Lemma 15.23.13. Let $R$ be a Noetherian ring. Let $\varphi : M \to N$ be a map of $R$-modules. Assume $M$ is finite and that for every prime $\mathfrak p$ of $R$ one of the following happens

$M_\mathfrak p \to N_\mathfrak p$ is an isomorphism, or

$\text{depth}(M_\mathfrak p) \geq 2$ and $\mathfrak p \not\in \text{Ass}(N)$.

Then $\varphi $ is an isomorphism.

**Proof.**
By Lemma 15.23.12 we see that $\varphi $ is injective. Let $N' \subset N$ be an finitely generated $R$-module containing the image of $M$. Then $\text{Ass}(N_\mathfrak p) = \emptyset $ implies $\text{Ass}(N'_\mathfrak p) = \emptyset $. Hence the assumptions of the lemma hold for $M \to N'$. In order to prove that $\varphi $ is an isomorphism, it suffices to prove the same thing for every such $N' \subset N$. Thus we may assume $N$ is a finite $R$-module. In this case, $\mathfrak p \not\in \text{Ass}(N) \Rightarrow \text{depth}(N_\mathfrak p) \geq 1$, see Algebra, Lemma 10.63.18. Consider the short exact sequence

\[ 0 \to M \to N \to Q \to 0 \]

defining $Q$. Looking at the conditions we see that either $Q_\mathfrak p = 0$ in case (1) or $\text{depth}(Q_\mathfrak p) \geq 1$ in case (2) by Algebra, Lemma 10.72.6. This implies that $Q$ does not have any associated primes, hence $Q = 0$ by Algebra, Lemma 10.63.7.
$\square$

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