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The following generalization of Hilbert 90 can be found in Serre's Corps Locaux (Chap. X, §1, ex.2, p.160 of the French edition), see also this question:

Theorem: If $L|K$ is a finite Galois extension and $R$ is a finite-dimensional $K$-algebra, then $H^1(L|K, (R \otimes_K L)^{\times})) = \{1\}$.

I am wondering about the following possible converse: if $L|K$ is a finite Galois extension and $R$ is a finite-dimensional $L$-algebra, (edit) equipped with a $\mathrm{Gal}(L|K)$-action (as an algebra, i.e. the action respects products) for which $H^1(L|K, R^{\times}) = \{1\}$ holds, does we necessarily have $R = R^{\mathrm{Gal}(L|K)} \otimes_K L$? If this fails, I'm curious if there are "natural" hypotheses that would suffice to apply known results of Galois descent.

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    $\begingroup$ What do you mean by $H^1(L|K, R^\times)$? Do you assume that ${\rm Gal}(L|K)$ already acts in $R$? $\endgroup$
    – Mikhail Borovoi
    Commented Mar 21 at 13:08
  • $\begingroup$ Yes, sorry, edited to fix that. I mean that $R$ is a finite-dimensional $L$-algebra equipped with a $\mathrm{Gal}(L|K)$-action (as an algebra) satisfying H90. $\endgroup$
    – Moinsdeuxcat
    Commented Mar 21 at 15:23

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Let $R$ be a finite dimensional $L$-algebra with an action of ${\rm Gal}(L/K)$ such that $\sigma (\lambda .r)=\sigma (\lambda )\sigma (r)$, $\lambda\in L$, $r\in R$, $\sigma\in {\rm Gal}(L/K)$. Then if $R_o$ is the $K$-algebra $R^{{\rm Gal}(L/K)}$, the natural algebra homomorphism $R_o \otimes L\longrightarrow R$ is an isomorphism of $L$-algebras. You do not need to assume that a cohomology set vanishes! Indeed the natural algebra homomorphism is bijective by Galois descent for vector spaces, which is a consequence of Hilbert $90$ : $H^1 (L/K , {\rm GL}(n,L))=\{ 1\}$.

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    $\begingroup$ This assumes the inclusion $L \to R$ is Galois-equivariant, which is of course reasonable to assume but not quite explicitly stated in the question. Without this assumption, I guess the answer is negative for silly reasons: You can choose an algebra which contains two different fields $L, L'$ which happen to have the same Galois group, and let the Galois group of $L$ act via the natural action of the Galois group of $L'$. Then cohomology will vanish by generalized Hilbert 90 for $L'$ but descent will not hold. $\endgroup$
    – Will Sawin
    Commented Mar 21 at 19:24

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