Let $\sigma \in Aut(K)$ have infinite order and $F = \mathcal{F}(\sigma)$. Show that if $K/F$ is algebraic, then $K$ is normal over $F$.

Let $$K$$ be a field, and suppose that $$\sigma \in Aut(K)$$ has infinite order. Let $$F$$ be the fixed field of $$\sigma$$. If $$K/F$$ is algebraic, show that $$K$$ is normal over $$F$$.

I have to use

Definition. If $$K$$ is a field extension of $$F$$, then $$K$$ is normal over $$F$$ if $$K$$ is splitting field of a set of polynomials over $$F$$

Criteria for normality:

Proposition. If $$K$$ is algebraic over $$F$$, then the following statements are equivalent:

1. The field $$K$$ is normal over $$F$$
2. If $$M$$ is an algebraic closure of $$K$$ and if $$\tau: K \to M$$ is an $$F$$-homomorphism, then $$\tau(K)=K$$.
3. If $$F \subset L \subset K \subset N$$ are fields and if $$\sigma: L \to N$$ is an $$F$$-homomorphism, then $$\sigma(L) \subset K$$, and there is a $$\tau \in Gal(K/F)$$ with $$\tau|_{L} = \sigma$$
4. For any irreducible $$f(x) \in F[x]$$, if $$f$$ has a root in $$K$$, then $$f$$ splits over $$K$$.

I'm not sure what to do, I'm trying to use the statement 2.

We know that $$\mathcal{F}(\sigma) = F$$, so $$\sigma(F) = F$$. Let $$M$$ be an algebraic closure of $$K$$. We know that $$\sigma$$ is an $$F$$-automorphism, in particular, $$\sigma$$ is an $$F$$-homomorphism... $$\sigma: K \to M$$ would not it be a $$F$$-homomorphism?

Seems very simple, I imagine I'm not seeing something, because I didn't use the hypothesis $$\sigma$$ has infinite order. Thanks for the help!

• It’s certainly true in the familiar case that $\sigma$ is of finite order. I think that the reason for mentioning infiniteness is to point out that the proposition is more general than usually realized. – Lubin May 6 '18 at 4:50

Let $P$ be an irreducible polynomial $\in F[X]$ which has a root $x\in K$ and whose higher coefficient is $1$, we denote by $x,\sigma(x),...,\sigma^n(x)$ the orbit of $x$, remark that $\sigma^i(x)\in K$ since $\sigma\in Aut(X)$, write $Q=(X-x)(X-\sigma(x))..(X-\sigma^n(x))$, we have $Q^{\sigma}=Q$ implies that $Q\in F[X]$, we deduce that $Q$ divides $P$ since $\sigma^i(x)$ is a root of $P$ since $P^{\sigma}=P$, since $P$ is irreducible, $P=Q$.
• I have some questions. 1 - What $Q^{\sigma}$? 2 - The orbit of $x$ would not to be infinity? – Corrêa May 5 '18 at 16:15
• If $Q=a_0+a_1X+..+a_pX^p$, $Q^{\sigma}=\sigma(a_0)+..\sigma(a_p)X^p$. – Tsemo Aristide May 5 '18 at 16:16