TITLE: Finitely generated modules in exact sequence
QUESTION [9 upvotes]: For $A$-modules and homomorphisms $0\to M'\stackrel{u}{\to}M\stackrel{v}{\to}M''\to 0$ is exact. Prove if $M'$ and $M''$ are fintely generated then $M$ is finitely generated.

REPLY [3 votes]: $\require{begingroup} \begingroup$
$\def\coker{\operatorname{Coker}}$
Since $M',M''$ are finitely generated, we can pick surjections $s_{M'}:R^m\to M'$ and $s_{M''}:R^n\to M''$. Now consider the following commutative diagram:
$\require{AMScd}$
\begin{CD}
0 @>>>R^m @>>> R^m\oplus R^n @>>> R^n@>>>0 \\
@. @VV{s_{M'}}V @. @VV{s_{M''}}V\\
0 @>>>M' @>{u}>>M @>{v}>> M'' @>>> 0\\
\end{CD}
where the vertical arrows are the surjections. Then as $M\to M''$ is a surjection and $R^n$ is projective (since free), we get map $w:R^n\to M$ so that the composition $v\circ w$ is the same as the surjection $s_{M''}:R^n\to M''$. We can then define a map from $s_M:R^{m+n}=R^m\oplus R^n\to M$ as follows: send $(x,y)\mapsto (u\circ s_{M'})(x) + w(y)$.
Now by snake lemma, we have exact sequence $\coker{s_{M'}}\to\coker{s_M}\to\coker{s_{M''}}$. Then $\coker{s_{M'}}, \coker{s_{M''}}$ are trivial by assumption, so is $\coker{s_{M}}$.
$\endgroup$