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Consider a morphism of commutative rings $h\colon R\rightarrow S$. This gives rise to a functor $h^*\colon{\sf Mod}(R)\rightarrow{\sf Mod}(S)$, called scalar extension by means of $h$. This functor has a right adjoint, hence it commutes with inductive limits. We may ask now whether or not $h^*$ commutes with projective limits.

Clearly, $h^*$ is left exact if and only if $h$ is flat. Therefore (and using some general nonsense), $h^*$ commutes with projective limits if and only if $h$ is flat and $h^*$ commutes with infinite products. Flatness of $h$ does not imply that $h^*$ commutes with infinite products. So, the question is as follows:

Are there some conditions on a morphism of rings $h\colon R\rightarrow S$ that ensure that the scalar extension functor $h^*$ commutes with infinite products?

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    $\begingroup$ This holds if $S$ is a finitely presented $R$-module. $\endgroup$ – abx Mar 25 '15 at 13:42
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    $\begingroup$ $h^*$ is misleading because it's covariant, you should write it $h_*$ $\endgroup$ – YCor Mar 25 '15 at 14:52
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    $\begingroup$ A necessary condition is that $S$ is a finitely generated $R$-module. Indeed, suppose that $h_*$ commutes with taking the $S$-fold product $M\mapsto M^S$ (where $S$ is just viewed as a set!). Then $S^S$ is generated by $h(R)^S$ as an $S$-module. In particular, we can write $\mathrm{id}_S=\sum_{i=1}^ks_if_i$ with $f_i\in h(R)^S$. Thus $s=\sum_{i=1}^kf_i(s)s_i$ for all $s\in S$. This means that $s_1,\dots,s_k$ generates $S$ as an $R$-module. $\endgroup$ – YCor Mar 25 '15 at 15:03
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    $\begingroup$ Thanks, @abx! The "if" in your statement may in fact be replaced by an "iff" - see Bourbaki, A.X.1 Exercice 18. (Or, for a proof, T.Y.Lam, Lectures on modules and rings, Proposition 4.44.) $\endgroup$ – Fred Rohrer Mar 25 '15 at 18:56
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    $\begingroup$ I guess the upper star is because the functor is really 'pullback' of qcoh sheaves along the induced morphism of spectra. I agree about the 'variance' argument, but if you write $h_*$ I immediately think of a pushforward.. $\endgroup$ – Mattia Talpo Apr 6 '15 at 23:02
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(This was answered in the comments, essentially by abx.)

The scalar extension functor by means of $h\colon R\rightarrow S$ commutes with infinite products if and only if $S$, considered as an $R$-module by means of $h$, is of finite presentation.

A proof can be found in T.Y.Lam, Lectures on modules and rings, Proposition 4.44. See also Bourbaki, A.X.1 Exercice 18 for additional information.

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    $\begingroup$ Dear @Fred Rohrer, A ring map of finite presentation is one which makes the target into a finitely presented algebra over the source (this is common usage in commutative ring theory). It would be clearer probably to say that $h$ makes the target into a finitely presented module over the source. $\endgroup$ – Keenan Kidwell May 14 '15 at 19:42
  • $\begingroup$ Dear @Keenan, thank you - I edited my answer accordingly. $\endgroup$ – Fred Rohrer May 14 '15 at 19:48

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