# Over a principal ideal domain $R$, a submodule $N$ of a finitely-generated $R-$module M is finitely-generated.

Let $M$ be a finitely generated module on a PID, then $M=L_R(\left \{ b_1,b_2,..., b_n\right \})$. Let $N$ be a submodule of $M$, so $N=\left \langle n \right \rangle$ because we are in a PID ... Why it is necessary that $M$ be finitely generated, it is not enough that we are in a PID and already every submodule is generated by a single element And is thus finitely generated?

• Why should $N$ be generated by a single element? Ideals of $R$ are generated by a single element. There exist modules that cannot be generated by a single element: for instance, $R^2$. – egreg Mar 12 '17 at 16:34

## 1 Answer

Why is it necessary that $M$ be finitely generated?

Because $M$ is a submodule of itself. So, if every submodule of $M$ is finitely generated, $M$ itself is finitely generated.

Is it not enough that we are in a PID and already every submodule is generated by a single element?

Of course not!

The simplest example of a PID is a field. Let $V$ be a three-dimensional vector space over $\mathbb{R}$. Then $V$ has infinitely many subspaces which cannot be generated by a single element.

The definition of PID tells you that ideals of $R$ are generated by a single element.

In particular, every ideal is finitely generated, so $R$ is Noetherian. Over a Noetherian ring, submodules of finitely generated submodules are finitely generated.