Fields

[kathrynag]

1. Let K be a finite field of characteristic p > 0. Show that the map f : K--->K given
by f(a) = a^p is surjective, hence if B is in K, then N = l^p for some element l of K.

2. Let K be a finite field with q = p^n elements (p a prime). Show that if f(x) in K[x] and
l is a root of f(x) in some extension F of K, then l^q is also a root of f(x).

3. Let f(x) in Zp[x] be irreducible of degree m. Show that if f(x) | x^p^n? x, then m | n.

4. Let K be a field with p^m elements and let l in K. Show that l^p^mk= l for every
positive integer k.




Ideas:
1. Since K is a finite field, f maps K to itself, f is surjective iff f is injective.
So I need to show f is injective, I'm assuming?

2.I began by writing f(x) = a_mx^m+a_m?1x^m?1+· · ·+a_1x+a_0. I was thinking that might help somehow.

3.Let f(x) | x^p^n? x and let K be the field of p^n elements. Then x^p^n? x,
and hence f(x), splits over K. I get this, but then I don't know where to go from here.

4.I think we can use induction somehow?

 

[daon]

1) what is N? B is an element of K?
2) yes, so plug in l. as its a root, you get f(l)=0. Then apply g(x) = x^q to the value f(l). It is an automorphism, so...
3) this is actually an if-and-only-if. x^p^n - x is a separable polynomial and has splitting field Z_p^n over Z_p[x] (it actually IS that field). Hence f(x) has all of its roots in Z_p^n, and you may conclude that the splitting field for f(x) is a subfield of this. You can take it from here...
4) induction works. the nonzero field elements form a group under multiplication. the order of that group is p^m - 1

 

[kathrynag]

1) That N should have been B
2)Do you mean do x^q=0 when you say apply g(x) to the value f(l)

 

[daon]

No, g(f(l))=g(0), since f(l)=0. But, g is an automorphism, and so fixes 0.

For the other one, any map of finite dimensional vector spaces of the same dimension satisfy "injective iff surjective". You can show the kernel contains only 0.

 

[kathrynag]

Let f(x) | x^p^n? x and let K be the field of p^n elements. Then x^p^n? x,
and hence f(x), splits over K. Let a in K be a root of f(x). That's where I get stuck. Like I know that implies f(a)=0, but I don't know where to take it from there.

 

[kathrynag]

For 4, using induction,
I start with k=1
I get a^p^m
I don't know how that would show a^p^m=a
Then I use assume true for k=n
Show true for k=n+1
a^p^(n+1)=a^(p^n*p)

 

[daon]

Okay, so F(a) is a subfield of K. What is [F(a):F]?

There is only one step to do for the induction... a^(p^(n+1)) = (a^p^n)^p = ...

Plus you know that if m divides n then Fp^m is a subfield of Fp^n. So the automorphism x -> x^p^n fixes Fp^m. That gives you your base case.