Lemma 10.82.12. Let $R \to S$ be a ring map. Let $M \to M'$ be a map of $S$-modules. The following are equivalent

1. $M \to M'$ is universally injective as a map of $R$-modules,

2. for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R$-modules,

3. for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R$-modules,

4. for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak q$ in $R$, and

5. for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak m$ in $R$.

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

$(M \otimes _ R N)_{\mathfrak q} = M_{\mathfrak q} \otimes _ R N = M_{\mathfrak q} \otimes _{R_{\mathfrak p}} N_{\mathfrak p}.$

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. $\square$

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