cohomology

# Contents

## 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 $K$ be a finite Galois extension of a field $k$, with a cyclic Galois group $G = \langle g \rangle$ of order $n$. Regard the multiplicative group $K^\ast$ as a $G$-module. Then the group cohomology of $G$ with coefficients in $K^\ast$ – the Galois cohomology – satisfies

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

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

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

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

$D \coloneqq g - 1,$
$\,$
$N \coloneqq 1 + g + g^2 + \ldots + g^{k-1}$

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

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

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

$\beta^N = \beta^{1 + g + \ldots + g^{n-1}} = \beta \cdot \beta^g \cdot \ldots \beta^{g^{n-1}}$

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

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

$1 + \beta g + \beta^{1+g}g^2 + \ldots + \beta^{1 + g + \ldots + g^{n-2}}g^{n-1}$

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

$\alpha = \theta + \beta \theta^g + \beta^{1+g}\theta^{g^2} + \ldots + \beta^{1 + g + \ldots + g^{n-2}}\theta^{g^{n-1}}$

is non-zero. Using the fact that $N(\beta) = 1$, one may easily calculate that $\beta \alpha^g = \alpha$, 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 $K$ as a $G$-module, we have

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

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

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

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

$\alpha \coloneqq \frac1{Tr(\theta)}(\beta g(\theta) + (\beta + g(\beta))g^2(\theta) + \ldots + (\beta + g(\beta) + \ldots + g^{n-2}(\beta))g^{n-1}(\theta).$

One may then calculate that

$\array{ \alpha - g(\alpha) & = & \frac1{Tr(\theta)}(\beta g(\theta) + \beta g^2(\theta) + \ldots + \beta g^{n-1}(\theta) - (g(\beta) + \ldots + g^{n-1}(\beta))\theta \\ & = & \frac1{Tr(\theta)}(\beta g(\theta) + \beta g^2(\theta) + \ldots + \beta g^{n-1}(\theta) + \beta \theta) \\ & = & \beta }$

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

## 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 $K$ be a field, let $G$ be a monoid, and let $\chi_1, \ldots, \chi_n \colon G \to K^\ast$ be distinct monoid homomorphisms. Then the functions $\chi_i$, considered as functions valued in $K$, are $K$-linearly independent.

###### Proof

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

(1)$a_1 \chi_1 + \ldots + a_n \chi_n = 0$

where $a_i \in K$, and assume $n$ is as small as possible. In particular, no $a_i$ is equal to $0$, and $n \geq 2$. Choose $g \in G$ such that $\chi_1(g) \neq \chi_2(g)$. Then for all $h \in G$ we have

$a_1 \chi_1(g h) + \ldots + a_n \chi_n(g h) = 0$

so that

(2)$a_1 \chi_1(g) \chi_1 + \ldots + a_n \chi_n(g)\chi_n = 0.$

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

$(a_2\frac{\chi_2(g)}{\chi_1(g)} - a_2)\chi_2 + \ldots = 0$

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

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