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\begin{document}

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\section*{Hilbert&#39;s Theorem 90}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{cohomology}{}\paragraph*{{Cohomology}}\label{cohomology}

[[!include cohomology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{independence_of_characters}{Independence of characters}\dotfill \pageref*{independence_of_characters} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A theorem in [[Galois cohomology]] due to [[David Hilbert]].

\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

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

\begin{theorem}
\label{mult90}\hypertarget{mult90}{}
\textbf{(Hilbert)} Suppose $K$ be a [[finite set|finite]] [[Galois extension]] of a [[field]] $k$, with a [[cyclic group|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

\begin{displaymath}
H^1(G, K^\ast) = 0
  \,.
\end{displaymath}
\end{theorem}
Before embarking on the proof, we recall from the article \href{http://ncatlab.org/nlab/show/projective+resolution#CohomologyOfCyclicGroups}{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:

\begin{displaymath}
\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
\end{displaymath}
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 $D$, $N$:

\begin{displaymath}
D \coloneqq g - 1,
\end{displaymath}
\begin{displaymath}
\,
\end{displaymath}
\begin{displaymath}
N \coloneqq 1 + g + g^2 + \ldots + g^{k-1}
\end{displaymath}
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

\begin{displaymath}
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
\end{displaymath}
\begin{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

\begin{displaymath}
\beta^N = \beta^{1 + g + \ldots + g^{n-1}} = \beta \cdot \beta^g \cdot \ldots \beta^{g^{n-1}}
\end{displaymath}
which is precisely the \emph{[[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 \ref{XYZ} 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, $K$-[[linear independence|linearly independent]]. It follows in particular that

\begin{displaymath}
1 + \beta g + \beta^{1+g}g^2 + \ldots + \beta^{1 + g + \ldots + g^{n-2}}g^{n-1}
\end{displaymath}
is not identically zero, and therefore there exists $\theta \in K^\ast$ such that the element

\begin{displaymath}
\alpha = \theta + \beta \theta^g + \beta^{1+g}\theta^{g^2} + \ldots + \beta^{1 + g + \ldots + g^{n-2}}\theta^{g^{n-1}}
\end{displaymath}
is non-zero. Using the fact that $N(\beta) = 1$, one may easily calculate that $\beta \alpha^g = \alpha$, as was to be shown.

\end{proof}
Now we give the additive version of Hilbert's theorem 90:

\begin{theorem}
\label{add90}\hypertarget{add90}{}
Under the same hypotheses given in \ref{mult90}, and regarding the additive group $K$ as a $G$-module, we have

\begin{displaymath}
H^1(G, K) = 0
  \,.
\end{displaymath}
\end{theorem}
\begin{proof}
\begin{displaymath}
Tr(\alpha) \coloneqq N \cdot \alpha = \alpha + g(\alpha) + \ldots + g^{n-1}(\alpha).
\end{displaymath}
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

\begin{displaymath}
\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).
\end{displaymath}
One may then calculate that

\begin{displaymath}
\itexarray{
\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 
}
\end{displaymath}
where in the second line we used $Tr(\beta) = 0$.

\end{proof}
\hypertarget{independence_of_characters}{}\subsection*{{Independence of characters}}\label{independence_of_characters}

The next result may be thought of as establishing ``independence of characters'' (where ``[[group character|characters]]'' are valued in the [[multiplicative group]] of a field):

\begin{lemma}
\label{XYZ}\hypertarget{XYZ}{}
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 $K$-[[linear independence|linearly independent]].

\end{lemma}
\begin{proof}
A single $\chi \colon G \to K^\ast$ obviously forms a [[linear independence|linearly independent]] set. Now suppose we have an [[equation]]

\begin{displaymath}
a_1 \chi_1 + \ldots + a_n \chi_n = 0
\end{displaymath}
where $a_i \in K$, and assume $n$ is as small as possible. In particular, no $a_i$ is [[equality|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

\begin{displaymath}
a_1 \chi_1(g h) + \ldots + a_n \chi_n(g h) = 0
\end{displaymath}
so that

\begin{displaymath}
a_1 \chi_1(g) \chi_1 + \ldots + a_n \chi_n(g)\chi_n = 0.
\end{displaymath}
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

\begin{displaymath}
(a_2\frac{\chi_2(g)}{\chi_1(g)} - a_2)\chi_2 + \ldots = 0
\end{displaymath}
which is a [[contradiction]].

\end{proof}
A corollary of this result is an important result in its own right, the \textbf{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!''?)

\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[Kummer theory]]

\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

e.g.

\begin{itemize}%
\item [[Günter Tamme]], section II 4.3 of \emph{[[Introduction to Étale Cohomology]]}

\end{itemize}
[[!redirects Hilbert Theorem 90]] [[!redirects Hilbert theorem 90]]

[[!redirects Hilbert's theorem 90]]



\end{document}