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:



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> 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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Extension of scalars and projective limits
*(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.


