Hilbert's theorem 90 for a finite cyclic extension

For the forward direction from 2 to 1, suppose $b=\beta \mathrm{/}\sigma (\beta )$ for some $\beta \in K^{\times }$. Since the embeddings of $K\mathrm{/}F$ are $1,\sigma ,\dots ,{\sigma }^{n-1}$, [L1] gives $N_{K\mathrm{/}F}(b)={\prod }_{i=0}^{n-1}\frac{{\sigma }^i(\beta )}{{\sigma }^{i+1}(\beta )}=1,$ the numerator and denominator cancelling cyclically because ${\sigma }^n=1$.

For the converse, assume $N_{K\mathrm{/}F}(b)=1$. For $0\le i<n$, set $c_i:={\prod }_{j=0}^{i-1}{\sigma }^j(b),$ with the empty product $c_0=1$. Then the distinct automorphisms $1,\sigma ,\dots ,{\sigma }^{n-1}$ restrict to distinct characters $K^{\times }\to K^{\times }$, so [L2] implies that the $K$-linear operator $T(x):={\sum }_{i=0}^{n-1}{c}_i{\sigma }^i(x)$ is not identically zero. Choose $x\in K$ with $\beta :=T(x)\mathrm{\ne }0$.

The coefficients satisfy $c_{i+1}=c_i{\sigma }^i(b)$ for $0\le i<n-1$, and $c_n=N_{K\mathrm{/}F}(b)=1$. Therefore $\sigma (c_i)=b^{-1}{c}_{i+1}(0\le i<n-1),\sigma (c_{n-1})=b^{-1}\mathrm{.}$ Applying $\sigma$ to $\beta =T(x)$ and re-indexing the sum gives $\sigma (\beta )=b^{-1}\beta \mathrm{.}$ Hence $b=\beta \mathrm{/}\sigma (\beta )$.

Steps 1.1 and 2.1 prove the equivalence.