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!

  • $\begingroup$ 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. $\endgroup$
    – Lubin
    May 6, 2018 at 4:50

1 Answer 1


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, $P=Q$.

  • $\begingroup$ I have some questions. 1 - What $Q^{\sigma}$? 2 - The orbit of $x$ would not to be infinity? $\endgroup$
    – Lucas
    May 5, 2018 at 16:15
  • 1
    $\begingroup$ If $Q=a_0+a_1X+..+a_pX^p$, $Q^{\sigma}=\sigma(a_0)+..\sigma(a_p)X^p$. $\endgroup$ May 5, 2018 at 16:16

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