Re: Pari query

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.

Regards Harvey

On Wed, 10 Nov 2021 at 12:01, Bill Allombert <Bill.Allombert@math.u-bordeaux.fr> wrote:

On Wed, Nov 10, 2021 at 11:25:28AM +0000, Harvey Rose wrote:
> Dear Pari expert
> I am trying to use factorff(x,a,p) and I keep getting error
> messages. For example I am wanting to factorise the polynomial
> t^5+t^3+1 modulo 2 over the finite field of order 32 generated
> by the polynomial t^5+t^2+1. I am assuming that the field has
> basis 1 and a, say, and the non-zero elements of the field can
> be represented by the powers a^n for n between 1 and 31, or as a
> vector space of dim 5 with basis elements (1,a, a^2,a^3,a^4).
>
> When I try the factorff function I keep getting the error message
> : incorrect priority in factormod, variable x <=t. or something
> very similar. I have tried using Mod(-,2) as coefficients of the
> polynomial x and/or polynomial p to no avail. So when you have a
> spare bit of time I would welcome any help here.

You need to use distinct variables for defining the polynomial and the
field, and the field variable must have lower priority than the
polynomial variable (x always have the highest priority).

So for example you can do
? factorff(x^5+x^3+1,a^5+a^2+1,2)
%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]
? liftall(%)
%3 =
[x+(a^3+1),1;x+(a^3+a+1),1;x+(a^3+a^2+a+1),1;x+(a^4+a),1;x+(a^4+a^3+a^2+a+1),1]

Alternatvely, you can first define a finite field generator with
a=ffgen((a^5+a^2+1)*Mod(1,2),'a)
and then do
factor(t^5+t^3+1,a)
%42 =
[t+(a^3+1),1;t+(a^3+a+1),1;t+(a^3+a^2+a+1),1;t+(a^4+a),1;t+(a^4+a^3+a^2+a+1),1]
This way you can escape the priority rule.

Cheers,
Bill.