Lemma 63.15.5. Let $P$ be a finite projective $A[G]$-module and $M$ a $\Lambda [G]$-module, finite projective as a $\Lambda$-module. Then $P \otimes _ A M$ is a finite projective $\Lambda [G]$-module, for the structure induced by the diagonal action of $G$.

Proof. For any $\Lambda [G]$-module $N$ one has

$\mathop{\mathrm{Hom}}\nolimits _{\Lambda [G]}\left(P \otimes _ A M, N\right)= \mathop{\mathrm{Hom}}\nolimits _{A[G]}\left(P, \mathop{\mathrm{Hom}}\nolimits _{\Lambda }(M, N)\right)$

where the $G$-action on $\mathop{\mathrm{Hom}}\nolimits _{\Lambda }(M, N)$ is given by $(g\cdot \varphi )(m) = g \varphi (g^{-1} m)$. Now it suffices to observe that the right-hand side is a composition of exact functors, because of the projectivity of $P$ and $M$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).