While studying Field Theory by Pete L. Clark ($Theorem$ $8.30$) , I came across
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.
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.
Best Answer
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 $Aut(K/F)$.