I am trying to check if the following statement is true
Let $R$ be a commutative ring and $M$ a $R-$module. Suppose that $G$ is a group acting on $M$ by $R-$linear automorphisms. Let $N$ be a projective $R-$module. Suppose that $G$ acts on $M\otimes_{R}N$ by the formula $$ g(m\otimes n)=g(m)\otimes n $$ for all $g\in G, m\in M$ and $n\in N$. Then $$ (M\otimes_{R}N)^G=M^{G}\otimes_{R} N. $$
Here is what I believe that should be a good solution of the claim:
Using the projectivity of $N$, let $P$ be a $R-$module such that $N\oplus P\cong R^{\oplus S}$ for some set $S$. The isomorphisms $$ M\otimes_{R}(N\oplus P)\cong M\otimes_{R}R^{\oplus S}\cong M^{\oplus S} $$ are $G-$equivariant where we define the actions $G$ on $M\otimes_{R}(N\oplus P)$ and $M\otimes_{R}P$ in analogous way to the one on $M\otimes_{R}N$. Taking $G-$invariants in the above isomorphisms we obtain $$ (M^{G})^{\oplus S}= (M^{\oplus S})^G\cong \big(M\otimes_{R}(N\oplus P)\big)^G \cong (M\otimes_{R}N)^G\oplus (M\otimes_{R}P)^G $$ Using the flatness of $N$ and $P$ over $R$ we have that $M^G\otimes_{R}N$ and $M^G\otimes_{R}P$ are contained in $M\otimes_{R}N$ and $M\otimes_{R}P$, respectively. It is also clear that $M^G\otimes_{R}N\subseteq (M\otimes_{R}N)^G$ and $M^G\otimes_{R}P\subseteq (M\otimes_{R}P)^G$ hold by definition. Hence $$ (M^{G})^{\oplus S}= (M^G \otimes N)\oplus (M^G \otimes P)\subseteq (M\otimes_{R}N)^G \oplus (M\otimes_{R}P)^G=(M^{G})^{\oplus S} $$ Therefore all of the inclusions must be equalities and the result follows.
Can anyone tell me if there is anything wrong with it?