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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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    I think it's correct to say that since $-\otimes_S F$ is an adjoint, and is exact, then it preserves projectives.2012-12-20
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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.

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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.