TITLE: $M$ finitely generated if submodule and quotient are finitely generated.
QUESTION [15 upvotes]: I know this must be easy. I'm new to modules, so perhaps I must be missing  something important.

Let $M$ be an $R$-module. Show that if there exists a submodule $N$ such that $N$ and $M/N$ are finitely generated, then $M$ is finitely generated.

If the exact sequence $0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0 $ (where the second and third arrows are inclusion and projection respectively) splits then $M$ would be isomorphic to the direct sum of $M$ and $M/N$ and the conclusion would follow (I think). But I can't show that it splits.
Thank you.

REPLY [28 votes]: 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 $j=1,\dots n$