The Stacks project

Proof. Let $N$ be an $R$-module. Let $\mathfrak q$ be a prime of $S$ lying over the prime $\mathfrak p$ of $R$. Then we have

Moreover, the same thing holds for $M'$ and localization is exact. Also, if $N$ is an $R_{\mathfrak p}$-module, then $N_{\mathfrak p} = N$. Using this the equivalences can be proved in a straightforward manner.

For example, suppose that (5) holds. Let $K = \mathop{\mathrm{Ker}}(M \otimes _ R N \to M' \otimes _ R N)$. By the remarks above we see that $K_{\mathfrak m} = 0$ for each maximal ideal $\mathfrak m$ of $S$. Hence $K = 0$ by Lemma 10.23.1. Thus (1) holds. Conversely, suppose that (1) holds. Take any $\mathfrak q \subset S$ lying over $\mathfrak p \subset R$. Take any module $N$ over $R_{\mathfrak p}$. Then by assumption $\mathop{\mathrm{Ker}}(M \otimes _ R N \to M' \otimes _ R N) = 0$. Hence by the formulae above and the fact that $N = N_{\mathfrak p}$ we see that $\mathop{\mathrm{Ker}}(M_{\mathfrak q} \otimes _{R_{\mathfrak p}} N \to M'_{\mathfrak q} \otimes _{R_{\mathfrak p}} N) = 0$. In other words (4) holds. Of course (4) $\Rightarrow$ (5) is immediate. Hence (1), (4) and (5) are all equivalent. We omit the proof of the other equivalences. $\square$