$K$ is normal over $F$. Let $K$ be a field and suppose that $\sigma \in \operatorname  {Aut}(K)$ has infinite order. Let $F$ be the fixed field of $\sigma$. If $K/F$ is algebraic, show that $K$ is normal over $F$.
Note: $F=\{x \in K| \sigma(x)=x \}.$
I tried it this way, if $\sigma \in \operatorname  {Aut}(K)$ has infinite order, then  there exists $\alpha \in K$ such that $\sigma^{n}(\alpha) \neq \alpha \forall n \in \mathbb{N}$. 
Now take $\alpha \in K-F$, then
$\alpha,\sigma(\alpha), \sigma^2(\alpha), ....$ are all distinct and $f(\alpha)=0$ as $K$ is algebraic over $F$. So $f(\alpha)=0, f(\sigma(\alpha)) =0, f(\sigma^2(\alpha)) =0, ...$ that is $f$ has infinitely many roots! And I got stuck...
Thanks for any help!
 A: Let $x$ be an element of $K-F$, you have $\sigma(x)\neq x$. Since $K$ is an algebraic extension of $F$, there exists an irreducible polynomial $P\in F[X]$ such that $P(x)=0$. You also have $P(\sigma(x))=0$ since the coefficients of $P$ are invariant by $P$. This implies that the orbit of $x$ by $\sigma$ is finite. Write $x,\sigma(x),....,\sigma^n(x)$ this orbit. $Q(X)=(X-x)(X-\sigma^1(x))...(X-\sigma^n(x))$ is invariant by $\sigma$ and divides $P$ since $P(\sigma^i(x))=0, i\leq n$. This implies that $Q(X)$ is in $F[X]$; since $P$ is the minimal polynomial of $x$, you deduce that $Q=P$. Henceforth $K$ is a normal extension of $F$ since all the roots of $P$ are in $K$.
A: Your argument shows that for any $\alpha\in K$, there can be only finitely many different values $\sigma^n(\alpha)$ for $n\in\mathbb{Z}$, since they must all be roots of the minimal polynomial of $\alpha$ over $F$.  However, this doesn't give a contradiction: all you know is that there does not exist any $n$ such that $\sigma^n(\alpha)=\alpha$ for all $\alpha$.  However, for any fixed $\alpha$, there might exist an $n$ (which depends on $\alpha$) such that $\sigma^n(\alpha)=\alpha$.  You just can't have the same $n$ work for all $\alpha$ at once.
Still, what you have done is a good first step.  To show that $K$ is normal over $F$, you want to show that if $\alpha\in K$, then the minimal polynomial of $\alpha$ over $F$ splits in $K$.  So fix $\alpha\in K$.  By the argument you have given, there are only finitely many different values $\sigma^n(\alpha)$.  Let $L$ be the subfield of $K$ generated by $F$ and all the different values of $\sigma^n(\alpha)$.  Then $L$ is finite over $F$, and $\sigma$ maps $L$ to itself.  You can now apply the version of what you're trying to prove for finite extensions (which is standard Galois theory) to finish.
