Back to Algebra

After a short combinatorical hiatus, I’m back to algebra. After writing about what you can do with Dedekind domains, I’m going to talk about why you should care – in particular, why rings of integers of number fields are Dedekind domains. But first, I’m going to define number fields more rigorously.

First, if K is a field and L is a field that contains K, we can regard L as a vector space over K. Vector spaces over fields are algebraic structures that mimic structures like R^2, the set of all points of the form (x, y) where x and y are any real numbers. Vector spaces are abelian groups under addition, and have another operation, scalar multiplication, that has the properties we expect the operation of multiplying (x, y) by z to get (xz, yz) to have.

The important thing about vector spaces is that they have a dimension. I’m just going to talk about finite dimensions, but infinite-dimensional spaces exist. If K is any field, then a vector space of the form K^n, the set of ordered n-tuples with elements from K, has dimension n. A bit more formally, if we can find elements x1, x2… x(n) that span the space, that is, every element in the space can be written as k1x1 + k2x2 + … + k(n)x(n), then we say the dimension is n or less.

I don’t want to get too much into linear algebra here, so I’ll just note four important things about the dimension n, without proof. Any set of size smaller than n doesn’t span the space, pretty much by definition. A set of size n that spans the space is linearly independent, that is, if k1x1 + k2x2 + … + k(n)x(n) = 0 then k1 = k2 = … = k(n) = 0. A set of size greater than n is always linearly dependent. And by labeling x1 as (1, 0, …, 0), x2 as (0, 1, 0, …, 0), etc., and x(n) as (0, …, 0, 1), we can just regard the space as K^n.

A number field K is a field that contains Q and has finite dimension over it. Not all fields containing Q have finite dimension; R has infinite dimension, since every finite subset of the infinite set {SQRT(2), SQRT(3), SQRT(5), SQRT(7), SQRT(11)…} is linearly independent over Q, and we can make it as big as we like.

If K is a number field, then every element of K is algebraic over Q; that is, every k in K is the root of some polynomial with coefficients in Q (even in Z, since we can multiply by all the denominators of the coefficients). If the dimension is n, then the set {1, k, k^2, …, k^n} is linearly dependent, which gives us a polynomial over Q that k satisfies. If k is any element of a field containing Q, then k is algebraic iff Q(k), the smallest field containing Q and k, has finite dimension over Q. In fact, I know two proofs that every number field K is of the form Q(k) for some k, but one of them requires too much theory and the other is so tedious that to me it’s no better than “Because I say so.”

So far, I talked about number fields. But in every number field, there’s something called the ring of integers, which simply consists of all elements that satisfy monic polynomials over Z. For example, if K = Q(SQRT(2)) = {a + bSQRT(2): a and b are in Q}, then the ring of integers, O(K), is just Z[SQRT(2)] = {a + bSQRT(2): a and b are in Z}. But if K = Q(SQRT(5)), then since (1+SQRT(5))/2 satisfies the monic polynomial x^2 + x – 1 = 0, O(K) needs to be bigger, Z[(1+SQRT(5))/2] = {a + b(1+SQRT(5))/2: a and b are in Z} (it contains SQRT(5), which is just -1 + 2*(1+SQRT(5))/2).

I know it’s dense, so I’ll hold off the proof that O(K) is actually a ring for next time. Then I’ll show that as a group, O(K) is the same as Z^n (where n is the same as the dimension of K over Z), which will help prove it’s a Dedekind domain.