Finitely generated modules in exact sequence

Solution 1:

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 $u(x_1),\dots,u(x_n), y_1,\dots,y_m$.

Solution 2:

$\require{begingroup} \begingroup$ $\def\coker{\operatorname{Coker}}$

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 $\coker{s_{M}}$. $\endgroup$