# Cohomology and tensor product

Let $G$ be a profinite group, $A$ a free $\mathbb{Z}_p$-module of finite rank with a continuous action of $G$ and $B$ any $\mathbb{Z}_p$-module (I am not supposing it to be free), with the trivial action of $G$.

I am looking at continuous cohomology groups. I tought that $H^i(G,A \otimes B)$ would be isomorphic to $H^i(G,A) \otimes B$, which is false in general.

Is there a condition on $B$ making this statement true ?

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You certainly need $B$ to be flat. I think this condition should also be sufficient. So let's assume $B$ is flat. If $B$ is finitely generated, this means it's free already. If $B$ is not finitely generated, I don't exactly know what happens. If tensoring with $B$ turns injective ${\mathbb Z}_pG$-modules to acyclic ones, the property follows because of the (in that case split) Grothendieck spectral-sequence, but I don't know if that is the case. – Anton Nov 20 '10 at 19:02
To amplify Anton's "you certainly need $B$ to be flat" comment: consider the case where $A$ is $\mathbf{Z}_p$, $G$ is $1+p\mathbf{Z}_p$ acting on $A$ by multiplication, and $B=\mathbf{Z}/p\mathbf{Z}$. Then even $H^0$ does not commute with the tensor product. – Kevin Buzzard Nov 20 '10 at 22:07
I am pretty aware that the reduction mod p of the cohomology is not the cohomology of the reduction mod p, which is quite strange to me... – A M Nov 20 '10 at 23:19