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Consider a morphism of commutative rings $h\colon R\rightarrow S$. This yields the two functors $h_*\colon{\sf Mod}(S)\rightarrow{\sf Mod}(R)$ (scalar restriction) and $h^*\colon{\sf Mod}(R)\rightarrow{\sf Mod}(S)$ (scalar extension), and $h^*$ is left adjoint to $h_*$. The unit of this adjunction is for an $R$-module $M$ given by the morphism $$\rho_h(M)\colon M\rightarrow h_*(h^*(M)),\;x\mapsto x\otimes 1_S.$$

It is easy to see the following:

a) $\rho_h$ is an isomorphism if and only if $h$ is bijective;

b) $\rho_h$ is an epimorphism if and only if $h$ is surjective;

c) If $\rho_h$ is a monomorphism then $h$ is injective.

It is furthermore clear that the converse of c) need not be true. So, we may ask the following question:

For which (necessarily injective) morphisms of rings $h\colon R\rightarrow S$ is the unit $\rho_h$ of the adjunction $(h^*,h_*)$ a monomorphism?

An example of a class of such morphisms are those whose source is absolutely flat.

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    $\begingroup$ Such morphisms are called pure, there is an extensive literature about them. Faithfully flat morphisms are pure, but there are more pure morphisms: see this question and its answers. $\endgroup$
    – abx
    Mar 11, 2015 at 20:59
  • $\begingroup$ Dear @abx, may I ask you to add your comment as an answer? It seems that knowing the terminology essentially answers my question... $\endgroup$ Mar 11, 2015 at 21:34
  • $\begingroup$ OK , just done. $\endgroup$
    – abx
    Mar 12, 2015 at 3:57

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Such morphisms are called pure, there is an extensive literature about them. Faithfully flat morphisms are pure, but there are more pure morphisms: see this question and its answers. – abx 6 hours ago

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