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 $G=⟨g⟩$ of order $n$. Regard the multiplicative group ${K}^{*}$ as a $G$-module. Then the group cohomology of $G$ with coefficients in ${K}^{*}$ – the Galois cohomology – satisfies

${H}^{1}\left(G,{K}^{*}\right)=0\phantom{\rule{thinmathspace}{0ex}}.$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 $ℤ$ as trivial $G$-module:

$\dots \stackrel{N}{\to }ℤG\stackrel{D}{\to }ℤG\stackrel{N}{\to }ℤG\stackrel{D}{\to }ℤG\to ℤ\to 0$\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 $ℤG\to ℤ$ 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 $ℤG$, also denoted $D$, $N$:

$D≔g-1,$D \coloneqq g - 1,
$\phantom{\rule{thinmathspace}{0ex}}$\,
$N≔1+g+{g}^{2}+\dots +{g}^{k-1}$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 ${H}^{n}\left(G,A\right)$ (in the case $A={K}^{*}$), by taking cohomology of the induced cochain complex

$0\to {\mathrm{hom}}_{ℤG}\left(ℤ,A\right)\to {\mathrm{hom}}_{ℤG}\left(ℤG,A\right)\to {\mathrm{hom}}_{ℤG}\left(ℤG,A\right)\to \dots$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 $\sigma \in ℤG$ 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 $N\in ℤG$ is

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

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

By lemma 1 below, the homomorphisms $1,g,\dots ,{g}^{n-1}:{K}^{*}\to {K}^{*}$ 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}+\dots +{\beta }^{1+g+\dots +{g}^{n-2}}{g}^{n-1}$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 $\theta \in {K}^{*}$ such that the element

$\alpha =\theta +\beta {\theta }^{g}+{\beta }^{1+g}{\theta }^{{g}^{2}}+\dots +{\beta }^{1+g+\dots +{g}^{n-2}}{\theta }^{{g}^{n-1}}$\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 $N\left(\beta \right)=1$, 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

${H}^{1}\left(G,K\right)=0\phantom{\rule{thinmathspace}{0ex}}.$H^1(G, K) = 0 \,.

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

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

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

$\alpha ≔\frac{1}{\mathrm{Tr}\left(\theta \right)}\left(\beta g\left(\theta \right)+\left(\beta +g\left(\beta \right)\right){g}^{2}\left(\theta \right)+\dots +\left(\beta +g\left(\beta \right)+\dots +{g}^{n-2}\left(\beta \right)\right){g}^{n-1}\left(\theta \right).$\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

$\begin{array}{ccc}\alpha -g\left(\alpha \right)& =& \frac{1}{\mathrm{Tr}\left(\theta \right)}\left(\beta g\left(\theta \right)+\beta {g}^{2}\left(\theta \right)+\dots +\beta {g}^{n-1}\left(\theta \right)-\left(g\left(\beta \right)+\dots +{g}^{n-1}\left(\beta \right)\right)\theta \\ & =& \frac{1}{\mathrm{Tr}\left(\theta \right)}\left(\beta g\left(\theta \right)+\beta {g}^{2}\left(\theta \right)+\dots +\beta {g}^{n-1}\left(\theta \right)+\beta \theta \right)\\ & =& \beta \end{array}$\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 $\mathrm{Tr}\left(\beta \right)=0$.

## 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 ${\chi }_{1},\dots ,{\chi }_{n}:G\to {K}^{*}$ be distinct monoid homomorphisms. Then the functions ${\chi }_{i}$, considered as functions valued in $K$, are $K$-linearly independent.

###### Proof

A single $\chi :G\to {K}^{*}$ obviously forms a linearly independent set. Now suppose we have an equation

(1)${a}_{1}{\chi }_{1}+\dots +{a}_{n}{\chi }_{n}=0$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 $n\ge 2$. Choose $g\in G$ such that ${\chi }_{1}\left(g\right)\ne {\chi }_{2}\left(g\right)$. Then for all $h\in G$ we have

${a}_{1}{\chi }_{1}\left(gh\right)+\dots +{a}_{n}{\chi }_{n}\left(gh\right)=0$a_1 \chi_1(g h) + \ldots + a_n \chi_n(g h) = 0

so that

(2)${a}_{1}{\chi }_{1}\left(g\right){\chi }_{1}+\dots +{a}_{n}{\chi }_{n}\left(g\right){\chi }_{n}=0.$a_1 \chi_1(g) \chi_1 + \ldots + a_n \chi_n(g)\chi_n = 0.

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

$\left({a}_{2}\frac{{\chi }_{2}\left(g\right)}{{\chi }_{1}\left(g\right)}-{a}_{2}\right){\chi }_{2}+\dots =0$(a_2\frac{\chi_2(g)}{\chi_1(g)} - a_2)\chi_2 + \ldots = 0