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2017-10-03, 14:19   #11
Dr Sardonicus

Feb 2017
Nowhere

132·23 Posts

Quote:
 Originally Posted by carpetpool I would like to announce one result involving the field K = Q(\zeta_{23}). **Conjecture/Observation** 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
;-D

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