Thread: 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 F such that x⁷=1 and
x^k 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) x^k 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?

Thanks in advance.

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 u^k is 7/(gcd(k,7)) we see that |u^k| = 7 for k = 1,2,3,4,5,6, and only u⁷ = 1 has order 7/7 = 1.

So I guess I'm right with my solution, eh Deveno?

Thanks