The Stacks project

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. $\square$