The Stacks project

Proof. Assume $u$ is injective. Our goal is to show $u$ is universally injective. First we choose a module $Q$ such that $N \oplus Q$ is free. On considering the map $N \oplus Q \to M \oplus Q$ we see that it suffices to prove the lemma in case $N$ is free. In this case $N$ is a directed colimit of finite free $R$-modules. Thus we reduce to the case that $N$ is a finite free $R$-module, say $N = R^{\oplus n}$. We prove the lemma by induction on $n$. The case $n = 0$ is trivial.

Let $u : R^{\oplus n} \to M$ be an injective module map with $M$ projective. Choose an $R$-module $Q$ such that $M \oplus Q$ is free. After replacing $u$ by the composition $R^{\oplus n} \to M \to M \oplus Q$ we see that we may assume that $M$ is free. Then we can find a direct summand $R^{\oplus m} \subset M$ such that $u(R^{\oplus n}) \subset R^{\oplus m}$. Hence we may assume that $M = R^{\oplus m}$. In this case $u$ is given by a matrix $A = (a_{ij})$ so that $u(x_1, \ldots , x_ n) = (\sum x_ i a_{i1}, \ldots , \sum x_ i a_{im})$. As $u$ is injective, in particular $u(x, 0, \ldots , 0) = (xa_{11}, xa_{12}, \ldots , xa_{1m}) \not= 0$ if $x \not= 0$, and as $R$ has property (P) we see that $a_{11}R + a_{12}R + \ldots + a_{1m}R = R$. Hence see that $R(a_{11}, \ldots , a_{1m}) \subset R^{\oplus m}$ is a direct summand of $R^{\oplus m}$, in particular $R^{\oplus m}/R(a_{11}, \ldots , a_{1m})$ is a projective $R$-module. We get a commutative diagram

with split exact rows. Thus the right vertical arrow is injective and we may apply the induction hypothesis to conclude that the right vertical arrow is universally injective. It follows that the middle vertical arrow is universally injective. $\square$