Finite Fields

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Homework Statement



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.

Homework Equations





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
 

Answers and Replies

  • #2
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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.
 
  • #3
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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?
 
  • #4
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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?

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).
 
  • #5
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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

????
 
  • #6
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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

????

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]??
 
  • #7
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Well Xb=1

so then ca=1 has a=b
 
  • #8
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Well Xb=1

so then ca=1 has a=b

No, a and b are fixed. they can be anything!! They don't need to equal each other...
 
  • #9
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Ok I am thoroughly confused and getting frustrated...
 

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