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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

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