# What does it mean when the extension of scalars is free?

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 $$1$$?

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.
• 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 $S$. Sep 29 '20 at 18:46
• 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! Sep 29 '20 at 18:49
• @Nate: again, with $\varphi : R \to S$ faithfully flat either of the above hypotheses it already follows that $M$ is invertible and hence projective. Sep 29 '20 at 21:25