Bill Allombert on Thu, 10 Oct 2013 20:45:23 +0200

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Re: Degree extension

On Thu, Oct 10, 2013 at 05:57:06PM +0200, wrote:
> Dear All,
> I am Laura, a new member of the list of the
> users of PARI.


> I have a question. I have the roots of
> some polynomials of the form y^2=x^3+Ax+B.
> For example p:=y^2=x^3+2*x+5.
> I call a, b and c the roots of p.
> I would like to calculate the degree
> of the number field
> Q(sqrt(a-b),sqrt(a-c),sqrt(b-c),sqrt(-1)).
> Is it possible with PARI?

Yes, this is possible 
Let P=x^3+A*x+B be your polynomial
and K=Q(sqrt(a-b),sqrt(a-c),sqrt(b-c),sqrt(-1))

1) Build the Galois closure of P as follow:
  (This depends on the Galois group of P, but this will work in both case)

2) Compute the roots of P in the field Q[X]/P:
  Now the roots are given by a=N[1], b=N[2], c=N[3]
  in term of a root alpha of S.

3) Compute the minimal polynomial of a-b as follow
  By Galois theory, it has the same degree as S.

4) The minimal polynomial of sqrt(a-b) is a factor of Mab(x^2)

5) Build the tensor product Q[x]/Msab \otimes Q(sqrt(-1))
MsabI = polcompositum(Msab,x^2+1)[1]

6) Factor Msab over L=Q[X]/MsabI:
  This will be given in term of a root beta of MsabI.

7.1) If Msab split in linear factor, then K=L and the degree
is poldegree(MsabI). This can be checked with

7.2) otherwise, we have to go one step higher. Set 

In that case K=Q[X]/MsabcI and the degree is poldegree(MsabcI)