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Old 2017-10-03, 14:19   #11
Dr Sardonicus
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Feb 2017

132·23 Posts

Originally Posted by carpetpool View Post
I would like to announce one result involving the field K = Q(\zeta_{23}).


For a prime p = 1 (mod 23), p is a norm of a principal ideal if and only if there are 3 solutions to x^3-x-1 = 0 (mod p).
Let K be the field of n-th roots of unity. Let Cl(K) be the ideal class group of K, h its order. What you seek is a defining polynomial for the Hilbert Class Field of K. This will have degree h over K. Among the many fascinating properties of L/K, if P is a prime ideal of K, then POL splits into h distinct ideals (so [apart from possible extraneous factors of the polynomial discriminant] the polynomial splits into distinct linear factors over OK/P precisely when P is principal.

For n = 23, it turns out that the quadratic subfield k = Q(sqrt(-23)) already has class number 3, and (fortuitously) the class number does not increase when you extend up to K.

It is well known that the cubic polynomial f = x^3 - x - 1 defines the Hilbert Class Field L of k. Accordingly, by Class Field Theory, everything you would want to be true about factorizations of f mod p, is true. The prime ideal P in Ok is principal if and only if f splits completely in the residue field Ok/P. If -23 is a quadratic non-residue (mod p), then f has one linear and one quadratic factor (mod p); P = pOk is prime, and Ok/P is the field of p^2 elements.

If -23 is a quadratic residue (mod p), then pOk = PP' where P and P' are of degree 1; the residue fields Ok/P and Ok/P' both have p elements. In this case, f (mod p) remains irreducible if P and P' are non-principal, but splits into linear factors (mod p) if P and P' are principal.

If w^2 - w + 6 = 0, then P is principal when P = x + y*w for integers x and y; or p = x^2 + x*y + 6*y^2, or (simple argument) p = X^2 + 23*Y^2 for integers X and Y.

Fortuitously, K has the same class number as K, and the join KL has degree 3 over K, so f serves as a defining polynomial for the Hilbert Class Field of K as well as of k.

Unfortunately, this approach doesn't work when K is the field of 29th roots of unity. In this case, all the proper subfields of K have class number 1. But K itself has class number 8, so a defining polynomial for its Hilbert Class Field L has degree 8 over K. The class group is "elementary Abelian," the direct product of 3 cyclic groups of order 2. It is therefore possible (in theory) to determine L/K as the join of quadratic extensions of K, which could be done using Pari's rnfkummer(). Alas, I'm too lazy to try to work out the necessary subgroups of the class group

In general, determining the Hilbert Class Field of a cyclotomic field is... difficult.
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