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 *x*1, *x*2… *x*(*n*) that span the space, that is, every element in the space can be written as *k*1*x*1 + *k*2*x*2 + … + *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 *k*1*x*1 + *k*2*x*2 + … + *k*(*n*)*x*(*n*) = 0 then *k*1 = *k*2 = … = *k*(*n*) = 0. A set of size greater than *n *is always linearly dependent. And by labeling *x*1 as (1, 0, …, 0), *x*2 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* + *b*SQRT(2): *a* and *b* are in **Q**}, then the ring of integers, O(K), is just **Z**[SQRT(2)] = {*a* + *b*SQRT(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.