# Prove module $M$ is finitely generated if $N$ and $M/N$ are finitely generated

Let $R$ be a ring with $1$ and $N$ be a submodule of an $R$-module $M$.

Prove that $M$ is finitely generated if $N$ and $M/N$ are finitely generated.

There are two definitions of "finitely generated" for me to use:

1. $M$ is finitely generated if $\exists k\in \mathbb{Z}^+, m_1,m_2,\cdots,m_k\in M$ such that $M=Rm_1+Rm_2+\cdots+Rm_k$.
2. $M$ is finitely generated if there exists a surjective $R$-module homomorphism $\phi : \delta\rightarrow M$, where $\delta$ is free of finite rank.

Any suggestion on which definition will make the proof easier?

• I am new to modules. Using first definition, we have $N=Rn_1+\cdots+Rn_k$ and $M/N=Rm_1N+\cdots+Rm_lN$. How to get $M$ from this? – user130916 Apr 7 '14 at 23:26
• Let's pick an arbitrary $m\in M$, and let $\overline m \in M/N$ be the image of $m$ in $M/N$. Then $m = \overline m + n$ for some $n$. (in the last sentence $\overline m$ is $m$ "reduced modulo $N$") – chriseur Apr 7 '14 at 23:28