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

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

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

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

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

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

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

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

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

$-\sigma (y){\sigma }^2(z)-(\sigma (y)+{\sigma }^2(y)){\sigma }^3(z)-\mathrm{\cdots }-(\sigma (y)+\mathrm{\cdots }+{\sigma }^{n-2}){\sigma }^{n-1}(z)+yz$

$=$$y\mathrm{Tr}(z),$

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