Associated Topics || Dr. Math Home || Search Dr. Math

### Abstract Algebra and Frobenius Automorphism

Date: 04/14/2004 at 08:52:58
From: Sebastian
Subject: Abstract algebra, Frobenius automorphism

Hi Dr. Math.

I'm having difficulty with one of the problems in my textbook, and I
hope you can give me some help.  Here's the problem:

Let E be a finite field of order p^n.

(a) Show that the Frobenius automorphism s_p has order n.

(b) Deduce from part (a) that G(E/Z_p) is cyclic of order n with
generator s_p.

I think this problem is very hard, so this is the only idea I can come
up with:

We have |E| = p^n. By a theorem, I know a \in E is a zero of the
polynomial x^{p^n} - x \in Z_p.  Then we have a^{p^n} - a = 0 <==>
a^{p^n} = a.  Because a^{p^n} = a, I can apply s_p on both sides,
getting s_p (a^{p^n}) = s_p (a).

By function-composition s_p * s_p * s_p ... *s_p (n times), we get
s_p^n.  Applying this on a, we get s_p^n (a) = a^{p^n}, and hence
s_p^n (a) = (a) = I(a). ==> s_p^n = I, where I is the identity
automorphism.  And this means s_p has order n. (??)

Am I going in the right direction?

Sebastian




Date: 04/30/2004 at 10:39:29
From: Doctor Nitrogen
Subject: Re: Abstract algebra, frobenius automorphism

Hi, Sebastian:

You can use this argument for your proof of (a) and (b) above.

Any element from the field Z_p is left fixed by the Frobenius
automorphism s_p, as, by Fermat's Little Theorem, a_p is congruent to
a modulo p. So if a is an element from Z_p,

s_p(a) = a^p = a (modulo p).

You might start by showing that there is a subfield of E, call it

E_(s_p),

which has all its elements left fixed by s_p, and that it is
isomorphic to Z_p, that is, show that

E_(s_p) = Z_p (up to isomorphism).

If E is a finite extension of E_(s_p), and if this subfield of E is
identified with Z_p, you can then argue that E is a separable
extension of Z_p.

Now

|E| = p^n,

and

|Z_p| = p,

so the degree of E over Z_p (or, [E : Z_p]) is equal to n.  This
suggests E is the splitting field of

x^p^n - x

over Z_p.  This polynomial has at most p^n zeros, and the smallest
power of s_p that will leave all the p^n elements of E fixed is n
(why?).  But s_p is an element of G(E/Z_p), and

|G(E/Z_p)| = [E : Z_p] = n.

This implies s_p generates the elements of G(E/Z_p), which means
G(E/Z_p) is cyclic.  Make this informal argument more formal, and you
will have your proof for both (a) and (b).

What I have done, really, is used an argument found on page 469 in the
text:

"A First Course in Abstract Algebra", John B. Fraleigh, Addison-
Wesley, 1999, ISBN# 0-201-33596-4.

Can you get this book?

Another way you might prove that G(E/Z_p) is cyclic is by showing
that, if E is isomorphic to GF(p^n), then G(E/Z_p) is isomorphic to
the group (Z_n, +), and since Z_n is cyclic, G(E/Z_p) must be cyclic.
(If you get the Fraleigh book, then look at Example 9.6.8 on page
469).

Another helpful book is: "Introduction to finite fields and their
aplications," revised edition, R. Lidl, H. Niederreiter, Cambridge
University Press, ISBN# 0-521-46094-8.

Hope this helped answer the questions you had concerning your
mathematics problem.  You are welcome to return to The Math
Forum/Doctor Math whenever you have any math related questions.

- Doctor Nitrogen, The Math Forum
http://mathforum.org/dr.math/

Associated Topics:
College Linear Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search