**Proof.**
Proof of (1). If $x_1, \ldots , x_ n$ are generators of $M_1$ and $y_1, \ldots , y_ m \in M_2$ are elements whose images in $M_3$ are generators of $M_3$, then $x_1, \ldots , x_ n, y_1, \ldots , y_ m$ generate $M_2$.

Part (3) is immediate from the definition.

Proof of (5). Assume $M_3$ is finitely presented and $M_2$ finite. Choose a presentation

\[ R^{\oplus m} \to R^{\oplus n} \to M_3 \to 0 \]

By Lemma 10.5.2 there exists a map $R^{\oplus n} \to M_2$ such that the solid diagram

\[ \xymatrix{ & R^{\oplus m} \ar[r] \ar@{..>}[d] & R^{\oplus n} \ar[r] \ar[d] & M_3 \ar[r] \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 } \]

commutes. This produces the dotted arrow. By the snake lemma (Lemma 10.4.1) we see that we get an isomorphism

\[ \mathop{\mathrm{Coker}}(R^{\oplus m} \to M_1) \cong \mathop{\mathrm{Coker}}(R^{\oplus n} \to M_2) \]

In particular we conclude that $\mathop{\mathrm{Coker}}(R^{\oplus m} \to M_1)$ is a finite $R$-module. Since $\mathop{\mathrm{Im}}(R^{\oplus m} \to M_1)$ is finite by (3), we see that $M_1$ is finite by part (1).

Proof of (4). Assume $M_2$ is finitely presented and $M_1$ is finite. Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M_2 \to 0$. Choose a surjection $R^{\oplus k} \to M_1$. By Lemma 10.5.2 there exists a factorization $R^{\oplus k} \to R^{\oplus n} \to M_2$ of the composition $R^{\oplus k} \to M_1 \to M_2$. Then $R^{\oplus k + m} \to R^{\oplus n} \to M_3 \to 0$ is a presentation.

Proof of (2). Assume that $M_1$ and $M_3$ are finitely presented. The argument in the proof of part (1) produces a commutative diagram

\[ \xymatrix{ 0 \ar[r] & R^{\oplus n} \ar[d] \ar[r] & R^{\oplus n + m} \ar[d] \ar[r] & R^{\oplus m} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 } \]

with surjective vertical arrows. By the snake lemma we obtain a short exact sequence

\[ 0 \to \mathop{\mathrm{Ker}}(R^{\oplus n} \to M_1) \to \mathop{\mathrm{Ker}}(R^{\oplus n + m} \to M_2) \to \mathop{\mathrm{Ker}}(R^{\oplus m} \to M_3) \to 0 \]

By part (5) we see that the outer two modules are finite. Hence the middle one is finite too. By (4) we see that $M_2$ is of finite presentation.
$\square$

## Comments (1)

Comment #691 by Anfang Zhou on

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