Re: Pari query

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Wed, Nov 10, 2021 at 03:00:01PM +0000, Harvey Rose wrote:
> Dear Bill        Many thanks for your prompt reply.  I have a printed
> version (2.4.2) of the documentation
> which I have not updated to save paper!!  One problem is that in the 2.4.2
> version the order of the
> variables in factorff  is (x,p,a)  whereas yours is (x,a,p) - p is the base
> prime and a is the defining
> polynomial of the finite field.  I should at least download a copy of the
> latest documentation, being
> old-fashioned I still prefer paper versions, it also helps with eye-strain.

Ah sorry, you are right, traditional GP use factorff(x,p,a), so it is

? factorff(x^5+x^3+1,2,a^5+a^2+1)
%2 =
[Mod(Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2))*x+Mod(Mod(1,2)*a^3+Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2)),1;Mod(Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2))*x+Mod(Mod(1,2)*a^3+Mod(1,2)*a+Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2)),1;Mod(Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2))*x+Mod(Mod(1,2)*a^3+Mod(1,2)*a^2+Mod(1,2)*a+Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2)),1;Mod(Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2))*x+Mod(Mod(1,2)*a^4+Mod(1,2)*a,Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2)),1;Mod(Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2))*x+Mod(Mod(1,2)*a^4+Mod(1,2)*a^3+Mod(1,2)*a^2+Mod(1,2)*a+Mod(1,2),Mod(1,2)*a^5+Mod(1,2)*a^2+Mod(1,2)),1]

With the latest version, both works, but this function is a bit
deprecated: we advise to use ffgen() to create finite field elements,
which are less cumbersome.

Cheers,
Bill.