# proof of additive form of Hilbert’s theorem 90

Set $n=[L:K]$.

First, let there be $x\in L$ such that $y=x-\sigma(x)$. Then

 $\operatorname{Tr}(y)=(x-\sigma(x))+(\sigma(x)-\sigma^{2}(x))+\cdots+(\sigma^{n% -1}(x)-\sigma^{n}(x))=0$

because $x=\sigma^{n}(x)$.

Now, let $\operatorname{Tr}(y)=0$. Choose $z\in L$ with $\operatorname{Tr}(z)\neq 0$. Then there exists $x\in L$ with

 $x\operatorname{Tr}(z)=y\sigma(z)+(y+\sigma(y))\sigma^{2}(z)+\cdots+(y+\sigma(y% )+\cdots+\sigma^{n-1}(y))\sigma^{n-1}(z).$

Since $\operatorname{Tr}(z)\in K$ we have

 $\sigma(x)\operatorname{Tr}(z)=\sigma(y)\sigma^{2}(z)+(\sigma(y)+\sigma^{2}(y))% \sigma^{3}(z)+\cdots+(\sigma(y)+\cdots+\sigma^{n-2})\sigma^{n-1}(z)+(\sigma(y)% +\cdots+\sigma^{n-1}(y))\sigma^{n}(z).$

Now remember that $\operatorname{Tr}(y)=0$. We obtain

 $\displaystyle(x-\sigma(x))\operatorname{Tr}(z)$ $\displaystyle=$ $\displaystyle y\sigma(z)+(y+\sigma(y))\sigma^{2}(z)+\cdots+(y+\sigma(y)+\cdots% +\sigma^{n-1}(y))\sigma^{n-1}(z)$ $\displaystyle-\sigma(y)\sigma^{2}(z)-(\sigma(y)+\sigma^{2}(y))\sigma^{3}(z)-% \cdots-(\sigma(y)+\cdots+\sigma^{n-2})\sigma^{n-1}(z)+yz$ $\displaystyle=$ $\displaystyle y\operatorname{Tr}(z),$

so $y=x-\sigma(x)$, as we wanted to show.

Title proof of additive form of Hilbert’s theorem 90 ProofOfAdditiveFormOfHilbertsTheorem90 2013-03-22 15:21:25 2013-03-22 15:21:25 mathwizard (128) mathwizard (128) 4 mathwizard (128) Proof msc 12F10 msc 11R32