**Proof.**
The first statement is immediate from the definition. During the rest of the proof we will use the results of Lemma 10.5.3 without further mention.

Let $\varphi : N \to M$ satisfy the assumptions of (2). Suppose that $N$ is generated by $x_1, \ldots , x_ n$. By Definition 10.90.1 the kernel $K$ of the induced map $R^{\oplus n} \to M$, $e_ i \mapsto \varphi (x_ i)$ is of finite type. Hence $\mathop{\mathrm{Ker}}(\varphi )$ which is the image of the composition $K \to R^{\oplus n} \to N$ is of finite type. This proves (2).

Let $\varphi : N \to M$ satisfy the assumptions of (3). By (2) the kernel of $\varphi $ is of finite type and hence by (1) it is coherent.

With the same hypotheses let us show that $\mathop{\mathrm{Coker}}(\varphi )$ is coherent. Since $M$ is finite so is $\mathop{\mathrm{Coker}}(\varphi )$. Let $\overline{x}_ i \in \mathop{\mathrm{Coker}}(\varphi )$. We have to show that the kernel of the associated morphism $\overline{\Psi } : R^{\oplus n} \to \mathop{\mathrm{Coker}}(\varphi )$ is finite. Choose $x_ i \in M$ lifting $\overline{x}_ i$. Choose additionally generators $y_1, \ldots , y_ m$ of $\mathop{\mathrm{Im}}(\varphi )$. Let $\Phi : R^{\oplus m} \to \mathop{\mathrm{Im}}(\varphi )$ using $y_ j$ and $\Psi : R^{\oplus m} \oplus R^{\oplus n} \to M$ using $y_ j$ and $x_ i$ be the corresponding maps. Consider the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & R^{\oplus m} \ar[d]_\Phi \ar[r] & R^{\oplus m} \oplus R^{\oplus n} \ar[d]_\Psi \ar[r] & R^{\oplus n} \ar[d]_{\overline{\Psi }} \ar[r] & 0 \\ 0 \ar[r] & \mathop{\mathrm{Im}}(\varphi ) \ar[r] & M \ar[r] & \mathop{\mathrm{Coker}}(\varphi ) \ar[r] & 0 } \]

with exact rows. By Lemma 10.4.1 we get an exact sequence $\mathop{\mathrm{Ker}}(\Psi ) \to \mathop{\mathrm{Ker}}(\overline{\Psi }) \to 0$. Since $\mathop{\mathrm{Ker}}(\Psi )$ is a finite $R$-module, we see that $\mathop{\mathrm{Ker}}(\overline{\Psi })$ is finite.

Statement (4) follows from (3).

Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $R$-modules. It suffices 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 $x_1, \ldots , x_ n$ be finitely many elements of $M_2$. We have to show that the module of relations $K$ between them is finite. Consider the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & 0 \ar[r] \ar[d] & \bigoplus _{i = 1}^{n} R \ar[r] \ar[d] & \bigoplus _{i = 1}^{n} R \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 } \]

with obvious notation. By the snake lemma we get an exact sequence $0 \to K \to K_3 \to M_1$ where $K_3$ is the module of relations among the images of the $x_ i$ in $M_3$. Since $M_3$ is coherent we see that $K_3$ is a finite module. Since $M_1$ is coherent we see that the image $I$ of $K_3 \to M_1$ is coherent. Hence $K$ is the kernel of the map $K_3 \to I$ between a finite module and a coherent module and hence finite by (2).
$\square$

## Comments (1)

Comment #8089 by Steven Sam on

There are also: