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# Difference of finite fields

I'm trying to write a funcion that included a for cycle.

This cycle should take all the elements in the set GF(16)\GF(4).

However I'm having some probelms with this, I tried using .difference but this way doesn't work as intended, the resulting set ends up having 14 elements instead of 12 as I want.

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## 2 Answers

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We need an explicit embedding of GF(4) into GF(16). Viewing GF(16) as an extension of GF(4), we can create a list of elements from GF(16)\GF(4) as follows:

F.<z> = GF(4)                             # GF(4)
R.<x> = F[]                                # polynomials in x over GF(4)
p = R.irreducible_element(2)     # some irreducible polynomial over GF(4) of degree 2
K = R.quotient_ring(p, 'a')          # GF(16) as a factor ring GF(4)[x] / (p(x))
[t for t in K if t.lift().degree() > 0]


It gives the following 12 elements:

[z*a,
z*a + z,
z*a + z + 1,
z*a + 1,
(z + 1)*a,
(z + 1)*a + z,
(z + 1)*a + z + 1,
(z + 1)*a + 1,
a,
a + z,
a + z + 1,
a + 1]


where $\{0,1,z,1+z\}$ form GF(4) (and polynomials of degree $\leq 0$ in R), and a is a zero of the polynomial p ($=x^2 + (z + 1)x + 1$ in the above example).

P.S. You may find this answer also relevant to your question.

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## Comments

Nice, However mine was only an example, I need to do this for different extensions that may have different dimensions so I can't force polynomial to be of degree 2

You can specify the needed degree d for polynomial p as p = R.irreducible_element(d).

At the end I settled for this solution :

  K=[]
def FieldM(q, n):
k.<t> = GF(q^n)
Frob = k.frobenius_endomorphism()
for a in k:
K.append(a)
for l in divisors(n):
if l!=n:
for a in k:
if a==power(Frob, l)(a):
if a in K:
K.remove(a)


Where we have the function power defined as such

    def power(func, n):
def pow(x, i=n):
return func(pow(x, i-1)) if i > 0 else x
return pow

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Asked: 2021-02-16 17:43:25 +0200

Seen: 63 times

Last updated: Feb 20