The Stacks project

Proof. By Algebra, Theorem 10.85.4 the module $M$ is free. If we show the result holds for every finitely generated direct summand of $M$, then the lemma follows. Hence we may assume that $M$ is finite free. Write $N = \mathop{\mathrm{colim}}\nolimits _ i N_ i$ as a directed colimit of finite free modules, see Algebra, Theorem 10.81.4. Note that $u : M \to N$ factors through $N_ i$ for some $i$ (as $M$ is finite free). Denote $u_ i : M \to N_ i$ the corresponding $R$-module map. As $\overline{u}$ is injective we see that $\overline{u_ i} : M/\mathfrak mM \to N_ i/\mathfrak mN_ i$ is injective and remains injective on composing with the maps $N_ i/\mathfrak mN_ i \to N_{i'}/\mathfrak mN_{i'}$ for all $i' \geq i$. As $M$ and $N_{i'}$ are finite free over the local ring $R$ this implies that $M \to N_{i'}$ is a split injection for all $i' \geq i$. Hence for any $R$-module $Q$ we see that $M \otimes _ R Q \to N_{i'} \otimes _ R Q$ is injective for all $i' \geq i$. As $- \otimes _ R Q$ commutes with colimits we conclude that $M \otimes _ R Q \to N_{i'} \otimes _ R Q$ is injective as desired. $\square$