nLab Hilbert's Theorem 90

Idea

A type of theorem in Galois cohomology due to David Hilbert.

Statements

There are several versions referred to as Hilbert’s theorem 90.

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

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

Proof

For the following, note (see here), that if $G = C_n$ is a finite cyclic group of order $n$, then there is a projective resolution of $\mathbb{Z}$ as a 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 that follow refer to this resolution as a means of defining $H^n(G; A)$ (in the case $A = K^\ast$), by forming the cochain cohomology of the induced cochain complex

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

With that understood, let now $\sigma \in \mathbb{Z}G$ 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.

Additive version

Theorem

Under the same hypotheses as in Thm. , and regarding the additive group $K$ as a $G$-module, we have

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

Proof

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 lemma below, there exists $\theta$ such that $Tr(\theta) \neq 0$; notice that $Tr(\theta)$ belongs to the ground field $k$ since $g \cdot N = N$. Put

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

One may then calculate that

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

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

Modern version

Theorem

Suppose $K$ is a finite Galois extension of a field $k$ with Galois group $G$. Regard the general linear group $GL(n,K)$ as a $G$-module. Then

$$H^1\big( G; GL(n,K) \big) \;=\; 0 \,.$$

(Gille & Szamuely 2017 p. 29)

Independence of characters

The next result establishes the lemma of “independence of characters” used in the above proofs (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 theorem.

Related concepts

• Kummer theory

References

• Günter Tamme, section II 4.3 of: Introduction to Étale Cohomology, Springer (1994) [doi:10.1007/978-3-642-78421-7]

The modern version, Thm. , of Hilbert’s Theorem 90, explained as part of a general theory of Galois descent:

• Philippe Gille, Tamás Szamuely, p. 29 of: Central Simple Algebras and Galois Cohomology, Cambridge U. Press, (2017) [doi:10.1017/CBO9780511607219, pdf]

An attempt to understand the Galois descent approach more conceptually:

• John Baez: Group cohomology and homotopy fixed points, The n-Category Café. [web]

Version in symmetric monoidal categories:

• Ahmed Saleh Al-Rawashdeh, Bachuki Mesablishvili, Hilbert’s Theorem 90 in monoidal categories, Journal of Algebra 602 (2022) [doi:10.1016/j.jalgebra.2022.02.013]