Lemma 10.82.13. Let $\varphi : A \to B$ be a ring map. Let $S \subset A$ and $S' \subset B$ be multiplicative subsets such that $\varphi (S) \subset S'$. Let $M \to M'$ be a map of $B$-modules.

1. If $M \to M'$ is universally injective as a map of $A$-modules, then $(S')^{-1}M \to (S')^{-1}M'$ is universally injective as a map of $A$-modules and as a map of $S^{-1}A$-modules.

2. If $M$ and $M'$ are $(S')^{-1}B$-modules, then $M \to M'$ is universally injective as a map of $A$-modules if and only if it is universally injective as a map of $S^{-1}A$-modules.

Proof. You can prove this using Lemma 10.82.12 but you can also prove it directly as follows. Assume $M \to M'$ is $A$-universally injective. Let $Q$ be an $A$-module. Then $Q \otimes _ A M \to Q \otimes _ A M'$ is injective. Since localization is exact we see that $(S')^{-1}(Q \otimes _ A M) \to (S')^{-1}(Q \otimes _ A M')$ is injective. As $(S')^{-1}(Q \otimes _ A M) = Q \otimes _ A (S')^{-1}M$ and similarly for $M'$ we see that $Q \otimes _ A (S')^{-1}M \to Q \otimes _ A (S')^{-1}M'$ is injective, hence $(S')^{-1}M \to (S')^{-1}M'$ is universally injective as a map of $A$-modules. This proves the first part of (1). To see (2) we can use the following two facts: (a) if $Q$ is an $S^{-1}A$-module, then $Q \otimes _ A S^{-1}A = Q$, i.e., tensoring with $Q$ over $A$ is the same thing as tensoring with $Q$ over $S^{-1}A$, (b) if $M$ is any $A$-module on which the elements of $S$ are invertible, then $M \otimes _ A Q = M \otimes _{S^{-1}A} S^{-1}Q$. Part (2) follows from this immediately. $\square$

There are also:

• 2 comment(s) on Section 10.82: Universally injective module maps

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).