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Hilbert's Theorem 90

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Idea

A theorem in Galois cohomology due to David Hilbert.

Statement

There are actually two versions of Hilbert’s theorem 90, one multiplicative and the other additive. We begin with the multiplicative version.

Theorem

(Hilbert) Suppose KK be a finite Galois extension of a field kk, with a cyclic Galois group G=gG = \langle g \rangle of order nn. Regard the multiplicative group K *K^\ast as a GG-module. Then the group cohomology of GG with coefficients in K *K^\ast – the Galois cohomology – satisfies

H 1(G,K *)=0. H^1(G, K^\ast) = 0 \,.

Before embarking on the proof, we recall from the article projective resolution that if G=C nG = C_n is a finite cyclic group of order nn, then there is a projective resolution of \mathbb{Z} as trivial GG-module:

NGDGNGDG0\ldots \stackrel{N}{\to} \mathbb{Z}G \stackrel{D}{\to} \mathbb{Z}G \stackrel{N}{\to} \mathbb{Z}G \stackrel{D}{\to} \mathbb{Z}G \to \mathbb{Z} \to 0

where the map G\mathbb{Z}G \to \mathbb{Z} is induced from the trivial group homomorphism G1G \to 1 (hence is the map that forms the sum of all coefficients of all group elements), and where DD, NN are multiplication by special elements in G\mathbb{Z}G, also denoted DD, NN:

Dg1,D \coloneqq g - 1,
\,
N1+g+g 2++g k1N \coloneqq 1 + g + g^2 + \ldots + g^{k-1}

The calculations in the proof that follows implicitly refer to this resolution as a means to defining H n(G,A)H^n(G, A) (in the case A=K *A = K^\ast), by taking cohomology of the induced cochain complex

0hom G(,A)hom G(G,A)hom G(G,A)0 \to \hom_{\mathbb{Z}G}(\mathbb{Z}, A) \to \hom_{\mathbb{Z}G}(\mathbb{Z}G, A) \to \hom_{\mathbb{Z}G}(\mathbb{Z}G, A) \to \ldots
Proof

Let σG\sigma \in \mathbb{Z}G be an element of the group algebra, and denote the action of σ\sigma on an element βK\beta \in K by exponential notation β σ\beta^\sigma. The action of the element NGN \in \mathbb{Z}G is

β N=β 1+g++g n1=ββ gβ g n1\beta^N = \beta^{1 + g + \ldots + g^{n-1}} = \beta \cdot \beta^g \cdot \ldots \beta^{g^{n-1}}

which is precisely the norm N(β)N(\beta). We are to show that if N(β)=1N(\beta) = 1, then there exists αK\alpha \in K such that β=α/g(α)\beta = \alpha/g(\alpha).

By lemma 1 below, the homomorphisms 1,g,,g n1:K *K *1, g, \ldots, g^{n-1}: K^\ast \to K^\ast are, when considered as elements in a vector space of KK-valued functions, KK-linearly independent. It follows in particular that

1+βg+β 1+gg 2++β 1+g++g n2g n11 + \beta g + \beta^{1+g}g^2 + \ldots + \beta^{1 + g + \ldots + g^{n-2}}g^{n-1}

is not identically zero, and therefore there exists θK *\theta \in K^\ast such that the element

α=θ+βθ g+β 1+gθ g 2++β 1+g++g n2θ g n1\alpha = \theta + \beta \theta^g + \beta^{1+g}\theta^{g^2} + \ldots + \beta^{1 + g + \ldots + g^{n-2}}\theta^{g^{n-1}}

is non-zero. Using the fact that N(β)=1N(\beta) = 1, one may easily calculate that βα g=α\beta \alpha^g = \alpha, as was to be shown.

Now we give the additive version of Hilbert’s theorem 90:

Theorem

Under the same hypotheses given in 1, and regarding the additive group KK as a GG-module, we have

H 1(G,K)=0. H^1(G, K) = 0 \,.

The trace of an element αK\alpha \in K is defined by

Tr(α)Nα=α+g(α)++g n1(α).Tr(\alpha) \coloneqq N \cdot \alpha = \alpha + g(\alpha) + \ldots + g^{n-1}(\alpha).

We want to show that if Tr(β)=0Tr(\beta) = 0, then there exists αK\alpha \in K such that β=αg(α)\beta = \alpha - g(\alpha). By the theorem on linear independence of characters (following section), there exists θ\theta such that Tr(θ)0Tr(\theta) \neq 0; notice Tr(θ)Tr(\theta) belongs to the ground field kk since gN=Ng \cdot N = N. Put

α1Tr(θ)(βg(θ)+(β+g(β))g 2(θ)++(β+g(β)++g n2(β))g n1(θ).\alpha \coloneqq \frac1{Tr(\theta)}(\beta g(\theta) + (\beta + g(\beta))g^2(\theta) + \ldots + (\beta + g(\beta) + \ldots + g^{n-2}(\beta))g^{n-1}(\theta).

One may then calculate that

αg(α) = 1Tr(θ)(βg(θ)+βg 2(θ)++βg n1(θ)(g(β)++g n1(β))θ = 1Tr(θ)(βg(θ)+βg 2(θ)++βg n1(θ)+βθ) = β\array{ \alpha - g(\alpha) & = & \frac1{Tr(\theta)}(\beta g(\theta) + \beta g^2(\theta) + \ldots + \beta g^{n-1}(\theta) - (g(\beta) + \ldots + g^{n-1}(\beta))\theta \\ & = & \frac1{Tr(\theta)}(\beta g(\theta) + \beta g^2(\theta) + \ldots + \beta g^{n-1}(\theta) + \beta \theta) \\ & = & \beta }

where in the second line we used Tr(β)=0Tr(\beta) = 0.

Independence of characters

The next result may be thought of as establishing “independence of characters” (where “characters” are valued in the multiplicative group of a field):

Lemma

Let KK be a field, let GG be a monoid, and let χ 1,,χ n:GK *\chi_1, \ldots, \chi_n \colon G \to K^\ast be distinct monoid homomorphisms. Then the functions χ i\chi_i, considered as functions valued in KK, are KK-linearly independent.

Proof

A single χ:GK *\chi \colon G \to K^\ast obviously forms a linearly independent set. Now suppose we have an equation

(1)a 1χ 1++a nχ n=0a_1 \chi_1 + \ldots + a_n \chi_n = 0

where a iKa_i \in K, and assume nn is as small as possible. In particular, no a ia_i is equal to 00, and n2n \geq 2. Choose gGg \in G such that χ 1(g)χ 2(g)\chi_1(g) \neq \chi_2(g). Then for all hGh \in G we have

a 1χ 1(gh)++a nχ n(gh)=0a_1 \chi_1(g h) + \ldots + a_n \chi_n(g h) = 0

so that

(2)a 1χ 1(g)χ 1++a nχ n(g)χ n=0.a_1 \chi_1(g) \chi_1 + \ldots + a_n \chi_n(g)\chi_n = 0.

Dividing equation 2 by χ 1(g)\chi_1(g) and subtracting from it equation 1, the first term cancels, and we are left with a shorter relation

(a 2χ 2(g)χ 1(g)a 2)χ 2+=0(a_2\frac{\chi_2(g)}{\chi_1(g)} - a_2)\chi_2 + \ldots = 0

which is a contradiction.

A corollary of this result is an important result in its own right, the normal basis heorem.

(Will write this out later. I am puzzled that all the proofs I’ve so far looked at involve determinants. What happened to the battle cry, “Down with determinants!”?)

References

e.g.

Revised on November 25, 2013 06:17:49 by Urs Schreiber (89.204.153.236)