If P is a Projective R Module and Q is an Injective R module then $P \otimes_{R} Q$ is Injective

Question

I am doing some basics on Protective, flat and Injective but I have no idea how to proceed for this one.

Any help is appreciated!

Answer 1

I guess your ring $R$ is commutative.

Since $P$ is projective, it is a direct summand of a free module, say $P\oplus P'=R^{(X)}$ is a direct sum of copies of $R$. Since $$ R^{(X)}\otimes_RQ\cong(P\otimes_R Q)\oplus(P'\otimes_R Q) $$ you just need to prove the statement for $P$ free. You also have $$ R^{(X)}\otimes_RQ\cong(R\otimes_RQ)^{(X)}\cong Q^{(X)} $$ so you just need to prove that any direct sum of copies of $Q$ is injective. This is not true in general, but it is if $R$ is Noetherian.

If $P$ is finitely generated, then $X$ can be chosen finite, so the assumption $R$ being Noetherian is not needed.

A module $Q$ is called $\Sigma$-injective if every direct sum of copies of $Q$ is injective. Injective module over non Noetherian rings need not be $\Sigma$-injective. See this paper by Guil Asensio, Jain and Srivastava for references.