is the tensor product of projective modules again projective?

Question

Let $R$ be a commutative ring and let $A_1$ and $A_2$ be (not necessarily commutative) $R$-algebras. Under which conditions on $A_1$ and $A_2$ is the following true: For every projective $A_1$-module $P_1$ and every projective $A_2$-module $P_2$ we have that $P_1\otimes_R P_2$ is projective as a $A_1\otimes_R A_2$-module?

For instance, what about the case when $A_1$ and $A_2$ are projective over $R$? Or else, does it help if $Tor_R(A_1,A_2)$ vanishes?

I am also looking for counterexamples for the general case.

Answer 1

Since $P_1$ is projective there exists $Q_1$ $A_1$-module and an isomorphism $$ P_1\oplus Q_1 = A_1^{\oplus I_1}$$ for some index set $I_2$. Analogously there exists $Q_2$ $A_2$-module and an index set $I_2$ such that $$ P_2\oplus Q_2 = A_2^{\oplus I_2} $$ Tensoring the two previous relations $$P_1 \otimes P_2 \oplus (P_1\otimes Q_2 \oplus Q_1\otimes P_2 \oplus Q_1\otimes Q_2) = A_1^{\oplus I_1}\otimes A_2^{\oplus I_2} = (A_1\otimes A_2)^{\oplus I_1\times I_2}$$ So $P_1\otimes P_2$ is a direct summand of a free $A_1\otimes A_2$-module and so it is projective.

Answer 2

Recall that $P$ is projective iff $\text{Hom}(P, -)$ is exact. We have

$$\text{Hom}_{A_1 \otimes A_2}(P_1 \otimes P_2, -) \cong \text{Hom}_{A_1}(P_1, \text{Hom}_{A_2}(P_2, -))$$

and a composition of exact functors is exact.

Answer 3

Another way of putting the same thing : embed $P_i$ in a free $A_i$-module $L_i$ with a $A_i$-linear retraction $r_i: L_i\rightarrow P_i$. Then $P_1\otimes _RP_2$ embeds into $L_1\otimes _RL_2$ with a $\ (A_1\otimes _RA_2)$-linear retraction $r_1\otimes r_2: L_1\otimes _RL_2\rightarrow P_1\otimes _RP_2$.