Hilbert's theorem 90 ($H^1(K,{\overline{K}}^{\ast })=0$)

For a finite Galois extension $L/K$, multiplicative Hilbert theorem 90 says $H^1(\mathrm{Gal}(L/K),L^{\ast })=0$. A multiplicative cocycle has the form $c_{\sigma }=\sigma (b)/b$. One proof forms a nonzero sum $A={\sum }_{\sigma }{c}_{\sigma }\sigma (a)$, using linear independence of field automorphisms. Its transformation law is $\tau A=c_{\tau }^{-1}A$, so $b=A^{-1}$ gives $c_{\tau }=\tau (b)/b$. Passing to a separable closure gives $H^1(G_K,K_s^{\times })=0$. For $m$ prime to the characteristic of a field, the multiplicative Kummer sequence then identifies $H^1(G_K,{\mu }_m)$ with $K^{\times }/(K^{\times }{)}^m$. The characteristic restriction makes the power map surjective on the separable closure.

For an unramified extension of non-Archimedean local fields, let $\varphi$ generate its cyclic Galois group. The ordinary Hilbert theorem 90 supplies $b\in L^{\times }$ with $x=\varphi (b)/b$. A common uniformizer $\pi \in K$ gives $y={\pi }^{-v_L(b)}b\in {\mathcal{O}}_L^{\times }$, retaining the same quotient because $\varphi$ fixes $\pi$. The reverse implication follows by telescoping the field norm.