The Stacks project

Proof. By assumption there exists a nonzero element $x \in R$ such that for every $f \in \mathfrak m$ we have $f^ n x = 0$. Say $I = (f_1, \ldots , f_ r)$. Then $x$ is in the kernel of $R \to \bigoplus R_{f_ i}$. Hence we see that there exists a nonzero $y \in R$ such that $f_ i y = 0$ for all $i$, see Algebra, Lemma 10.24.4. As $y \in \text{Ann}_ R(I)$ we win. $\square$

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. $\square$

Proof. The implication (1) $\Rightarrow $ (2) is Lemma 15.15.3. It is clear that (3) and (4) are equivalent. We have (2) $\Rightarrow $ (3), (4) by Algebra, Lemma 10.82.4. Part (5) is a special case of (4). Assume (5). Let $I = (a_1, \ldots , a_ n)$ be a proper finitely generated ideal of $R$. As $I \not= R$ we see that $R \to R^{\oplus n}$, $x \mapsto (xa_1, \ldots , xa_ n)$ is not a split injection. Hence it has a nonzero kernel and we conclude that $\text{Ann}_ R(I) \not= 0$. Thus (1) holds. $\square$

Proof. If any matrix coefficient of $\varphi $ is not in $\mathfrak m$, then we apply Algebra, Lemma 10.102.2 to write $\varphi $ as the sum of $1 : R \to R$ and a map $\varphi ' : R^{m-1} \to R^{n-1}$. It is easy to see that the lemma for $\varphi '$ implies the lemma for $\varphi $. Thus we may assume from the outset that all the matrix coefficients of $\varphi $ are in $\mathfrak m$.

Suppose $\varphi $ is injective. We may assume $m > 0$. Let $\mathfrak q \in \text{WeakAss}(R)$ so that $R_\mathfrak q$ is an auto-associated ring. Then $\varphi $ induces a injective map $R_\mathfrak q^ m \to R_\mathfrak q^ n$ which is universally injective by Lemmas 15.15.2 and 15.15.3. Thus $\varphi : \kappa (\mathfrak q)^ m \to \kappa (\mathfrak q)^ n$ is injective. Hence the rank of $\varphi \bmod \mathfrak q$ is $m$ and $I(\varphi \otimes \kappa (\mathfrak q))$ is not the zero ideal. Since $m$ is the maximum rank $\varphi $ can have, we conclude that $\varphi $ has rank $m$ as well (ranks of matrices can only drop after base change). Hence $I(\varphi ) \cdot \kappa (\mathfrak q) = I(\varphi \otimes \kappa (\mathfrak q))$ is not zero. Thus $I(\varphi )$ is not contained in $\mathfrak q$. Thus none of the weakly associated primes of $R$ are weakly associated primes of the $R$-module $\text{Ann}_ R I(\varphi )$. Thus $\text{Ann}_ R I(\varphi )$ has no weakly associated primes, see Algebra, Lemma 10.66.4. It follows from Algebra, Lemma 10.66.5 that $\text{Ann}_ R I(\varphi )$ is zero.

Conversely, assume (2). The rank being $m$ implies $n \geq m$. Write $I(\varphi ) = (f_1, \ldots , f_ r)$ which is possible as $I(\varphi )$ is finitely generated. By Algebra, Lemma 10.15.5 we can find maps $\psi _ i : R^ n \to R^ m$ such that $\psi \circ \varphi = f_ i \text{id}_{R^ m}$. Thus $\varphi (x) = 0$ implies $f_ i x = 0$ for $i = 1, \ldots , r$. This implies $x = 0$ and hence $\varphi $ is injective.

For the equivalence of (1) and (3) in the Noetherian local case we refer to Algebra, Proposition 10.102.9. If the ring $R$ is Noetherian but not local, then the reader can deduce it from the local case; details omitted. Another option is to redo the argument above using associated primes, using that there are finitely many of these, using prime avoidance, and using the characterization of nonzerodivisors as elements of a Noetherian ring not contained in any associated prime. $\square$

Proof. Let $\varphi ^ t : R^ n \to R^ n$ be the transpose of $\varphi $. The lemma claims that $\varphi ^ t$ is injective. With notation as in Lemma 15.15.6 we see that the rank of $\varphi ^ t$ is $n$ and that $I(\varphi ) = I(\varphi ^ t)$. Thus we conclude by the equivalence of (1) and (2) of the lemma. $\square$