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Let $G$ be a group and let $K/k$ be a field extension. Suppose that $V$ is a $kG$-module, and let $V_K = K \otimes_k V$ be the $KG$-module given by changing the base field.

Is it true that $H^n(G,V_K) \cong K \otimes_k H^n(G,V)$?

If no, what about with some extra assumptions (for example $G$ is finite and $k$ is finite)?

My main motivation is in computing some cohomology/ext groups for $KG$-modules, where $K$ is algebraically closed and $G$ is finite. When $k$ is a finite field, there are ways to compute these with GAP for example.

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  • $\begingroup$ (Especially David Sprehn's answer, but note that he means $H^0(1,R)=R$, not $H^n(1,R)$) $\endgroup$
    – Dave Benson
    Commented Aug 27, 2023 at 19:03
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    $\begingroup$ @DaveBenson: Yes it does, I didn't see that question before. Thank you! $\endgroup$
    – testaccount
    Commented Aug 28, 2023 at 5:01

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There is a commutative diagram $\require{AMScd}$ \begin{CD} k[G]\text{-Mod} @>{-\otimes_{k[G]}K[G]}>> K[G]\text{-Mod}\\ @V{-^G}VV @V{-^G}VV\\ k\text{-Vect} @>{-\otimes_kK}>> K\text{-Vect}, \end{CD} i.e., $(K[G]\otimes_{k[G]}V)^G\simeq K\otimes_kV^G$, since $K$ is a free $k$-vector space (and hence $K[G]\simeq K\otimes_kk[G]$ is a free $k[G]$-module). Now the horizontal arrows are exact, so the derived functor of the composition $k[G]\text{-Mod}\to K\text{-Vect}$ is isomorphic to both $H^i(G,V\otimes_{k[G]}K[G])$ and $H^i(G,V)\otimes_kK$.

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  • $\begingroup$ Thanks! There is also an alternative argument using the Kunneth formula, in a question linked in the comments. $\endgroup$
    – testaccount
    Commented Aug 28, 2023 at 5:03

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