I was reading about $F$-purity and $F$-splittings, when I came across then following statement which I can't prove:
Definition: Let $R$ be a commutative ring with identity, and $M,N$ be $R$-modules. An $R$-linear map $\alpha : M\rightarrow N$ is called pure if for any $R$-module $W$ the map ${\alpha \otimes id}: {M\otimes_R W}\rightarrow N\otimes_R W$ is injective.
My question: Let $\alpha :R\rightarrow S$ be an extension of rings. If $S$ is faithfully flat as an $R$-module, then prove that $\alpha$ is pure.
Thanks in advance.
P.S. Can someone suggest some good reference for pure and split extensions, where the proofs of the results have been done nicely?
Let $N$ be an $R$-module. We have to show that $N \to N \otimes_R S$ is injective. As $S$ is faithfully flat as an $R$-module, it suffices to prove this after tensoring with $S$. Hence it suffices to show that $N \otimes_R S \to N \otimes_R S \otimes_R S$, $n \otimes s \mapsto n \otimes 1 \otimes s$ is injective. This is true because there is a section, namely, $n \otimes s \otimes s' \mapsto n \otimes ss'$.