# 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?

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You are right: if $F$ is free, say $F=R^{\oplus I}$, then $M\otimes_RF$ is just $M^{\oplus I}$, which is not free in general. –  Matemáticos Chibchas Jul 6 at 20:44
Is $R$ assumed commutative? –  Edward ffitch Jul 6 at 20:57
Yes, I think so. –  Nishant Jul 6 at 21:00
The question seems to have been edited by someone else because I was asking for a proof of a false result. –  Nishant Jul 7 at 16:17
@user26857 Hmm, well I certainly didn't put any part of my question in a box... –  Nishant Jul 7 at 18:06

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.

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Wait, so then how can tensoring with a flat module preserve projective resolutions? –  Nishant Jul 6 at 21:26
Well, I would say it does not, since, in that case, it should preserve projectives as well (because if $P$ is projective $0\to P\stackrel{id}{\to}P\to 0$ is a projective resolution of $P$)... –  Marco Vergura Jul 6 at 21:40
Well, it was in the specific case where $R\to S$ is a homomorphism such that when $S$ is considered as an $R$-module in the obvious way, then if $S$ is flat, tensoring with $R$ sends $R$-projective resolutions to $S$-projective resolutions. –  Nishant Jul 6 at 21:52
Yes, indeed: in that case it is actually true! –  Marco Vergura Jul 6 at 21:53
Well, essentially it is because, in that case, if $P$ is projective as an $R$ module and it is a direct summand of $R^{(X)}$ for some set $X$, then $P\otimes_{R} S$ is a direct summand of $R^{(X)}\otimes_{R} S\simeq S^{(X)}$, so it is a projective $S-$module. Therefore, if $P_{\bullet}\to M\to 0$ is a projective resolution of the $R$ module $M$, then tensoring with $S$ gives a $S-$projective resolution... –  Marco Vergura Jul 6 at 22:03
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.