Re: Why is nfinit(f).zk not the same as nfbasis(f) ??

[Karim Belabas <Karim.Belabas@math.u-bordeaux1.fr>]

* Jeroen Demeyer [2010-08-19 21:30]:
> With the svn version of PARI, I get:
> 
> gp> f = x^3 + x^2 - 2*x + 8
> %1 = x^3 + x^2 - 2*x + 8
> gp> nfbasis(f)
> %2 = [1, x, 1/2*x^2 - 1/2*x]
> gp> nfinit(f).zk
> %3 = [1, 1/2*x^2 + 1/2*x - 1, x]
> 
> Is it normal to get different results here?  I agree that a basis is not
> canonical, but it's not clear to me why nfbasis() uses a different
> algorithm than nfinit().

It is normal. nfinit() calls [the internal function underlying]
nfbasis(), then spends some time finding a good LLL-reduced basis.

Cheers,

    K.B.
--
Karim Belabas, IMB (UMR 5251)  Tel: (+33) (0)5 40 00 26 17
Universite Bordeaux 1          Fax: (+33) (0)5 40 00 69 50
351, cours de la Liberation    http://www.math.u-bordeaux1.fr/~belabas/
F-33405 Talence (France)       http://pari.math.u-bordeaux1.fr/  [PARI/GP]
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