Assuming that a field F of order (p prime) contains an element whose minimal polynomial over 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 is an element of F whose minimal polynomial over has degree . call this polynomial then now any fixes elements of and so it's completely
determined by and so there are at most possibilities for
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.