Additive Hilbert 90: trace zero is the image of $\alpha \mapsto \alpha -\sigma (\alpha )$

For the forward direction from 2 to 1, suppose $b=\alpha -\sigma (\alpha )$. Summing the conjugates gives ${\mathrm{Tr}}_{K\mathrm{/}F}(b)={\sum }_{i=0}^{n-1}{\sigma }^i(\alpha -\sigma (\alpha ))={\sum }_{i=0}^{n-1}({\sigma }^i(\alpha )-{\sigma }^{i+1}(\alpha ))=0,$ again by telescoping and ${\sigma }^n=1$.

For the converse, assume ${\mathrm{Tr}}_{K\mathrm{/}F}(b)=0$. By [F1] and [L1], choose $c\in K$ with ${\mathrm{Tr}}_{K\mathrm{/}F}(c)=1$. Define $\alpha :={\sum }_{i=0}^{n-1}({\sum }_{j=0}^{i-1}{\sigma }^j(b)){\sigma }^i(c),$ where the inner sum is $0$ for $i=0$.

Put $s_i={\sum }_{j=0}^{i-1}{\sigma }^j(b)$, so $s_0=0$, $s_n={\mathrm{Tr}}_{K\mathrm{/}F}(b)=0$, and $s_i=b+\sigma (s_{i-1})$ for $1\le i\le n-1$, while $0=s_n=b+\sigma (s_{n-1})$. Applying $\sigma$ to the coefficients as well gives $\sigma (\alpha )={\sum }_{i=0}^{n-1}\sigma (s_i){\sigma }^{i+1}(c)=\sigma (s_{n-1})c+{\sum }_{i=1}^{n-1}\sigma (s_{i-1}){\sigma }^i(c)\mathrm{.}$ Therefore every coefficient of $\alpha -\sigma (\alpha )$ is $b$, so $\alpha -\sigma (\alpha )=b{\sum }_{i=0}^{n-1}{\sigma }^i(c)=b{\mathrm{Tr}}_{K\mathrm{/}F}(c)=b\mathrm{.}$

Steps 1.1 and 2.1 prove the equivalence.