## If N and M/N are finitely generated as R-modules, then so is M

Let $R$ be a ring with 1, and suppose $M$ is a left $R$-module with submodule $N$. Suppose further that $N$ and $M/N$ are finitely generated. Prove that $M$ is finitely generated.

Say $M/N = (a_i + N \ |\ 1 \leq i \leq m)$ and $N = (b_j \ |\ 1 \leq j \leq n)$. We claim that $M = (a_i,b_j \ |\ 1 \leq i \leq m, 1 \leq j \leq n)$. Certainly $(\supseteq)$ holds. Now suppose $m \in M$. Note that $m + N = \sum_i r_i \cdot (a_i + N)$ $= (\sum_i r_i \cdot a_i) + N$. Now $m - \sum_i r_i \cdot a_i \in N$, so that $m - \sum_i r_i \cdot a_i = \sum_j s_j b_j$. Then $m = \sum_i r_i \cdot a_i + \sum_j s_j \cdot b_j \in (a_i, b_j \ |\ 1 \leq i \leq m, 1 \leq j \leq n)$. In particular, $M$ is finitely generated.