ProofOfHilbertTheorem90 2005-05-31 09:20:53 2007-11-28 19:28:54 Proof Hilbert Theorem 90 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 \begin{align*} \sigma(b)&=\sum_{\tau\in G}\sigma(a(\tau)\tau(c))\\ &=\sum_{\tau\in G}\sigma(a(\tau))(\sigma\tau)(c)\\ &=\sum_{\tau\in G}a(\sigma)^{-1}a(\sigma\tau)(\sigma\tau)(c), \end{align*} since $a$ is a cocycle, i.e. $a(\sigma\tau)=a(\sigma)\sigma(a(\tau))$. Then we get \begin{align*} \sigma(b)&=a(\sigma)^{-1}\sum_{\tau\in G}a(\sigma\tau)(\sigma\tau)(c)\\ &=a(\sigma)^{-1}b. \end{align*} 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 \begin{align*} \tilde{x}([i+k])&=\prod_{0\leq j\leq i+k-1}\sigma^j(x)\\ &=\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)\\ &=\tilde{x}([i])\sigma^i(\tilde{x}([j])). \end{align*} Therefore $\tilde{x}$ is a cocycle. Because of Hilberts Theorem 90, there exists $y\in L^*$, such that $x=\tilde{x}([1])=y\sigma(y)^{-1}$.