# Thread: Order of this automorphism group

1. ## 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.

2. Originally Posted by alakazam

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 $\sigma(\alpha).$

3. 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 $\alpha$ 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)?

4. 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.