# $M$ finitely generated if submodule and quotient are finitely generated.

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.

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Can you do it for vector spaces? –  Mariano Suárez-Alvarez Jan 18 '12 at 2:06
By the way, the sequence does not split in general! It is therefore good that you were not able to prove it does :) –  Mariano Suárez-Alvarez Jan 18 '12 at 4:03
@Mariano: thank you. What's an example that it doesn't split? –  Weltschmerz Jan 18 '12 at 12:35
@Dylan: Thank you. The first map, though, is not inclusion... –  Weltschmerz Jan 18 '12 at 22:22

Take a generating set $\{\bar z_1, \ldots, \bar z_m\}$ for $M/N$. If $x$ is an element of $M$, then there exist $a_1, \ldots, a_n \in R$ such that the image $\bar x$ of $x$ in $M/N$ is $\bar x = a_1\bar z_1 + \cdots + a_n\bar z_n.$ Since the map on the right is surjective, you can choose lifts $\{z_1, \ldots, z_m\}$ in $M$ of the aforementioned generating set. Now, the element $x' = a_1z_1 + \cdots + a_nz_n$ has the same image as $x$ in $M/N$. Do you see a way to use the exactness in the middle?

Note that the sequence need not split: consider the $\mathbf Z$-submodule $2\mathbf Z$ of $\mathbf Z$.

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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 $j=1,\dots n$

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