$_RM_S$ is R-projective and $_SN$ is S-projective then $_RM \otimes_S N$ is R-projective.

Question

I want to prove that if $_RM_S$ is R-projective and $_SN$ is S-projective then $_RM \otimes_S N$ is R-projective.

Projectivity of $_RM$ and $_SN \implies M \oplus K \cong R^{(I)}$ and $N \oplus L \cong S^{(J)}$

My question is: I want to construct $R^{(I)} \otimes_S S^{(J)} \cong (R \otimes S)^{(I \times J)}$ but are $K$ and $R^{(I)}$ S-Modules? How do I show that?

Maybe I see why $R^{(I)}$ is an S-Module (induced by $M_S$?), but $K$?

If in fact they are, the prove is clear.

$(R \otimes_S S)^{(I \times J)} \cong R^{(I)} \otimes_S S^{(J)} = (M \oplus K) \otimes_S (N \oplus L) = (M \otimes_S N) \oplus ((M \otimes_SL) \oplus (¿¿K \otimes_S (N \oplus K)??)) $

so $(M \otimes_S N)$ is a direct summand of a free R-module and hence R-projective.

Thanks in advance.

Answer 1

Since $_SN$ is projective, there is $N'$ such that $N\oplus N'\cong S^{(I)}$ is a free module. Then we can tensor: $$ M\otimes_SS^{(I)}\cong (M\otimes_SN)\oplus(M\otimes_SN') $$ as $R$-modules. Now $$ M\otimes_SS^{(I)}\cong M^{(I)} $$ as $R$-modules. Since $M$ is projective, every direct power of it is projective as well. So $M\otimes_SN$ is a direct summand of a projective module.