# proof of Hilbert Theorem 90

Remember that two cocycles $a,a^{\prime}\colon G\to L^{*}$ 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

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

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

 $\alpha\colon L\to L,c\mapsto\sum_{\tau\in G}a(\tau)\tau(c).$

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

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

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

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

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 $\tilde{x}\colon\mathbb{Z}/n\mathbb{Z}\to L^{*}$ by

 $\tilde{x}([i])=\prod_{0\leq j\leq i-1}\sigma^{j}(x),$

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

 $\displaystyle\tilde{x}([i+k])$ $\displaystyle=\prod_{0\leq j\leq i+k-1}\sigma^{j}(x)$ $\displaystyle=\left(\prod_{0\leq j\leq i-1}\sigma^{j}(x)\right)\sigma^{i}\left% (\prod_{0\leq j\leq k-1}\sigma^{j}(x)\right)$ $\displaystyle=\tilde{x}([i])\sigma^{i}(\tilde{x}([j])).$

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 ProofOfHilbertTheorem90 2013-03-22 15:19:27 2013-03-22 15:19:27 mathcam (2727) mathcam (2727) 8 mathcam (2727) Proof msc 11R32 msc 11S25 msc 11R34