Lemma 10.82.14. Let $R$ be a ring and let $M \to M'$ be a map of $R$-modules. If $M'$ is flat, then $M \to M'$ is universally injective if and only if $M/IM \to M'/IM'$ is injective for every finitely generated ideal $I$ of $R$.

**Proof.**
It suffices to show that $M \otimes _ R Q \to M' \otimes _ R Q$ is injective for every finite $R$-module $Q$, see Theorem 10.82.3. Then $Q$ has a finite filtration $0 = Q_0 \subset Q_1 \subset \ldots \subset Q_ n = Q$ by submodules whose subquotients are isomorphic to cyclic modules $R/I_ i$, see Lemma 10.5.4. Since $M'$ is flat, we obtain a filtration

of $M' \otimes _ R Q$ by submodules $M' \otimes _ R Q_ i$ whose successive quotients are $M' \otimes _ R R/I_ i = M'/I_ iM'$. A simple induction argument shows that it suffices to check $M/I_ i M \to M'/I_ i M'$ is injective. Note that the collection of finitely generated ideals $I'_ i \subset I_ i$ is a directed set. Thus $M/I_ iM = \mathop{\mathrm{colim}}\nolimits M/I'_ iM$ is a filtered colimit, similarly for $M'$, the maps $M/I'_ iM \to M'/I'_ i M'$ are injective by assumption, and since filtered colimits are exact (Lemma 10.8.8) we conclude. $\square$

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