Show that the minimal polynomial over $F$ of each element of $E$ has only one distinct root.

Question

Let $K$ be a finite normal extension of $F$ and let $E$ be the fixed field of the group of all $F-$automorphisms of $K$. Show that the minimal polynomial over $F$ of each element of $E$ has only one distinct root.

Answer 1

Let $\alpha \in E$. Let $f$ be the minimal polynomial of $\alpha$ over $F$ (which exists as $F \subseteq E$ is a finite extension). Suppose that $f$ has a root distinct from $\alpha$, call it $\beta$. As $K$ is a normal extension of $F$ we have that $\beta \in K$.

Now obviously sending $\alpha$ to $\beta$ is an $F$-isomorphism between $F(\alpha)$ and $F(\beta)$. Call it $\phi$. We'll try to extend $\phi$ to an $F$-isomorphism of $K$. Now as $K$ is a normal extension of $F$, it is a normal extension of $F(\alpha)$, too. Hence it is the splitting field of some polynomial $f(x) \in F(\alpha)[x]$. Also it is the splitting field of $\phi(f)(x) \in F(\beta)[x]$. Now, by an easy exercise (Theorem $5.1.6$ in David Cox's "Galois Theory") we have an $F$ - automorphism of $K$, call it $\overline{\phi}$, s.t. $\overline{\phi}$ restricted on $F(\alpha)$ is $\phi$

Finally, as $F(\alpha) \subseteq E$ we have that $\overline{\phi}(\alpha) = \alpha$ and thus $\phi(\alpha) = \alpha$. A contradiction, as by construction $\phi(\alpha) = \beta$ and $\beta \not = \alpha$. Therefore we conclude that $f$ has a unique distinct root. Hence the proof.

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On the other side, opposite to what has been mentioned in the comments it is not true, that $E = F$. Indeed, take $F = \mathbb{F}_2(t)$ and $K = \mathbb{F}_2(\sqrt{t})$. This is obviously a finite extension and also normal, being of degree $2$. However the only $F$-automorphism of $K$ is the identity. To conclude this, note that $\sqrt{t}$ has to be sent to a root of its minimal polynomial over $F$, namely $X^2-t$. However it is the only root of $X^2-t$. So any $F$-automorphism of $K$ fixes $F$ and $\sqrt{t}$, which in other words fixes $K$ itself, hence it's the identity one. From here it's not hard to conclude that $E=K \not = F$.

As you may have noticed, the obstruction for the claim $E=F$ to hold is the fact that $K$ isn't a separable extension over $F$. Indeed, otherwise the extension would be Galois and the claim $E=F$ holds by the Fundamental Theorem of Galois Theory.