nLab Hilbert's Theorem 90

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 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 , 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

$$a_1 \chi_1 + \ldots + a_n \chi_n = 0$$

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

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

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!”?)

Related concepts

• Kummer theory

References

e.g.

• Günter Tamme, section II 4.3 of Introduction to Étale Cohomology