I'm dealing with the following problem:
Let $K$ be a field and suppose that $\sigma\in\mbox{Aut}(K)$ has infinite order. Let $F$ be the fixed field of $\sigma$. If $K/F$ is algebraic, show that $K$ is normal over $F$.
${\it Proof. }$ Let $\alpha\in K$, I want to show that $f(x)=\mbox{min}(F,\alpha)$ splits over $K$. Let $i\in \mathbb{N}$, notice that $f(\sigma^i(\alpha))=\sigma^i(f(\alpha))=\sigma^i(0)=0$, thus $\sigma^i(\alpha)$ is a root of $f(x)$ for all $i\in \mathbb{N}$. Consider $n$ the minimal integer such that $\sigma^n(\alpha)=\alpha$ ($n$ exists since we have a finite number of roots of $f$), then we can factor $f$ over $K$ as $f(x)=(x-\alpha)(x-\sigma(\alpha))...(x-\sigma^n(\alpha))g(x)$ where $g(x)\in K$, moreover, since $f=\sigma(f)$, we have
$$(x-\alpha)(x-\sigma(\alpha))...(x-\sigma^n(\alpha))g(x)=(x-\alpha)(x-\sigma(\alpha))...(x-\sigma^n(\alpha))\sigma(g(x))$$
and therefore $\sigma(g)=g$, thus $g(x)\in F[x]$, this implies that the factorization of $f$ is a factorization over $F$, hence $g(x)=1$ since $f$ is irreducible over $F$ and so $f(x)=(x-\alpha)(x-\sigma(\alpha))...(x-\sigma^n(\alpha))$, i.e., $f$ splits over $K$.
Is this proof correct? If that's the case... Where did I use the infinite order of $\sigma$? Because if the proof is correct, it seems I've proved something like "There are not irreducible polynomials of order $n$ in a fixed field of an automorphism of order less than $n$". Thanks for your answers.