Results 1 to 3 of 3

Math Help - Finite fields - Roots of unity

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    111
    Thanks
    1

    Lightbulb 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?

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,312
    Thanks
    692

    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.
    Last edited by Deveno; November 19th 2012 at 12:48 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2009
    Posts
    111
    Thanks
    1

    Re: Finite fields - Roots of unity

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

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 8th 2012, 09:48 PM
  2. Distinct Bases for Finite Vector Spaces over Finite Fields
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 29th 2011, 12:31 PM
  3. Roots of unity.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: January 9th 2010, 06:05 PM
  4. primitive pth root of unity and cyclotomic fields
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 14th 2009, 10:32 PM
  5. nth roots of unity ???
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 7th 2009, 07:44 PM

Search Tags


/mathhelpforum @mathhelpforum