Lemma 15.15.6. Let $(R, \mathfrak m)$ be a local ring. Suppose that $\varphi : R^ m \to R^ n$ is a map of finite free modules. The following are equivalent

$\varphi $ is injective,

the rank of $\varphi $ is $m$ and the annihilator of $I(\varphi )$ in $R$ is zero.

If $R$ is Noetherian these are also equivalent to

the rank of $\varphi $ is $m$ and either $I(\varphi ) = R$ or it contains a nonzerodivisor.

Here the rank of $\varphi $ and $I(\varphi )$ are defined as in Algebra, Definition 10.102.5.

**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.
$\square$

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