Tensor product of a projective module over an arbitrary ring still projective?

If $S$ is any ring, and $P$ is a projective $S$ module, and $Q$ is any $S$ module, then will $Q \otimes_{S} P$ be a projective $S$-module?

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No. Let $S$ be a ring with nonprojective module $Q$. Then $Q\otimes_S S\cong Q$ is not projective.

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Thank you, do you think it would be true if $Q$ was also $S$-projective?, since ive seen this result when $S$ is commutative but not for an arbitrary ring. –  user39947 Dec 20 '12 at 15:31
It is true if $Q$ is projective. The point is that if $P$ is just a flat module, then $-\otimes_S P$ is an exact functor. Because of that, it will send projectives to projectives. Thus if $Q$ were projective, then $Q\otimes P$ would be also. $P$ just needs to be flat: of course, if it is projective it is a fortiori flat. –  rschwieb Dec 20 '12 at 15:34
Oh ok yeah I see, thanks. The question came up because I wanted to see if for Q a finitely generated free $\mathbb{Z}$-module that is also a $S$ module, the above was true. –  user39947 Dec 20 '12 at 15:46
At least I thought that exact functors do that :) I'm worried that I'm recalling it incorrectly. Still, it definitely works for two projective modules: you can use the characterization of a projective module as a submodule of a free module to assemble an argument using the same trick in my solution. –  rschwieb Dec 20 '12 at 15:46
I think it's correct to say that since $-\otimes_S F$ is an adjoint, and is exact, then it preserves projectives. –  rschwieb Dec 20 '12 at 15:50
No, let $S = P = \mathbf Z$ and $Q = \mathbf{Z}/ 2 \mathbf{Z}$. $P$ is free, hence projective, but $P \otimes Q \simeq \mathbf{Z}/2\mathbf{Z}$ is not.
No. Take $S =P = \Bbb{Z}$ and $Q = \Bbb{Z}$. Then $\Bbb{Z}$ as a $\Bbb{Z}$ - module is free and hence projective. But the tensor product $\Bbb{Z} \otimes_{\Bbb{Z}} \Bbb{Q} \cong \Bbb{Q}$ is isomorphic to $\Bbb{Q}$ which is not projective as a $\Bbb{Z}$ - module.