Algebraic Numbers, The Geometry of Numbers

The Geometry of Numbers

This theorem, which is the foundation for much of algebraic number theory, is due to Minkowski. (biography) It places certain constraints on discriminants and class groups of number fields. Needless to say, it's clever!

Let S be the ring of integers in a number field E/Q, and let H be an ideal in S. Embed S into complex space, and real space, as described in the previous section. Yes, you really need to understand these lattices before you proceed.

Build a generalized octahedron W in complex space as follows. The space that contains W has r1 real dimensions and r2 complex dimensions. The sum of the absolute values of the real coordinates, plus twice the sum of the norms of the complex coordinates, must be less than t. We are interested in the volume of such a shape. This is an exercise in multiple integration.

To start, the volume scales as tn, so we can set t = 1, and multiply by tn later.

Concentrate on 1 section, with all coordinates positive, and multiply by 2n later.

Run through the real coordinates first. The volume of a real simplex has already been computed. If the sum of the coordinates is bounded by x, in k dimensions, the volume is xk/k!. Move to the first complex coordinate and apply this to the quarter unit circle in the complex plane with radius ½. In polar coordinates, the integrand becomes r×(1-2r)k/k!. Integrate by parts, as r runs from 0 to ½, and θ runs from 0 to π/2, and get π/(8×(k+2)!).

Do this for each of the r2 complex coordinates and the volume of our region is (π/8)r2/n!. Multiply this by (2t)n to get the volume of W.

View the same space as n dimensional real space, and apply the standard embedding of S into Rn. Thus the lattice and the octahedron are superimposed.

Let v be the covolume of the lattice. If H is an ideal, its covolume becomes v×|H|. With this in mind, set t to the nth root of n!×(8/π)r2×v×|H|. Now W has a volume of 2n×v×|H|.

Since W is closed, convex, and symmetric, apply the point lattice theorem, and some nonzero x in H has its image in W. The sum of the norms of the conjugates of x is bounded by t. Their average is bounded by t/n.

The arithmetic mean bounds the geometric mean, hence the product of these norms, which is the norm of x, is bounded by (t/n)n. Substitute for t and get this.

|x| ≤ n!/nn × (8/π)r2 × v×|H|

Remember that v is the covolume of a modified lattice. The original lattice in complex space comes from the n field isomorphisms of E. This matrix, when multiplied by its transpose, gives the discriminant of the extension E/Q. This is often easier to determine. So rewrite the equation in terms of d, the discriminant of E/Q.

|x| ≤ n!/nn × (4/π)r2 × sqrt(d)×|H|

The coefficient on |H| becomes the constant l that proves the class group is finite.

Applications

Adjoin sqrt(-5) to Z. In an earlier section we found l = 10.472. Let's see what happens here. Use the standard basis of 1 and sqrt(-5). Build the 2×2 matrix of trace of bibj. The antidiagonal products are pure imaginary, and their trace is 0. The diagonal products are 1 and -5, and these are doubled for the trace. The determinant, or discriminant, is -20. Alternatively, the discriminant is the square of the difference between the two conjugates ±sqrt(-5). Remember that n = 2 and r2 = 1. Plug and chug, and l = 2.847. We know that this extension is not a ufd. It has a nontrivial class group. A generator for this group must have index 2, and must lie over 2. There can be only one totally ramified prime ideal. Therefore the class group is Z2. If you skip ahead to quadratic extensions, you'll find that the prime lying over 2, which generates the class group, is generated by 2 and 1+sqrt(-5).

Consider Z adjoin sqrt(2). The difference between the two conjugates is 2×sqrt(2), and the discriminant is 8. With n = 2 and r2 = 0, l = 1.414. A proper ideal cannot have index less than 2, hence this extension is a pid.

Adjoin sqrt(3) and get l = 1.732, another pid.

Adjoin (1+sqrt(5))/2 and get l = 1.118, another pid.

Discriminant is at Least 3

Let the integral ring S act as its own ideal. Select any nonzero x in S and its norm is an integer in R. The conjugates of x are all nonzero, hence |x| is at least 1. If the discriminant is too low, there is some x with |x| < 1, and that is impossible. Set |x| = 1 and solve for d, to see how small d can be.

sqrt(d) ≥ nn/n! × (π/4)r2

If n = 1 the extension is trivial. The basis is 1, the trace is 1, and the discriminant is 1. Assume n > 1.

Pair up i with n-i in n!. The product of these two factors is the square of their geometric mean. This only increases if we use the square of the arithmetic mean, or (n/2)2. Rais this to the n/2 for the number of pairs. Then there's another n at the end, and if n is even, there's an n/2 in the middle. The denominator n! is at most nn/2n-1.

sqrt(d) ≥ 2r1 × πr2 / 2

Since π is less than 4, this is smallest when the embeddings are all complex. Thus d is at least (π/2)2, or 2.467. Since d is an integer, the discriminant is at least 3.

By adjoining (1+sqrt(-3))/2, the discriminant can indeed be 3.

If T is an integral domain, and a free R module of rank n, let E be its fraction field, and extend T up to S, the integral ring. Since T is a subring of S it creates a sublattice, with an even larger covolume. This translates to a larger discriminant. Thus the discriminant of T is at least 3, and at least 12 if T is a proper subring of S, i.e. if T is not integrally closed. For example, adjoin sqrt(5) to Z. This is a subring of the pid that was described earlier. Its discriminant is 20, whereas the discriminant of the pid, produced by adjoining (1+sqrt(5))/2, is 5.