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?


It's really a preference. If this is your first exposure to modules, and you like doing things concretely, 1. would be better.

If you know about short exact sequences, then 2. will give you a more "hands off" solution.

  • $\begingroup$ 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? $\endgroup$ – user130916 Apr 7 '14 at 23:26
  • 1
    $\begingroup$ 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 $N$") $\endgroup$ – chriseur Apr 7 '14 at 23:28

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