Lemma 38.8.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map, and $N$ an $S$-module. Assume
$R$ is a Noetherian ring,
$S$ is a Noetherian ring,
$N$ is a finite $S$-module, and
for any finite $R$-module $Q$, any $\mathfrak q \in \text{Ass}_ S(Q \otimes _ R N)$ satisfies $IS + \mathfrak q \not= S$.
Then the map $N \to N^\wedge $ of $N$ into the $I$-adic completion of $N$ is universally injective as a map of $R$-modules.
Proof. We have to show that for any finite $R$-module $Q$ the map $Q \otimes _ R N \to Q \otimes _ R N^\wedge $ is injective, see Algebra, Theorem 10.82.3. As there is a canonical map $Q \otimes _ R N^\wedge \to (Q \otimes _ R N)^\wedge $ it suffices to prove that the canonical map $Q \otimes _ R N \to (Q \otimes _ R N)^\wedge $ is injective. Hence we may replace $N$ by $Q \otimes _ R N$ and it suffices to prove the injectivity for the map $N \to N^\wedge $.
Let $K = \mathop{\mathrm{Ker}}(N \to N^\wedge )$. It suffices to show that $K_{\mathfrak q} = 0$ for $\mathfrak q \in \text{Ass}(N)$ as $N$ is a submodule of $\prod _{\mathfrak q \in \text{Ass}(N)} N_{\mathfrak q}$, see Algebra, Lemma 10.63.19. Pick $\mathfrak q \in \text{Ass}(N)$. By the last assumption we see that there exists a prime $\mathfrak q' \supset IS + \mathfrak q$. Since $K_{\mathfrak q}$ is a localization of $K_{\mathfrak q'}$ it suffices to prove the vanishing of $K_{\mathfrak q'}$. Note that $K = \bigcap I^ nN$, hence $K_{\mathfrak q'} \subset \bigcap I^ nN_{\mathfrak q'}$. Hence $K_{\mathfrak q'} = 0$ by Algebra, Lemma 10.51.4. $\square$
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