Since Ben is being frivolous I’m going to try to pick up the math blogging. I’m finally back in Berkeley (after a delightful afternoon at coblogger David’s new apartment in Boston).
The last week at Mathcamp I taught a class on quadratic imaginary number theory. While preparing for this class (and with a bunch of hints from Jared Weinstein) I finally understand the relationship between two “different” uses of Minkowski’s theorem in elementary algebraic number theory. Warning, this is a bit detailed for a blog post.
The first use appears on the Ross problem sets. We want to show that any rational integer p for which -1 is a quadratic residue modulo p is of the form . Since -1 is a quadratic residue, there exists x such that . So look at the lattice of points (where the congruence is componentwise). It is easy to see that for any lattice point here we have . In order to get we need only find a nonzero lattice point inside the circle . Since the area of this circle is while four times the are of the fundamental domain is 4p, Minkowski’s theorem guarantees the existence of such a lattice point.
The other use of Minkowski’s theorem I had in mind is the famous Minkowski bound. The usual statement of this (for quadratic imaginary number fields) is that in the ring of integers R of the number field any ideal class contains an ideal of norm smaller than B (where B is up to an annoying factor of 2). I didn’t want to talk about ideal classes, but if you restrict your attention to trying to distinguish number rings with unique factorization, then you can use a more elementary statement of the Minkowski bound. Namely, if every rational prime smaller than B factors into primes (not just irreducibles) in R, then R has unique factorization. (That is, if every ideal of small norm is principle, then all ideals are principle.)
Let’s relate this to the previous use of Minkowski’s theorem. First we need some way to see when rational integers factor into a product of primes in R. The number field is the splitting field of some polynomial f (either or ). Let z be a root of f. Since, we see that p is prime in R if and only if f is prime in . Since polynomial rings over a field have unique factorization and since f has degree 2, we see that p is prime if and only if f doesn’t have a root in .
On the other hand, suppose that p has a factor q in R. Then is a quotient of . Since it is a nontrivial quotient of a ring with elements it must have p elements, and is thus a field. Hence q is prime.
Combining the last two paragraphs we see that, in order to prove unique factorization in R we need only show that if f has a nontrivial root in then p is the norm of something in R. This is again a question about finding a certain nonzero lattice point. We again build a lattice of points with . However, we cannot guarantee a lattice point of norm smaller than 2p anymore, since d could be large.
Here induction saves the day! Suppose that p is the smallest rational prime causing a problem. Now suppose that we have a lattice point with for k smaller than p. Since k is smaller than p, we see that k factors as a product of primes in R. With a little work this actually implies that there is another element whose norm is p (basically you cancel out the prime factors of k from both sides). Hence, we only need to find a lattice point with . Now the area of the ellipse grows quadratically, while the area of the fundamental domain grows linearly. Hence if the first problem p is larger than the Minkowski bound then there’s no problem at all!
Explicitly, suppose that , then the area of the ellipse is while the area of the fundamental domain is 4p. Thus, the Minkowski bound is . The formula in the other case is not much trickier.