# Finite fields - Roots of unity

• November 19th 2012, 11:45 AM
MAX09
Finite fields - Roots of unity
I'd like to know if my solution to the following problem is correct

"Let F be a field of 8 elements and A = {x $\epsilon$ F such that x7=1 and
xk $\neq$ 1 for all natural numbers k<7}"

Find the no. of elements of A.

MY SOLUTION

I used the following result.

The nth roots of unity form a group which is cyclic and it'll have 'd' elements
where d = g.c.d (n, |F|-1).

Here n = 7 and |F|-1 =7 which means there are d = g.c.d(7,7) = 7

But since the question wants to simply find numbers such that
i) x7 = 1 and
ii) xk $\neq$1 for any k<7,

the actual answer should be one less than 7, as we gotta remove the
multiplicative identity 1 from this list of elements.

I guess the answer has to be 6.

Am I right in my conclusion?

• November 19th 2012, 12:25 PM
Deveno
Re: Finite fields - Roots of unity
the number of generators of a cyclic group of order d is φ(d), the euler totient of d. if d = p, a prime number, then φ(p) = p-1 (so yes, you are correct).

(assumed as given: the group of units of a finite field is cyclic).

***********

let's make a finite field of order 8. to do this, we need an irreducible cubic in Z2[x]. let's try f(x) = x3+x+1:

f(0) = 1
f(1) = 1, so f has no roots in Z2, so f has no linear factors (and thus is irreducible) in Z2[x].

so let's form F = Z2(u), where u is a root of f. by definition, we have: u3 = -u-1 = u+1 (since in any field of characteristic 2, -a = a for all a).

let's go ahead and list all possible elements of F = {0,1,u,u+1,u2,u2+u,u2+1,u2+u+1}.

(viewing F as the vector space Z2xZ2xZ2, with (ordered) basis {u2,u,1} these would be:

F = {(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,1,0),( 1,0,1),(1,1,1)} if you prefer).

let's explicitly find a generator for F* (which will prove it's cyclic of order 7). we'll try u first:

u2 ≠ 1 (u is not of order 2, trivial).
u3 = u+1 (not of order 3)
u4 = u(u3) = u(u+1) = u2+u (not of order 4)
u5 = (u2)(u3) = u2(u+1) = u3+u2 = u2+u+1 (not of order 5)
u6 = (u3)2 = (u+1)2 = u2+1 (using the frobenius morphism, and also: not of order 6)
u7 = u(u6) = u(u2+1) = u3+u = (u+1)+u = 1 <---u is of order 7.

since the order of uk is 7/(gcd(k,7)) we see that |uk| = 7 for k = 1,2,3,4,5,6, and only u7 = 1 has order 7/7 = 1.
• November 19th 2012, 12:43 PM
MAX09
Re: Finite fields - Roots of unity
So I guess I'm right with my solution, eh Deveno?

Thanks