Re: polredabs() again

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

On Thu, 3 Apr 2003, Igor Schein wrote:

> On Fri, Apr 04, 2003 at 12:18:34AM +0200, Karim BELABAS wrote:
> [deleted]
> > On the other hand, the current specification of polredabs is quite useless.
> > There's no application whatsoever for a "polynomial of absolute smallest
> > T2-norm". It's not even guaranteed to have minimal discriminant, or to
> > yield smallest coefficients. The only one I can see is to give a
> > pseudo-canonical representative for the field (this helps table builders,
> > less isomorphism tests...)
>
> An appropriate observation.  Here's an example:
>
> ? p1=x^40-444796*x^30+91848351606*x^20-11536862712826396*x^10+672749994932560009201;
> time = 0 ms.
> ? p2=polredabs(p1)
> time = 10,160 ms.
> %5 = x^40 - 45*x^38 + 1325*x^36 - 32390*x^34 + 714250*x^32 -
> 12409705*x^30 + 186413550*x^28 - 2499178250*x^26 + 29415089025*x^24 -
> 274143015750*x^22 + 2260522643650*x^20 - 16155295928625*x^18 +
> 89060060373750*x^16 - 251196632524250*x^14 + 670266070790625*x^12 -
> 1618446757085125*x^10 + 2840715854153125*x^8 - 389148315762500*x^6 +
> 53298286370000*x^4 - 7270708200000*x^2 + 922368160000
>
> Anybody in his right mind, I believe, would prefer p1 over p2 as a
> representative for this field.

I might prefer p2:

(00:25) gp > poldisc(p2)/poldisc(p1)*1.
time = 70 ms.
%1 = 2.227868806881974253905227025 E-46

    Karim.
-- 
Karim Belabas                     Tel: (+33) (0)1 69 15 57 48
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