*Disclaimer:**This post is a quick discussion of some number theory related things; it is not meant to be rigorous. Throughout, I take to be a base field and to be an extension of ; towards the end of the post will* *represent the Hilbert Class Field of .*

Recently, I’ve spent a lot of time thinking about (non-)unique factorisation. This is a topic that, on some level, most people are familiar with; even at school, the Fundamental Theorem of Arithmetic is at the very least alluded to.

Fairly early on in some algebra course, we’ll see examples of rings which are no longer unique factorisation domains (UFDs); a classic example being , where ; all these numbers on the right hand side are irreducible and non-associate, so unique factorisation fails. For the rest of this post, I’m mainly going to be considering number fields and their associated ring of integers ; I am a number theorist after all!

Later in this post, I’ll discuss some ways we can deal with this when it is an issue, but first I’ll discuss some aspects of unique factorisation that are discussed a little less. Perhaps you don’t really care what irreducible factorisation of a number you pick, you only care about the number of elements in this factorisation; it seems (to me at least) that could be quite safe. Perhaps all irreducible factorisations have the same number of elements in the product? Wishful thinking, doesn’t work.

If the field has class number one, we’re fine. We can show that has class number one if and only if is a UFD. If has class number 2, we’re also fine, but beyond this, we run into problems (expect a post soon-ish about class numbers, the Class Number One problem is a favourite of mine!).

Theorem 1All irreducible factorisations of have the same length (number of irreducibles in the factorisation) if and only if has class number 1 or 2.

So, can we deal with this? Kind of, depending on what we care about. People familiar with the subject will have been screaming about ideals from the get go. For a number field , the ring of integers is a Dedekind Ring, and we have the following result.

Theorem 2Let be a dedekind ring. Every ideal of other than and has a factorisation into nonzero prime ideals of which is unique up to the order of the factors.

What we’re saying is, if we don’t mind thinking about as a principal ideal rather than an element, we are sorted! We have a unique factorisation into prime ideals. This actually “solves” the mystery of where the irreducible factorisations come from. Considering the previous example in , we find that and , giving us one factorisation from the prime ideal decomposition. The other comes from noting , and .

The problem here is these ideals aren’t principal, so unique factorisation doesn’t flow down to the level of elements. Perhaps we can come up with a different approach. Could we perhaps, given number field , find a finite extension such that is a unique factorisation domain (i.e. has class number one)? It turns out, while this is sometimes possible, it’s not true in general. Shafarevich and Golod were able to show that there are number fields with no such extension.

If we reduce what we’re asking for, maybe we can get a partial solution. Instead of asking for unique factorisation in our larger field, perhaps we ask that all ideals of are principal in ; that is, is principal in . This turns out to be possible; For every number field we can find its Hilbert Class Field , the maximal abelian unramified extension of . Then by the Principal Ideal Theorem, all ideals of is principal in .

To continue the example we’ve been working on, consider with . I claim that is the Hilbert Class Field of ; we can show this by considering the prime ideals lying above as generates the class group of . After some calculation we find that and which are principal as we desired. Similarly to before, by regrouping these, we can obtain the two factorisations of 6 we have previously seen.

We still have problems here; the elements generating these principal ideals in general need not be themselves irreducible. Thus, we don’t obtain a unique factorisation into irreducibles on the level of elements still, as commented earlier. However, it is interesting to note that any irreducible factorisation in the base field comes from regrouping the products of the generators of the principal ideals we get in the Hilbert Class Field.

The Hilbert Class Field has other nice properties, my favourite being that the degree of the extension is equal to the Class Number of . This (kind of) goes some way to explaining why some people refer to the class number as a measure of “how close” the ring of integers of a number field is to being a UFD.

For further, more detailed reading, I recommend this paper; in future posts I’ll be going into more detail on things mentioned on this page, in particular class numbers/groups and class field theory. Any issues spotted or questions, please comment or tweet me!