# Finitely generated modules over PID's

I'm trying to proof a statement about finitely generated modules over PID's, but I'm not sure if the statement is even true and if so how to continue my proof. I do however strongly suspect that the statement is true.

The statement is as follows: let $0 \to A \to B \to C \to 0$ be an exact sequence of $R$ modules, where $R$ is a principal ideal domain. Then if $B$ is finitely generated, then $A$ and $C$ are also finitely generated.

My attempt at a proof is as follows: because $R$ is a PID, every ideal is finitely generated. We can easily find an isomorphism between the product ring $R^{n}$ and $B$ where $n$ is the cardinality of a smallest set that generates $B$.

It's obvious that therefore, every sub module of $B$ corresponds with a unique sub module of $R^{n}$. Now note that sub modules of $R^{n}$ correspond with products of ideals in $R$, so therefore every sub module is finitely generated. (A cute, but probably irrelevant, result is that we can also show that there is also no infinite strictly increasing chain of submodule inclusions).

I now want to proof that $A$ and $C$ can be associated to some submodules of $R^{n}$, but I'm having some difficulties with the fact that my first arrow need not be injective and my second not surjective, since our sequence is not short.

• If $0 \to A \to B \to C \to 0$ is an exact sequence, then it is a short exact sequence by definition. Or do you mean $A \to B \to C$ without the zeros? – lhf Nov 25 '16 at 15:03
• @2015 I'm not really sure how the structure theorem over PID would help here, could you elaborate on that? – pokemonfan Nov 25 '16 at 19:10

A finitely generated module is a module $M$ for which there exists an $R$-module surjection $R^n\to M$, with $n$ a positive integer.

From there, assuming that $B$ is finitely generated, it follows that any quotient of $B$ is finitely generated. Hence $C$ is finitely generated.

Proving that $A$ is finitely generated amounts to proving that any submodule of a finitely generated module is finitely generated. You seem to already know the following result: Any submodule of $R^n$ is finitely generated. You can use it as follows: If $f:R^n\to B$ is a surjection (it exists since $B$ is finitely generated), then since $A$ is a submodule of $B$, consider the preimage $f^{-1}(A)$ of $A$ by $f$. It is a submodule of $R^n$, hence it is finitely generated. Since $A$ is a quotient of $f^{-1}(A)$, it is also finitely generated.

Now, I would like to discuss some sentences of you post which attracted my attention.

1. We can easily find an isomorphism between the product ring $R^n$ and $B$ where $n$ is the cardinality of a smallest set that generates $B$.

This is not true; it would mean that $B$ is a free module. What we have is a surjection from $R^n$ to $B$. (To see an example where there is no isomorphism, take $R= \mathbb{Z}$ and $B=\mathbb{Z}/2\mathbb{Z}$.)

1. Now note that sub modules of $R^n$ correspond with products of ideals in $R$

This is also untrue: look at the submodule $\{(m,m) \ | \ m\in \mathbb{Z}\}$ of the $\mathbb{Z}$-module $\mathbb{Z}^2$. What is true is that ideals of the ring $R^n$ are products of ideals of $R$.

1. I'm having some difficulties with the fact that my first arrow need not be injective and my second not surjective, since our sequence is not short.

If you have an exact sequence of the form $0\to A\stackrel{g}{\to} B$, then $g$ is automatically injective (compute its kernel, and see that it is zero). Similarly, in an exact sequence $B\stackrel{h}{\to} C \to 0$, the morphism $h$ is automatically surjective.

Therefore your exact sequence is a short exact sequence, as pointed out by lhf in the comments.