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.



is the tensor product of projective modules again projective?
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.


