Let $R$ be a ring, $M$ an $R$-module and $U$ a submodule of $M$.
Show that, if $U$ and $M/U$ are finitely generated, then $M$ is finitely generated aswell.
I thought to maybe show this by taking (finite) generating systems of both $U$ and $M/U$, and to construct a new generating system, consisting of the previous ones. But I'm not entirely sure if this really always works. Thanks in advance!
Edit: If $(u_1, ..., u_n)$ is a finite generating system for $U$ and $(v_1, ..., v_m)$ a finite generating system of $M/U$, does this already implicate that the system $(u_1, ..., u_n, v_1, ..., v_m)$ generates $M$? Each $m \in M$ can be written as $m' + u$, with $m' \in M/U, u \in U$, and $m'$ can be generated via $(u_1, ..., u_n)$, and $u$ via $(v_1, ..., v_m)$, does this already implicate that our new system is indeed a finite generating system of $M$? Or is there more to it?
Your edit is basically the right idea but notationally there's a problem. You can't say every $m \in M$ can be written as $m' + u$ with $m' \in M/U$ and $u \in U$ because $M/U$ is not a submodule of $M$.
What you want to say is that $(v_1 + U, \ldots, v_m + U)$ are generators for $M/U$ so given $m \in M$ you first push it $M/U$ to get $m + U$. Then write $$m + U = c_1(v_1 + U) + \cdots + c_m(v_m + U) = (c_1v_1 + \cdots + c_mv_m) + U$$ so that $m - (c_1v_v + \cdots + c_mv_m) \in U$. Now that it's in $U$ write it as a linear expression involving the $u_i$ and then solve that expression for $m$.