Cubic Extensions

From: Michael Blanc (mblanc_at_znet.com)
Date: 06/24/04 

Date: Thu, 24 Jun 2004 17:00:41 +0000 (UTC)

Write f(x) = x^3 + Ax^2 + Bx + C, a polynomial in a single indeterminate
with rational
integral coefficients. Let Disc(f) = -4A^3C + A^2B^2 + 18ABC - 4B^3 - 27C^2,
the
discriminant of f(x), an integer.

Denote the square-free part of Disc(f) by d. The roots of f(x) will generate
a field that
contains sqrt(d). I wish to parameterize families of cubic polynomials by d,
and proceed
as follows.

Step 1. Substitute -dA for C in f(x):

Now Disc(f) = 4dA^4 + A^2B^2 - 18dA^2B - 4B^3 - 27d^2A^2 =
4dA^4 + A^2[B^2 - 18dB - 27d^2] - 4B^3.

Step 2. Substitute -dk^2 for B, and solve for k:

Disc(f) becomes 4dA^4 + A^2[d^2k^4+ 18d^2k^2 - 27d^2] + 4d^3k^6 =
d (4A^4 + dA^2 [k^4+ 18k^2 - 27] + 4d^2k^6 ).

Solving k^4 - 8k^3 + 18k^2 - 27 = (k-3)^3(k+1), gives two
parameterizations:

x^3 + Ax^2 - dx - dA = (x + A)(x^2 - d) is always reducible, but

x^3 + Ax^2 - 9dx - dA seems usually is not to be.

It occurs as an interesting question, whether all cubic extensions of the
rational
numbers can be enumerated as splitting fields for polynomials of this second
type. That
is, first sort the extensions by quadratic subfield, then index each family
by a single
parameter A. Even better if that coefficient A can be restricted to integer
values, and
still account for all the field extensions.

I recall that a few years ago something like this was published in Holland
for the case
d=1, comprising the abelian cubic extensions of Q. And ofcourse the other
extensions are
each cyclic over some Q(sqrt(d)), so perhaps a possible result like that can
be easily
extended. I have yet to find this paper.

Any leads?

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