Ask Your Question

hds's profile - activity

2020-09-16 21:42:04 +0100 received badge  Famous Question (source)
2018-03-14 14:05:07 +0100 received badge  Notable Question (source)
2018-03-14 14:05:07 +0100 received badge  Popular Question (source)
2014-01-13 05:32:33 +0100 commented answer Representing finite field elements in terms of subfield elements

Fantastic! Thanks Luca, that's exactly what I needed.

2014-01-13 05:31:49 +0100 marked best answer Representing finite field elements in terms of subfield elements

This is not supported yet. Your best option, for the moment, is use linear algebra:

sage: M = matrix([vector(w4^i*w2^j) for i in range(2) for j in range(2)])
sage: M
[1 0 0 0]
[0 1 1 0]
[0 1 0 0]
[0 0 1 1]
sage: vector(a) * M^-1
(1, 0, 1, 1)

which, of course, says $a$ equals $1 + w_4(1+w_2)$.

2014-01-13 05:31:49 +0100 received badge  Scholar (source)
2014-01-13 05:16:49 +0100 received badge  Supporter (source)
2014-01-11 15:36:15 +0100 asked a question Representing finite field elements in terms of subfield elements

I am trying to do something along the lines of magma's Eltseq function between two finite fields of the same characteristic. I'm hoping that this will be easier now that the conway polynomial work has been included in sage.

Given an element of a finite field, I would like to be able to get that element as a polynomial in the generator of the finite field, but with coefficients in a non-prime subfield.

K.<w2> = FiniteField(2^2, conway=True, prefix="w")
L.<w4> = FiniteField(2^4, conway=True, prefix="w")
a = (w2+1)*w4 + 1 # a = w4^3 + w4^2 + w4 + 1

# Currently we can do this to get a polynomial with coefficients in the prime subfield:
a.polynomial().list() # returns [1, 1, 1, 1]

# I would like this same, but for any subfield:
a.polynomial(K).list() # would return [1, w2+1]

Does anything like this exist? I've been wracking my brain trying to work out how to do this in an efficient manner, but I can't come up with anything.

Any help in this, even just pointing me in the right direction, would be very much appreciated.