1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finite Fields

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Let q=pm and let F be a finite field with qn elements. Let K={x in F: xq=x}

    (a) Show that K is a subfield of F with at most q elements.

    (b) Show that if a and b are positive integers, and a divides b, then Xa-1 divides Xb-1

    i. Conclude that q-1 divides qn-1 (in Z), and therefore
    ii. Xq-X divides Xqn (in F[X])

    (c) Use the fact that Xq-X divides Xqn-X, and the fact that every element of F is a root of Xqn-X, to show that K has exactly q elements.

    2. Relevant equations

    3. The attempt at a solution

    (a) I have completed this part. Any element of F satisfies Xq-X so it was easy to show that a+b, ab, a-b, a/b are in K.

    (b) since a divides b => b=a*s for some s a positive integer, but I don't see the connection with that in showing Xa-1 divides Xb-1
  2. jcsd
  3. Aug 9, 2011 #2
    For (b). Try to show that the roots of [itex]X^a-1[/itex] are also roots of [itex]X^b-1[/itex].
    Note that the roots are roots of unity, and they form a cyclic group.
  4. Aug 9, 2011 #3
    So because a and b are real numbers, the roots of unity are 1 and -1, right? So this implies Xa-1 divides Xb-1?
  5. Aug 9, 2011 #4
    No, there can be more roots. But they lie in a field extension. In general, [itex]X^a-1[/itex] has exactly a roots (counting multiplicity).
  6. Aug 9, 2011 #5
    Xa-1 has a roots
    Xb-1 has b roots

    but since a divides b b=a*s for some positive integer s

    so set a=a*s => s=1 so the number of roots in each equation is the same

  7. Aug 9, 2011 #6
    Not really, that makes little sense.

    Take a root c. It holds that [itex]c^a=1[/itex]. Can you prove that c is a root of [itex]X^b-1[/itex]??
  8. Aug 9, 2011 #7
    Well Xb=1

    so then ca=1 has a=b
  9. Aug 9, 2011 #8
    No, a and b are fixed. they can be anything!! They don't need to equal each other...
  10. Aug 9, 2011 #9
    Ok I am thoroughly confused and getting frustrated...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook