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 July 14th, 2009, 12:33 PM #1 Newbie   Joined: Jul 2009 Posts: 1 Thanks: 0 Automorphism in Finite Fields I need help on the following question: Let o be the automorphism in F_q in Prop II 1.5. Prove that the set of elements left fixed by o^j is the field F_k, k = p^d where d = gcd(j,f). Prop II 1.5: Let F_q be the finite field of q = p^f elements, and let o be the map that sends every element to its p-th power: o(a) = a^p. Then o is an automorphism of the field F_q (a 1-1 map of the field to itself which preserves addition and multiplcation). The elements of F_q which are kept fixed by o are precisely the elements of the prime field F_p. The f-th power of (and no lower power) the map o is the identity map.
 August 21st, 2009, 01:06 PM #2 Senior Member   Joined: Dec 2008 Posts: 160 Thanks: 0 Re: Automorphism in Finite Fields For every m, divisor of n, there is a unique Field $F_{p^m}$ that is a subfield of the $F_{p^n}$ To prove this, look at the enclosures of the roots of the polynomials that splits in those fields. As well looking at the map of the roots we can establish that the more common dividers j has with f - the more roots stay fixed. Consider this: $a^j$ has order $\frac{f}{gcd(f, j)}$ this will help to show that $F_{p^{gcd(f, j)}}$ is fixed

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