Abstract Algebra and Frobenius AutomorphismDate: 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/ |
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