Integral Bases

At last, I’m going to show that every ring of integers of a number field has the additive structure of Z^n, where n is the degree of the field over Q. To do that, I’ll use the discriminant. The point is to show that the ring has an integral basis, that is, a set of n elements that are linearly independent over Q and that span the whole ring over Z.

If K = Q(a) is a number field of degree n and a is an algebraic integer, then consider the disc(a) = disc{1, a, a^2, …, a^(n-1)}. The underlying set is linearly independent over Q, so disc(a) != 0. Since |disc{x1, x2, … x(n)}| is a non-negative integer, and we can choose the x(j)’s to make the discriminant nonzero, we can talk about the least positive value of the discriminant, say D. Then, there exists a set X = {x1, x2, … x(n)} whose discriminant has absolute value D.

Since disc(X) is nonzero, the elements of X are a Q-basis for K. In particular, every element of O(K) can be written as a linear combination of them over Q in a unique way. It’s then enough to show that all coefficients in the linear combination will actually be in Z.

But suppose on the contrary that y is an element of O(K) that has a proper rational coefficient, say the x1 coefficient. Denote it by p/q, where q > 1. We can write p as p‘ + kq where p‘ and k are integers and 0 < p‘ < q. Then y‘ = ykx1 has x1 coefficient p‘/q, where 0 < p‘/q < 1.

Now, consider disc(Y), Y = {y‘, x2, x3, …, x(n)}. If the matrices corresponding to the sets X and Y are A and B respectively, then we can say B is obtained from A by multiplying the first column by p‘/q and adding some rational multiples of the other columns. By the rules of determinants, |B| = (p‘/q)|A|, so disc(Y) = (p‘/q)*disc(X), so that 0 < disc(Y) < disc(X). But that contradicts the minimality of disc(X), so the element y is nonexistent, and the set X is in fact an integral basis.

Although K always has a rational basis of the form {1, a, a^2, …, a^(n-1)}, this doesn’t necessarily hold for O(K). There’s an explicit counterexample, but it requires tedious computations to prove what is in fact an integral basis over just the one ring.

Since O(K) has an integral basis, we can write it additively like Z^n, that is as n-tuples of integers, with integral basis (1, 0, 0, …, 0), (0, 1, 0, …, 0), …, (0, 0, 0, …, 1). This says nothing about the multiplicative structure of O(K), except that if k is in Z, then k*(a1, a2, …, a(n)) = (k*a1, k*a2, …, k*a(n)). So the ideal generated by k consists of n-tuples whose values are all divisible by k. In particular, O(K)/(k) has size k^n, which is finite.

Furthermore, if b is any nonzero element of O(K), then b divides some integer k, namely N(b). To see why, note that N(b)/b is the quotient of two nonzero elements of K, so it’s in K. Further, N(b) is the product of all conjugates of b, so N(b)/b is the product of all conjugates of b, except b itself. The conjugates are all algebraic integers, so N(b)/b is in B. Then it’s in O(K), which is just the intersection of K and B.

Now, let I be any nonzero ideal of O(K). I contains some principal ideal, which will then contain some ideal generated by an integer, (k). So O(K)/I will have at most as many elements as O(K)/(k) – in other words, it’ll have finitely many elements.

An ideal I of a ring R is prime iff R/I is an integral domain. It’s maximal iff R/I is a field. And if R is a finite integral domain, it’s a field (proof later today). So if I in O(K) is prime, it’s maximal.

In addition, any increasing sequence of ideals of O(K) will induce a decreasing sequence of the numbers of elements of the quotient rings. So it’ll have to terminate, since otherwise we’ll get quotient rings with a negative number of elements, which is absurd.

So O(K) is Noetherian and has no non-maximal prime ideals except (0). To see that it’s integrally closed, recall the proof that B is a ring, which uses the fact that a is an algebraic integer iff the set {1, a, a^2, …} is finitely generated over Z. This works over every ring: a is algebraic over O(K) iff {1, a, a^2, …} is finitely generated over O(K).

But suppose that a in K is algebraic over O(K). Then {1, a, a^2, …} is generated by, say, {1, a, a^2, …, a^m}. Additively, we can then embed O(K)[a] into K^m. But that embeds it into Z^mn, so the set {1, a, a^2, …} is finitely generated over Z, and a is in B. That shows that a is in O(K), so that O(K) is integrally closed.

One Response to Integral Bases

  1. […] The proof that every ring of integers of a number field has an integral basis relies on two ideas: first, that O(K) has n elements that are linearly independent over Q (where n is as usual the degree of K), and second, that if such a set is not an integral basis then we can find another set of smaller discriminant. Part two works for any nonzero ideal of O(K) – just do a global search and replace and write I instead of O(K). […]

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: