If $P$ is a projective module and $M$ is a flat module, both over some commutative ring $R$, then how do you prove that $M\otimes_R P$ is flat?
I tried using the fact that $P$ is the direct summand of a free module, but that line of reasoning only works of $M$ tensored with a free module gives another free module, which I don't think is the case. Is it? If not, how can I prove that projectiveness is preserved?
Best Answer
Let $M, N$ be $R-$modules. Then the following holds.
$$ F'':=F\otimes_{R}F' $$ is free (tensor product of free modules) and $M\otimes_{R}N$ is a direct summand of $F''$ (as tensor product commutes with direct sums).
(I am being a little fast here, I admit, but these are all pretty standard results that you can easily find also on the Net).
Thus, in your case, you get that $M\otimes_{R}P$ is certainly flat and that is all you can say about it. In particular, $M\otimes_{R}P$ is not projective in general: take $M$ as a flat module which is not projective (for example $\mathbb{Q}$ seen as a $\mathbb{Z}-$module) and $P$ as $R$ (which is projective as a module over itself), so that $M\otimes_{R}R\simeq M$, which is not projective by assumption.