$M$ finitely generated $\nRightarrow$ $N$ and $M/N$ finitely generated

34 Views Asked by At

In an algebra lecture we looked at the following lemma:

Let $R$ be a ring, $M$ an $R$-module and $N \subset M$ a submodule. Then

$N$ and $M/N$ are finitely generated $\implies$ $M$ is also finitely generated.

I understand the proof for this lemma and it makes sense to me. 

Our professor also mentioned the converse:

$M$ finitely generated $\implies$ $M/N$ finitely generated

$M$ finitely generated + $R$ noetherian $\implies$ $N$ and $M/N$ finitely generated

Here I don't understand why we don't have:

$M$ finitely generated $\implies$ $N$ and $M/N$ finitely generated

Since $N$ is a submodule of $M$ and therefore $(N,+)$ is a subgroup of $(M,+)$, I would have intuitively said that $N$ also must be finitely generated. But this must be false since we need the condition that $R$ is noetherian for that to be the case.

I would greatly appreciate any help.