Re: Shorter way to convert polynomial in x^3 to x?

[Richard Heylen <richardheylen@gmail.com>]

Thanks Bill and Karim,

I'm glad to be taught new stuff about gp/pari.

Richard

On 16 November 2013 23:28, Bill Allombert
<Bill.Allombert@math.u-bordeaux1.fr> wrote:
> On Sat, Nov 16, 2013 at 10:50:32PM +0100, Karim Belabas wrote:
>> * Richard Heylen [2013-11-16 22:43]:
>> > ? f=factor(elldivpol(ellinit([0,1]),13))[1,1]
>> > %32 = 13*x^12 + 52*x^9 + 1536*x^6 + 1024*x^3 + 256
>> > ? Pol(vector(poldegree(f)/3+1,i,polcoeff(f,poldegree(f)-3*i+3)))
>> > %33 = 13*x^4 + 52*x^3 + 1536*x^2 + 1024*x + 256
>> >
>> >
>> > I tried using subst(f,x,y^(1/3)) but to no avail. Is there a short way
>> > of doing this?
>>
>> (22:50) gp > substpol(f,x^3,x)
>> %2 = 13*x^4 + 52*x^3 + 1536*x^2 + 1024*x + 256
>
> Eventually Richard idea can be made to work:
> ? lift(subst(f,x,Mod(x,x^3-y)))
> %9 = 13*y^4+52*y^3+1536*y^2+1024*y+256
>
> (i.e. y^(1/3) can be denoted by Mod(x,x^3-y) in GP)
>
> Cheers,
> Bill.
>