## Fun with Fractional Ideals

Note: despite what the title might make you think, I do go out.

After establishing that each Dedekind domain has unique factorization into prime ideals, let’s look at a concrete example. Since principal ideal domains are somewhat pointless here – you already know how to multiply positive rational numbers, for example (at least, I hope you do) – I’ll use the simplest example of a Dedekind domain that doesn’t have unique factorization: Z[SQRT(-5)]. Elements of the ring are of the form a+bs, where a and b are integers, and s is my shorthand notation for SQRT(-5) for the purposes of this post.

The fraction field of Z[s] is simply Q[s], since if a and b are rational numbers, then 1/(a+bs) = (abs)/(abs)(a+bs) = (abs)/(a^2+5b^2) and a^2+5b^2 is rational.

Every fractional ideal of Z[s] is either principal, or of the form a(2, 1+s). There is a fairly easy proof for it, but it requires more theory as background than I’m willing to clutter this blog with, so now you’ll have to make do with “Because I said so.”

For example, you may recall that (3, 1+s) is another ideal of the ring. To express it in the above form, first note that (3, 1+s)(2, 1+s) = (6, 2+2s, 3+3s, 2s-4) = (1+s), and (2, 1+s)(2, 1+s) = (4, 2+2s, 2s-4) = (2). So (3, 1+s) = (1+s)/(2, 1+s) = ((1+s)/2)(2/(2, 1+s)) = ((1+s)/2)(2, 1+s).

We can explicitly calculate the inverse of (2, 1+s), but it’s enough to know that it exists for the above calculation to work. Either way, (2)/(2, 1+s) = (2, 1+s), so (1)/(2, 1+s) = (1, (1+s)/2).

Another thing we can do with ideals, in a way, is add them. If A and B are (fractional) ideals, then A+B is the ideal of all sums (a+b), where a is in A and b in B; this is fairly clearly the smallest ideal containing both A and B. It’s a good idea to express A and B as products of prime ideals, and see how A+B will look in terms of these prime ideals.

To do that, it’s useful to introduce a new notation: if P1, P2, P3… P(n) are the distinct prime ideals dividing A, then we write A = (P1^a1)(P2^a2)…(P(n)^a(n)). If A is proper fractional, we can write it as B/(a) where B is integral, and then subtract the powers in (a) from those in B; then the a(i)’s can be positive or negative (or zero, but then we can cancel). We can now abandon the cumbersome Q = P^(-1) notation and use both P and Q to denote integral prime ideals.

If A is the product of P(i)^a(i)’s and B is the product of P(i)^b(i)’s, where the a(i)’s and b(i)’s can be zero, then A and B are both contained in C, the product of the P(i)^c(i)’s, where c(i) = min{a(i), b(i)}. So A+B is a subset of C. So what’s left to show is that C = A+B, or, equivalently, A/C + B/C = (A+B)/C = R.

A/C and B/C are both integral ideals of R, which are coprime: the powers they have in the P(i) decomposition are a(i)-c(i) and b(i)-c(i) respectively, which can’t be positive at the same time by the definition of c(i). But A/C + B/C is an integral ideal of R, which means it must have some decomposition into prime ideals. If any prime ideal Q divides A/C + B/C, then it divides A/C and B/C, which is impossible. So A/C + B/C is not contained in any prime ideal, which means it’s not contained in any maximal ideal, which means it’s the whole ring.

### 4 Responses to Fun with Fractional Ideals

1. suzie says:

upon reading this, i am now depressed.
yes, you do go out. that was you at the PEP party as the ringleader of feminist ladies, eh? congrats on being one of two guys present. that wus cool.

2. Alon Levy says:

Well, you know I go out… but Gordo, who made fun of me a few weeks ago for posting links on Saturday night instead of getting drunk or something, appears not to.

And yeah, that was me. It kind of depresses me that so few males care about abortion. I mean, I was pretty discombobulated when I realized that there actually were people who didn’t live in third-world backwaters who had moral problems with abortion.

Anyway, why did this post depress you?

3. suzie says: