Proving that tensoring a projective module with a flat module gives a projective module? 
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?
 A: Projective modules are always flat. So if $P$ is a projective $R$-module, $M$ is a flat $R$-module and 
 $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact sequence of $R$-modules, then, as $P$ is flat, $0 \rightarrow P\otimes A \rightarrow P\otimes B \rightarrow P\otimes C \rightarrow 0$ is exact, so by flatness of $M$, $0 \rightarrow M \otimes (P\otimes A) \rightarrow M \otimes (P\otimes B) \rightarrow M \otimes (P\otimes C) \rightarrow 0$ is exact. But this exact sequence is (naturally) isomorphic to:
$$0 \rightarrow (M \otimes P)\otimes A \rightarrow (M \otimes P)\otimes B \rightarrow (M \otimes P)\otimes C \rightarrow 0$$
hence $M \otimes P$ is flat, as required.
A: Let $M, N$ be $R-$modules. Then the following holds.


*

*If $M$ and $N$ is flat, then so is $M\otimes_{R}N$: see related question here.

*If $M$ and $N$ are projective, then so is $M\otimes_{R} N$. Indeed, writing $M\oplus M'=F,\ N\oplus N'=F'$, for free $R-$modules $F,\ F'$, one has that


$$
F'':=F\otimes_{R}F'
$$
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).


*

*Projective modules are flat. This is because free modules are flat and direct summands of flat modules are flat. (Another possible reason for this, if you are used to the language, is the following. Projective modules are projective objects in the category $R-\mathbf{Mod}$ of modules over $R$. Hence, they are $F-$acyclic for each (additive) right exact functor $F\colon R-\mathbf{Mod}\to \mathscr{D}$ into an arbitrary abelian category $\mathscr{D}$. In particular, a projective module $P$ is $(-\otimes P)-$acyclic, thus flat, by definition of flatness.)


(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.
