 3 (mod 4) intensively. Let

x : σ′ ⟶ x(σ′)

be the map of G′_{0}/G′_{1} onto ℤ /k′ ℤ. If λ′_{0} ≡ −1 (mod k′) and x((σ′_{0})^{2}) = \frac{k′}{2} we showed, incidentally, that (14.12) is valid. If λ′_{0} ≡ −1 (mod k′), x((σ′_{0})^{2}) = 0, and the action of H′_{0} on S(K′/L′) is reducible we saw that p^{ℓ} is a square p^{2ℓ′} and that the left side of (L) is

(−1)^{\frac{p^{ℓ′} −1}{2}} .

But the field with p^{ℓ′} elements must contains the k′th roots of unity and k′ ≡ 0 (mod 4). Thus

p^{ℓ′} − 1 ≡ 0 (mod 4)

and (14.12) is again valid. If k′ ≥ 8,

λ′_{0} ≡ \frac{k′}{2} − 1 (mod k′)

and the action of H′_{0} on S(K′/L′) is reducible, the left side of (14.12) is

(−1)^{\frac{p^{ℓ′} −1}{4}} .

This time

p^{ℓ′} − 1 ≡ 0 (mod 8).

To complete the proof of the lemma we show that in the case under consideration the action of H′_{0} and S(K′/L′) or, what is the same, the action of H_{0}^{∗} on V is reducible. If not the field generated over the prime field by the k′th roots of unity has p^{ℓ} elements. Thus

p^{ℓ} ≡ 1 (mod 4).

However as we have observed repeatedly, the number of elements in ϕ is congruent to 3 modulo 4. Thus ℓ is even. Let ℓ = 2ℓ′. Either p^{ℓ′} −1 or p^{ℓ′} +1 is congruent to 2 modulo 4. If p^{ℓ′} +1 ≡ 2 (mod 4) then k′ divides p^{ℓ′} − 1 because

\frac{p^{ℓ} − 1}{k′} = (\frac{p^{ℓ′} − 1}{k′}) (p^{ℓ′} + 1)

is even. Since k′ cannot divide p^{ℓ′} − 1 we have

p^{ℓ′} ≡ 3 (mod 4)

and ℓ′ is odd. Indeed it is 1 but that does not matter. Since k divides p^{ℓ} − 1, the kth roots of unity are contained in the field with p^{ℓ} elements. Adjoining them to ϕ = O_{F} /𝔓_{F} we obtain a quadratic extension because 4 does not divide ℓ. Therefore if σ belongs to G_{0}/G_{1}

θ_{0}(σ) = θ_{0}(σ)^{σ_{0}^{−2}} = θ_{0}(σ)^{λ(σ_{0}^{2})}

so that

λ(σ_{0}^{2}) ≡ 1 (mod k).

This contradicts the assumption that G/G_{0} acts faithfully on G_{0}/G_{1}.

Returning to the proof of (14.5), we suppose first that H_{0} is not contained in H^{∗} so that the action of G/G_{0} on G_{0}/G_{1} is faithful. Because of Lemma 14.9 the identity (14.5) is equivalent to

ν_{ϕ} (∏_{μ∈T} N_{F_{μ}/F} γ_{μ}) = 1.

If μ belongs to T but not to T′ then F_{μ} = L and, by Lemma 14.8,

ν_{ϕ}(N_{F_{μ}/F} γ_{μ}) = 1.

If μ belongs to T′ then G_{μ} is contained in H^{∗}C so that F_{μ} contains F′. Moreover we do not change F_{μ} if we replace K/F by K′/F′. Let ϖ_{F′} generate 𝔓_{F′} and take ϖ_{F} = N_{F′/F} ϖ_{F′}. If E′ is the fixed-field of H^{∗} we may suppose that

ϖ_{F′} = N_{E′/F′}ϖ_{E′}

and that

ϖ_{E} = N_{E′/E}ϖ_{E′}.

Then

ϖ_{F} = N_{E/F} ϖ_{E}

as required. Let

N_{F_{μ}/F} , ϖ_{F_{μ}}^{t} = γ′_{μ}ϖ^{\frac{t[Fμ:F′]}{k′}} .

Then

γ_{μ} = N_{F′/F} γ′_{μ}.

Since F′/F is ramified γ_{μ} is a square in U_{F} and (14.5) is proved. To prove (14.6) we have to show that

ν_{ϕ}(−1)^{Σ_{μ∈T} [ϕ_{μ}:ϕ]} = ν_{ϕ′}(−1)^{Σ_{μ∈T′} [ϕ_{μ}:ϕ′]} = 1.

But \frac{p^{ℓ}−1}{k′} is even and this follows from the simultaneous validity of (14.5′′) and (14.6′′).

We have yet to treat the case that \frac{q−1}{k} is even and H_{0} is contained in H^{∗}. Then F′/F is unramified and k′ = k. Suppose first of all that \frac{p^{ℓ}−1}{k} is also even. Then

\frac{q − 1}{2k} = (\frac{p^{ℓ} − 1}{k}) (\frac{p^{ℓ} + 1}{2})

is even. H_{0} is contained in H^{∗} and H is generated by σ_{0} and H_{0}. Consequently σ_{0} is not contained in H^{∗} and

σ_{0}ρ_{0}σ_{0}^{−1} = σ (\frac{k}{2} ) ρ_{0}.

Since ρ_{0} = σ(x_{0})τ_{0}

(μ_{0} − 1)x_{0} ≡ \frac{k}{2} (mod k)

if μ_{0} = λ(σ_{0}). If

y_{0} = x (σ_{0}^{[G:G_{0}]})

and m is the greatest common divisor of y_{0} and k then by the definition of x_{0} the greatest common divisor of x_{0} and k is \frac{m}{2}. Therefore \frac{k}{m} is the greatest common divisor of μ_{0} −1 and k. In particular m < k. The order of σ_{0} in H is

\frac{k}{m} [G : G_{0}].

Therefore [G : G_{0}] divides \frac{k}{2m} [G : G_{0}] and H^{∗} contains a cyclic subgroup of order

\frac{k}{2m} [G : G_{0}].

If σ is the element of order 2 in this subgroup, then σ belongs to H_{0} and π^{∗}(σ) does not have 1 as an eigenvalue. Thus no non-zero element of V is fixed by any element of this cyclic subgroup and

p^{ℓ} − 1 ≡ 0 (mod \frac{k}{2m} [G : G_{0}]) .

In particular [G : G_{0}] divides p^{ℓ} − 1 and

a′′ = (\frac{p^{ℓ} − 1}{k}) (\frac{p^{ℓ} − 1}{[G : G_{0}]})

is even. As before ν_{ϕ}(γ_{μ}) = 1 if μ belongs to T and F_{μ} = L. If F_{μ} ≠ L then μ belongs to T′ and G_{μ} lies in H^{∗}C so that F_{μ} contains F′. In the present situation F′/F is unramified and we may take ϖ_{F′} = ϖ_{F}. If

N_{F_{μ}/F′}ϖ_{F_{μ}}^{t} = γ′_{μ}ϖF^{\frac{t[Fμ:F′]}{k}}

then

N_{F_{μ}/F} ϖ_{F_{μ}}^{t} = (N_{F′/F} γ′_{μ}) ϖF^{\frac{t[Fμ:F]}{k}} .

The identity (14.5) reduces to

ν_{ϕ} (∏_{μ∈T′} N_{F′/F} γ′_{μ}) = (−1)^{a′}

or

ν_{ϕ′} (∏_{μ∈T′} γ′_{μ}) = (−1)^{a′}.

Since ϕ′ is a quadratic extension of ϕ the number ν_{ϕ′}(−1) is 1 and this relation is equivalent to (14.5′′). To prove (14.6) we have to show that

ν_{ϕ}(−1)^{Σ_{μ∈T} [ϕ_{μ}:ϕ]} = 1.

This is clear because 2 divides each of the degrees [ϕ_{μ} : ϕ].

Finally we have to suppose that \frac{p^{ℓ}−1}{k} is odd. Since [ϕ′ : ϕ] = 2 the relation (14.4′′) amounts to

(−1)^{a′+1} = 1.

Again

ν_{ϕ} (∏_{μ∈T} γ_{μ}) = ν_{ϕ′} (∏_{μ∈T′} γ′_{μ}) .  (14.13)

If μ belongs to T′ and σ ≠ 1 belongs to G_{μ} then some power of σ will equal ρ_{0}. Since

\frac{p^{ℓ} − 1}{k} = ∑_{μ∈T′} \frac{[F_{μ} : F′]}{k}

is odd and

\frac{[F_{μ} : F′]}{k}

is a power of 2 there is at least one μ in T′ for which [F_{μ} : F′] = k. Then G_{μ} must contain an element of the form σ(z_{0})σ_{0}^{2}. Then

ρ_{0} = σ(x_{0})τ_{0} = (σ(z_{0})σ_{0}^{2})^{\frac{1}{4} [G:G_{0}]} = σ ((\frac{μ_{0}^{\frac{1}{2} [G:G_{0}]} − 1}{μ_{0}^{2} − 1})^{z_{0}})^{τ_{0}} .

Thus

(\frac{μ_{0}^{\frac{1}{2} [G:G_{0}]} − 1}{μ_{0}^{2} − 1}) z_{0} ≡ x_{0} (mod k).

Let

\frac{1}{4} [G : G_{0}] = 2^{b}.

Since

μ_{0}^{2} ≡ 1 (mod 4)

and, as before, the greatest common divisor of x_{0} and k is \frac{m}{2} if the greatest common divisor of y_{0} and k is m, we infer that

\frac{μ_{0}^{\frac{1}{2} [G:G_{0}]} − 1}{μ_{0}^{2} − 1} = ∏_{j=1}^{b} · \frac{μ_{0}^{2^{j+1}} − 1}{μ_{0}^{2^{j}} − 1} = ∏_{j=1}^{b} μ_{0}^{2^{j}} + 1

is multiplicatively congruent to

\frac{[G : G_{0}]}{4}

modulo 2 and that the greatest common divisor of z_{0} and k is

\frac{2m}{[G : G_{0}]}.

In particular \frac{[G:G_{0}]}{2} divides m. z_{0} is odd if and only if

m = \frac{1}{2} [G : G_{0}].

If μ_{0} ≡ 1 (mod 4) the order of μ_{0} in the group of units of ℤ /k ℤ is m because as we observed when treating the case that \frac{p^{ℓ}−1}{k} is even, the greatest common divisor of μ_{0} − 1 and k is \frac{k}{m}. However

μ_{0}^{\frac{1}{2} [G:G_{0}]} ≡ λ(τ_{0}) ≡ 1 (mod k)

and in this case m divides \frac{1}{2} [G : G_{0}]. Thus

m = \frac{1}{2} [G : G_{0}]

if μ_{0} ≡ 1 (mod 4).

We shall define a sequence of fields F^{(i)}, L^{(i)}, K^{(i)}, 1 ≤ i ≤ n. n is an integer to be specified. We will have F^{(i)} ⊆ L^{(i)} ⊆ K^{(i)} and K^{(i)}/F^{(i)} and L^{(i)}/F^{(i)} will be Galois. Let G^{(i)} = 𝔊(K^{(i)}/F^{(i)}) and C^{(i)} = 𝔊 (L^{(i)}/F^{(i)}). There will be a subgroup H^{(i)} of G^{(i)} such that H^{(i)} ≠ {1}, H^{(i)} ∩ C^{(i)} = {1}, and G^{(i)} = H^{(i)}C^{(i)}. C^{(i)} will be a non-trivial abelian normal subgroup of G^{(i)} which is contained in every other non-trivial normal subgroup. H^{(n)} will be abelian but H^{(i)} will be non-abelian if i < n. Moreover k^{(i)} = [H_{0}^{(i)} : 1] will be at least 4 for all i and k^{(i)} will equal 2k^{(i+1)} if i < n. If x is an isomorphism of H_{0}^{(i)} with ℤ /k^{(i)} ℤ and σ belongs to H^{(i)} let

x(στσ^{−1}) = λ^{(i)}(σ)x(τ).

Then λ^{(i)}(σ) will be congruent to 1 modulo 8 if i < n. If q^{(i)} is the number of elements in C^{(i)} then

\frac{q^{(i)} − 1}{k^{(i)}}

will be odd.

F′ and K′ have already been defined. L′ is just the fixed field of C′.

\frac{q′ − 1}{k′} = \frac{p^{ℓ} − 1}{k}

is odd. If σ′ in H′ is the image of σ in H^{∗}C then

λ′(σ′) ≡ λ(σ) (mod k).

Since σ is a square modulo H_{0}

λ′(σ′) ≡ 1 (mod k).

If F^{(i)}, L^{(i)}, and K^{(i)} have been defined and H^{(i)} is not abelian we can define F^{(i+1)}, L^{(i+1)}, K^{(i+1)} by the process we used to pass from F, L, K to F′, L′, K′. We have seen that if

\frac{q^{(i)} − 1}{k^{(i)}}

is odd then

\frac{q^{(i+1)} − 1}{k^{(i+1)}}

is also odd and that

k^{(i)} = 2k^{(i+1)}.

We have also seen that k^{(i)} ≥ 8 if H^{(i)} is not abelian. If H^{(i)} is abelian we take n = i.

When we pass from the ith stage to the (i + 1)th we break up T^{(i)}, the analogue of T, into T^{(i+1)} and a complementary set U^{(i)}. We may think of T^{(i)} as lying in T. If σ_{0}^{(i)} generates H^{(i)} modulo H_{0}^{(i)} then

λ′(σ_{0}^{(i)}) ≡ 1 (mod 8).

We saw that this implies that U^{(i)} has an even number of elements. If μ belongs to U^{(i)} then F_{μ} is equal to L^{(i)}. Thus we may suppose that

ν_{ϕ′} (∏_{μ∈U^{(i)}} γ′_{μ}) = 1.

Moreover L^{(i)}/F^{(i)} is non-abelian and therefore L^{(i)}/F^{(i)} is not totally ramified. Thus μ is not in U^{(i)} if [F_{μ} : F′] = k.

Since L^{(n)}/F^{(n)} is abelian the isotropy group in H^{(n)} of any μ in T^{(n)} is trivial so that F_{μ} = L^{(n)} for such μ. Since

∑_{μ∈T′} \frac{[F_{μ} : F′]}{k} ≡ ∑_{μ∈T^{(n)}} \frac{[L^{(n)} : F′]}{k} (mod 2).

There are an odd number of elements in T^{(n)} and

[L^{(n)} : F′] = k.

Choose z_{0} so that σ(z_{0})σ_{0}^{2} lies in 𝔊(L/L^{(n)}). It then fixes each μ in T^{(n)}.

Since L^{(n)}/F′ must be totally ramified there is a δ in U_{F} such that

N_{L^{(n)}/F} ϖ_{L^{(n)}} = δϖ_{F}^{2} .

The right side of (14.13) is equal to ν_{ϕ}(δ). L^{(n)} is contained in L. Choose w_{0} in W_{L/F} so that τ_{L/F} (w_{0}) = ϖ_{F}. We may suppose that σ_{0} has been chosen to be σ(w_{0}). Let L_{0} be the fixed field of H_{0}. Choose u_{0} in W_{L/L_{0}} so that σ(u_{0}) = σ(z_{0}) and so that τ_{L/L_{0}}(u_{0}) is a unit. Clearly z_{0} is even if and only if τ_{L/L_{0}}(u_{0}) or

N_{L_{0}/F} (τ_{L/L_{0}}(u_{0})) = τ_{L/F} (u_{0})

is a square. Since σ(z_{0})σ_{0}^{2} lies in 𝔊(L/L^{(n)}),

u_{0}w_{0}^{2}

lies in W_{L/L^{(n)}}. We may take

ϖ_{L^{(n)}} = τ_{L/L^{(n)}} (u_{0}w_{0}^{2}).

Then

N_{L^{(n)}/F} (ϖ_{L^{(n)}}) = τ_{L/F} (u_{0}w_{0}^{2}) = τ_{L/F} ϖ_{F}^{2}

and δ = τ_{L/F} (u_{0}) is a square if and only if z_{0} is even.

Since (−1)^{a′+1} = 1 the relation (14.5) amounts to

(1)^{a′′−1} ν_{ϕ}(−1)^{\frac{q−1}{2k}} = (−1)^{z_{0}}.

(14.6) is equivalent to (14.5) because each [ϕ_{μ} : ϕ] = 2[ϕ_{μ} : ϕ′] is even. Recall that

a′′ = (\frac{p^{ℓ} − 1}{k}) (\frac{p^{ℓ} − 1}{[G : G_{0}]}) ≡ \frac{p^{ℓ} − 1}{[G : G_{0}]} (mod 2)

and that

\frac{q − 1}{2k} = (\frac{p^{ℓ} − 1}{k}) (\frac{p^{ℓ} + 1}{2}) ≡ \frac{p^{ℓ} + 1}{2} (mod 2).

If

μ_{0} ≡ 1 (mod 4)

then ν_{ϕ}(−1) = 1 and, as we observed earlier, z_{0} is odd. We have to show that a′′ is even. We showed before that H^{∗} has to contain a cyclic subgroup of order \frac{k}{2m} [G : G_{0}] and that \frac{k}{2m} [G : G_{0}] has to divide p^{ℓ} − 1. But \frac{k}{m} is the greatest common divisor of μ_{0} − 1 and k. Since 4 divides μ_{0} − 1 and k it divides \frac{k}{m} and 2[G : G_{0}] divides p^{ℓ} − 1. Thus a′′ is even.

If

μ_{0} ≡ 3 (mod 4)

then ν_{ϕ}(−1) = −1. Moreover k > 2 so that p^{ℓ} ≡ 1 (mod 4) and

\frac{q − 1}{2k} ≡ \frac{p^{ℓ} + 1}{2} ≡ 1 (mod 2).

We have to show that a′′ is odd if

m = \frac{1}{2} [G : G_{0}]

and even otherwise. But μ_{0} ≡ 3 (mod 4) so that \frac{k}{m} = 2 and m = \frac{k}{2} . Thus [G : G_{0}] = 2m if and only if [G : G_{0}] = k. If [G : G_{0}] = k then

a′′ ≡ \frac{p^{ℓ} − 1}{k} (mod 2)

is odd. Otherwise 2[G : G_{0}] divides k and a′′ is even.

Lemma 14.3 is now completely proved, so we turn to Lemma 14.4. In the proof of both Lemma 14.4 and 14.5, we will combine the induction assumption with Lemma 15.1 which is stated and proved in paragraph 15, the following paragraph. Suppose F ⊆ F′ ⊆ L and F′/F is cyclic of prime degree ℓ. Let 𝔊(K/F′) be H′C where H′ ⊆ H and let E′ be the fixed field of H′. Then E′/E is cyclic of prime order ℓ. If S(F′/F) is the set of characters of C_{F} /N_{F′/F} C_{F′} then

S(E′/E) = {ν_{E/F} | ν_{F} ∈ S(F′/F)}.

From Lemma 15.1 we see that for any quasi-character χ_{F},

Ind(W_{K/E}, W_{K/E′}, χ_{E′/E}) ≃ ⊕_{ν_{F} ∈S(F′/F)} ν_{E/F} χ_{E/F} .

Therefore

Ind(W_{K/F} , W_{K/E′}, χ_{E′/E}) ≃ ⊕_{ν_{F}} Ind(W_{K/F} , W_{K/E}, ν_{E/F} χ_{E/F})

which is equivalent to

⊕_{ν_{F}} {(⊕_{μ∈T} Ind(W_{K/F} , W_{K/F_{μ}}, μ′ ν_{F_{μ}/F} χ_{F_{μ}/F})) ⊕ ν_{F} χ_{F} }.

If T′ is a set of representatives for the non-trivial orbits of H′ in S(K/L) then

Ind(W_{K/F′}, W_{K/E′}, χ_{E′/F}) = σ

is equivalent to

(⊕_{μ∈T′}Ind(W_{K/F′}, W_{K/F′_{μ}}, μ′ χ_{F′_{μ}/F})) ⊕ χ_{F′/F} .

Moreover

Ind(W_{K/F} , W_{K/F′}, σ) ≃ Ind(W_{K/F} , W_{K/E′}, χ_{E′/E}).

Applying the induction assumption to L/F we see that

{∏_{ν_{F}} Δ(ν_{F} , χ_{F} , ψ_{F})} {∏_{ν_{F}} ∏_{μ∈T} Δ(μ′ν_{F_{μ}/F} χ_{F_{μ}/F} , ψ_{F_{μ}/F})λ(F/F, ψ_{F})}

is equal to

{Δ(χ_{F′/F} , ψ_{F′/F})λ(F′/F, ψ_{F})} {∏_{μ∈T′} Δ(μ′χ_{F′_{μ}/F} , ψ_{F′_{μ}/F})λ(F′_{μ}/F, ψ_{F}).  (14.14)

The application is legitimate because the fields F′, F_{μ}, and F′_{μ} all lie between F and L. By Lemma 4.5

λ(F′_{μ}/F, ψ_{F}) = λ(F′_{μ}/F′, ψ_{F′/F}) λ(F′/F, ψ_{F})^{[F′_{μ}:F′]}.

Chapter 14 187 Also

λ(F′/F, ψ_{F}) {∏_{μ∈T′} λ(F′/F, ψ_{F})^{[F′_{μ}:F′]}} = λ(F′/F, ψ_{F})^{[E′:F′]}.

Since the fields F′ and F′_{μ} lie between F′ and K we can apply the induction assumption to K/F′ to see that (14.14) is equal to the product of

λ(F′/F, ψ_{F})^{[E′:F′]}

and

Δ(χ_{E′/F} , ψ_{E′/F}) λ(E′/F′, ψ_{F′/F}).

Applying the induction assumption to K/E we see that

Δ(χ_{E′/F} , ψ_{E′/F})

is equal to

{∏_{ν_{F} ∈S(F′/F)} Δ(ν_{E/F} χ_{E/F} , ψ_{E/F})} λ(E′/E, ψ_{E/F})^{−1}.

We conclude that the quotient

∏_{ν_{F}} \{\frac{Δ(ν_{F} χ_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′ν_{F_{μ}/F} χ_{F_{μ}/F} , ψ_{F_{μ}/F}}{Δ(ν_{E/F} χ_{E/F} , ψ_{E/F})} \}  (14.15)

is independent of χ_{F}. Taking χ_{F} to be trivial we see that it equals

∏_{ν_{F}} \{\frac{Δ(ν_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′ ν_{F_{μ}/F} , ψ_{F_{μ}/F})}{Δ(ν_{E/F} , ψ_{E/F})}\}.  (14.16)

It is easily seen that the complex conjugate of Δ(ν_{F} , ψ_{F}) is

ν_{F} (−1) Δ(ν_{F}^{−1}, ψ_{F}).

Thus

Δ(ν_{F} , ψ_{F}) Δ(ν_{F}^{−1}, ψ_{F}) = ν_{F} (−1).

If ℓ is odd the right side is 1. Since

Δ(1, ψ_{F}) = 1

and ν_{F} ≠ ν_{F}^{−1} if ℓ is odd, the product

∏_{ν_{F} ∈S(L/F)} Δ(ν_{F} , ψ_{F}) = 1.

For the same reasons

∏_{ν_{F} ∈S(L/F)} Δ(ν_{E/F} , ψ_{E/F}) = 1.

However, if ℓ is 2

Δ(ν_{F} , ψ_{F}) = Δ(ν_{F}^{−1}, ψ_{F})

has square ±1 and is therefore a fourth root of unity. Thus

∏_{ν∈S(L/F)} Δ(ν_{F} , ψ_{F}) ∼ ∏_{ν∈S(L/F)} Δ(ν_{E/F} , ψ_{E/F}) ∼_{2} 1.

On the other hand, m(μ′) = t + 1 ≥ 2 while m(ν_{F_{μ}/F}) ≤ 1. Thus Lemma 9.5 shows that

Δ(μ′ν_{F_{μ}/F} , ψ_{F_{μ}/F}) ∼_{ℓ} Δ(μ′, ψ_{F_{μ}/F}).

Thus the expression (14.16) and therefore the expression (14.15) is equal to

η {∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})}^{ℓ}

where η ∼_{ℓ} 1.

If m(χ_{F}) is 0 or 1, Lemma 14.4 is a consequence of Lemma 14.2. We suppose therefore that m(χ_{F}) ≥ 2. In this case Lemma 9.5 implies that

∏_{ν_{F}} Δ(ν_{F} χ_{F} , ψ_{F}) ∼_{ℓ} Δ(χ_{F} , ψ_{F})^{ℓ}

and that

∏_{ν_{F}} Δ(ν_{E/F} χ_{E/F} , ψ_{E/F}) ∼_{ℓ} Δ(χ_{E/F} , ψ_{E/F})^{ℓ}.

We also saw in the beginning of the paragraph that, in all cases, m(μ′χ_{F_{μ}/F}) ≥ 2. Thus

Δ(μ′ν_{F_{μ}/F} χ_{F_{μ}/F} , ψ_{F_{μ}/F}) ∼_{ℓ} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}).

Putting these facts together we see that if

σ{Δ (χ_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F})}^{ℓ}

is equal to

{Δ(χ_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})}^{ℓ}

then σ ∼_{ℓ} 1. Since σ = ρ^{ℓ} we conclude that

ρ ∼_{ℓ} 1.

Finally we have to prove Lemma 14.5. Let F′ be the fixed field of H_{1}C and let L′ be the fixed field of H_{2}C. Let E′ be the fixed field of H_{1} and let K′ be the fixed field of H_{2}. Let P be a set of representatives for the orbits under 𝔊(L/F) of the characters in S(L/L′). If ν is one of these representatives, let H_{ν}H_{2}C with H_{ν} and H_{1} be its isotropy group and let F_{ν} be the fixed field of H_{ν}H_{2}C. Applying the induction assumption and Lemma 15.1 to the extension L/F we see that

Δ(χ_{F′/F} , ψ_{F′/F}) ρ (F′/F, ψ_{F})

is equal to

∏_{ν∈P} Δ(ν′χ_{F_{ν}/F} , ψ_{F_{ν}/F}) λ(F_{ν}/F, ψ_{F}).  (14.17)

Let

R = {ν ∈ P | F_{ν} = F }

and let S be the complement of R in P. R consists of the elements of S(L/L′) fixed by each element of 𝔊(L/F). It is a subgroup of S(L/L′) and its order r must therefore be a power of ℓ. The expression (14.17) may be written as

{∏_{ν∈R} Δ(ν′χ_{F} , ψ_{F})} {∏_{ν∈S} Δ(ν′χ_{F_{ν}/F} , ψ_{F_{ν}/F}) λ(F_{ν}/F, ψ_{F})}.

If F is replaced by E and F′ by E′ then P is replaced by

{ν_{K′/L′} | ν = ν_{L′} ∈ P }.

Also F_{ν} is replaced by E_{ν}, the fixed field of H_{ν}H_{2}, and ν′ is replaced by ν′_{E_{ν}/F_{ν}}. Applying the induction assumption to K/E we see that

Δ(χ_{E′/F} , ψ_{E′/F}) λ(E′/E, ψ_{E/F})

is equal to the product of

{∏_{ν∈R} Δ(ν′_{E/F} χ_{E/F} , ψ_{E/F} }

and

{∏_{ν∈S} Δ(ν′_{E_{ν}/F_{ν}} χ_{E_{ν}/F_{ν}} , ψ_{E_{ν}/F}) λ(E_{ν}/E, ψ_{E/F} }.

This equality will be referred to as relation (14.18).

To derive this equality we have used not only the induction assumption but also Lemma 15.1 which implies that

Ind(W_{K/E}, W_{K/E′}, χ_{E′/F})

is equivalent to

{⊕_{R} Ind(W_{K/E}, W_{K/E}, ν′_{E/F} χ_{E/F})} ⊕ {⊕_{S} Ind(W_{K/E}, W_{K/E_{ν}} , ν′_{E_{ν}/F_{ν}} χ_{E_{ν}/F})}.

Thus

Ind(W_{K/F} , W_{K/E′}, χ_{E′/F})

will be equivalent to the direct sum of

⊕_{R} Ind(W_{K/F} , W_{K/E}, ν′_{E/F} χ_{E/F})

and

⊕_{S} Ind(W_{K/F} , W_{K/E_{ν}} , ν′_{E_{ν}/F_{ν}} χ_{E_{ν}/F}).

If ν is in R we can apply Lemma 15.1 to see that

Ind (W_{K/F} , W_{K/E}, ν′_{E/F} χ_{E/F})

is equivalent to

{⊕_{μ∈T} Ind(W_{K/F} , W_{K/F_{μ}}, μ′ν′_{F_{μ}/F} χ_{F_{μ}/F})} ⊕ ν′χ_{F} .

We can obtain

Ind (W_{K/F} , W_{K/F_{ν}} , ν′_{E_{ν}/F_{ν}} χ_{E_{ν}/F})

by first inducing from W_{K/E_{ν}} to W_{K/F_{ν}} and then from W_{K/F_{ν}} to W_{K/F}.

If T_{ν} is a set of representatives for the orbits of S(K/L) under the action of 𝔊(K/F_{ν}) and F_{ν,μ} is the fixed field of the isotropy group of μ in T_{ν} then, by Lemma 15.1 again,

Ind (W_{K/F_{ν}} , W_{K/E_{ν}} , ν′_{E_{ν}/F_{ν}} χ_{E_{ν}/F})

is equivalent to

⊕_{T_{ν}} Ind (W_{K/F_{ν}} , W_{K/F_{ν,μ}}, μ′ν′_{F_{ν,μ/Fν}} χ_{F_{ν},μ/F}).

Since [K : F_{ν}] < [K : F] if ν belongs to S, we can apply the induction assumption to see that

Δ(ν′_{E_{ν}/F_{ν}} χ_{E_{ν}/F} , ψ_{E_{ν}/F}) λ(E_{ν}/F_{ν}, ψ_{F_{ν}/F})

is equal to

∏_{μ∈T_{ν}} Δ(μ′ν′_{F_{ν,μ}/F_{ν}} χ_{F_{ν,μ}/F} , ψ_{F_{ν}/F}) λ(F_{ν,μ}/F_{ν}, ψ_{F_{ν}/F}).

This equality will be referred to as relation (14.19).

It also follows that

Ind(W_{K/F} , W_{K/E_{ν}} , ν′_{E_{ν}/F_{ν}} χ_{E_{ν}/F})

is equivalent to

⊕_{μ∈T_{ν}} Ind(W_{K/F} , W_{K/F_{ν,μ}}, μ′ν′_{F_{ν,μ}/F_{ν}} χ_{F_{ν,μ}/F}).

The fields F_{ν} and F_{ν,μ} all lie between F and L. Thus we have expressed

Ind(W_{K/F} , W_{K/E′}, χ_{E′/F})  (14.20)

as a direct sum of terms of the form

Ind(W_{K/F} , W_{K/M}, χ_{M})  (14.21)

where M lies between F and L. Moreover such a representation is in fact a representation of W_{K/F} obtained by inflating a representation of W_{L/F}, namely, by inflating

Ind(W_{L/F} , W_{L/M}, χ_{M}).

Thus any other expression of (14.20) as a sum of representations of the form (14.21) will lead, by an application of the induction assumption to L/F, to an identity between the numbers Δ(χ_{M}, ψ_{M/F}).

To obtain another such expression, we observe that the representation (14.20) can be obtained by first inducing from W_{K/E′} to W_{K/F′} and then from W_{K/F′} to W_{K/F}. If T′ is a set of representatives for the orbits of non-trivial characters in S(K/L) under the action of 𝔊(K/F′) and F′_{μ} is the fixed field of the isotropy group in 𝔊(K/F′) of μ in T′ then

Ind(W_{K/F′}, W_{K/E′}, χ_{E′/F})

is equivalent to

{⊕_{μ∈T′} Ind(W_{K/F′}, W_{K/F′_{μ}}, μ′χ_{F′_{μ}/F} } ⊕ χ_{F′/F} .

Thus (14.20) is equivalent to the direct sum of

Ind(W_{K/F} , W_{K/F′}, χ_{F′/F})

and

⊕_{μ∈T′} Ind(W_{K/F} , W_{K/F′_{μ}}, μ′ χ_{F′_{μ}/F}).

We shall describe the resultant identity in a moment. We first apply the induction assumption to the extension K/F′ to see that

Δ(χ_{E′/F} , ψ_{E′/F}) λ(E′/F′, ψ_{F′/F})

is equal to

Δ(χ_{F′/F} , ψ_{F′/F}) ∏_{μ∈T′} Δ(μ′χ_{F′_{μ}/F} , ψ_{F′_{μ}/F}) λ(F′_{μ}/F′, ψ_{F′/F}).

This equality will be relation (14.22).

The two expressions for the representation (14.20) lead to the conclusion that the product of

∏_{ν∈R} Δ(ν′χ_{F} , ψ_{F})  (14.23)

and

∏_{ν∈R} ∏_{μ∈T} Δ(μ′ν′_{F_{μ}/F} χ_{F_{μ}/F} , ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F})  (14.24)

and

∏_{ν∈S} ∏_{μ∈T_{ν}} Δ(μ′ν′_{F_{ν,μ}/F_{ν}} χ_{F_{ν,μ}/F} , ψ_{F_{ν,μ}/F}) λ(F_{ν,μ}/F, ψ_{F})  (14.25)

is equal to the product of

Δ(χ_{F′/F} , ψ_{F′/F}) λ(F′/F, ψ_{F})

and

∏_{μ∈T′} Δ(μ′χ_{F′_{μ}/F} , ψ_{F′_{μ}/F}) λ(F′_{μ}/F, ψ_{F}).

Applying relation (14.22) and Lemma 4.5 we see that the second of these two products is equal to

Δ(χ_{E′/F} , ψ_{E′/F}) λ(E′/F′, ψ_{F′/F}) λ(F′/F, ψ_{F})^{[E′:F′]}.

According to the relation (14.18) this expression is the product of

{∏_{ν∈R} Δ(ν′_{E/F} χ_{E/F} , ψ_{E/F})} {∏_{ν∈S} Δ(ν′_{E/F} χ_{E/F} , ψ_{E/F})}

and

∏_{ν∈S} λ(E_{ν}/E, ψ_{E/F})

and

λ(E′/E, ψ_{E/F})^{−1} λ(E′/F′, ψ_{F′/F}) λ(F′/F, ψ_{F})^{[E′:F′]}.  (14.26)

Equating this final product to the product of (14.23), (14.24), and (14.25) and then making certain cancellations by means of (14.19), we see that the product of (14.23) and (14.24) and

∏_{ν∈S} ∏_{μ∈T_{ν}} λ^{−1}(F_{ν,μ}/F_{ν}, ψ_{F_{ν}/F}) λ(F_{ν,μ}/F, ψ_{F})

is equal to the product of

∏_{ν∈R} Δ(ν′_{E/F} χ_{E/F} , ψ_{E/F})

and

∏_{ν∈S} λ^{−1}(E_{ν}/F_{ν}, ψ_{F_{ν}/F}) λ(E_{ν}/E, ψ_{E/F})

and the expression (14.26).

In particular, the expression

∏_{ν∈R} \{\frac{Δ(ν′χ_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′ν′_{F_{μ}/F} χ_{F_{μ}/F} , ψ_{F_{μ}/F})}{Δ(ν′_{E/F} χ_{E/F} , ψ_{E/F})}\}

is independent of χ_{F}. Taking χ_{F} to be trivial we see that

∏_{ν∈R} \{\frac{Δ(ν′χ_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′ν′_{F_{μ}/F} χ_{F_{μ}/F} , ψ_{F_{μ}/F})}{Δ(ν′_{E/F} χ_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′ν′_{F_{μ}/F} , ψ_{F_{μ}/F})}\}

is equal to

∏_{ν∈R} \frac{Δ(ν′, ψ_{F})}{Δ(ν′_{E/F} , ψ_{E/F})}.

The set

R′ = {ν′ | ν ∈ R}

is a group of characters of C_{F} or of H. Regarded as characters of H the elements of R′ are just those characters which are trivial on H_{1}. As a group R′ is cyclic and its order is a power of ℓ. The argument used in the proof of Lemma 14.4 shows that

∏_{ν∈R} Δ(ν′, ψ_{F}) ∼_{ℓ} 1

and

∏_{ν∈R} Δ(ν′_{E/F} , ψ_{E/F}) ∼_{ℓ} 1.

If m(χ_{F}) is 0 or 1, Lemma 14.5 is a consequence of Lemma 14.2. We may as well suppose therefore that m(χ_{F}) > 1. If ν belongs to R then ν′ is 1 on N_{L/F} C_{L}. Therefore m(ν′), as well as m(ν′_{F_{μ}/F}) and m(ν′_{E/F}) is at most 1. We saw in the beginning of this paragraph that m(χ_{E/F}) would also be at least 2. We also saw that m(μ′χ_{F_{μ}/F}) would be either t + 1 or ψ_{F_{μ}/F} (m − 1) + 1. In any case it is at least 2. Also m(μ′) = t + 1 is at least 2. Lemma 9.5 therefore implies the following relations:

Δ(ν′χ_{F} , ψ_{F}) ∼_{ℓ} Δ(χ_{F} , ψ_{F})
Δ(ν′_{E/F} χ_{E/F} , ψ_{E/F} ∼_{ℓ} Δ(χ_{E/F} , ψ_{E/F})
Δ(μ′ν′_{F_{μ}/F} χ_{F_{μ}/F} , ψ_{F_{μ}/F} ∼_{ℓ} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F})
Δ(μ′ν′_{F_{μ}/F} , ψ_{F_{μ}/F}) ∼_{ℓ} Δ(μ′, ψ_{F_{μ}/F}).

We conclude finally that

\{\frac{Δ(χ_{F} , ψ_{F}) ∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F})}{Δ(χ_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})}\}^{r} ∼_{ℓ} 1

if r is the number of elements in R. The lemma follows.

Chapter Fifteen.

Another Lemma

Suppose K/F is normal and G = 𝔊(K/F). Suppose H is a subgroup of G and C is an abelian normal subgroup of G. Let E be the fixed field of H and L that of C. If μ is a character of C and h belongs to H, define μ^{h} by

μ^{h}(c) = μ(hch^{−1}).

The set of characters of C may be identified with S(K/L). If α belongs to C_{L}

μ^{h}(α) = μ(h(α)).

The set of elements in S(K/L) which are trivial on H ∩ C is invariant under H. Let T be a set of representatives for the orbits of H in this set. If μ ∈ T let H_{μ} be the isotropy group of μ, let G_{μ} = H_{μ}C and let F_{μ} be the fixed field of G_{μ}. Define a character μ′ of G_{μ} by

μ′(hc) = μ(c)

if h ∈ H_{μ} and c ∈ C. μ′ may be regarded as a character of C_{F_{μ}}.

Lemma 15.1

If χ_{F} is a quasi-character of C_{F} then

ρ = Ind(W_{K/F} , W_{K/E}, χ_{E/F})

is equivalent to

⊕_{μ∈T} Ind(W_{K/F} , W_{K/F_{μ}}, μ′χ_{F_{μ}/F}).

Let G′ = HC and let F′ be the fixed field of G′. F′ is contained in E and in the fields F_{μ}. Because of the transitivity of the induction process it is enough to show that

Ind(W_{K/F′}, W_{K/E}, χ_{E/F})

is equivalent to

⊕_{μ∈T} Ind(W_{K/F′}, W_{K/F_{μ}}, μ′χ_{F_{μ}/F}).

If

χ′_{F′} = χ_{F′/F}

then

χ_{E/F} = χ′_{E/F′}

and

χ_{F_{μ}/F} = χ′_{F_{μ}/F′}.

Consequently we may suppose, with no loss of generality, that F′ is F.

If K′ is the fixed field of H ∩ C and ν ∈ S(K′/L) let φ_{ν} be the function on W_{K/F} defined by

φ_{ν}(hc) = χ_{F} (τ_{K/F} (hc)) ν(τ_{K/L}(c))

for h in W_{K/E}, c in W_{K/L}. ρ acts on the space of all functions φ on W_{K/F} satisfying

φ(hg) = χ_{F} (τ_{K/F} (h)) φ(g)

for all h in W_{K/E} and all g in W_{K/F}. The set

{φ_{ν} | ν ∈ S(K′/L)}

is a basis for this space. Clearly

ρ(c)φ_{ν} = χ_{F} (τ_{K/F} (c)) ν(τ_{K/L}(c))φ_{ν}

if c belongs to W_{K/L} and

ρ(h)φ_{ν} = χ_{F} (τ_{K/F} (h))φ_{ν′},

with ν′ = ν^{h^{−1}}, if h belongs to W_{K/E}. Thus if R is an orbit of H in S(K′/L)

⊕_{ν∈R}ℂ_{φ_{ν}} = V

is an invariant subspace.

Let μ be the element common to T and R and consider

σ = Ind(W_{K/F} , W_{K/F_{μ}}, μ′χ_{F_{μ}/F}).

If W_{K/F} is the disjoint union

⋃_{i=1}^{r} W_{K/F_{μ}}h_{i}

and if φ_{i}(w) = 0 unless

w ∈ W_{K/F_{μ}}h_{i}

while

φ_{i}(wh_{i}) = μ′χ_{F_{μ}/F} (τ_{K/F_{μ}}(w))

for w in W_{K/F_{μ}} then

{φ_{i} | 1 ≤ i ≤ r}

is a basis for the space U on which σ acts. If ν_{i} = μ^{h_{i}} and if λ is the map from U to V which sends φ_{i} to χ_{F}^{−1}(τ_{K/F} (h_{i}))φ_{ν_{i}} then, as one verifies easily,

λσ(w) = ρ(w)λ

for all w in W_{K/F}. The lemma follows.

The lemma has a corollary. Lemma 15.2

If Theorem 2.1 is valid for K/F then

Δ(χ_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})

is equal to

∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}).

If Theorem 2.1 is valid

Δ(χ_{E/F} , ψ_{E/F}) λ(E/F, ψ_{F})

is equal to

∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}).

Taking χ_{F} = 1, we see that

λ(E/F, ψ_{F}) = ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}).

Substituting this into the first equality and cancelling the non-zero factor

∏_{μ∈T} λ(F_{μ}/F, ψ_{F})

we obtain the lemma.

To define the λ-function we shall need the following lemma.

Lemma 15.3

Suppose Theorem 2.1 is valid for all Galois extensions K_{1}/F_{1} with F ⊆ F_{1} ⊆ K_{1} ⊆ K and [K_{1} : F_{1}] < [K : F]. Then

Δ(χ_{E/F} , ψ_{E/F}) ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F})

is equal to

∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}).

The conclusion of this lemma is the same as that of the previous one. There is however a critical difference in the assumptions.

Let F′ be the fixed field of HC. If

ψ′_{F′} = ψ_{F′/F}

then for all separable extensions E′ of F′

ψ′_{E′/F′} = ψ_{E′/F} .

If [K : F′] < [K : F] the relation of the lemma is a consequence of the induction assumption and the previous lemma. We thus suppose that F = F′ and G = HC.

Suppose in addition that there is a subgroup C_{1} of C, which is neither C nor {1}, whose normalizer contains H. C_{1} is then a normal subgroup of G. Let F_{1} be the fixed field of HC_{1} and L_{1} the fixed field of C_{1}. Lemma 15.1 applies to the extension K/F_{1}. Thus there are fields A_{1}, …, A_{r} lying between F_{1} and L_{1} and quasi-characters χ_{A_{1}}, …, χ_{A_{r}} such that

Ind(W_{K/F_{1}}, W_{K/E}, χ_{E/F})

Chapter 15 197 is equivalent to

⊕_{i=1}^{r} Ind(W_{K/F_{1}}, W_{K/A_{i}}, χ_{A_{i}}).

The induction assumption then implies that

Δ(χ_{E/F} , ψ_{E/F}) λ(E/F_{1}, ψ_{F_{1}/F})  (15.1)

is equal to

∏_{i=1}^{r} Δ(χ_{A_{i}}, ψ_{A_{i}/F}) λ(A_{i}/F_{1}, ψ_{F_{1}/F}).  (15.2)

Inducing the first of these two representations from W_{K/F_{1}} to W_{K/F}, we obtain

Ind(W_{K/F} , W_{K/E}, ψ_{E/F}).

Thus

⊕_{μ∈T} Ind(W_{K/F} , W_{K/F_{μ}}, μ′χ_{F_{μ}/F})  (15.3)

is equivalent to

⊕_{i=1}^{r} Ind(W_{K/F} , W_{K/A_{i}}, χ_{A_{i}}).  (15.4)

We recall that there exist surjective homomorphisms

τ_{K/F,L_{1}/F} : W_{K/F} ⟶ W_{L_{1}/F}
τ_{K/A_{i},L_{1}/A_{i}} : W_{K/A_{i}} ⟶ W_{L_{1}/A_{i}}
τ_{K/F_{μ},L_{1}/F_{μ}} : W_{K/F_{μ}} ⟶ W_{L_{1}/F_{μ}}

whose kernels are all equal to the commutator subgroup W_{K/L_{1}}^{c} of W_{K/L_{1}}. Moreover the diagrams W_{K/A_{i}} ⟶ W_{L_{1}/A_{i}}

↓ ↓
W_{K/F} ⟶ W_{L_{1}/F}

and

W_{K/F_{μ}} ⟶ W_{L_{1}/F_{μ}}
↓ ↓
W_{K/F} ⟶ W_{L_{1}/F}

may be supposed commutative. Since W_{K/L_{1}}^{c} lies in the kernel of χ_{A_{i}} and μ′χ_{F_{μ}/F} the equivalence of (15.3) and (15.4) amounts to the equivalence of

⊕_{μ∈T} Ind(W_{L_{1}/F} , W_{L_{1}/F_{μ}}, μ′χ_{F_{μ}/F})

and

⊕_{i=1}^{r}Ind(W_{L_{1}/F} , W_{L_{1}/A_{i}}, χ_{A_{i}}).

The induction assumption applied to the extension L_{1}/F implies that

∏_{i=1}^{r} Δ(χ_{A_{i}}, ψ_{A_{i}/F}) λ(A_{i}/F, ψ_{F})

is equal to

∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}).

It also implies that

λ(A_{i}/F, ψ_{F}) = λ(A_{i}/F_{1}, ψ_{F_{1}/F}) λ(F_{1}/F, ψ_{F})^{[A_{i}:F_{1}]}.

Since

∑_{i}[A_{i} : F_{1}] = [E : F_{1}]

we infer from the equality of (15.1) and (15.2) that

Δ(χ_{E/F} , ψ_{E/F}) λ(E/F_{1}, ψ_{F_{1}/F}) λ(F_{1}/F, ψ_{F})^{[E:F_{1}]}

is equal to

∏_{μ∈T} Δ(μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}).

Taking χ_{F} = 1 to find the value of

λ(E/F_{1}, ψ_{F_{1}/F}) λ(F_{1}/F, ψ_{F})^{[E:F_{1}]}

and then substituting the result into the equation and cancelling the common factors we obtain the assertion of the lemma.

Now suppose that H contains a normal subgroup H_{1} ≠ {1} which lies in the centralizer of C. H_{1} is a normal subgroup of G if, as we are assuming, G = HC. K_{1}, the fixed field of H_{1}, contains E and all the fields F_{μ}. Lemma 15.1 together with the argument just applied to L_{1} shows that

Ind(W_{K_{1}/F} , W_{K_{1}/E}, χ_{E})

is equivalent to

⊕_{μ∈T} Ind(W_{K_{1}/F} , W_{K_{1}/F_{μ}}, μ′χ_{F_{μ}/F}).

In this case the assertion of the lemma follows from the induction assumption applied to K_{1}/F.

We have finally to suppose that G = HC, C contains no proper subgroup invariant under H, and H contains no normal subgroup lying in the centralizer of C. In particular H ∩ C = {1}. If Z is the centralizer of C then Z = (Z ∩ H)C and Z ∩ H is a normal subgroup of H. Consequently Z = C. If D is a normal subgroup of G and D does not contain C then

D ∩ C = {1}.

This implies that D is contained in Z. Thus D is contained in C and D = {1}. If H ≠ {1} the assertion of the lemma is that of the third and fourth main lemmas. If H = {1} then G = C and C is cyclic of prime order so that the assertion is that of the first main lemma.

Chapter Sixteen.

Definition of the λ Functions

In this and the next three paragraphs, we take a fixed Galois extension K/F, assume that Theorem 2.1 is valid for all Galois extensions K′/F′ with F ⊆ F′ ⊆ K′ ⊆ K and [K′ : F′] < [K : F], and prove that it is valid for K/F itself. The first step is to define and establish some simple properties of the function which will serve as the λ-function. Lemma 16.1

Suppose

E/F′ ⟶ λ(E/F′, ψ_{F′})

is a weak λ-function on P_{0}(K′/F′). If σ ∈ 𝔊(K′/F′) let

E^{σ} = {σ^{−1}(α) | α ∈ E}.

Then

λ(E^{σ}/F′, ψ_{F′}) = λ(E/F′, ψ_{F′}).

If μ is a character of 𝔊(K/E) let μ^{σ} be the character of 𝔊(K/E^{σ}) defined by

μ^{σ}(ρ) = μ(σρσ^{−1}).

According to Lemma 13.2

Δ(μ^{σ}, ψ_{E^{σ}/F′}) = Δ(μ, ψ_{E/F′}).

The representation

Ind(𝔊(K′/F′), 𝔊(K′/E), μ)

acts on the space U of functions φ on 𝔊(K′/F′) satisfying

φ(ρτ) = μ(ρ) φ(τ)

for all τ in 𝔊(K′/F′) and all ρ in 𝔊(K′/E). The map φ ⟶ ψ with

ψ(τ) = φ(στ)

is a 𝔊(K′/F′) isomorphism of U with the space on which

Ind(𝔊(K′/F′), 𝔊(K′/E^{σ}), μ^{σ})

acts. Thus the two representations are equivalent.

If

⊕_{i=1}^{r}Ind(𝔊(K′/F′), 𝔊(K′/E_{i}), μ_{i})

is equivalent to

⊕_{j=1}^{s}Ind(𝔊(K′/F′), 𝔊(K′/F_{j}), ν_{j})

then

⊕_{i=1}^{r}Ind(𝔊(K′/F′), 𝔊(K′/E_{i}^{σ}), μ_{i}^{σ})

is equivalent to

⊕_{j=1}^{s}Ind(𝔊(K′/F′), 𝔊(K′/F_{j}^{σ}), ν_{j}^{σ})

and, with the conventions of the fourth paragraph,

∏_{i=1}^{r}(χ_{E_{i}^{σ}}, ψ_{E_{i}^{σ}/F′}) λ(E_{i}^{σ}/F′, ψ_{F′})

is equal to

∏_{j=1}^{s} Δ(χ_{F_{j}^{σ}}, ψ_{F_{j}^{σ}/F′}) λ(F_{j}^{σ}/F′, ψ_{F′}).

Since

Δ(χ_{F_{j}^{σ}}, ψ_{F_{j}^{σ}/F′}) = Δ(χ_{F_{j}}, ψ_{F_{j}/F′})

and

Δ(χ_{E_{i}^{σ}}, ψ_{E_{i}^{σ}/F′}) = Δ(χ_{E_{i}}, ψ_{E_{i}/F′}).

We conclude that

∏_{i=1}^{r} Δ(χ_{E_{i}}, ψ_{E_{i}/F′}) λ(E_{i}^{σ}/F′, ψ_{F′})

is equal to

∏_{j=1}^{s} Δ(χ_{F_{j}}, ψ_{F_{j}/F′}) λ(F_{j}^{σ}/F′, ψ_{F′}).

In other words

E/F′ ⟶ λ(E^{σ}/F′, ψ_{F′})

is a weak λ-function on P_{0}(K′/F′). Lemma 16.1 follows from the uniqueness of such functions.

We return to the problem of defining a λ-function on P_{0}(K/F). Choose a non-trivial abelian normal subgroup C of G = 𝔊(K/F) and let L be the fixed field of C. If E is any field lying between F and K let H be the corresponding subgroup of G. Choose the set T of characters and the fields F_{μ} as in the previous paragraph. Since F_{μ} ⊆ L the numbers λ(F_{μ}/F, ψ_{F}) are defined.

Lemma 16.2

Suppose F ⊆ E ⊆ K_{1} ⊂_{≠} K with K_{1}/F normal so that λ(E/F, ψ_{F}) is defined. Then

λ(E/F, ψ_{F}) = ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}).

Let K_{1} be the fixed field of H_{1}. If H_{1} ∩C ≠ {1} we may enlarge K_{1} and replace H_{1} by H_{1} ∩C. Thus we may suppose that either H_{1} is contained in C or H_{1} ∩ C = {1}. In either case H_{1} is contained in the centralizer of C. We saw in the previous paragraph that under these circumstance

Ind(W_{K_{1}/F} , W_{K_{1}/E}, 1) ≃ ⊕_{μ∈T} Ind(W_{K_{1}/F} , W_{K_{1}/F_{μ}}, μ′).

Consequently

λ(E/F, ψ_{F}) = ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}).

In general, we define

λ(E/F, ψ_{F}) = ∏_{μ∈T} Δ(μ′, ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F})

if E/F is in P_{0}(K/F) T is, of course, not always uniquely determined. We may replace any μ in T by μ^{σ} with σ in H. Then H_{μ} and G_{μ} are replaced by σ^{−1}H_{μ}σ and σ^{−1}G_{μ}σ while F_{μ} is replaced by F_{μ}^{σ} and μ′ is replaced by (μ′)^{σ}. Since

Δ(μ′, ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}) = Δ((μ′)^{σ}, ψ_{F_{μ}^{σ}/F}) λ(F_{μ}^{σ}/F, ψ_{F})

the number λ(E/F, ψ_{F}) does not depend on T. A priori, it may depend on C but that is unimportant since C is fixed and, the uniqueness having been proved, we are interested only in the existence of a λ-function.

We shall need only one property of the function just defined.

Lemma 16.3

If F ⊆ E ⊆ E′ ⊆ K then
λ(E′/F, ψ_{F}) = λ(E′/E, ψ_{E/F}) λ(E/F, ψ_{F})^{[E′:E]}.

If E = F then

λ(E′/E, ψ_{E/F}) = λ(E′/F, ψ_{F})

and if E ≠ F

λ(E′/E, ψ_{E/F})

is the value of the λ-function of P(K/E), which is defined by assumption, at E′/E. Since

λ(F/F, ψ_{F}) = 1

the assertion is clear if E = F. It is also clear if E = E′.

Let E be the fixed field of H as before and let F′ be the fixed field of HC. We suppose that H ≠ G. Lemma 4.5 and the induction assumption imply that

λ(F_{μ}/F, ψ_{F}) = λ(F_{μ}/F′, ψ_{F′/F}) λ(F′/F, ψ_{F})^{[F_{μ}:F′]}.

The relation

[E : F′] = ∑ [F_{μ} : F′]

implies that

λ(E/F, ψ_{F}) = λ(E/F′, ψ_{F′/F}) λ(F′/F, ψ_{F})^{[E:F′]}.

There is a similar formula for λ(E′/F, ψ_{F}). If F′ ≠ F the induction assumption implies that

λ(E′/E, ψ_{E/F}) λ(E/F′, ψ_{F′/F})^{[E′:E]} = λ(E′/F′, ψ_{F′/F}).

Since

[E′ : F′] = [E′ : E] [E : F′]

the assertion of the lemma is proved simply by multiplying both sides of this equation by

λ(F′/F, ψ_{F})^{[E′:F′]}.

Now suppose that G = HC and H ∩C = {1}. Let E′ be the fixed field of H′ and let F′ be the fixed field of H′C = G′. Each character of H′ may be identified with a character of C_{E′}/N_{K/E′}C_{K} and each character of G′ may be identified with a character of C_{F′}/N_{K/F′}C_{K}. Any character χ_{E′} of H′ may be extended to a character χ_{F′} of G′ by setting

χ_{F′}(ρσ) = χ_{E′(ρ)}

if ρ ∈ H′ and σ ∈ C. Then

χ_{E′} = χ_{E′/F′}.

It follows from Lemma 15.1 that there are fields of F_{i}(E′), 1 ≤ i ≤ m(E′), lying between F′ and L and characters μ_{F_{i}(E′)} such that

Ind(W_{K/F′}, W_{K/E′}, χ_{E′})

is equivalent to

⊕_{i=1}^{m(E′)}Ind(W_{K/F′}, W_{K/F_{i}(E′)}, μ_{F_{i}(E′)}χ_{F_{i}(E′)/F′}).

If E ≠ E′ so that F ≠ F′ the induction assumption implies that

Δ(χ_{E′}, ψ_{E′/F}) λ(E′/F′, ψ_{F′/F})

is equal to

∏_{i=1}^{m(E′)} Δ(μ_{F_{i}(E′)}χ_{F_{i}(E′)/F′}, ψ_{F_{i}(E′)/F}) λ(F_{i}(E)/F′, ψ_{F′/F}).

We have seen that the lemma is valid for any pair E′, E for which HC ≠ G. In particular, it is valid for the pair E′, F′ and the pairs F_{i}(E′), F′. Multiplying the equality just obtained by

λ(F′/F, ψ_{F})^{[E′:F′]}

we see that

Δ(χ_{E′}, ψ_{E′/F}) λ(E′/F, ψ_{F})  (16.1)

is equal to

∏_{i=1}^{m(E′)} Δ(μ_{F_{i}(E′)}χ_{F_{i}(E′)/F′}, ψ_{F_{i}(E′)/F}) λ(F_{i}(E′)/F, ψ_{F}).  (16.2)

If F′ = F the equality of (16.1) and (16.2), for a suitable choice of the fields F_{i}(E′), results from Lemma 15.1, Lemma 15.3, and the definition of

λ(E′/F, ψ_{F}).

In any case the equality is valid for all fields lying between E and K.

Suppose E_{1}, …, E_{r}, E′_{1}, …, E′_{s} are such fields, χ_{E_{i}} is a character of C_{E_{i}}/N_{K/E_{i}}C_{K}, χ_{E′_{j}} is a character of C_{E′_{j}}/N_{K/E′_{j}}C_{K}, and

⊕_{i=1}^{r}Ind(W_{K/E}, W_{K/E_{i}}, χ_{E_{i}})

is equivalent to

⊕_{j=1}^{s}Ind(W_{K/E}, W_{K/E′_{j}}, χ_{E′_{j}}).

Then

∑_{i=1}^{r}[E_{i} : E] = ∑_{j=1}^{s}[E′_{j} : E]  (16.3)

and, by the transitivity of the induction process,

⊕_{i=1}^{r} ⊕_{k=1}^{m(E_{i})} Ind(W_{K/F} , W_{K/F_{k}(E_{i})}, μ_{F_{k}(E_{i})} χ_{F_{k}(E_{i})/F_{i}})

is equivalent to

⊕_{j=1}^{s} ⊕_{ℓ=1}^{m(E′_{j})} Ind(W_{K/F} , W_{K/F_{ℓ}(E′_{j})}, μ_{F_{ℓ}(E′_{j})} χ_{F_{ℓ}(E′_{j})/F′_{j}}).

If E_{i} is the fixed field of H_{i} and E′_{j} the fixed field of H′_{j} then F_{i} and F′_{j} are the fixed fields of H_{i}C and H′_{j}C. This equivalence and the induction assumption for L/F imply that

∏_{i=1}^{r} ∏_{k=1}^{m(E_{i})} Δ(μ_{F_{k}(E_{i})} χ_{F_{k}(E_{i})/F_{i}}, ψ_{F_{k}(E_{i})/F}) λ(F_{k}(E_{i})/F, ψ_{F})

is equal to

∏_{j=1}^{s} ∏_{ℓ=1}^{m(E′_{j})} Δ(μ_{F_{ℓ}(E′_{j})} χ_{F_{ℓ}(E′_{j})/F′_{j}}, ψ_{F_{ℓ}(E′_{j})/F}) λ(F_{ℓ}(E′_{j})/F, ψ_{F}).

This equality, the equality of (16.1) and (16.2), and the relation (16.3) imply that

∏_{i=1}^{r} Δ(χ_{E_{i}}, ψ_{E_{i}/F}) λ(E_{i}/F, ψ_{F}) λ(E/F, ψ_{F})^{−[E_{i}:E]}

is equal to

∏_{j=1}^{s} Δ(χ_{E′_{j}}, ψ_{E′_{j}/F}) λ(E′_{j}/F, ψ_{F}) λ(E/F, ψ_{F})^{−[E′_{j}:E]}.

Consequently

E′ ⟶ λ(E′/F, ψ_{F}) λ(E/F, ψ_{F})^{−[E′:E]}

is a weak λ-function on P_{0}(K/E). The lemma of uniqueness implies that

λ(E′/F, ψ_{F}) λ(E/F, ψ_{F})^{−[E′:E]} = λ(E′/E, ψ_{E/F}).

This is, of course, the assertion of the lemma.

At this point, we have proved the lemma when various supplementary conditions are satisfied. Before proving it, in general, we make an observation. Suppose

F ⊆ E ⊆ E′ ⊆ E′′ ⊆ K

and the assertion of the lemma is valid for E′′/E′ and E′/E. Then

λ(E′′/F, ψ_{F}) = λ(E′′/E′, ψ_{E′/F}) λ(E′/F, ψ_{F})^{[E′′:E′]}

and

λ(E′/F, ψ_{F}) = λ(E′/E, ψ_{E/F}) λ(E/F, ψ_{F})^{[E′:E]}.

Moreover, by induction,

λ(E′′/E, ψ_{E/F}) = λ(E′′/E′, ψ_{E′/F}) λ(E′/E, ψ_{E/F})^{[E′′:E′]}.

The assertion for E′′/E is obtained by substituting the second relation in the first and simplifying according to the third.

If the lemma is false in general, chose amongst all the extensions in P(K/F) for which it is false one E′/E for which [E′ : E] is a minimum. Let E be the fixed field of H and E′ that of H′. According to the previous discussion G = HC, H ∩ C ≠ {1}, and there are no fields lying between E and E′. If H′ ∩ C = H ∩ C, which is a normal subgroup of G, the fields F, E, and E′ are contained in the fixed field of H ∩ C and the assertion is a consequence of the induction assumption. Thus H′ is a proper subgroup of H′(H ∩ C). Because there are no intermediate fields H = H′(H ∩ C).

As we have seen there are fields E_{1}, …, E_{r} lying between E and the fixed field K_{1} of H ∩ C and characters μ_{E_{1}}, …, μ_{E_{r}} such that

Ind(W_{K/E}, W_{K/E′}, 1)

is equivalent to

⊕_{i=1}^{r}Ind(W_{K/E}, W_{K/E_{i}}, μ_{E_{i}}).

Then

λ(E′/E, ψ_{E/F}) = ∏_{i=1}^{r} Δ(μ_{E_{i}}, ψ_{E_{i}/E}) λ(E_{i}/E, ψ_{E/F}).

By the induction assumption, applied to K_{1}/F,

λ(E_{i}/E, ψ_{E/F}) λ(E/F, ψ_{F})^{[E_{i}:E]} = λ(E_{i}/F, ψ_{F}).

Thus

λ(E′/E, ψ_{E/F}) λ(E/F, ψ_{F})^{[E′:E]}

is equal to

∏_{i=1}^{r} Δ(μ_{E_{i}}, ψ_{E_{i}/E}) λ(E_{i}/F, ψ_{F}).  (16.4)

Moreover, by the transitivity of the induction process,

Ind(W_{K/F} , W_{K/E′}, 1)  (16.5)

is equivalent to

⊕_{i=1}^{r}Ind(W_{K/F} , W_{K/E_{i}}, μ_{E_{i}}).  (16.6)

On the other hand, there are fields F_{1}, …, F_{s} contained in L and characters ν_{F_{1}}, …, ν_{F_{s}} such that (16.5) is equivalent to

⊕_{j=1}^{s}Ind(W_{K/F} , W_{K/F_{j}}, ν_{F_{j}})  (16.7)

and such that, by definition,

λ(E′/F, ψ_{F}) = ∏_{j=1}^{s} Δ(ν_{F_{j}}, ψ_{F_{j}/F}) λ(F_{j}/F, ψ_{F}).  (16.8)

Since the representations (16.6) and (16.7) are equivalent the induction assumption, applied to K_{1}/F, shows that (16.4) is equal to the right side of (16.8). This is a contradiction.

Chapter Seventeen.

A Simplification.

We shall use the symbol Ω to denote an orbit in the set of quasi-characters of C_{K} under the action of 𝔊(K/F) or, what is the same, under the action of W_{K/F} on C_{K} by means of inner automorphisms. If χ_{K} is a quasi-character of C_{K} its orbit will be denoted Ω(χ_{K}). If ρ is a representation of W_{K/F} the restriction of ρ to C_{K} is the direct sum of one-dimensional representations. Let S(ρ) be the collection of quasi-characters to which these one-dimensional representations correspond.

Suppose

ρ = Ind(W_{K/F} , W_{K/E}, χ_{E}).

Let W_{K/F} be the disjoint union

⋃_{i=1}^{m} W_{K/E}w_{i}.

Define the function φ_{i} by

φ_{i}(ww_{j}) = 0  w ∈ W_{K/E}, j ≠ i
φ_{i}(ww_{i}) = χ_{E}(τ_{K/E}w) w ∈ W_{K/E}.

{φ_{1}, …, φ_{m}} is a basis for the space of functions of which ρ acts. If a ∈ C_{K} then

ww_{j}a = w(w_{j}aw_{j}^{−1})w_{j}

and w_{j}aw_{j}^{−1} belongs to C_{K} which, of course, lies in W_{K/E}. Thus

ρ(a)φ_{i} = χ_{E}(τ_{K/E}(w_{i}aw_{i}^{−1}))φ_{i} = χ_{K/E}^{σ_{i}}(a)φ_{i}

if σ_{i} is the image of w_{i} in 𝔊(K/F). Thus

S(ρ) = Ω(χ_{K/E}).

Suppose E_{1}, …, E_{r}, E′_{1}, …, E′_{s} lie between F and K, χ_{E_{i}} is a quasi-character of E_{i}, and χ_{E′_{j}} is a quasi-character of E′_{j}. Let

ρ_{i} = Ind(W_{K/F} , W_{K/E_{i}}, χ_{E_{i}})

and let

ρ′_{j} = Ind(W_{K/F} , W_{K/E′_{j}}, χ_{E′_{j}}).

Suppose ρ_{i} acts on V_{i} and ρ′_{j} acts on V′_{j}. The direct sum of the representations ρ_{i} acts on

V = ⊕_{i=1}^{r}V_{i}

and the direct sum of the representations ρ′_{j} acts on

V′ = ⊕_{j=1}^{s}V′_{j}.

Let

V_{Ω} = ⊕_{{i | χ_{K/E_{i}} ∈Ω}}V_{i}
V′_{Ω} = ⊕_{{i | χ_{K/E′_{j}} ∈Ω}}V′_{j}.

Any isomorphism of V with V′ which commutes with the action of W_{K/F} takes V_{Ω} to V′_{Ω}.

If χ_{K/E_{i}} ∈ Ω(χ_{K}) there is a σ in 𝔊(K/F) such that χ_{K} = χ_{K/E_{i}}^{σ}. Then
ρ_{i} ≃ Ind(W_{K/F} , W_{K/E_{i}^{σ}}, χ_{E_{i}}^{σ})

and

Δ(χ_{E_{i}}, ψ_{E_{i}/F}) λ(E_{i}/F, ψ_{F}) = Δ(χ_{E_{i}}^{σ}, ψ_{E_{i}^{σ}/F}) λ(E_{i}^{σ}/F, ψ_{F}).

We conclude that Theorem 2.1 is a consequence of the following lemma.

Lemma 17.1

Suppose χ_{K} is a quasi-character of C_{K}. Suppose E_{1}, …, E_{r}, E′_{1}, …, E′_{s} lie between F and K, χ_{E_{i}} is a quasi-character of C_{E_{i}}, χ_{E′_{j}} is a quasi-character of C_{E′_{j}}, and

ρ = ⊕_{i=1}^{r}Ind(W_{K/F} , W_{K/E_{i}}, χ_{E_{i}})

is equivalent to

ρ′ = ⊕_{j=1}^{s} Ind(W_{K/F} , W_{K/E′_{j}}, χ_{E′_{j}}).

If χ_{K/E_{i}} = χ_{K/E′_{j}} = χ_{K} for all i and j then

∏_{i=1}^{r} Δ(χ_{E_{i}}, ψ_{E_{i}/F}) λ(E_{i}/F, ψ_{F})

is equal to

∏_{j=1}^{s} Δ(χ_{E′_{j}}, ψ_{E′_{j}/F}), λ(E′_{j}/F, ψ_{F}).

Let F (χ_{K}) be the fixed field of the isotropy group of χ_{K}. Let ρ act on V and let ρ′ act on V′. Let

V (χ_{K}) = {v ∈ V | ρ(a)v = χ_{K}(a)v for all a in C_{K}}.

Define V′(χ_{K}) in a similar fashion. It is clear that any isomorphism of V with V′ which commutes with the action of W_{K/F} takes V (χ_{K}) to V′(χ_{K}). The group W_{K/F (χ_{K})} leaves both V (χ_{K}) and V′(χ_{K}) invariant and its representations on these two spaces are equivalent.

Let

Ind(W_{K/F} , W_{K/E_{i}}, χ_{E_{i}})

acts on V_{i} and define V_{i}(χ_{K}) in the obvious manner. Then

V (χ_{K}) = ⊕_{i=1}^{r}V_{i}(χ_{K}).

Chapter 17 207 Defining V′_{j} and V′_{j}(χ_{K}) in a similar manner, we have

V′(χ_{K}) = ⊕_{j=1}^{s}V′_{j}(χ_{K}).

It is clear that the representation of W_{K/F (χ_{K})} on V_{i}(χ_{K}) is equivalent to

Ind(W_{K/F (χ_{K})}, W_{K/E_{i}}, χ_{E_{i}}).

Thus

⊕_{i=1}^{r}Ind(W_{K/F (χ_{K})}, W_{K/E_{i}}, χ_{E_{i}})

is equivalent to

⊕_{j=1}^{s}Ind(W_{K/F (χ_{K})}, W_{K/E′_{j}}, χ_{E′_{j}}).

If F (χ_{K}) ≠ F the assertion of the lemma follows from the induction assumption and Lemma 16.3.

Chapter Eighteen.

Nilpotent Groups.

In this paragraph we prove Lemma 17.1 assuming that F = F (χ_{K}) and that G = 𝔊(K/F) is nilpotent.

Lemma 18.1

Suppose D is a normal subgroup of G of prime order ℓ which is contained in the center of G. Let M be the fixed field of D. Suppose F ⊆ E ⊆ K and χ_{E} is a quasi-character of C_{E}. Suppose also that F (χ_{K/E}) = F.

(a) There are fields F_{1}, …, F_{r} contained in M and quasi-characters χ_{F_{1}}, …, χ_{F_{r}} such that χ_{K/F_{i}} =

χ_{K/E} and such that

Ind(W_{K/F} , W_{K/E}, χ_{E})

is equivalent to

⊕_{i=1}^{r}Ind(W_{K/F} , W_{K/F_{i}}, χ_{F_{i}}).

(b) If Theorem 2.1 is valid for all Galois extensions K′/F′ in P(K/F) with [K′ : F′] < [K : F]

then

Δ(χ_{E}, ψ_{E/F}) λ(E/F, ψ_{F})

is equal to

∏_{i=1}^{r} Δ(χ_{F_{i}}, ψ_{F_{i}/F}) λ(F_{i}/F, ψ_{F}).

We prove the lemma by induction on [K : F]. Let H be the subgroup of G corresponding to E; let G′ = HD and let F′ be the fixed field of G′. If F′ ≠ F the induction assumption implies that there are fields F_{1}, …, F_{r} contained in M and quasi-characters χ_{F_{1}}, …, χ_{F_{r}} such that χ_{K/F_{i}} = χ_{K/E} for each i and such that

Ind(W_{K/F} , W_{K/E}, χ_{E})

is equivalent to

⊕_{i=1}^{r}Ind(W_{K/F′}, W_{K/F_{i}}, χ_{F_{i}}).

The first part of the lemma follows from the transitivity of the induction process. The second part follows from Lemma 16.3 and the assumed validity of Theorem 2.1 for the extension K/F′.

We suppose now that G = HD. Suppose that H contains a normal subgroup H_{1} of G which is different from {1} and suppose that, if K_{1} is the fixed field of H_{1}, F (χ_{K_{1}/E}) = F. If M_{1} is the fixed field of H_{1}D then, according to the induction assumption, there are fields F_{1}, …, F_{r} contained in M_{1} and quasi-characters χ_{F_{1}}, …, χ_{F_{r}} such that

χ_{K_{1}/F_{i}} = χ_{K_{1}/E}

and such that

Ind(W_{K_{1}/F} , W_{K_{1}/E}, χ_{E})

is equivalent to

⊕_{i=1}^{r}Ind(W_{K_{1}/F} , W_{K_{1}/F_{i}}, χ_{F_{i}}).

It follows immediately that

χ_{K/F_{i}} = χ_{K/E}

and that

Ind(W_{K/F} , W_{K/E}, χ_{E})

is equivalent to

⊕_{i=1}^{r}Ind(W_{K/F} , W_{K/F_{i}}, χ_{F_{i}}).

The equality of (b) is a consequence of the assumed validity of Theorem 2.1 for K_{1}/F.

We assume now that G = HD and that if H_{1} is a normal subgroup of H different from {1} with fixed field K_{1} the field F (χ_{K_{1}/E}) is not F. If w_{1} belongs to W_{K/E} and w_{2} belongs to W_{K/M} then

w_{1}w_{2}w_{1}^{−1}w_{2}^{−1} ∈ C_{K}.

Let χ_{K} = χ_{K/E}. Since F (χ_{K}) = F

χ_{K}(w_{1}w_{2}w_{1}^{−1}w_{2}^{−1}) = χ_{K}(w_{2}w_{1}^{−1}w_{2}^{−1}w_{1}) = χ_{K}(w_{1}^{−1}w_{2}^{−1}w_{1}w_{2}) = χ_{K}(w_{2}^{−1}w_{1}w_{2}w_{1}^{−1}).

Denote the common value of the expressions by ω(w_{1}, w_{2}). Then ω(v_{1}w_{1}, w_{2}) is equal to

χ_{K}(v_{1}w_{1}w_{2}w_{1}^{−1}v_{1}^{−1}w_{2}^{−1}) = χ_{K}(w_{2}^{−1}w_{1}w_{2}w_{1}^{−1}v_{1}^{−1}w_{2}^{−1}v_{1}w_{2}).

The right side is

ω(v_{1}, w_{2}) ω(w_{1}, w_{2}).

In the same way ω(w_{1}, v_{2}w_{2}) is

χ_{K}(w_{1}v_{2}w_{2}w_{1}^{−1}w_{2}^{−1}v_{2}^{−1}) = χ_{K}(w_{1}^{−1}v_{2}^{−1}w_{1}v_{2}w_{2}w_{1}^{−1}w_{2}^{−1}w_{1})

which equals

ω(w_{1}, v_{2}) ω(w_{1}, w_{2}).

If either w_{1} or w_{2} belong to C_{K}, we have

ω(w_{1}, w_{2}) = 1.

Thus, for each w_{2},

w_{1} ⟶ ω(w_{1}, w_{2})

is a homomorphism of H = W_{K/E}/C_{K} into ℂ^{∗} and, for each w_{1},

w_{2} ⟶ ω(w_{1}, w_{2})

is a homomorphism of D = W_{K/M}/C_{K} into ℂ^{∗}. If w belongs to W_{K/F} then

ω(ww_{1}w^{−1}, ww_{2}w^{−1}) = ω(w_{1}, w_{2}).

Thus there is a normal extension K_{1} containing E such that

W_{K/K_{1}} = {w_{1} | ω(w_{1}, w_{2}) = 1 for all w_{1} ∈ W_{K/M}}.

But F (χ_{K_{1}/E}) will be F so that K_{1} must be K.

It follows immediately that H is isomorphic to a subgroup of the dual group of D. Thus H = {1} or H is cyclic of order ℓ. In either case H must lie in the centralizer of D so that E/F is normal and 𝔊(E/F) is isomorphic to D. If H = {1} then χ_{E} may be extended from C_{E} = C_{K} to a quasi-character of W_{E/F}. In other words, there is a quasi-character χ_{F} of C_{F} such that χ_{E} = χ_{E/F}. Then

Ind(W_{K/F} , W_{K/E}, χ_{E})

is equivalent to

⊕_{μ_{F} ∈S(E/F)} Ind(W_{K/F} , W_{K/F} , μ_{F} χ_{F}).

Suppose H ≠ {1}. Since W_{K/M}/C_{K} is cyclic there is a quasi-character χ_{M} of C_{M} such that χ_{K} = χ_{K/M}. If w_{1} belongs to W_{K/E} let χ_{w_{1}} be the character of W_{K/M} or, what is the same, of C_{M} defined by

χ_{w_{1}}(w_{2}) = ω(w_{1}, w_{2})

and if w_{2} belongs to W_{K/M} let

χ_{w_{2}}(w_{1}) = ω(w_{1}, w_{2}).

Clearly

{χ_{w_{1}} | w_{1} ∈ W_{K/E}} = S(K/M)

and

{χ_{w_{2}} | w_{2} ∈ W_{K/M}} = S(K/E).

If σ_{1} is the image of w_{1} in H and σ_{2} the image of w_{2} in D then

χ_{E}^{σ_{2}}(w_{1}) = χ_{E}(w_{2}w_{1}w_{2}^{−1}w_{1}^{−1}w_{1}) = χ_{w_{2}}^{−1}(w_{1}) χ_{E}(w_{1})

and

χ_{M}^{σ_{1}}(w_{2}) = χ_{M}(w_{1}w_{2}w_{1}^{−1}w_{2}^{−1}w_{2}) = χ_{w_{1}}(w_{2}) χ_{M}(w_{2}).

Let W_{K/F} be the disjoint union

⋃_{i=1}^{ℓ} W_{K/E}v_{i}

with v_{i} in W_{K/M}. Define the function φ_{i} on W_{K/F} by

φ_{i}(wv_{j}) = 0

if w ∈ W_{K/E} and j ≠ i and by

φ_{i}(wv_{i}) = χ_{E}(w)

if w ∈ W_{K/E}. Then

{φ_{i} | 1 ≤ i ≤ ℓ}

is a basis for the space U on which

Ind(W_{K/F} , W_{K/E}, χ_{E})

acts. Let ψ_{i}, 1 ≤ i ≤ ℓ be the function W_{K/F} defined by

ψ_{i}(w_{2}w_{1}) = χ_{M}(w_{2})χ_{E}^{σ(v_{i})}(w_{1})

if w_{1} belongs to W_{K/E} and w_{2} belongs to W_{K/M}. Here σ(v_{i}) is the image of v_{i} in 𝔊(K/F). It is necessary, but easy, to verify that ψ_{i} is well-defined. The collection

{ψ_{i} | 1 ≤ i ≤ ℓ}

is a basis for the space V on which

Ind(W_{K/F} , W_{K/M}, χ_{M})

acts. It is easily verified that the isomorphism of U with V which sends χ_{M}(v_{i})φ_{i} to ψ_{i} is an isomorphism. Thus

Ind(W_{K/F} , W_{K/E}, χ_{E}) ≃ Ind(W_{K/F} , W_{K/M}, χ_{M}).

This takes care of the first part of the lemma.

Whether H = {1} or not,

Ind(W_{K/F} , W_{K/E}, 1)

is equivalent to

⊕_{μ_{F} ∈S(E/F)} Ind(W_{K/F} , W_{K/F} , μ_{F}).

If H ≠ 1 we may apply Theorem 2.1 to E/F to see that

λ(E/F, ψ_{F}) = ∏_{μ_{F} ∈S(E/F)} Δ(μ_{F} , ψ_{F}).

If H = {1} this equality is just the definition of the left side. In this case the second part of the lemma asserts that

Δ(χ_{E}, ψ_{E/F}) ∏_{μ_{F} ∈S(E/F)} Δ(μ_{F} , ψ_{F})  (18.1)

is equal to

∏_{μ_{F} ∈S(E/F)} Δ(μ_{F} χ_{F} , ψ_{F})

where χ_{E} = χ_{E/F}. This is a consequence of the first main lemma. If H ≠ {1}, Theorem 2.1 applied to M/F, shows that

λ(M/F, ψ_{F}) = ∏_{μ_{F} ∈S(M/F)} Δ(μ_{F} , ψ_{F})

and the second part of the lemma asserts that (18.1) is equal to

Δ(χ_{M}, ψ_{M/F}) ∏_{μ_{F} ∈S(M/F)} Δ(μ_{F} , ψ_{F}).

This is a consequence of the second main lemma.

A non-trivial nilpotent group always contains a subgroup D satisfying the conditions of the previous lemma. Lemma 17.1 is clear if K = F. If K ≠ F and 𝔊(K/F) is nilpotent it is a consequence of the following lemma.

Lemma 18.2

Suppose K/F is normal and Theorem 2.1 is valid for all normal extensions K′/F′ in P(K/F) with [K′ : F′] < [K : F]. Suppose F ⊆ M ⊂_{≠} K and M/F is normal. Suppose E_{1}, …, E_{r},

E′_{1}, …, E′_{s} lie between F and M, χ_{E_{i}} is a quasi-character of C_{E_{i}}, χ_{E′_{j}} is a quasi-character of C_{E′_{j}}, and

⊕_{i=1}^{r}Ind(W_{K/F} , W_{K/E_{i}}, χ_{E_{i}})

is equivalent to

⊕_{j=1}^{s}Ind(W_{K/F} , W_{K/E′_{j}}, χ_{E′_{j}}).

Then

∏_{i=1}^{r} Δ(χ_{E_{i}}, ψ_{E_{i}/F}) λ(E_{i}/F, ψ_{F})

is equal to

∏_{j=1}^{s} Δ(χ_{E′_{j}}, ψ_{E′_{j}/F}) λ(E′_{j}/F, ψ_{F}).

The representation

Ind(W_{K/F} , W_{K/E_{i}}, χ_{E_{i}})

can be obtained by inflating the representation

Ind(W_{M/F} , W_{M/E_{i}}, χ_{E_{i}})

from W_{M/F} to W_{K/F}. A similar remark applies to the representations induced from the χ_{E′_{j}}. Thus

⊕_{i=1}^{r}Ind(W_{M/F} , W_{M/E_{i}}, χ_{E_{i}})

is equivalent to

⊕_{j=1}^{s}Ind(W_{M/F} , W_{M/E′_{j}}, χ_{E′_{j}}).

Applying Theorem 2.1 to the extension M/F we obtain the lemma.

Chapter Nineteen.

Proof of the Main Theorem.

We shall first prove Lemma 17.1 when there is a quasi-character χ_{F} of C_{F} such that χ_{K} = χ_{K/F}. Implicit in the statement of the following lemma as in that of Lemma 17.1, is the assumption that Theorem 2.1 is valid for all pairs K′/F′ in P(K/F) for which [K′ : F′] < [K : F]. Recall that we have fixed a non-trivial abelian normal subgroup C of G = 𝔊(K/F) and that L is its fixed field.

Lemma 19.1

Suppose F ⊆ E ⊆ K, χ_{F} is a quasi-character of C_{F} , χ_{E} is a quasi-character if C_{E}, and χ_{K/E} = χ_{K/F}. There are fields F_{1}, …, F_{r} contained in L and quasi-characters χ_{F_{i}}, 1 ≤ i ≤ r, such that χ_{K/F_{i}} = χ_{K/F},

Ind(W_{K/F} , W_{K/E}, χ_{E})

is equivalent to

⊕_{i=1}^{r}Ind(W_{K/F} , W_{K/F_{i}}, χ_{F_{i}})

and

Δ(χ_{E}, ψ_{E/F}) λ(E/F, ψ_{F})

is equal to

∏_{i=1}^{r} Δ(χ_{F_{i}}, ψ_{F_{i}/F}) λ(F_{i}/F, ψ_{F}).

We prove the lemma by induction on [K : F]. Let E be the fixed field of H and let F′ be the fixed field of HC. If F′ ≠ F then, by induction, there are fields F_{1}, …, F_{r} lying between F′ and L and quasi-characters χ_{F_{1}}, …, χ_{F_{r}} such that χ_{K/F_{i}} = χ_{K/F} and

Ind(W_{K/F′}, W_{K/E}, χ_{E})

is equivalent to

⊕_{i=1}^{r}Ind(W_{K/F′}, W_{K/F_{i}}, χ_{F_{i}}).

In this case the lemma follows from the transitivity of the induction process, the assumed validity of Theorem 2.1 for K/F′ and Lemma 16.3.

We suppose henceforth that G = HC. There is a character θ_{E} in S(K/E) such that χ_{E} = θ_{E}χ_{E/F}. θ_{E} may be regarded as a character of H. If H ∩ C = {1} we may define a character θ_{F} of G by setting

θ_{F} (hc) = θ_{E}(h)

if h is in H and c is in C. θ_{F} may be regarded as a character of C_{F} and θ_{E} = θ_{E/F}. Replacing χ_{F} by θ_{F} χ_{F} we suppose that χ_{E} = χ_{E/F}. Then in the notation of Lemma 15.1, we may take

{F_{1}, …, F_{r}} = {F_{μ} | μ ∈ T }

and if F_{i} = F_{μ},

χ_{F_{i}} = μ′χ_{F_{μ}/F} .

The assertions of the lemma are consequences of Lemmas 15.1 and 15.3.

We suppose now not only that G = HC but also that H ∩ C ≠ {1}. Let S be the set of characters in S(K/L) whose restriction to H ∩ C agrees with the restriction of θ_{E}. S is invariant under the action of H on S(K/L). If ν belongs to S let φ_{ν} be the function on W_{K/F} defined by

φ_{ν}(wv) = χ_{E}(w) χ_{L/F} (v) ν(v)

if w is in W_{K/E} and v is in W_{K/L}. ν is a character of C and may therefore be regarded as a character of W_{K/L} or of C_{L}. It is easy to verify that φ_{ν} is well-defined. If

ρ = Ind(W_{K/F} , W_{K/E}, χ_{E})

then

{φ_{ν} | ν ∈ S}

is a basis for the space of functions on which ρ acts. If w belongs to W_{K/E}

ρ(w)φ_{ν} = χ_{E}(w)φ_{ν′}

with ν′ = ν^{σ^{−1}} is σ is the image of w in 𝔊(K/F). If v belongs to W_{K/L}

ρ(v)φ_{ν} = χ_{L/F} (v) ν(v)φ_{ν}.

Thus if R is an orbit in S under the action of H, the space

V_{R} = ∑_{ν∈R} ℂ_{φ_{ν}}

is invariant under W_{K/F} and ρ is the direct sum of its restrictions to the spaces V_{R}.

If μ belongs to R let H_{μ} be the isotropy group of μ, let G_{μ} = H_{μ}C, and let F_{μ} be the fixed field of G_{μ}. Extend μ to a character μ′ of G_{μ} by setting

μ′(hc) = θ_{E}(h) μ(c)

if h is in H_{μ} and c is in C. μ′, which is easily seen to be well-defined, may be regarded as a character of W_{K/F_{μ}} of C_{F_{μ}}. Let W_{K/F} be the disjoint union

⋃_{i=1}^{s} W_{K/F_{μ}}w_{i}

with w_{i} in W_{K/E} and let σ_{i} be the image of w_{i} in 𝔊(K/F). Let φ_{i} be the function of W_{K/F} defined by

φ_{i}(ww_{j}) = 0  w ∈ W_{K/F_{μ}}, j ≠ i
φ_{i}(ww_{i}) = μ′(w)χ_{F_{μ}/F} (w)  w ∈ W_{K/F_{μ}}.

The collection

{φ_{i} | 1 ≤ i ≤ s}

is a basis for the space V_{μ} on which the representation

σ_{μ} = Ind(W_{K/F} , W_{K/F_{μ}}, μ′χ_{F_{μ}/F})

acts. Let

ψ_{i} = χ_{E}(w_{i})φ_{i}.

If w belongs to W_{K/L}

σ_{μ}(w)ψ_{i} = μ^{σ_{i}}(w) χ_{L/F} (w)ψ_{i}.

If w belongs to W_{K/E} and w_{j}w = vw_{i} with v in W_{K/F_{μ}} then

σ_{μ}(w)ψ_{i} = χ_{E}(w)ψ_{j}.

Thus the isomorphism of V_{μ} with V_{R} which takes ψ_{i} to φ_{μ^{σ_{i}}} commutes with the action of W_{K/F}. If T is a set of representatives for the orbits in S

ρ ≃ ⊕_{μ∈T} σ_{μ}.

If K_{1} is the fixed field of H ∩ C then K_{1}/F is normal and ρ is the inflation to W_{K/F} of

Ind(W_{K_{1}/F} , W_{K_{1}/E}, χ_{E})  (19.1)

and σ_{μ} is the inflation of

Ind(W_{K_{1}/F} , W_{K_{1}/F_{μ}}, μ′χ_{F_{μ}/F}).

Thus the representation (19.1) is equivalent to

⊕_{μ∈T} Ind(W_{K_{1}/F} , W_{K_{1}/F_{μ}}, μ′χ_{F_{μ}/F}).

Applying Theorem 2.1 to K_{1}/F we see that

Δ(χ_{E}, ψ_{E/F}) λ(E/F, ψ_{F})

is equal to

∏_{μ∈T} (μ′χ_{F_{μ}/F} , ψ_{F_{μ}/F}) λ(F_{μ}/F, ψ_{F}).

If there is a quasi-character χ_{F} such that χ_{K} = χ_{K/F}, Lemma 17.1 follows from Lemma 18.2 and the lemma just proved. To complete the proof of Theorem 2.1 we have to prove Lemma 17.1 when F = F (χ_{K}), G is not nilpotent, and there is no quasi-character χ_{F} of C_{F} such that χ_{K} = χ_{K/F}. In this case none of the fields E_{1}, …, E_{r}, E′_{1}, …, E′_{s} is equal to F and Theorem 2.1 may be applied to K/E_{i} and K/E′_{j}.

Lemma 19.2

Suppose A and B lie between F and K. Suppose χ_{A} and χ_{B} are quasi-characters of C_{A} and C_{B} respectively. There are fields A_{1}, …, A_{m} lying between A and K, fields B_{1}, …, B_{m} lying between B and K, elements σ_{1}, …, σ_{m} in G, and quasi-characters χ_{A_{1}}, …, χ_{A_{m}}, χ_{B_{1}}, …, χ_{B_{m}} such that B_{i} = A_{i}^{σ_{i}}, χ_{B_{i}} = χ_{A_{i}}^{σ_{i}}, and such that the tensor product

Ind(W_{K/F} , W_{K/A}, χ_{A}) ⊗ Ind(W_{K/F} , W_{K/B}, χ_{B})

is equivalent to

⊕_{i=1}^{m}Ind(W_{K/F} , W_{K/A_{i}}, χ_{A_{i}})

and to

⊕_{i=1}^{m}Ind(W_{K/F} , W_{K/B_{i}}, χ_{B_{i}}).

Let

ρ = Ind(W_{K/F} , W_{K/A}, χ_{A})
σ = Ind(W_{K/F} , W_{K/B}, χ_{B}).

Let α be the restriction of σ to W_{K/A} and β the restriction of ρ to W_{K/B}. By Lemma 2.3

ρ ⊗ σ ≃ Ind(W_{K/F} , W_{K/A}, χ_{A} ⊗ α)

and

ρ ⊗ σ ≃ Ind(W_{K/F} , W_{K/B}, χ_{B} ⊗ β).

Let W_{K/F} be the disjoint union

⋃_{i=1}^{m} W_{K/A}w_{i}W_{K/B}.

If U_{i} is the space of functions in U, the space on which ρ acts, which are zero outside of the double coset W_{K/A}w_{i}W_{K/B} then U_{i} is invariant under β. Define the field B_{i} by demanding that

W_{K/B_{i}} = W_{K/B} ∩ w_{i}^{−1}W_{K/A}w_{i}.

If σ_{i} is the image of w_{i} in 𝔊(K/F) let χ′_{B_{i}} be the restriction of χ_{A}^{σ_{i}} to W_{K/B_{i}}. If U′_{i} is the space of functions on which

Ind(W_{K/B}, W_{K/B_{i}}, χ′_{B_{i}})

acts, the map of U_{i} to U′_{i} which sends φ to the function φ′ defined by

φ′(w) = φ(w_{i}w)

if w is in W_{K/B} is an isomorphism which commutes with the action of W_{K/B}. Thus

β ≃ ⊕_{i=1}^{m}Ind(W_{K/B}, W_{K/B_{i}}, χ′_{B_{i}})

and, if χ_{B_{i}} = χ_{B_{i}/B} χ′_{B_{i}},

χ_{B} ⊗ β ≃ ⊕_{i=1}^{m}Ind(W_{K/B}, W_{K/B_{i}}, χ_{B_{i}}).

Similar considerations apply if the roles of A and B are interchanged. The double coset decomposition becomes

⋃_{i=1}^{m} W_{K/B}w_{i}^{−1}W_{K/A}

and

W_{K/A_{i}} = W_{K/A} ∩ w_{i}W_{K/B}w_{i}^{−1} = w_{i}W_{K/B_{i}}w_{i}^{−1}.

Thus B_{i} = A_{i}^{σ_{i}}. It is also clear that χ_{B_{i}} = χ_{A_{i}}^{σ_{i}}.

To complete the proof of Lemma 17.1 we use Brauer’s theorem in the following form. There are fields F_{1}, …, F_{n} lying between F and K such that 𝔊(K/F_{k}) is nilpotent for each k, characters χ_{F_{k}} of C_{F_{k}}/N_{K/F_{k}}C_{K}, and integers m_{1}, …, m_{n} such that

1 ≃ ⊕_{k=1}^{n}m_{k} Ind(W_{K/F} , W_{K/F_{k}}, χ_{F_{k}}).

Since we are assuming that G is not nilpotent none of the F_{k} are equal to F and we may apply Theorem 2.1 to each of the extensions K/F_{k}.

We shall apply the previous lemma with A = E_{i}, B = F_{k} and with A = E′_{j}, B = F_{k}. m will be denoted by m(ik) or m′(jℓ). A_{ℓ} will be denoted by E_{ikℓ} or E′_{jkℓ} and B_{ℓ} will be denoted by F_{ikℓ} or F′_{jkℓ}. Observe that

Δ(χ_{E_{ikℓ}}, ψ_{E_{ikℓ}/F} λ(E_{ikℓ}/F, ψ_{F})  (19.2)

is equal to

Δ(χ_{F_{ikℓ}}, ψ_{F_{ikℓ}/F}) λ(F_{ikℓ}/F, ψ_{F})  (19.3)

and that

Δ(χ_{E′_{jkℓ}}, ψ_{E′_{jkℓ}/F}) λ(E′_{jkℓ}/F, ψ_{F})  (19.4)

is equal to

Δ(χ_{F′_{jkℓ}}, ψ_{F′_{jkℓ}/F}) λ(F′_{jkℓ}/F, ψ_{F}).  (19.5)

χ_{E_{i}} may be regarded as a one-dimensional representation of W_{K/E_{i}} and as such is equivalent to

⊕_{k=1}^{n} ⊕_{ℓ=1}^{m(ik)} m_{k} Ind(W_{K/E_{i}}, W_{K/E_{ikℓ}}, χ_{E_{ikℓ}}).

Therefore

1 = ∑_{k=1}^{n} ∑_{ℓ=1}^{m(ik)} m_{k} [E_{ikℓ} : E_{i}]

and

Δ(χ_{E_{i}}, ψ_{E_{i}/F})

is equal to

∏_{k=1}^{n} ∏_{ℓ=1}^{m(ik)}{Δ(χ_{E_{ikℓ}}, ψ_{E_{ikℓ}/F}) λ(E_{ikℓ}/E_{i}, ψ_{E_{i}/F})}^{m_{k}}.

Multiplying both of these expressions by λ(E_{i}/F, ψ_{F}), we see that

Δ(χ_{E_{i}}, ψ_{E_{i}/F}) λ(E_{i}/F, ψ_{F})  (19.6)

is equal to

∏_{k=1}^{n} ∏_{ℓ=1}^{m(ik)}{Δ(χ_{E_{ikℓ}}, ψ_{E_{ikℓ}/F}) λ(E_{ikℓ}/F, ψ_{F})}^{m_{k}}.  (19.7)

The same argument establishes that

Δ(χ_{E′_{j}}, ψ_{E′_{j}/F}) λ(E′_{j}/F, ψ_{F})  (19.8)

is equal to

∏_{k=1}^{n} ∏_{ℓ=1}^{m′(jk)}{Δ(χ_{E′_{jkℓ}}, ψ_{E′_{jkℓ}/F}) λ(E′_{jkℓ}/F, ψ_{F})}^{m_{k}}.  (19.9)

We are trying to show that the product over i of the expressions (19.6) is equal to the product over j of the expressions (19.8). It will be enough to show that the product of the expressions (19.7) is equal to the product of the expressions (19.9).

The representations

⊕_{i=1}^{r} ⊕_{ℓ=1}^{m(ik)} Ind(W_{K/F_{k}}, W_{K/F_{ikℓ}}, χ_{F_{ikℓ}})

and

⊕_{j=1}^{s} ⊕_{ℓ=1}^{m′(jk)} Ind(W_{K/F_{k}}, W_{K/F′_{jkℓ}}, χ_{F′_{jkℓ}})

are equivalent. Therefore

∑_{i=1}^{r} ∑_{ℓ=1}^{m(ik)}[F_{ikℓ} : F_{k}] = ∑_{j=1}^{s} ∑_{ℓ=1}^{m′(jk)}[F′_{jkℓ} : F_{k}].

Denote the common value of these expressions by N(k). Moreover

∏_{i=1}^{r} ∏_{ℓ=1}^{m(ik)} Δ(χ_{F_{ikℓ}}, ψ_{F_{ikℓ}/F}) λ(F_{ikℓ}/F_{k}, ψ_{F_{k}/F})

is equal to

∏_{j=1}^{s} ∏_{ℓ=1}^{m′(jk)} Δ(χ_{F′_{jkℓ}}, ψ_{F′_{jkℓ}/F}) λ(F′_{jkℓ}/F, ψ_{F_{k}/F}).

Multiplying both of these expressions by

λ(F_{k}/F, ψ_{F})^{N(k)}

we see that

∏_{i=1}^{r} ∏_{ℓ=1}^{m(ik)} Δ(χ_{F_{ikℓ}}, ψ_{F_{ikℓ}/F}) λ(F_{ikℓ}/F, ψ_{F})  (19.10)

is equal to

∏_{j=1}^{s} ∏_{ℓ=1}^{m′(jk)} Δ(χ_{F′_{jkℓ}}, ψ_{F′_{jkℓ}/F}) λ(F′_{jkℓ}/F, ψ_{F}).  (19.11)

Because of the equality of (19.2) and (19.3) the product over i of the expressions (19.7) is equal to the product over k of of the m_{k}th powers of the expressions (19.10). The product over j of the expressions (19.9) is equal to the product over k of the m_{k}th powers of the expressions (19.11). Lemma 17.1, and with it Theorem 2.1, is now completely proved.

Chapter Twenty.

Artin L-functions.

Suppose ω is an equivalence class of representations of the Weil group of the non-archimedean local field F. Let K be a Galois extension of F and let σ be a representation of W_{K/F} in the class ω. Suppose σ acts on V. Let V^{0} be the subspace of V fixed by every element of W_{K/F}^{0} . Since W_{K/F}^{0} is a normal subgroup of W_{K/F} the space V^{0} is invariant under W_{K/F} and on V^{0} we get a representation σ^{0}. Since W_{K/F}^{0} = τ_{K/F}^{−1} (u_{F}^{0} ) the class of σ^{0} depends only on w. σ^{0} breaks up into the direct sum of 1-dimensional representations corresponding to unramified generalized characters μ_{1}, …, μ_{r} of C_{F}. We set

L(s, w) = ∏_{i=1}^{r} \frac{1}{1 − μ_{i}(π_{F}) |π_{F} |^{s}} .

This we take as the local function. It is clear that when w is one-dimensional the present definition agrees with that of the introduction and that of ω = ω_{1} ⊕ ω_{2}. Then

L(s, ω) = L(s, ω_{1}) ⊕ L(s, ω_{2}).

Suppose F ⊆ E ⊆ K, ρ is a representation of W_{K/E}, and

σ = Ind(W_{K/F} , W_{K/E}, ρ).

We have to show that if θ is the class of ρ then

L(s, ω) = L(s, θ).

Let ρ act on W. Then V is the space of functions f on W_{K/F} with values in W which satisfy

f(uv) = ρ(u)f(v)

for u in W_{K/E} and v in W_{K/F}. If f lies in V_{0} and u lies in W_{K/E}^{0} then

ρ(u)f(v) = f(uv) = f(vv^{−1}uv) = f(v)

because v^{−1} lies in W_{K/F}^{0} . Thus f takes values in W^{0}. In other words, we may as well assume that W = W^{0}. Indeed we may as well go further and assume that W = W^{0} has dimension one.

Let N_{E/F} π_{E} = επ_{F}^{f} where ε is a unit and choose w_{0} in W_{K/F} so that τ_{K/F} w_{0} = π_{F}. Then w^{f} = u_{0}v_{0} with u_{0} in W_{K/F}^{0} and v_{0} in W_{K/E} such that τ_{K/E}v_{0} = π_{E}. Clearly, V^{0} consists of the functions f with values in W which satisfy f(uw) = f(w) for u in W_{K/F}^{0} and f(uw) = μ(τ_{K/E}u)f(w) if ρ is associated to the generalized character μ of C_{E}. Take as basis of V^{0} the functions φ_{0}, …, φ_{f−1} defined by

φ_{i}(uvw_{0}^{j}) = μ(τ_{K/E}v)δ_{i}^{j}x

where x is a non-zero vector in W, u belongs to W_{K/F}^{0} , v belongs to W_{K/E}, 0 ≤ j < f, and δ_{i}^{j} is Kronecker’s delta. The matrix of σ(w_{0}) with respect to this basis is

A = ( \begin{matrix}0 & · & · & · & μ(τ_{K/E}v_{0}) \\ 1 & 0 & & & · \\ & 1 & ... & & · \\ & & ... & ... & · \\ & 0 & & 1 & 0 \end{matrix} )

and

L(s, ω) = \frac{1}{det(I − A |π_{F} |^{s})} = \frac{1}{1 − μ(π_{E}) |π_{F} |^{f_{s}}} = L(s, θ)

since |π_{F} |^{f} = |π_{E}|.

For archimedean fields we proceed in a different manner. If we write ω, as we may, as a sum of irreducible representations the components are unique up to order. If ω = ∑_{i=1}^{r} ⊕ω_{i}, we will have to have

L(s, ω) = ∏_{i=1}^{r} L(s, ω_{i}).

Thus it is a matter of defining L(s, ω) for irreducible ω. If ω is one-dimensional this was done in the introduction. If ω is not one-dimensional then F must be ℝ. Let σ be a representation of W_{ℂ/ℝ} in the class ω. W_{ℂ/ℝ} is an extension of the group of order 2 by ℂ^{∗}. Let W_{ℂ/ℝ} = ℂ^{×} ∪ w_{0}ℂ^{×}. If σ acts on V there is a non-zero vector x in V and a generalized character μ of ℂ^{×} such that σ(a)x = μ(a)x for all a in ℂ^{×}. Then the space spanned by {x, σ(w_{0})x} is invariant and therefore all of V. Since V is not one-dimensional σ(w_{0})x is not a multiple of x. Notice that σ(a)σ(w_{0})x = σ(w_{0})σ(w_{0}^{−1}aw_{0})x = μ(\overline{a})σ(w_{0})x. If

μ(z) = |z|^{r} \frac{z^{m}\overline{z}^{n}}{|z|^{\frac{m+n}{2}}}

with m + n ≥ 0, mn = 0 we set

L(s, ω) = 2(2π)^{−(s+r+ \frac{m+n}{2})} Γ (s + r + \frac{m + n}{2}) .

The initial choice of μ is of course not uniquely determined. However if μ_{0} is one choice the only other choice is the character a → μ_{0}(\overline{a}). Thus the resulting local L-function is independent of the choice.

The only point to be checked is that the local L-function behaves properly under induction. We have to verify that if ρ is a representation of ℂ^{∗} = W_{ℂ/ℂ} in the class θ and

σ = Ind(W_{ℂ/ℝ}, W_{ℂ/ℂ}, ρ)

is in the class w then L(s, w) = L(s, θ). We may as well assume that ρ is irreducible and therefore one-dimensional. Let it correspond to the generalized character ν. If σ is irreducible we could choose the generalized character μ above to be ν and the equality of the two L-functions becomes a matter of definition. If σ is irreducible it breaks up into the sum of two one-dimensional representatives. It follows easily that ν(a) = ν(\overline{a}) for all a. Thus ν is of the form ν(a) = |a|^{r} and

L(s, θ) = 2(2π)^{−(s+r)} Γ(s + r).

If μ_{ℝ} = μ is the generalized character x → |x|^{r} of ℝ^{×} then ν = μ_{ℂ/ℝ} and, as we saw in chapter 10, the representation σ is equivalent to the direct sum of the one-dimensional representations corresponding to μ and to μ′ where μ′(x) = sgnxμ(x). Thus

L(s, ω) = \{π^{− \frac{1}{2} (s+r)} Γ (\frac{s + r}{2} )\}\{π^{− \frac{1}{2} (s+r+1)} Γ (\frac{s + r + 1}{2})\}.

The required result is thus a consequence of the familiar duplication formula

2^{2z−1} Γ(z) Γ(z + 1/2) = π^{1/2} Γ(2z).

If F is a global field and ω is an equivalence class of representations of the Weil group of F, we define as in the introduction, the global L-function to be

L(s, ω) = ∏_{𝔭} L(s, ω_{𝔭}).

I repreat that the product is taken over all primes, including those at infinity. It is not difficult to see that the product converges in a half-plane Re s > c. One need only verify it for ω irreducible. Choose a Galois extension K of F so that there is a representation σ of W_{K/F} in the class ω. The restriction of σ to C_{K} is equivalent to the direct sum of 1-dimensional representations corresponding to generalized characters μ^{(1)}, …, μ^{(r)} of C_{K}. For each i and j there is a σ in 𝔊(K/F) such that μ^{j}(a) ≡ μ^{i}(σ(a)). Then |μ_{i}^{−1}μ_{j}(a)| = |μ^{i}(a^{−1}σ(a))| = 1 because a^{−1}σ(a) belongs to the compact group of i idèle classes of norm 1. Let |μ′(a)| = |a|^{r}. Let ν_{F} be the generalized character a → |a|^{r} of C_{F}. Replacing σ by ν_{F}^{−1} ⊗ σ we replace L(s, w_{𝔭}) by L(s − r, w_{𝔭}) and μ^{(i)} by |μ^{(i)}|^{−1}μ^{(i)}. Thus we may as well suppose that all μ^{(i)} are ordinary characters. Since C_{K} is of finite index in W_{K/F} the eigenvalues of σ(w) will all have absolute 1 for any w in W_{K/F} and at any non-archimedean prime the local L-function will be of the form

∏_{i=1}^{s} \frac{1}{1 − α_{i} |π_{F_{𝔭}}|^{s}}

with s ≤ dim w and |α_{i}| = 1, 1 ≤ i ≤ s. The required result follows from the well-known fact that

∏^{1} \frac{1}{1 − |π_{F_{𝔭}}|^{s}}

converges from Re s > 1. This product is taken only over the non-archimedean primes.

Chapter Twenty-one.

Proof of the Functional Equation.

Choose a non-trivial character ψ_{F} of 𝔸_{F} /F. Before we can write down the factor appearing in the functional equation of the global L-function we have to verify that ε(s, ω_{γ}, ψ_{F_{γ}}) = 1 for all but a finite number of γ.

Let ω be realized as a representation σ of W_{K/F} and let the restriction of σ to C_{K} be equivalent to the direct sum of 1-dimensional representations corresponding to the generalized characters μ^{(1)}, …, μ^{(r)}. All but finitely many primes 𝔭 will satisfy the following conditions.

(i) 𝔭 is non-archimedean. (ii) n(ψ_{F_{𝔭}}) = 1. (iii) 𝔭 does not ramify in K.

(iv) m(μ_{𝔓}^{(i)}) = 0 for all 𝔓 dividing 𝔭 and all i.

Choose one such 𝔭 and let 𝔓 divide 𝔭. Corresponding to the map K/F ⟶ K_{𝔓}/F_{𝔭} is a map φ_{𝔭} : W_{K_{𝔓}/F_{𝔭}} ⟶ W_{K/F}. ω_{𝔭} is the class of σ_{𝔭} = σ ∘ φ_{𝔭}. The kernel of σ_{𝔭} contains U_{K_{𝔓}}. Since K_{𝔓}/F_{𝔭} is unramified the quotient of W_{K_{𝔓}/F_{𝔭}} by U_{K_{𝔓}} is abelian and σ_{𝔭} is the direct sum of one-dimensional representations. Let them correspond to the generalized characters ν_{𝔭}^{(1)}, …, ν_{𝔭}^{(r)} of C_{F_{𝔭}}. Since τ_{K_{𝔓}/F_{𝔭}} takes U_{K_{𝔓}} onto U_{F_{𝔭}} each of these characters is unramified. Thus

ε(s, ω_{𝔭}, ψ_{F_{𝔭}}) = ∏_{i=1}^{r} Δ (α_{F_{𝔭}}^{s− \frac{1}{2}} ν_{𝔭}^{(i)}, ψ_{F_{𝔭}}) = 1.

If ψ′_{F} is another non-trivial character of 𝔸_{F} /F there is a β in F^{∗} such that ψ′_{F} (x) ≡ ψ_{F} (βx). According to Lemma 5.1

ε(s, ω, ψ_{F_{𝔭}}) = α_{F_{𝔭}}^{s− \frac{1}{2}} (β)det ω_{𝔭}(β) ε(s, ω, ψ_{F_{𝔭}}).

Since

∏_{𝔭} α_{F_{𝔭}}(β)^{s− \frac{1}{2}} det ω_{𝔭}(β) = |β|^{s− \frac{1}{2}} det ω(β) = 1

the function

ε(s, ω) = ∏_{𝔭} ε(s, ω_{𝔭}, ψ_{F_{𝔭}})

is indeed independent of ψ_{F}.

We can infer from Tate’s thesis not only that L(s, ω) is meromorphic in the whole complex plane if ω is one-dimensional but also that it satisfies the functional equation

L(s, ω) = ε(s, ω) L(1 − s, ω̃)

if ω̃ is contragredient to ω. As is well-known, Lemma 2.2 then implies that L(s, ω) is meromorphic in the whole complex plane for any ω. In any case, Theorem B is true for one-dimensional ω and, granting this, we have to establish it in general.

First we need a lemma.

Lemma 21.1

Suppose F is a global field, K is a Galois extension of F, E is a field lying between F and K, χ is a generalized character of C_{E} and

σ = Ind(W_{K/F} , W_{K/E}, χ).

If ω is the class of σ and, for each prime 𝔮 of E, χ_{𝔮} is the restriction of χ to C_{E_{𝔮}} then for each prime 𝔭 of F

ε(s, ω_{𝔭}, ψ_{F_{𝔭}}) = ∏_{𝔮|𝔭}{ε(s, χ_{𝔮}, ψ_{E_{𝔮}/F_{𝔭}}) ρ (E_{𝔮}/F_{𝔭}, ψ_{F_{𝔭}})}.

Let 𝔓 be a prime of K dividing 𝔭. The first step is to find a set of representatives for the double cosets W_{K/E} w W_{K_{𝔓}/F_{𝔭}}. Since C_{K} ⊆ W_{K/E} is a normal subgroup of W_{K/F} we can factor out C_{K} and merely find a set of representatives for the double cosets

𝔊(K/E)σ 𝔊(K_{𝔓}/F_{𝔭}).

Let 𝔓_{1}, …, 𝔓_{r} be the primes of K dividing 𝔭 and let 𝔓_{1} divide 𝔮_{i} in E. 𝔊(K/F) is the disjoint union

⋃_{i=1}^{r} σ_{i}𝔊(K_{𝔓}/F_{𝔭})

where σ_{i}(𝔓) = 𝔓_{i}. If σ_{i} and σ_{j} belong to the same double coset 𝔮_{i} = 𝔮_{j}. Conversely, if 𝔮_{i} = 𝔮_{j} there is a ρ in 𝔊(K/E) such that ρ(𝔓_{i}) = 𝔓_{j}. Then ρσ_{i}(𝔓) = σ_{j}(𝔓) and

ρσ_{i} ∈ σ_{j} 𝔊(K_{𝔓}/F_{𝔭}).

Thus we may write 𝔊(K/F) as the disjoint union

⋃_{τ∈S} 𝔊(K/E)τ 𝔊(K_{𝔓}/F_{𝔭})

so that if 𝔓 divides 𝔮 in E the collection {τ(𝔮) | τ ∈ S} is the collection of distinct primes in E dividing 𝔭.

For each τ in S choose a representative w(τ) in W_{K/F}. For each τ in S the restriction of σ to W_{K_{𝔓}/F_{𝔭}} leaves invariant the space of functions f on the double coset W_{K/E}w(τ)W_{K_{𝔓}|F_{𝔭}} which satisfy f(vw) = χ(τ_{K/E}(v))f(w) for all v in W_{K/E}. The representation of W_{K_{𝔓}/F_{𝔭}} on this space is equivalent to

Ind(W_{K_{𝔓}/F_{𝔭}}, W_{K_{𝔓}/E_{𝔮^{τ}}^{τ}} , χ′_{𝔮^{τ}}

if E^{τ} = τ^{−1}(E) and

χ′_{𝔮^{τ}} (a) = χ(τ(a)).

Thus

ε(s, ω_{𝔮}, ψ_{F_{𝔭}}) = ∏_{τ∈S} ε(s, χ′_{𝔮^{τ}} , ψ_{E_{𝔮^{τ}}^{τ} /F_{𝔭}}) ρ(E_{𝔮^{τ}}^{τ} /F_{𝔭}, ψ_{F_{𝔭}})

which is of course equal to

∏_{τ∈S} ε(s, χ_{𝔮}, ψ_{E_{𝔮}/F_{𝔭}}) ρ(E_{𝔮}/F_{𝔭}, ψ_{F_{𝔭}}).

We set

ρ(E/F) = ∏_{𝔮} ∏_{𝔮|𝔭} ρ(E_{𝔮}/F_{𝔭}, ψ_{F_{𝔭}}).

The preceding discussion together with Lemma 5.1 shows that it does not depend on ψ_{F}. However that does not really matter since we are about to show that for any choice of ψ_{F} it is 1. Observe first of all that the previous lemma implies immediately that if ω is the class of

σ = Ind(W_{K/F} , W_{K/E}, χ)

then

ε(s, ω) = ε(s, χ) ρ(E/F).

Given an arbitrary class ω realizable as a representation of W_{K/F} we can find fields E_{1}, …, E_{r} lying between F and K, generalized characters χ_{E_{1}}, …, χ_{E_{r}}, and integers m_{1}, …, m_{r} such that

∑_{i=1}^{r} ⊕m_{i} Ind(W_{K/F} , W_{K/E_{i}}, χ_{E_{i}})

is in the class ω. Then

ε(s, ω) = ∏_{i=1}^{r} {ε(s, χ_{E_{i}})^{m_{i}} ρ(E_{i}/F)^{m_{i}}}.

On the other hand

L(s, ω) = ∏_{i=1}^{r} L(s, χ_{E_{i}})^{m_{i}}

and

L(s, ω̃) = ∏_{i=1}^{r} L(s, χ_{E_{i}}^{−1})^{m_{i}}.

Since

L(s, χ_{E_{i}}) = ε(s, χ_{E_{i}}) L(1 − s, χ_{E_{i}}^{−1})

we have

L(s, ω) = ∏_{i=1}^{r} ε(s, χ_{E_{i}})^{m_{i}} L(1 − s, ω̃)

because ω̃ contains

∑_{i=1}^{r} ⊕m_{i} Ind(W_{K/F} , W_{K/E_{i}}, χ_{E_{i}}^{−1}).

Consequently

∏_{i=1}^{r} ε(s, χ_{E_{i}})^{m_{i}}

depends only on ω and not on the particular way it is written as a sum of induced representations. Thus

∏_{i=1}^{r} ρ(E_{i}/F)^{m_{i}}

also depends only on ω. We call it H(ω). It is clear that to prove Theorem B we have to show that H(ω) = 1 for all ω or, what is the same, that ρ(E/F) = 1 for all E and F.

Suppose F ⊆ E ⊆ E′. Denote the primes of F by 𝔭, those of E by 𝔮, and those of E′ by 𝔮′. Then

ρ(E′/F) = ∏_{𝔭} ∏_{𝔮′|𝔭} ρ(E′_{𝔮′}/F_{𝔭}, ψ_{F_{𝔭}}).

Apply Lemma 4.5 to see that the right side equals

∏_{𝔭} ∏_{𝔮|𝔭} ∏_{𝔮′|𝔮} {ρ(E′_{𝔮′}/E_{𝔮}, ψ_{F_{𝔮}/F_{𝔭}}) ρ(E_{𝔮}/F_{𝔭}, ψ_{F_{𝔭}})^{[E′_{𝔮}:E_{𝔮}]}}.

Since

∑_{𝔮′|𝔮} [E′_{𝔮′} : E_{𝔮}] = [E′ : E]

this may be written as

{∏_{𝔮} ∏_{𝔮′|𝔮} ρ(E′_{𝔮′}/E_{𝔮}, ψ_{F_{𝔮}/F_{𝔭}})} {∏_{𝔭} ∏_{𝔮|𝔭} ρ(E_{𝔮}/F_{𝔭}, ψ_{F_{𝔭}})}^{[E′:E]}

which is of course

ρ(E′/E) ρ(E/F)^{[E′:E]}.  (20.1)

Suppose E/F is an abelian extension and ω is the class of the representation of W_{E/F} induced from the trivial representation of C_{E} = W_{E/E}. Then H(ω) = ρ(E/F). On the other hand, ω is the direct sum of [E : F] one-dimensional representations; so H(ω) = ρ(F/F)^{[E:F]} = 1. It follows immediately not only that ρ(E/F) = 1 if E/F is abelian but also that ρ(E/F) = 1 if E can be obtained from F by a succession of abelian extensions. In particular if F ⊆ E ⊆ L and L/F is nilpotent ρ(E/F) = 1.

Observe that (20.1) together with Lemma 2.2 and the transitivity of induction imply that if ω is the class of

σ = Ind(W_{K/F} , W_{K/E}, ρ)

and θ is the class of ρ then

H(ω) = H(θ) ρ(E/F)^{dim θ}.

To complete the proof we will show that H(ω_{1} ⊗ ω_{2}) = H(ω_{2})^{dim ω_{1}} for all ω_{1} and ω_{2}. Taking ω_{2} = 1 we find H(ω_{1}) = 1. It is enough to prove the equality when ω_{1} and ω_{2} are both realizable as representations of W_{K/F} and there is a field E lying between E and K with 𝔊(K/E) nilpotent and a generalized character χ_{E} such that ω_{2} is the class of

Ind(W_{K/F} , W_{K/E}, χ_{E}).

Then H(ω_{2}) = ρ(E/F). If ρ is a representation in the restriction of ω_{1} to W_{K/E} then, by Lemma 2.3, ω_{1} ⊗ ω_{2} is the class of

Ind(W_{K/F} , W_{K/E}, ρ ⊗ χ_{E}).

Let θ be the class of ρ ⊗ χ_{E}. H(θ) is of the form

∏_{i=1}^{r} ρ(E_{i}/E)^{m_{i}}

where E ⊆ E_{i} ⊆ K and is therefore 1. Thus

H(ω_{1} ⊗ ω_{2}) = H(θ)ρ(E/F)^{dim θ} = ρ(E/F)^{dim ω_{1}}.

as required.

Chapter Twenty-two.

Appendix.

There is clearly not much to be said about the functions ε(s, ω, ψ_{F}) when F is archimedean. However for non-archimedean F their properties are more obscure. In this appendix we shall describe and prove some properties which were not needed in the proofs of the main theorems and so found no place in the main body of the paper but which will be used elsewhere.

The first step is to define the Artin conductor of ω. We follow a well-trodden path. If K is a finite Galois extension of the local field F then W_{K/F}^{0} contains U_{K} as a subgroup of finite index and is therefore compact. It is, in fact, a maximal compact subgroup of W_{K/F}. Choose that Haar measure dw on W_{K/F} which assigns the measure 1 to W_{K/F}^{0} . If f is a locally constant function on W_{K/F} and u is a non-negative real number set

f̂(u) = {∫_{W_{K/F}^{u}} dw}^{−1} ∫_{W_{K/F}^{u}} {f(1) − f(w)}dw.

Since W_{K/F}^{u} is an open subgroup of W_{K/F} it is meaningful to restrict dw to it. f̂(u) is bounded, continuous from the left, and 0 for u sufficiently large. Since W_{K/F}^{u} = W_{K/F}^{0} for 1 < u ≤ 0 we have f̂(u) = f̂(0) for such u. The integral

∫_{−1}^{∞} f̂(u)du

is well-defined.

There are some simple lemmas to be verified.

Lemma 22.1

Suppose F ⊆ K ⊆ L and L/F is also a Galois extension. Define g on W_{L/F} by g(w) = f_{(τ_{L/F} ,K/F} (w)). Then ĝ(u) = f̂(u) for all u.

This is immediate because by Lemma 6.16, τ_{L/F,K/F} maps W_{L/F}^{u} onto W_{K/F}^{u} for every u.

When we want to make the roles of K and F explicit we write f̂(u) = f̂_{K/F} (u).

Lemma 22.2

Suppose F ⊆ E ⊆ K and g is a function on W_{K/E} satisfying g(wzw^{−1}) = g(z) for all z and w in W_{K/E}. Regard W_{K/E} as a subgroup of W_{K/F} and set

f(w) = ∑_{z∈W_{K/E}\W_{K/F}} g(z^{−1}wz).

If N_{E/F} π_{E} is a unit times π_{F}^{f_{E/F}} and 𝔓_{E}^{δ_{E/F}} is the different of E/F

∫_{−1}^{∞} f̂_{K/F} (u)du = f_{E/F} ∫_{−1}^{∞} ĝ_{K/E}(u)du + f_{E/F} δ_{E/F} g(1).

Let dw_{K/F} be the normalized Haar measure on W_{K/F} and let dw_{K/E} be the normalized Haar measure on W_{K/E}. On W_{K/E}

dw_{K/E} = [W_{K/F}^{0} : W_{K/E}^{0}]dw_{K/F} .

Suppose at first that g(1) = 0. Denote also by g the function on W_{K/F} which equals the given g on W_{K/E} but is 0 outside of W_{K/E}. Then

f̂_{K/F} (u) = [W_{K/F} : W_{K/E}] ĝ_{K/F} (u).

Since W_{K/F}^{u} ∩ W_{K/E} = W_{K/E}^{v} if v = ψ_{E/F} (u)

ĝ_{K/F} (u) = −[W_{K/F}^{0} : W_{K/F}^{u}] ∫_{W_{K/F}^{u}} g(w)dw_{K/F}

= −\frac{[W_{K/F}^{0} : W_{K/F}^{u}]}{[W_{K/F}^{0} : W_{K/E}^{0}]} ∫_{W_{K/E}^{v}} g(w)dw_{K/E}

= \frac{1}{[W_{K/F}^{0} : W_{K/E}^{0}]} \frac{[W_{K/F}^{0} : W_{K/F}^{u}]}{[W_{K/E}^{0} : W_{K/E}^{v}]} ĝ_{K/E}(v).

Recall that

f_{E/F} = \frac{[W_{K/F} : W_{K/E}]}{[W_{K/F}^{0} : W_{K/E}^{0}]}.

Moreover

\frac{[W_{K/F}^{0} : W_{K/F}^{u}]}{[W_{K/E}^{0} : W_{K/E}^{v}]} = \frac{[W_{K/F}^{0} : W_{K/F}^{u} U_{K}^{0}]}{[W_{K/E}^{0} : W_{K/E}^{v} U_{K}^{0}]} · \frac{[U_{K}^{0} : U_{K}^{0} ∩ W_{K/F}^{u}]}{[U_{K}^{0} : U_{K}^{0} ∩ W_{K/E}^{v}]}

and U_{K}^{0} ∩ W_{K/F}^{u} = U_{K}^{0} ∩ W_{K/E}^{v}. By Lemma 6.11 the first term in this product is equal to

\frac{[G^{0} : G^{u}]}{[H^{0} : H^{v}]}

if G = 𝔊(K/F) and H = 𝔊(K/E). But

[G^{0} : G^{u}] = ψ′_{K/F} (u)

and

[H^{0} : H^{v}] = ψ′_{K/E}(v)

while

ψ′_{K/F} (u) = ψ′_{K/E}(v)ψ′_{E/F} (u).

Thus

∫_{−1}^{∞} f̂_{K/F} (u)du = f_{E/F} ∫_{−1}^{∞} ĝ_{K/E}(ψ_{E/F} (u)) ψ′_{E/F} (u)du
= f_{E/F} ∫_{−1}^{∞} ĝ_{K/E}(v)dv.

To complete the proof of the lemma, we have to show that if g(w) = 1 so that ĝ_{K/E}(u) ≡ 0 then

∫_{−1}^{∞} f̂_{K/F} (u)du = f_{E/F} δ_{E/F} .

In this case

f̂_{K/F} (u) = [G : H] − \frac{[G : H]}{[G^{0} : H^{0}]} \frac{[G^{0} : G^{u}]}{[H^{0} : H^{v}]}

if v = ψ_{E/F} (u). After some simple rearranging this becomes

\frac{[G : H]}{[G^{u} : 1]} \{[G^{u} : 1] − [H^{v} : 1]\} = \frac{[G : H]}{[G^{u} : 1]} \{([G^{u} : 1] − 1) − ([H^{v} : 1] − 1)\}.

The factor

\frac{[G : H]}{[G^{u} : 1]} = \frac{[G : G^{0}]}{[H : 1]} ψ′_{K/F} (u)

and, from paragraph IV.2 of [ ],

∫_{1}^{∞}([G^{u} : 1] − 1)ψ′_{K/F} (u)du = ∫_{−1}^{∞}([G_{x} : 1] − 1)dx = δ_{K/F}

while

∫_{−1}^{∞}([H^{v} : 1] − 1) ψ′_{K/F} (u)du = ∫_{−1}^{∞}([H^{v} : 1] − 1) ψ′_{K/E}(v)dv = δ_{K/E}.

Thus

∫_{−1}^{∞} f̂_{K/F} (u)du = \frac{[G : G^{0}]}{[H : 1]} (δ_{K/F} − δ_{K/E}) = f_{E/F} δ_{E/F}

because

δ_{K/F} = δ_{K/E} + [H_{0} : 1] δ_{E/F} .

Suppose ω is an equivalence class of representations of the Weil group of F and σ is a representation of W_{K/F} in the class of ω. Let f_{σ} be the character of σ. It follows from Lemma 22.1 that the value of

∫_{−1}^{∞} f̂_{σ}(u)du

depends only on ω and not on σ. We call it the order of ω and denote it by m(ω). Since f̂_{σ}(u) is clearly non-negative for all u and vanishes identically if and only if W_{K/F}^{0} is contained in the kernel of σ, the order m(ω) is always non-negative and equals zero if and only if the kernel of each realization σ of ω contains W_{K/F}^{0} .

Lemma 22.3

(a) If ω = ω_{1} ⊕ ω_{2} then m(ω) = m(ω_{1}) + m(ω_{2}).

(b) If

ω = Ind(W_{K/F} , W_{K/E}, ν)

then

m(ω) = f_{E/F} m(ν) = f_{E/F} δ_{E/F} dim ν.

(c) m(ω) is a non-negative integer.

The first property is immediate. The second is a consequence of Lemma 22.2. To verify the third we merely have to show that m(ω) is integral. If ω = μ ⊕ ν and the assertion is true for any two of μ, ν and ω it is true for the third. This observation, together with part (b) and Lemma 2.2, shows that it is enough to verify (c) when ω is the one-dimensional class corresponding to a generalized character χ_{F} of C_{F}. To do this we show that m(ω) = m(χ_{F}).

If f(a) = χ_{F} (a) for a in C_{F} = W_{F/F} then f̂(u) = f̂(m) for m − 1 < u ≤ m and

f̂(m) = [U_{F}^{0} : U_{F}^{m}] ∫_{U_{F}^{m}}{1 − χ_{F} (a)}da.

The right side is 1 if m < m(χ_{F}) and 0 if m ≥ m(χ_{F}). Thus

m(ω) = ∫_{−1}^{m(χ_{F})−1} du = m(χ_{F}).

The function ω ⟶ m(ω) is characterized by (a) and (b) together with the fact that m(ω) = m(χ_{F}) if ω is the class of χ_{F}.

Lemma 22.4

If ω is an equivalence class of representations of the Weil group of the non-archimedean local field F and ψ_{F} is a non-trivial additive character of F set m′(ω) = m(ω) + n(ψ_{F}) dim ω. There is a non-zero complex constant a(ω) such that, as a function of s,

ε(s, ω, ψ_{F}) = a(ω) |π_{F} |^{m′(ω)s}.

If ω = μ⊕ν and the lemma is true for any two of μ, ν, and ω, it is true for the third. Applying Lemma 2.2 we see that it is enough to verify it when ω contains a representation

Ind(W_{K/F} , W_{K/E}, χ_{E}).

Then

ε(s, ω, ψ_{F}) = Δ(α_{E}^{s− \frac{1}{2}} χ_{E}, ψ_{E/F}) ρ(E/F, ψ_{F}).

Clearly

Δ(α_{E}^{s− \frac{1}{2}} χ_{E}, ψ_{E/F}) = α_{E}^{s− \frac{1}{2}} (π_{E}^{m(χ_{E})+δ_{E/F}} π_{F}^{n(ψ_{F})}) Δ(χ_{E}, ψ_{E/F}).

But

α_{E} (π_{E}^{m(χ_{E})+δ_{E/F}} π_{F}^{n(ψ_{F})}) = α_{F} (N_{E/F} (π_{E}^{m(χ_{E})+δ_{E/F}} π_{F}^{n(ψ_{F})}))

and the argument on the right is the product of a unit and

π_{F}^{f_{E/F} (m(χ_{E})+δ_{E/F})+n(ψ_{F}) dim ω} = π_{F}^{m′(ω)}.

The lemma follows.

The next lemma is rather technical and to prove it we will have to use the notations and results of paragraphs 8 and 9.

Lemma 22.5

Let ω be an equivalence class of representations of the Weil group of the non-archimedean local field F and m_{1} a positive integer. There is a positive integer m_{2} such that if χ_{F} and μ_{1}, …, μ_{r} with r = dim ω, are generalized characters of C_{F} and m(χ_{F}) ≥ m_{2}, m(μ_{i}) ≤ m_{1}, 1 ≤ i ≤ r, while

∏_{i=1}^{r} μ_{i} = detω

then for any non-trivial additive character ψ_{F}

ε(s, χ_{F} ⊗ σ, ψ_{F}) = ∏_{i=1}^{r} ε(s, μ_{i}χ_{F} , ψ_{F}).

Choose, as a start, m_{2} ≥ 2m_{1}+1. If μ_{F} is a generalized character of C_{F} and m(μ_{F}) ≤ m_{1} while m(χ_{F}) ≥ m_{2} then m(μ_{F} χ_{F}) = m(χ_{F}) = m. Let n = n(ψ_{F}) and choose γ so that O_{F} γ = 𝔓_{F}^{m+n}. If β = β(χ_{F}) we may choose β(μ_{F} χ_{F}) = β. Appealing to Lemmas 8.1 and 9.4 we see that

ε(s, μ_{F} χ_{F} , ψ_{F}) = Δ(α_{F}^{s− \frac{1}{2}} μ_{F} χ_{F} , ψ_{F})

= (α_{F}^{s− \frac{1}{2}} μ_{F}) (\frac{γ}{β} ) Δ(χ_{F} , ψ_{F}).

In particular

∏_{i=1}^{r} ε(s, μ_{i}χ_{F} , ψ_{F}) = α_{F}^{r(s− \frac{1}{2} )} (\frac{γ}{β} ) detω (\frac{γ}{β} ) \{Δ(χ_{F} , ψ_{F})\}^{r}.

If ω = μ ⊕ ν then
ε(s, χ_{F} ⊗ ω, ψ_{F}) = ε(s, χ_{F} ⊗ μ, ψ_{F}) ε(s, χ_{F} ⊗ ν, ψ_{F})

and all three terms are different from zero. Thus if the lemma is true for two of μ, ν and ω it is true for the third. Using Lemma 2.2 once again, we see that it is enough to prove the lemma when there is an intermediate field E and a generalized character μ_{E} of C_{E} such that ω is the class of

Ind(W_{K/F} , W_{K/E}, μ_{E}).

Then χ_{F} ⊗ ω is the class of

Ind(W_{K/F} , W_{K/E}, μ_{E} χ_{E/F})

and

ε(s, χ_{F} ⊗ ω, ψ_{F}) = Δ(α_{E}^{s− \frac{1}{2}} μ_{E}χ_{E/F} , ψ_{E/F}) ρ(E/F, ψ_{F}).

There are two simple lemmas which we need before we can proceed further and we digress to prove them.

Lemma 22.6

Let E be a separable extension of F. If m is sufficiently large

ψ_{E/F} (m − 1) + 1 = m e_{E/F} − δ_{E/F}

if e_{E/F} is the index of ramification of F in E.

Suppose F ⊆ E ⊆ K where K/F is Galois and the assertion is true for K/F and K/E. Subtracting 1 from both sides of the equation, applying ψ_{K/E}, and then adding 1, we obtain the equivalent equation

ψ_{K/F} (m − 1) + 1 = ψ_{K/E}(m e_{E/F} − δ_{E/F} − 1) + 1.

By assumption, the left side equals

m e_{K/F} − δ_{K/F}

and the right side equals

(m e_{E/F} − δ_{E/F}) e_{K/E} − δ_{K/E}.

Since e_{K/F} = e_{K/E}e_{E/F} and δ_{K/F} = δ_{K/E} + e_{K/E}δ_{E/F} these two expressions are equal and we have only to prove the lemma for Galois extensions.

Suppose F ⊆ K ⊆ L and L/F and K/F are Galois. Suppose also that the lemma is true for L/K and K/F. Then

ψ_{L/F} (m − 1) + 1 = ψ_{L/F} (ψ_{K/F} (m − 1)) + 1
= ψ_{L/F} (m e_{K/F} − δ_{K/F} − 1) + 1
= (m e_{K/F} − δ_{K/F}) e_{L/K} − δ_{L/K}
= m e_{L/F} − δ_{L/F}

as before. Thus, if we use induction, we need only verify the lemma directly for a Galois extension K/F of prime degree.

We apply Lemma 6.3. If K/F is unramified, e_{K/F} = 1 and δ_{K/F} = 0 while ψ_{K/F} (m − 1) = m − 1; so the relation follows. If K/F is ramified there is an integer t such that δ_{K/F} = ([K : F] − 1)(t + 1) while ψ_{K/F} (m − 1) + 1 = [K : F]m − ([K : F] − 1)(t + 1) for m − 1 ≥ t. Since e_{K/F} = [K : F] the relation follows again.

If n = n(ψ_{F}) then
n′ = n(ψ_{E/F}) = n e_{E/F} + δ_{E/F} .

Thus if m is sufficiently large and m′ = ψ_{E/F} (m − 1) + 1

m′ + n′ = (m + n) e_{E/F}

and if O_{F} γ = 𝔓_{F}^{m+n} then O_{E}γ = 𝔓_{E}^{m′+n′}. We define

P_{E/F}^{∗} (x) = P_{E/F}^{∗} (x; γ, γ)

as in paragraph 8.

Lemma 22.7

If m_{1} is a given positive integer then for m sufficiently large

P_{E/F}^{∗} (x) ≡ x (mod 𝔓_{E}^{m_{1}}).

As in paragraph 8, let d be the integral part of \frac{m}{2}, d′ the integral part of \frac{m′}{2} , and let m = 2d+ε, m′ = 2d′ +ε′. P_{E/F}^{∗} (x) depends only on the residue of x modulo 𝔓_{F}^{d} and is only determined modulo 𝔓_{E}^{d′}. Recall that if

P_{E/F} (y) = N_{E/F} (1 + y) − 1

for y in P_{E}^{d′+ε′} then

ψ_{E/F} (\frac{P_{E/F}^{∗} (x)y}{γ} ) = ψ_{F} (\frac{x P_{E/F} (y)}{γ}) .

To show that P_{E/F}^{∗} (x) ≡ x (mod 𝔓_{E}^{m_{1}}) when m is sufficiently large, we have to show that

ψ_{F} (\frac{x P_{E/F} (y)}{γ}) = ψ_{E/F} (\frac{xy}{γ} )

for y in 𝔓_{E}^{m′−m_{1}}. To do this we show that

P_{E/F} (y) ≡ S_{E/F} (y) (mod 𝔓_{F}^{m})

when m is sufficiently large and y is in 𝔓_{E}^{m′−m_{1}}.

To put it another way, we have to show that if K/F is any Galois extension the assertion is true for all intermediate fields E. For this we use induction on [K : F] together with Lemma 3.3. There are three facts to verify:

(i) If E/F is a Galois extension of prime degree then

P_{E/F} (y) ≡ S_{E/F} (y) (mod 𝔓_{F}^{m})

when m is sufficiently large and y is in 𝔓_{E}^{m′−m_{1}}. (ii) Suppose F ⊆ E ⊆ K and K/F is Galois. Let G = 𝔊(K/F) and let E be the fixed field of

H. Suppose H ≠ {1} and G = HC where H ∩ C = {1} and C is a non-trivial abelian normal subgroup of G which is contained in every other non-trivial normal subgroup. If the induction assumption is valid

P_{E/F} (y) ≡ S_{E/F} (y) (mod 𝔓_{F}^{m})

when m is sufficiently large and y is in 𝔓_{E}^{m′−m_{1}}. (iii) Suppose F ⊆ E ⊆ E′ ⊆ K and m′′ = ψ_{E′/F} (m − 1) + 1. If, for any choice of m_{1},

P_{E/F} (y) ≡ S_{E/F} (y) (mod 𝔓_{F}^{m})

when m is sufficiently large and y is in 𝔓_{E}^{m′−m_{1}} and, for any choice of m′_{1},

P_{E′/E}(y) ≡ S_{E′/E}(y) (mod 𝔓_{E}^{m′})

when m, or m′, is sufficiently large and y is in 𝔓_{E′}^{m′′−m′_{1}} then, for any choice of m′_{1},

P_{E′/F} (y) ≡ S_{E′/F} (y) (mod 𝔓_{F}^{m})

if m is sufficiently large and y is in 𝔓_{E′}^{m′′−m′_{1}}.

We first verify (i) for E/F unramified. By paragraph V.2 of [12]

P_{E/F} (y) = N_{E/F} (1 + y) − 1 ≡ S_{E/F} (y) (mod 𝔓_{F}^{m})

if y belongs to 𝔓_{E}^{d′+ε′}. In this case m = m′ and we take m > 2m_{1} so that m′ − m_{1} > d′ + ε′. If E/F is ramified and of degree ℓ we again choose m sufficiently large that m′ > 2m_{1}. If m > t

\frac{2(m′ − m_{1}) + (ℓ − 1)(t + 1)}{ℓ} ≥ \frac{m′ + (ℓ − 1)(t + 1)}{ℓ} = m

so that by Chapter V of [12],

P_{E/F} (y) ≡ S_{E/F} (y) + N_{E/F} (y) (mod 𝔓_{F}^{m})

if y belongs to 𝔓_{E}^{m′−m_{1}}. t of course has its usual meaning. Since N_{E/F} (y) belongs to 𝔓_{F}^{m′−m_{1}} all we have to do is arrange that m′ − m_{1} ≥ m. Since

m′ − m_{1} = ℓm − (ℓ − 1) (t + 1) − m_{1}

and ℓ ≥ 2, this can certainly be done by choosing m sufficiently large.

To verify the second fact let L be the fixed field of C. We can assume that the required assertion is true for the extension K/L. Let ℓ = ψ_{L/F} (m − 1) + 1 and ℓ′ = ψ_{K/F} (m − 1) + 1. If m is sufficiently large and H_{0} is the inertial group of H

ℓ = [H_{0} : 1]m − ([H_{0} : 1] − 1)

and

ℓ′ = [H_{0} : 1]m′ − ([H_{0} : 1] − 1).

Thus

𝔓_{L}^{ℓ} ∩ F = 𝔓_{F}^{m}

and if ℓ_{1} = [H_{0} : 1]m_{1} then

𝔓_{K}^{ℓ′−ℓ_{1}} ∩ E = 𝔓_{E}^{m′−m_{1}}.

If m and therefore ℓ is sufficiently large

P_{K/L}(y) = N_{K/L}(1 + y) − 1 ≡ S_{K/L}(y) (mod 𝔓_{L}^{ℓ})

if y belongs to 𝔓_{K}^{ℓ′−ℓ_{1}}. Thus if y belongs to 𝔓_{E}^{m′−m_{1}}

P_{E/F} (y) = P_{K/L}(y) ≡ S_{K/L}(y) = S_{E/F} (y) (mod 𝔓_{F}^{m}).

Chapter 22 234

To verify the third fact we choose, once m′_{1} is given, m_{1} so that

S_{E′/E} (𝔓_{E′}^{−δ_{E′/E}−m′_{1}}) = 𝔓_{E}^{−m_{1}}.

If m is sufficiently large

m′′ = m′ e_{E′/E} − δ_{E′/E}

and if y belongs to 𝔓_{E′}^{m′′−m′_{1}}

S_{E′/E}(y) ∈ 𝔓_{E}^{m′−m_{1}}.

Taking in even larger if necessary, we have

P_{E′/F} (y) ≡ P_{E/F} (P_{E′/E}(y))
≡ P_{E/F} (S_{E′/E}(y))
≡ S_{E/F} (S_{E′/E}(y))
≡ S_{E′/F} (y) (mod 𝔓_{F}^{m}).

Returning to the proof of Lemma 22.5, we choose

β′ = β(χ_{E/F}) = P_{E/F}^{∗} (β).

If m(χ_{F}) and therefore m(χ_{E/F}) is sufficiently large,

Δ(α_{E}^{s− \frac{1}{2}} μ_{E} χ_{E/F} , ψ_{E/F}) = α_{E}^{s− \frac{1}{2}} (\frac{γ}{β′} ) μ_{E} (\frac{γ}{β′} ) Δ(χ_{E/F} , ψ_{E/F}).

Both β and β′ are units and therefore

α_{E}^{s− \frac{1}{2}} (\frac{γ}{β′} ) = α_{E}^{r− \frac{1}{2}} (\frac{γ}{β} ) = α_{F}^{s− \frac{1}{2}} (N_{E/F} (\frac{γ}{β} )) = α_{F}^{r(s− \frac{1}{2} )} (\frac{γ}{β} ) .

If m(χ_{F}) is sufficiently large

β′ ≡ β (mod 𝔓_{E}^{m(μ_{E})})

and μ_{E}(β′) = μ_{E}(β). In paragraph 5 we saw that

det ω (\frac{γ}{β} ) = μ_{E} (\frac{γ}{β} ) det ι_{E/F} (\frac{γ}{β} ) ,

if ι_{E/F} is the representation of W_{K/F} induced from the trivial representation of W_{K/E}. We are reduced to showing that

det ι_{E/F} (\frac{γ}{β} ) \{Δ(χ_{F} , ψ_{F})\}^{r} = Δ(χ_{E/F} , ψ_{E/F}) ρ(E/F, ψ_{F})  (22.1)

if m(χ_{F}) is sufficiently large. Of course r = [E : F].

What we do is show that for each Galois extension K/F the relation (22.1) is true for all fields E lying between K and F. For this we use induction on [K : F]. Let G = 𝔊(K/F) and let C be a non-trivial abelian normal subgroup of G. Let L be the fixed field of C. We saw in Chapter 13 that there are fields F_{1}, …, F_{s} lying between F and L and generalized characters μ_{1}, …, μ_{s} of C_{F_{1}}, …, C_{F_{s}} respectively such that

ι_{E/F} ≃ ⊕_{i=1}^{s}Ind(W_{K/F} , W_{K/F_{i}}, μ_{i}).

Then

χ_{F} ⊗ ι_{E/F} ≃ ⊕_{i=1}^{s}Ind(W_{K/F} , W_{K/E_{i}}, μ_{i} χ_{F_{i}/F})

and by Theorem 2.1, the Main Theorem, the right side of (22.1) is equal to

∏_{i=1}^{s} Δ(μ_{i} χ_{F_{i}/F} , ψ_{F_{i}/F}) ρ(F_{i}/F, ψ_{F}).

We just saw that if m(χ_{F}) is sufficiently large this is equal to

\{∏_{i=1}^{s} μ_{i} (\frac{γ}{β} )\} \{∏_{i=1}^{s} Δ(χ_{F_{i}/F} , ψ_{F_{i}/F}) ρ(F_{i}/F, ψ_{F})\}.

Since

∑_{i=1}^{s}[F_{i} : F] = [E : F]

we see upon applying the induction assumption to L/F that this equals

\{∏_{i=1}^{s} μ_{i} (\frac{γ}{β} ) det ι_{F_{i}/F} (\frac{γ}{β} )\} \{Δ(χ_{F} , ψ_{F})\}^{r}.

We complete the proof of (22.1) by appealing to Chapter 5 to see that

det ι_{E/F} = ∏_{i=1}^{s} μ_{i}det ι_{F_{i}/F} .

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