Re: bug in polredabs() (fwd)

[Igor Schein <igor@txc.com>]

On Fri, Dec 04, 1998 at 08:26:04PM -0500, Gerhard Niklasch wrote:
> In further response to
> > > Message-Id: <19981204193944.K14565@io.txc.com>
> > > Date: Fri, 4 Dec 1998 19:39:44 -0500
> > > From: Igor Schein <igor@txc.com>
> subsequent to
> > Message-Id: <199812050059.BAA14280@pchelwig1.mathematik.tu-muenchen.de>
> > Date: Sat, 5 Dec 1998 01:59:18 +0100 (MET)
> from yours truly:
> 
> > [...]
> > > So x^16+48 and x^16+3 generate the same number field,
> > 
> > No, they don't define the same field.
> 
> Cute.  x^8+3 and x^8+48 already define distinct fields (quadratic
> extensions of the same quartic field -- totally complex of discriminant
> 432 --) which, however, both have class number 1, the same regulator
> 24.0787745..., and the same order of the torsion unit subgroup (6th
> roots of unity).
> 
> > They define two distinct
> > fields which happen to have the same discriminant 2^48*3^15.
> 
> ...and the same class number (1), and presumably the same regulator,
> although this is hard to tell at default realprecision.  (The bnf[8][3]
> components differ by more than the computed regulators - almost 3%.)

Strange, I don't get 3% difference:

? bnfinit(x^8+48)[8][3]==bnfinit(x^8+3)[8][3]
%68 = 1

The equality is precision-independent, because I tried it at several
different \p.

A tiny difference between regulators can be attributed to round-off
arithmetic: 

? bnfinit(x^8+3)[8][2]-bnfinit(x^8+48)[8][2]
%76 = 1.75162308011204004200000000000 E-46

If you consider all polynomials of form f(m,n)=x^8+2^(4*m-4)*3^(2*n-1)
with m,n being positive integers, their number field are isomorphics
iff m is the same, i.e.
f(a,b) ~ f(c,d) iff a==c.

Igor