Infinite Galois extension

Question

While studying Field Theory by Pete L. Clark ($Theorem$ $8.30$) , I came across

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For an algebraic field extension $K/F,$ the following are equivalent: (i) $K^{Aut(K/F)} = F$. ("$K/F$ is Galois.") (ii) K is normal and separable.

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While studying this I am unable to understand , if $K$ is not normal , how fixed field of $K$ by $Aut(K/F)$ is affected .

Any hint would be welcome .Here , $K/F$ maybe infinite extension.

Answer 1

Let $x \in K$ not normal with minimal polynomial $P$ over $F$. Call $x_1, \cdots x_r$ its images by $Aut(K/F)$.

The polynomial $Q=(X-x_1)\cdots(X-x_r)$ divides $P$ in $K[X]$. Since $x$ is not normal, $Q$ can't be in $F[X]$, so there is at least one coefficient of $Q$ that is not in $F$ but by construction the coefficients of $Q$ are fixed by $Aut(K/F)$.

Answer 2

A classic example of a (finite) extension that is not normal: $K=\mathbb Q(\sqrt[3]{2})$ over $\mathbb Q$. Any automorphism of $K$ preserving $\mathbb Q$ must map $\sqrt[3]{2}$ into some root of $x^3-2=0$, but since the other two roots are not real numbers, and $K\subset\mathbb R$, it follows that this automorphism must map $\sqrt[3]{2}$ into itself, and so fixes the whole $K$.