Proving $Ind^{G}_{H}M:=M\bigotimes_{RH}RG$, where $M$ is projective, is projective.

Question

Let $H\leq G$ be a subgroup, $R$ a ring and $M$ be a projective $H$-module. Prove that the induced module $Ind^{G}_{H}M:=M\bigotimes_{RH}RG $ is projective.

Answer 1

This is a special case of the general result: let $P$ be a projective right $A$-module for some ring $A$, and let $B$ be a ring with a compatible left $A$-module structure, that is $a(bb')=(ab)b'$ for $a\in A$, $b$, $b'\in B$. Then $P\otimes_A B$ is a projective right $B$-module.

To prove this, note there's a projective $A$-module $Q$ with $P\oplus Q=F$ free. Then $(P\otimes B)\oplus(Q\otimes B)\cong F\otimes B$ is free over $B$ and so $P\otimes B$ is projective over $B$ (as a summand of a free module).