Thread: Order of this automorphism group

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.

Originally Posted by alakazam

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

Originally Posted by NonCommAlg

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

Woah slow down, why does F = K(a)?

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.