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.
So for example in the set of prime elements is . Now, since is a UFD, every element factors into a product of prime elements
Now, the case of interest is the even stronger case when is a PID:
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.
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.
This generalizes the notion of a Noetherian ring: a Noetherian ring is one for which is Noetherian as an -module.
2. The structure theorem
Our structure theorem takes two forms:
Proof: Factor each into prime factors (since is a UFD), then use the Chinese remainder theorem.
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
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 .
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.
So if , the matrix should take the form
and similarly when .
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 .
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.
Now, we can apply our structure theorem! I’ll just sketch proofs of these and let the reader fill in details.
Proof: In the structure theorem, .
Proof: View as a -module with action . By theorem for some polynomials , where . Write each block in the form described.
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.