In this post I’ll describe the structure theorem over PID’s which generalizes the following results:

- Finite dimensional vector fields over are all of the form ,
- The classification theorem for finitely generated abelian groups,
- The Frobenius normal form of a matrix,
- The Jordan decomposition of a matrix.

## 1. Some ring theory prerequisites

*Prototypical example for this section:* *.*

Before I can state the main theorem, I need to define a few terms for UFD’s, which behave much like : Our intuition from the case basically carries over verbatim. We don’t even need to deal with prime ideals and can factor elements instead.

**Definition 1**

If is a UFD, then is a **prime element** if is a prime ideal and . For UFD’s this is equivalent to the following property: if then either or is a unit.

So for example in the set of prime elements is . Now, since is a UFD, every element factors into a product of prime elements

**Definition 2**

We say **divides** if for some . This is written .

**Example 3** **(Divisibility in )**

The number is divisible by every element of . All other divisibility as expected.

**Ques 4**

Show that if and only if the exponent of each prime in is less than or equal to the corresponding exponent in .

Now, the case of interest is the even stronger case when is a PID:

**Proposition 5** **(PID’s are Noetherian UFD’s)**

If is a PID, then it is Noetherian and also a UFD.

*Proof:* The fact that is Noetherian is obvious. For to be a UFD we essentially repeat the proof for , using the fact that is principal in order to extract .

In this case, we have a Chinese remainder theorem for elements.

**Theorem 6** **(Chinese remainder theorem for rings)**

Let and be relatively prime elements, meaning . Then

*Proof:* This is the same as the proof of the usual Chinese remainder theorem. First, since we have for some and . Then we have a map

One can check that this map is well-defined and an isomorphism of rings. (Diligent readers invited to do so.)

Finally, we need to introduce the concept of a Noetherian -module.

**Definition 7**

An -module is **Noetherian** if it satisfies one of the two equivalent conditions:

- Its submodules obey the ascending chain condition: there is no infinite sequence of modules .
- All submodules of (including itself) are finitely generated.

This generalizes the notion of a Noetherian ring: a Noetherian ring is one for which is Noetherian as an -module.

**Ques 8**

Check these two conditions are equivalent. (Copy the proof for rings.)

## 2. The structure theorem

Our structure theorem takes two forms:

**Theorem 9** **(Structure theorem, invariant form)**

Let be a PID and let be any finitely generated -module. Then

for some satisfying .

**Corollary 10** **(Structure theorem, primary form)**

Let be a PID and let be any finitely generated -module. Then

where for some prime element and integer .

*Proof:* Factor each into prime factors (since is a UFD), then use the Chinese remainder theorem.

**Remark 11**

In both theorems the decomposition is unique up to permutations of the summands; good to know, but I won’t prove this.

## 3. Reduction to maps of free -modules

The proof of the structure theorem proceeds in two main steps. First, we reduce the problem to a *linear algebra* problem involving free -modules . Once that’s done, we just have to play with matrices; this is done in the next section.

Suppose is finitely generated by elements. Then there is a surjective map of -modules

whose image on the basis of are the generators of . Let denote the kernel.

We claim that is finitely generated as well. To this end we prove that

**Lemma 12** **(Direct sum of Noetherian modules is Noetherian)**

Let and be two Noetherian -modules. Then the direct sum is also a Noetherian -module.

*Proof:* It suffices to show that if , then is finitely generated. It’s unfortunately not true that (take ) so we will have to be more careful.

Consider the submodules

(Note the asymmetry for and : the proof doesn’t work otherwise.) Then is finitely generated by , \dots, , and is finitely generated by , \dots, . Let and let be elements of (where the ‘s are arbitrary things we don’t care about). Then and together generate .

**Ques 13**

Deduce that for a PID, is Noetherian.

Hence is finitely generated as claimed. So we can find another surjective map . Consequently, we have a composition

Observe that is the *cokernel* of the composition , i.e. we have that

So it suffices to understand the map well.

## 4. Smith normal form

The idea is now that we have reduced our problem to studying linear maps , which can be thought of as a generic matrix

for the standard basis , \dots, of and , \dots, of .

Of course, as you might expect it ought to be possible to change the given basis of such that has a nicer matrix form. We already saw this in *Jordan form*, where we had a map and changed the basis so that was “almost diagonal”. This time, we have *two* sets of bases we can change, so we would hope to get a diagonal basis, or even better.

Before proceeding let’s think about how we might edit the matrix: what operations are permitted? Here are some examples:

- Swapping rows and columns, which just corresponds to re-ordering the basis.
- Adding a multiple of a column to another column. For example, if we add times the first column to the second column, this is equivalent to replacing the basis
- Adding a multiple of a row to another row. One can see that adding times the first row to the second row is equivalent to replacing the basis

More generally, If is an invertible matrix we can replace with . This corresponds to replacing

(the “invertible” condition just guarantees the latter is a basis). Of course similarly we can replace with where is an invertible matrix; this corresponds to

Armed with this knowledge, we can now approach the following result.

**Theorem 14** **(Smith normal form)**

Let be a PID. Let and be free -modules and let be a linear map. Set .

Then we can select a pair of new bases for and such that has only diagonal entries , , \dots, and .

So if , the matrix should take the form

and similarly when .

**Ques 15**

Show that Smith normal form implies the structure theorem.

**Remark 16**

Note that this is not a generalization of Jordan form.

- In Jordan form we consider maps ; note that the source and target space are the
*same*, and we are considering one basis for the space . - In Smith form the maps are between
*different*modules, and we pick*two*sets of bases (one for and one for ).

**Example 17** **(Example of Smith normal form)**

To give a flavor of the idea of the proof, let’s work through a concrete example with the following matrix with entries from :

The GCD of all the entries is , and so motivated by this, we perform the **Euclidean algorithm on the left column**: subtract the second row from the first row, then three times the first row from the second:

Now that the GCD of is present, we move it to the upper-left by switching the two rows, and then kill off all the entries in the same row/column; since was the GCD all along, we isolate completely:

This reduces the problem to a matrix. So we just apply the Euclidean algorithm again there:

Now all we have to do is generalize this proof to work with any PID. It’s intuitively clear how to do this: the PID condition more or less lets you perform a Euclidean algorithm.

*Proof:* Begin with a generic matrix

We want to show, by a series of operations (gradually changing the given basis) that we can rearrange the matrix into Smith normal form.

Define to be any generator of the principal ideal .

**Claim 18** **(“Euclidean algorithm”)**

If and are entries in the same row or column, we can change bases to replace with and with something else.

*Proof:* We do just the case of columns. By hypothesis, for some . We must have now (we’re in a UFD). So there are and such that . Then

and the first matrix is invertible (check this!), as desired.

Let be the GCD of all entries. Now by repeatedly applying this algorithm, we can cause to appear in the upper left hand corner. Then, we use it to kill off all the entries in the first row and the first column, thus arriving at a matrix

Now we repeat the same procedure with this lower-right matrix, and so on. This gives the Smith normal form.

With the Smith normal form, we have in the original situation that

and applying the theorem to completes the proof of the structure theorem.

## 5. Applications

Now, we can apply our structure theorem! I’ll just sketch proofs of these and let the reader fill in details.

**Corollary 19** **(Finite-dimensional vector spaces are all isomorphic)**

A vector space over a field has a finite spanning set of vectors. Then for some , .

*Proof:* In the structure theorem, .

**Corollary 20** **(Frobenius normal form)**

Let where is a finite-dimensional vector space over an arbitrary field (not necessarily algebraically closed). Then one can write as a block-diagonal matrix whose blocks are all of the form

*Proof:* View as a -module with action . By theorem for some polynomials , where . Write each block in the form described.

**Corollary 21** **(Jordan normal form)**

Let where is a finite-dimensional vector space over an arbitrary field which is algebraically closed. Prove that can be written in Jordan form.

*Proof:* We now use the structure theorem in its primary form. Since is algebraically closed each is a linear factor, so every summand looks like for some .

*This is a draft of Chapter 15 of the Napkin.*