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