$M$ finitely generated if submodule and quotient are finitely generated.

Solution 1:

Perhaps this works?

Suppose $M/N$ is finitely generated with generators $a_1+N,\dots,a_m+N$, and $N$ is finitely generated with generators $b_1,\dots,b_n$. Take any $m\in M$. Then $m+N=\sum_{i=1}^m r_i(a_i+N)=\left(\sum_{i=1}^m r_ia_i\right)+N$ viewing $M/N$ as an $R$-module.

This implies $m-\sum_{i=1}^m r_ia_i\in N$, and thus $m-\sum_{i=1}^m r_ia_i=\sum_{j=1}^n s_jb_j$ for some $r_i,s_j\in R$. Finally, $m=\sum_{i=1}^m r_ia_i+\sum_{j=1}^n s_jb_j$. So $M$ is generated by $a_i, b_j$ for $i=1,\dots, m$ and $j=1,\dots n$

Solution 2:

Take a generating set $\{\bar z_1, \ldots, \bar z_m\}$ for $M/N$. If $x$ is an element of $M$, then there exist $a_1, \ldots, a_n \in R$ such that the image $\bar x$ of $x$ in $M/N$ is \[ \bar x = a_1\bar z_1 + \cdots + a_n\bar z_n. \] Since the map on the right is surjective, you can choose lifts $\{z_1, \ldots, z_m\}$ in $M$ of the aforementioned generating set. Now, the element \[ x' = a_1z_1 + \cdots + a_nz_n \] has the same image as $x$ in $M/N$. Do you see a way to use the exactness in the middle?

Note that the sequence need not split: consider the $\mathbf Z$-submodule $2\mathbf Z$ of $\mathbf Z$.