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In this post "ring" means "commutative ring with identity".

Background:

Let $\varphi : R \to S$ be a ring homomorphism. This gives us a functor $\varphi_!:= -\otimes_R S : \textbf{Mod}_R \to \textbf{Mod}_S$ often called the extension of scalars. This functor is left adjoint to the restriction of scalars, and therefore preserves colimits, and in particular direct sums. As a result, if $M$ is a free $R$-module, then $M \otimes_R S$ is a free $S$-module of the same rank.

Question:

I'm interested in the opposite question: What can be said about an $R$-module $M$ if $M \otimes_R S$ is a free $S$-module? What about the case when $M \otimes_R S$ is free of rank $1$?

Here's a possibly useful reframing of the question. The extension of scalars functor corresponds to the pullback of quasicoherent sheaves. The question then becomes, what can be said about a quasicoherent sheaf $\mathscr{F}$ on $\text{Spec}(R)$ if its pullback via $\text{Spec}(\varphi) : \text{Spec}(S) \to \text{Spec}(R)$ is free? What about the case when the pullback is free of rank $1$?

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In general nothing can be concluded: for example, if $S$ is a field then $M \otimes_R S$ is always free and this imposes no condition on $M$. (In this case $(-) \otimes_R S$ generally won't be faithful so it's not surprising that we lose information about $M$.)

If $\varphi : R \to S$ is faithfully flat then faithfully flat descent is available and a natural question to ask is whether $M \otimes_R S$ free implies that $M$ is locally free. According to the Stacks project this is false in general and an open question if we additionally assume that $S$ is finitely presented over $R$ (asked on MO with no answer), although apparently it holds if in addition $R$ is Noetherian or $M$ is finitely generated.

With either of these hypotheses, if $M \otimes_R S$ is free of rank $1$ then it follows that $M$ is locally free of rank $1$, or equivalently an invertible module / line bundle.

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    $\begingroup$ If your goal here is to prove that various characterizations of line bundles are equivalent you can get away with a bit less effort than this because you're free to choose many different values of $S$. $\endgroup$ Sep 29 '20 at 18:46
  • $\begingroup$ Thanks for the very helpful answer! Do you happen to know about the case when $M$ is a projective $R$-module (and $M \otimes_R S$ is free of rank $1$)? I should have put this in the original question! $\endgroup$ Sep 29 '20 at 18:49
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    $\begingroup$ @Nate: again, with $\varphi : R \to S$ faithfully flat either of the above hypotheses it already follows that $M$ is invertible and hence projective. $\endgroup$ Sep 29 '20 at 21:25

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