From: Igor Schein Subject: Re: Inverse Galois Theory question Date: Fri, 09 Apr 1999 14:30:20 GMT Newsgroups: sci.math Keywords: Dihedral Galois group Franz Lemmermeyer wrote: > Igor Schein wrote: >> >> Hi, >> >> is there a known method to generate a polynomial >> whose Galois group is Dihedral group of order N, >> for arbitrary N? > If you can find a complex quadratic number field > whose class group is cyclic of order N, then its > Hilbert class field is dihedral of order 2N over Q. > The construction is quite explicit, but the coefficients > have a tendency to explode as N gets larger. See Actually, this was my motivation for posting the question. I noticed ( using PARI ) exactly the same thing, that given a negative fundamental quadratic discriminant with class number N, the polynomial representing its Hilbert class field has Galois group D(N). > Cox, David A. > Primes of the form \$x^2+ny^2\$. > Fermat, class field theory and complex multiplication. > Other approaches: > Semin. Theor. Nombres Bordx., Ser. II 4, No.1, 141-153 (1992). > J. Number Theory 34, No.2, 153-173 (1990) > franz Thanks for the references, I'll check out the reasoning behind this fact. Igor ============================================================================== From: Franz Lemmermeyer Subject: Re: Inverse Galois Theory question Date: Fri, 09 Apr 1999 17:28:33 +0200 Newsgroups: sci.math Igor Schein wrote: > Actually, this was my motivation for posting the question. > I noticed ( using PARI ) exactly the same thing, that > given a negative fundamental quadratic discriminant with > class number N, the polynomial representing its Hilbert > class field has Galois group D(N). [...] > Thanks for the references, I'll check out the reasoning behind > this fact. That's "simple" if you know class field theory: the class field H of the quadratic field k has Galois group isomorphic to Z/N, and the base extension k/Q is cyclic and acts on Z/N via the Artin isomorphism. What this means is that the nontrivial isomorphism S of k/Q acts on a generator of Z/N in the same way as it acts on the class c generating the class group. But c^S = c^{-1}, since c^{S+1} = 1 (the norm of an ideal is principal since the base field Q has class number 1), and voila: the complete extension H/Q has the dihedral group. The fact that H/Q is normal follows at once from the characterization of H as the maximal unramified abelian extension of k: since H/k has these properties, so has the conjugate extension H'/k, hence H = H'. This is all in Cox. You might also want to look at Cohn, Introduction to the Construction of Class fields (Dover). franz