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.



Scalar restriction and scalar extension
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](https://mathoverflow.net/questions/152877/pure-morphisms-which-are-not-faithfully-flat) and its answers. – abx 6 hours ago 


