Cubic root in a finite field

[Irene]

Hi!

 I have the following:

p=3700001

Fp2=GF(p^2, 'b')

and I want to compute the cubic root of 344694*b + 1653339.

How can I do it? Because when I write (344694*b + 1653339)^(1/3) it gives me as result 1, and the same for every element that I consider in Fp2, or for every exponent 1/n.

[John Cremona]

Here is one way:

sage: p=3700001
sage: Fp2.<b>=GF(p^2)
sage: a = 344694*b + 1653339

Now either

sage: x = polygen(Fp2)
sage: (x^3-a).roots(multiplicities=False)
[401927*b + 661235, 259308*b + 3298074, 3038766*b + 3440693]

or just


sage: a.nth_root(3)
259308*b + 3298074
sage: a.nth_root(3,all=True)
[259308*b + 3298074, 3038766*b + 3440693, 401927*b + 661235]

Since the field contains the 3rd roots of unity there are three cube
roots (if any). I don't exactly know what the ^(1/3) promises, since
Sage calls pari and I have not looked up the pari documentation.

Since your output was 1 it looks suspiciously as if the 1/3 was
rounded before exponentiation. And since 1 is certainly not one of
the cube roots of a, I think you have found a bug!

John

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