let K be any field and $\sigma$ be an automorphism of of K where F is the fixed field of $\sigma$ and order of $\sigma$ is "s". now to prove is that [K:F] = s.
well if look at Gal(K/F) it contain all powers of $\sigma$ , so s elements are in it namely {$\sigma$, $\sigma^2$, $\sigma^3$,..., $\sigma^s$ = 1}. but also we know |Gal(K/F)| divides [K:F] and equal iff K/F is galois extension.
so my doubt here is first of all why Gal(K/F) is cyclic and second why K/F is Galois? if these two happen, we are through.
i know i am missing something trivial here. it has been a while i did Galois theory, actually this doubt came in non commutative rings. any hints or ideas or cf.
Fix $x\in K$, and let $\{x_0, \dots, x_{r-1}\}$ denote the orbit of $x$ under $\sigma$ for some $r\leq s$, where $x_i = \sigma^i(x)$. Then the polynomial $f(X) = \prod_{i < r} (X - x_i)\in F[X]$, since the coefficients of $f$ are symmetric polyomials in the $\sigma^i(x)$. Since all the $x_i\not\in F$ are distinct, $f$ is irreducible. It follows that $K/F$ is Galois (assuming $s$ is finite). Since $F$ is the fixed field of $G = \langle{\sigma}\rangle\subset \text{Gal}(K/F)$, we then have $[K:F] = \#G = s$.