proof of Hilbert Theorem 90

Remember that two cocycles $a,a^{\prime }:G\to L^{*}$ are called cohomologous, denoted by $a\sim a^{\prime }$, if there exists $b\in L^{*}$, such that $a^{\prime }(\tau )=ba(\tau )\tau (b^{-1})$ for all $\tau \in G$. Then

$$H^1(G,L^{*})=\{a:G\to L^{*}|a\text{ is a cocycle}\}/\sim .$$

Now let $a:G\to L^{*}$ be a cocycle. Then consider the map

$$\alpha :L\to L,c\mapsto \sum _{\tau \in G}a(\tau )\tau (c).$$

Since elements of the Galois group are linearly independent, $\alpha$ is not $0$. So we can choose $c\in L$, such that $b=\alpha (c)\ne 0$. Then for $\sigma \in G$ we have

$\sigma (b)$ $={\displaystyle \sum _{\tau \in G}}\sigma (a(\tau )\tau (c))$

$={\displaystyle \sum _{\tau \in G}}\sigma (a(\tau ))(\sigma \tau )(c)$

$={\displaystyle \sum _{\tau \in G}}a{(\sigma )}^{-1}a(\sigma \tau )(\sigma \tau )(c),$

since $a$ is a cocycle, i.e. $a(\sigma \tau )=a(\sigma )\sigma (a(\tau ))$. Then we get

$\sigma (b)$ $=a{(\sigma )}^{-1}{\displaystyle \sum _{\tau \in G}}a(\sigma \tau )(\sigma \tau )(c)$

$=a{(\sigma )}^{-1}b.$

Thus we have that $a(\sigma )=b\sigma {(b)}^{-1}$ is a 1-coboundary.

Now we prove the corollary. Denote the norm by $N$. Now if $x=\frac{y}{\sigma (y)}$, we have

$$N(x)=N\left(\frac{y}{\sigma (y)}\right)=\prod _{\tau \in G}\frac{\tau (y)}{\tau (\sigma (y))}=1.$$

Now let $N(x)=1$, $n=|G|$. Since $G$ is assumed cyclic, let $\sigma$ be a generator of $G$. $G$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. We define the map $\tilde{x}:\mathbb{Z}/n\mathbb{Z}\to L^{*}$ by

$$\tilde{x}([i])=\prod _{0\le j\le i-1}{\sigma }^j(x),$$

where $[i]$ denotes the class of $i\in \mathbb{Z}$ in $\mathbb{Z}/n\mathbb{Z}$. Since $N(x)=1$, $\tilde{x}$ is well defined. We have

$\tilde{x}([i+k])$ $={\displaystyle \prod _{0\le j\le i+k-1}}{\sigma }^j(x)$

$=\left({\displaystyle \prod _{0\le j\le i-1}}{\sigma }^j(x)\right){\sigma }^i\left({\displaystyle \prod _{0\le j\le k-1}}{\sigma }^j(x)\right)$

$=\tilde{x}([i]){\sigma }^i(\tilde{x}([j])).$

Therefore $\tilde{x}$ is a cocycle. Because of Hilberts Theorem 90, there exists $y\in L^{*}$, such that $x=\tilde{x}([1])=y\sigma {(y)}^{-1}$.

Title	proof of Hilbert Theorem 90
Canonical name	ProofOfHilbertTheorem90
Date of creation	2013-03-22 15:19:27
Last modified on	2013-03-22 15:19:27
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	8
Author	mathcam (2727)
Entry type	Proof
Classification	msc 11R32
Classification	msc 11S25
Classification	msc 11R34