# The Stacks Project

## Tag 0519

Lemma 10.5.3. Let $R$ be a ring. Let $$0 \to M_1 \to M_2 \to M_3 \to 0$$ be a short exact sequence of $R$-modules.

1. If $M_1$ and $M_3$ are finite $R$-modules, then $M_2$ is a finite $R$-module.
2. If $M_1$ and $M_3$ are finitely presented $R$-modules, then $M_2$ is a finitely presented $R$-module.
3. If $M_2$ is a finite $R$-module, then $M_3$ is a finite $R$-module.
4. If $M_2$ is a finitely presented $R$-module and $M_1$ is a finite $R$-module, then $M_3$ is a finitely presented $R$-module.
5. If $M_3$ is a finitely presented $R$-module and $M_2$ is a finite $R$-module, then $M_1$ is a finite $R$-module.

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 $$\text{Coker}(R^{\oplus m} \to M_1) \cong \text{Coker}(R^{\oplus n} \to M_2)$$ In particular we conclude that $\text{Coker}(R^{\oplus m} \to M_1)$ is a finite $R$-module. Since $\text{Im}(R^{\oplus m} \to M_1)$ 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 $$0 \to \text{Ker}(R^{\oplus n} \to M_1) \to \text{Ker}(R^{\oplus n + m} \to M_2) \to \text{Ker}(R^{\oplus m} \to M_3) \to 0$$ 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. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 381–400 (see updates for more information).

\begin{lemma}
\label{lemma-extension}
Let $R$ be a ring.
Let
$$0 \to M_1 \to M_2 \to M_3 \to 0$$
be a short exact sequence of $R$-modules.
\begin{enumerate}
\item If $M_1$ and $M_3$ are finite $R$-modules, then $M_2$ is a finite
$R$-module.
\item If $M_1$ and $M_3$ are finitely presented $R$-modules, then $M_2$
is a finitely presented $R$-module.
\item If $M_2$ is a finite $R$-module, then $M_3$ is a finite $R$-module.
\item If $M_2$ is a finitely presented $R$-module and $M_1$ is a
finite $R$-module, then $M_3$ is a finitely presented $R$-module.
\item If $M_3$ is a finitely presented $R$-module and $M_2$ is a finite
$R$-module, then $M_1$ is a finite $R$-module.
\end{enumerate}
\end{lemma}

\begin{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 $M_2$.

\medskip\noindent
Part (3) is immediate from the definition.

\medskip\noindent
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 \ref{lemma-lift-map} 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 \ref{lemma-snake}) we see that we get an isomorphism
$$\Coker(R^{\oplus m} \to M_1) \cong \Coker(R^{\oplus n} \to M_2)$$
In particular we conclude that $\Coker(R^{\oplus m} \to M_1)$
is a finite $R$-module. Since $\Im(R^{\oplus m} \to M_1)$
is finite by (3), we see that $M_1$ is finite by part (1).

\medskip\noindent
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 \ref{lemma-lift-map}
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.

\medskip\noindent
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
$$0 \to \Ker(R^{\oplus n} \to M_1) \to \Ker(R^{\oplus n + m} \to M_2) \to \Ker(R^{\oplus m} \to M_3) \to 0$$
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.
\end{proof}

Comment #691 by Anfang Zhou on June 14, 2014 a 11:40 pm UTC

Typo. In the proof of (4), it should be " $R^{\oplus k} \to R^{\oplus n} \to M_2$" , not" $R^{\oplus k} \to R^{\oplus n} \to M_1$".

There are also 5 comments on Section 10.5: Commutative Algebra.

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