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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

    micromass

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    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

    micromass

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    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

    micromass

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    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

    micromass

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    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...
     
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