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.

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    $\begingroup$ See Proposition 2.3 in Chapter IX of the book by Cartan-Eilenberg. $\endgroup$ – Mariano Suárez-Álvarez Apr 14 '14 at 22:58

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.

  • $\begingroup$ this is what I came up with too (sorry for not mentioning it), but the problem is, that it is not clear to me that why the given direct sum decomposition is valid as $A_1\otimes A_2$-modules. A priori it is only true as $R$-modules, which isn't enough. $\endgroup$ – user49605 Apr 14 '14 at 23:26
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    $\begingroup$ There's probably something I'm missing. Aren't all the direct summands $A_1\otimes A_2$-stable? $\endgroup$ – Denis Nardin Apr 14 '14 at 23:30
  • $\begingroup$ ok, looks like you are right $\endgroup$ – user49605 Apr 15 '14 at 7:53

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.

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    $\begingroup$ This is, in fact, the argument given by Cartan and Eilenberg. $\endgroup$ – Mariano Suárez-Álvarez Apr 15 '14 at 5:18

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 $r_1\otimes r_2: L_1\otimes _RL_2\rightarrow P_1\otimes _RP_2$.


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