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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, 2015 at 13:42
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    $\begingroup$ $h^*$ is misleading because it's covariant, you should write it $h_*$ $\endgroup$
    – YCor
    Mar 25, 2015 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, 2015 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.) $\endgroup$ Mar 25, 2015 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.. $\endgroup$ Apr 6, 2015 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. $\endgroup$ May 14, 2015 at 19:42
  • $\begingroup$ Dear @Keenan, thank you - I edited my answer accordingly. $\endgroup$ May 14, 2015 at 19:48
  • $\begingroup$ When you convert an answer in the comments to an actual answer, it's better to make the answer CW so you don't gain reputation from someone else's answer (that they thought wasn't significant enough to put as an answer, so instead put in the comments) $\endgroup$ Nov 24, 2023 at 13:24

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