Tensor product of a bimodule (projective as a right module) and a projective right module is projective

Question

Let R and S be rings. Let L be a projective R right module ($L_R$) and M be a R-S bimodule ($_RM_S$) such that M is projective as a S right module. I'm trying to prove that $$(L \otimes M)_S$$ is projective.

I've look the equivalent statements for a projective module but in all of them I can't see how $M_S$ being projective help me.

Do you have some ideas?

Thank you anyway.

Answer 1

Let's put the definition at the top for comparison:

Definition: Let $R$ be a ring. A right $R$-module $P$ is projective if, for each surjection $\pi : A \rightarrow B$ of right $R$-modules, and each map $f : P \rightarrow B$, there is a map $g : P \rightarrow A$ such that $\pi \circ g = f$.

Also, in problems like this, it is important to know the hom-tensor adjunction for bimodules, so I'll put that here too:

Theorem: Let $R$ and $S$. Let $M$ be a left $R$-module and a right $S$-module. Let $N$ be a left $R$-module and a right $S$-module. Let $L$ be a right $S$-module. Then there is an isomorphism of abelian groups $$ \text{Hom}_S ( M \otimes_R N, L) \cong \text{Hom}_R ( M, \text{Hom}_S (N, L))$$ Which, fixing $N$, is natural in $M$ and $L$.

In your setup, we have a projective right $R$-module $L$ and and an $R$-$S$ bimodule $M$ which is projective as a right $S$-module. Put $P = L \otimes_R M$, which is a right $S$-module. Let $\pi : A \rightarrow B$ be a surjection of right $S$-modules, and let $f : P \rightarrow B$ be an $S$-linear map.

Since $M$ is projective, the map $\rho : \text{Hom}_S (M, A) \rightarrow \text{Hom}_S (M, B)$ is surjective. By the Hom-Tensor adjunction, $f$ induces a map of right $R$-modules $L \rightarrow \text{Hom}_{S} (M, B)$. This map lifts along the surjection $\rho$ to a map $L \rightarrow \text{Hom}_S (M, A)$. By the hom-tensor adjunction, this induces a map of right $S$-modules $g : L \otimes_R M \rightarrow A$. I will leave it to you to check that $g$ is a lift for $f$, so that the definition above is satisfied.