Re: bug in polredabs() (fwd)

[Peter-Lawrence.Montgomery@cwi.nl]

Gerhard Niklasch <nikl@mathematik.tu-muenchen.de> writes

> In response to
> > Message-Id: <19981204193944.K14565@io.txc.com>
> > Date: Fri, 4 Dec 1998 19:39:44 -0500
> > From: Igor Schein <igor@txc.com>
> [...]
> > So x^16+48 and x^16+3 generate the same number field,
> 
> No, they don't define the same field.  They define two distinct
> fields which happen to have the same discriminant 2^48*3^15.
> 
> (If they did, then the field would also contain a 16th root of
> (-48)/(-3) = 16, or in other words, a 4th root of 2.  But it doesn't
> even contain a square root of 2 -- try to nffactor x^2-2 over the
> result of nfinit(polredabs(y^16+3)).  


     A field can contain a 16th root of 16 without a 4th root of 2.
It might instead have a fourth root of -2 or a square root of
+-1 +- sqrt(-1),

    I agree that the fields are not isomorphic.
If we try to factor x^16 + 48 over Q[y] where y^16 + 3 = 0,
we get only the four degree-4 factors of x^16 - 16 y^16.

    Peter Montgomery


                    GP/PARI CALCULATOR Version 2.0.11 (beta)
                             unknown 32-bit version
                (readline disabled, extended help not available)

                           Copyright (C) 1989-1998 by
          C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.

Send bug reports, suggestions and patches to pari@math.u-bordeaux.fr
Type ? for help.

   realprecision = 38 significant digits
   seriesprecision = 16 significant terms
   format = g0.38

parisize = 20000000, primelimit = 500000, buffersize = 30000
   echo = 1 (on)
? factornf(x^16+48,y^16+3)
%1 = 
[Mod(1, y^16 + 3)*x^4 + Mod(2*y^2, y^16 + 3)*x^2 + Mod(2*y^4, y^16 + 3) 1]

[Mod(1, y^16 + 3)*x^4 + Mod(-2*y^2, y^16 + 3)*x^2 + Mod(2*y^4, y^16 + 3) 1]

[Mod(1, y^16 + 3)*x^4 + Mod(-2*y^4, y^16 + 3) 1]

[Mod(1, y^16 + 3)*x^4 + Mod(2*y^4, y^16 + 3) 1]

? quit
Good bye!