Finite Fields Homework: Showing K has q Elements

[83956]

Homework Statement

Let q=p^m and let F be a finite field with qⁿ elements. Let K={x in F: x^q=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 X^a-1 divides X^b-1

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

(c) Use the fact that X^q-X divides X^qn-X, and the fact that every element of F is a root of X^qn-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 X^q-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 X^a-1 divides X^b-1

 

[micromass]

For (b). Try to show that the roots of $X^a-1$ are also roots of $X^b-1$.
Note that the roots are roots of unity, and they form a cyclic group.

 

[83956]

So because a and b are real numbers, the roots of unity are 1 and -1, right? So this implies X^a-1 divides X^b-1?

 

[micromass]

So because a and b are real numbers, the roots of unity are 1 and -1, right? So this implies X^a-1 divides X^b-1?

No, there can be more roots. But they lie in a field extension. In general, $X^a-1$ has exactly a roots (counting multiplicity).

 

[83956]

X^a-1 has a roots
X^b-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

?

 

[micromass]

X^a-1 has a roots
X^b-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 $c^a=1$. Can you prove that c is a root of $X^b-1$??

 

[83956]

Well X^b=1

so then c^a=1 has a=b

 

[micromass]

Well X^b=1

so then c^a=1 has a=b

No, a and b are fixed. they can be anything! They don't need to equal each other...

 

[83956]

Ok I am thoroughly confused and getting frustrated...