Re: some vexing cubic polynomials

From: Robin Chapman (rjc_at_ivorynospamtower.freeserve.co.uk)
Date: 06/09/04 

Date: Wed, 09 Jun 2004 07:58:42 +0100

Michael Blanc wrote:

> I'll need to lay out some machinery here, just to ask my question.
>
> A rational prime integer of the form p = 3k + 1, has an Eisenstein
> factorization:
>
> p = (A + B*w)(A + B*wbar), where A,B (and k) are integers,
>
> and w = (1/2)(-1 + sqrt(-3)), wbar = (1/2)(-1 - sqrt(-3)).
>
>
> Also, given some monic integral cubic polynomial f in a single
> indeterminant x,
>
> f(x) = x^3 + a*x^2 + b*x + c
>
> the discriminant of f is an integer D = a^2b^2 + 18abc - 4a^3c - 4b^3 -
> 27a^2,
> which will be a square iff either f(x) has three integer roots, or f(x)
> gives rise to
> an abelian extension of F over Q , i.e. galois with automorphism group of
> order 3.
>
> Further, such F is contained, it would appear, in a cyclotomic extension
> of Q generated by
> a primitive Nth root of one, PHI(N), where either:
>
> N is of the form pqr..., the product of finitely many distinct rational
> primes,
> each with residue 1 mod 3; or
> N = 9pqr..., nine times such a product.

Yes. Every abelian extension is contained in a cyclotomic field.
If the conductor is N, then 3 must divide phi(N)
(the order of the Galois group of Q(\zeta_N)). Thus N has a factor of
9 or a prime = 1 (mod 3). Some group theory shows that if F is contained
in Q(zeta_N) where 27|N or some prime p = 2 (mod 3) then F is contained
in a smaller cyclotomic field.

> Call the smallest such N for which PHI(N) contains F, the conductor of F.
>
> That's enough to at least state this question:
>
> For an abelian cubic extension F over Q, what is the _smallest_ value of
> D^1/2 for
> a monic irreducible integral cubic f(x) whose splitting field is F? I
> doubt that such a
> root-discriminant could be less than the conductor. I would like it to be
> equal to the
> conductor, but this does not seem to be always possible.

Let alpha denote a zero of f. The discriminant of F = Q(alpha) is
N^2 (this follows from the famous Fuhrerdiskriminantelproduktformel).
Then D will equal N^2 k^2 where k is the index of Z[alpha] in O_F
-- the ring of algebraic integers in F. So, there is an f with
discriminant N^2 if and only if O_F is monogenic: O_F = Z[alpha]
for some alpha in O_F. This is not always true. 

-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_