**Proof.**
The first statement is immediate from the definition.

Let $\varphi : N \to M$ satisfy the assumptions of (2). First, $\mathop{\mathrm{Im}}(\varphi )$ is finite, hence coherent by (1). In particular $\mathop{\mathrm{Im}}(\varphi )$ is finitely presented, so applying Lemma 10.5.3 to the exact sequence $0\to \mathop{\mathrm{Ker}}(\varphi )\to N\to \mathop{\mathrm{Im}}(\varphi )\to 0$ we see that $\mathop{\mathrm{Ker}}(\varphi )$ is finite. To prove that $\mathop{\mathrm{Coker}}(\varphi )$ is coherent, let $E \subset \mathop{\mathrm{Coker}}(\varphi )$ be a finite subomdule, and let $E'$ be its inverse image in $M$. From the exact sequence $0 \to \mathop{\mathrm{Ker}}(\varphi ) \to E' \to E \to 0$ and since $\mathop{\mathrm{Ker}}(\varphi )$ is finite we conclude by Lemma 10.5.3 that $E'\subset M$ is finite, hence finitely presented because $M$ is coherent. The same exact sequence then shows that $E$ is finitely presented, whence our claim.

Part (3) follows immediately from (1) and (2).

Let $0 \to M_1 \xrightarrow {i} M_2 \xrightarrow {p} M_3 \to 0$ be a short exact sequence of $R$-modules as in (4). It remains to prove that if $M_1$ and $M_3$ are coherent so is $M_2$. By Lemma 10.5.3 we see that $M_2$ is finite. Let $N_2 \subset M_2$ be a finite submodule. Put $N_3 = p(N_2) \subset M_3$ and $N_1 = i^{-1}(N_2) \subset M_1$. We have an exact sequence $0 \to N_1 \to N_2 \to N_3 \to 0$. Clearly $N_3$ is finite (as a quotient of $N_2$), hence finitely presented (as a finite submodule of $M_3$). It follows by Lemma 10.5.3 (5) that $N_1$ is finite, hence finitely presented (as a finite submodule of $M_1$). We conclude by Lemma 10.5.3 (2) that $M_2$ is finitely presented.
$\square$

## Comments (2)

Comment #8089 by Steven Sam on

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