# Finite Fields

1. Aug 9, 2011

### 83956

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. Aug 9, 2011

### micromass

Staff Emeritus
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.

3. Aug 9, 2011

### 83956

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. Aug 9, 2011

### micromass

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

5. Aug 9, 2011

### 83956

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. Aug 9, 2011

### micromass

Staff Emeritus
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 $X^b-1$??

7. Aug 9, 2011

### 83956

Well Xb=1

so then ca=1 has a=b

8. Aug 9, 2011

### micromass

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

9. Aug 9, 2011

### 83956

Ok I am thoroughly confused and getting frustrated...