Finitely generated modules in exact sequence 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.
 A: $\require{begingroup} \begingroup$
$\def\coker{\operatorname{Coker}}$
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 $\coker{s_{M}}$.
$\endgroup$
A: I would consider this as a very special case of the Horseshoe lemma and prove it like that. This is essentially the same prove as in Makoto Kato wrote down, but his is written down more elementary.
A: Suppose $M'$ is generated by $x_1,\dots,x_n$ and $M''$ is generated by $z_1,\dots,z_m$.
Let $v(y_i) = z_i$ for $i = 1,\dots,m$.
Let $x \in M$.
Then there exist $b_1,\dots,b_m \in A$ such that $v(x) = b_1z_1 + \cdots + b_mz_m$.
Then $v(x) = v(b_1y_1 + \cdots + b_my_m)$.
Hence $x - (b_1y_1 + \cdots + b_my_m) \in \operatorname{Ker}v$.
Since $\operatorname{Ker}v = \operatorname{Im}u$, there exist $a_1,\dots,a_n \in A$ such that $x - (b_1y_1 + \cdots + b_my_m) = a_1u(x_1) + \cdots + a_nu(x_n)$.
Hence $M$ is generated by $u(x_1),\dots,u(x_n), y_1,\dots,y_m$.
