# Finitely Generated Modules Over a PID

## The Basic Idea

We know what it means to have a module $M$ over a (commutative, say) ring $R$. We also know that if our ring $R$ is actually a field, our module becomes a vector space. But what happens if $R$ is "merely" a PID? Answer:

Recall from group theory that all finitely generated abelian groups can be classified up to isomorphism. So if you run into an abelian group of order 24, for instance, you know it's either going to have the same structure as $\mathbb{Z}/24$ or $\mathbb{Z}/3\oplus\mathbb{Z}/8$ or $\mathbb{Z}/3\oplus\mathbb{Z}/2\oplus \mathbb{Z}/4$ or $\mathbb{Z}/3\oplus\mathbb{Z}/2\oplus\mathbb{Z}/2\oplus\mathbb{Z}/2$. There are no other possibilities. In much the same way, we can classify all finitely generated modules over a PID. In fact, the result for abelian groups is a special case of this! (Notice $\mathbb{Z}$ is a PID and every abelian group is a $\mathbb{Z}$-module.)

Today we'll look at a proposition which, thanks to the language of exact sequences, is quite simple and from which the Fundamental Theorem of Finitely Generated Modules over a PID follows almost immediately. The information below is loosely based on section 12.1 of Dummit and Foote's *Abstract Algebra* as well as this article by K. Conrad.

## From English to Math

In short, the proposition says that every finitely generated module over a PID is comprised of a part which is free (i.e. has a basis) and a part which is not free:

Before the proof, let's look at two corollaries:

- If $n=0$, then $M$ is a torsion* module, i.e. $M=\text{Tor}(M)$. This is another way of saying $M$ is
*not*free: a basis for $M$ simply does not exist. Why? No finite subset $\mathcal{B}=\left\{m_1,\ldots,m_n\right\}$ of $M$ can be linearly independent! To see this, note that $M=\text{Tor}(M)$ means that for every $m_i\in\mathcal{B}\subset M$, there is a*nonzero*$r_i\in R$ such that $r_im_i=0$. So in particular $r_1m_1+\cdots+r_nm_n=0$ is a nontrivial relation.

- If $\text{Tor}(M)=0$ (and $n>0$) then $M$ is a free module. This is clear since $\text{Tor}(M)=0$ implies $M\cong R^n$ for some $n>0$. And indeed $R^n$ is a free module with bases $e_1,\ldots,e_n$ where $e_i=(0,\ldots,1,\ldots,0)$ (a 1 in the $i$th spot and zeros elsewhere).

To prove the proposition, we simply need to show the following is a split short exact sequence:

where $\text{Tor}(M)\hookrightarrow M$ is inclusion and $\pi : M \to M/\text{Tor}(M)$ is the natural projection map. The exactness of the sequence is immediate: $\text{Tor}(M)\hookrightarrow M$ is, of course, injective, $\pi$ is always surjective, and $\ker \pi=\text{Tor}(M)$ is indeed the image of the inclusion map. To see that the sequence splits, it's enough to prove $M/\text{Tor}(M)$ is free! Why? Because *all free modules are projective* and any time you have a projective module $P$, the exact sequence $0\to L\to M to P \to 0$, for any modules $L$ and $M$ always splits! (See proof here.) So let's show $M/\text{Tor}(M)$ is free. The following three facts will do it for us.

### Fact #1

**$M/\text{Tor}(M)$ is finitely generated.**

- This follows from the fact that $M$ is finitely generated and is actually true in general: quotients of finitely generated modules are finitely generated. (Here's the proof.)

### Fact #2

**$M/\text{Tor}(M)$ is torsion free.**

- Intuitively this makes sense: we've gotten rid of all the torsion elements by modding them out, so of course what's left over should be torsion free! (Here's the proof.)

### Fact #3

**If $R$ is a PID and $M$ is any finitely generated torsion free $R$-module, then $M$ is free.**

- This is precisely the second corollary following our main proposition above! (So at least you can believe it's true.) Of course to avoid a circular reasoning, we can prove it independently.

Now make the simple observation that Facts #1 and #2 allow us to apply Fact #3 to the $R$-module $M/\text{Tor}(M)$. So $M/\text{Tor}(M)$ is indeed free and, by our comments above, the sequence splits. Further since $M/\text{Tor}(M)$ is free it is isomorphic to $R^n$ for some $n\geq 0$. Thus

which completes the proof.

Before we close, let's quickly recall the Fundamental Theorem:

(Moreover, it can be shown that this expression is unique.) From our proposition above, it's clear that the Theorem is just one step away! One need only prove that

I won't include the proof here (it's about two pages long!), but you can find it in section 12.1, Theorem 4 of Dummit and Foote.

## So what's the takeaway here?

Using the machinery of homological algebra (e.g. short exact sequences) we were able to see that every finitely generated module over a PID is a direct sum of a free part and a not-free part. (And *that* reflects the fact that every element in the module either has torsion or it doesn't!) From there one obtains the Fundamental Theorem which says you can identify a finitely generated module over a PID simply by its **rank** (that's what the integer $n$ in the theorem is called) and the **invariant factors** - the sequence of the $a_i$'s with the divisibility property. Lastly, if we take our PID to be $\mathbb{Z}$ and our module $M$ to be an abelian group (i.e. a $\mathbb{Z}$-module), we obtain the familiar Fundamental Theorem of Finitely Generated Abelian Groups as a special case.

*Footnotes:*

* Quick reminder: the torsion submodule $\text{Tor}(M)$ of an $R$-module $M$ is defined to be the set of all elements $m\in M$ for which there exists a nonzero $r\in R$ so that $rm=0$.