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Math Help - Order of this automorphism group

  1. #1
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    Order of this automorphism group

    Assuming that a field F of order p^n (p prime) contains an element whose minimal polynomial over Z_{p} has degree n, show that the automorphism group of F has order at most n.
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  2. #2
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    Quote Originally Posted by alakazam View Post

    Assuming that a field F of order p^n (p prime) contains an element whose minimal polynomial over Z_{p} has degree n, show that the automorphism group of F has order at most n.
    well, the automorphism group of F is actually a cyclic group of order exactly n, but i guess you don't want to get there yet. to answer your question:

    suppose \alpha is an element of F whose minimal polynomial over K=\mathbb{Z}/p has degree n. call this polynomial f(x). then F=K(\alpha). now any \sigma \in \text{Aut}(F) fixes elements of K and so it's completely

    determined by \sigma(\alpha) and f(\sigma(\alpha))=\sigma(f(\alpha))=\sigma(0)= 0. so there are at most n possibilities for \sigma(\alpha).
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    well, the automorphism group of F is actually a cyclic group of order exactly n, but i guess you don't want to get there yet. to answer your question:

    suppose \alpha is an element of F whose minimal polynomial over K=\mathbb{Z}/p has degree n. call this polynomial f(x). then F=K(\alpha). now any \sigma \in \text{Aut}(F) fixes elements of K and so it's completely

    determined by \sigma(\alpha) and f(\sigma(\alpha))=\sigma(f(\alpha))=\sigma(0)= 0. so there are at most n possibilities for \sigma(\alpha).
    Woah slow down, why does F = K(a)?
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  4. #4
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    well, we know that F contains K(α), because it contains K = Zp and α. and because f is the minimal polynomial of α; 1, α, α^2,...,α^(n-1) are linearly independent (a linear combination of powers of α is a polynomial in α, see?).

    this means that |K(α)| = |K|^n = |Zp|^n = p^n = |F|, so K(α) = F.
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