The following is possibly true but I can't find a corresponding theorem:
If $E/F$ is the splitting field of some polynomial in $F$ and $F \subset K \subset E$ then:
$Gal(E/K)$ normal subgroup of $Gal(E/F)$ $\Leftrightarrow$ $K/F$ is a normal field extension
Is this true? I think saying that $E/F$ is the splitting field of some polynomial in $F$ is the same as saying $E/F$ is Galois and therefore the Galois correspondence theorem applies. But I'm not sure I see how the meaning of normality of a subgroup relates to the meaning of normality of a field extension. But maybe the above is wrong altogether?
In any case, many thanks for your help to clarify this.
Best Answer
You are confused about the Galois correspondence - normal subgroups $H$ of the Galois group $\text{Gal}(E/F)$ correspond to normal extensions $E^H/F$, where $E^H$ denotes the subfield of $E$ fixed by $H$. Note that $E$ being the splitting field of a polynomial in $F$ does not guarantee that $E/F$ is Galois. This is due to the fact that $E/F$ is Galois only when it is normal (i.e., is a compositum of some splitting fields) and separable.
However, I imagine the statement you intended was:
This is actually still true. It can be regarded as a salvaging of the Fundamental Theorem of Galois Theory in the case that $E/F$ is not necessarily separable. Here is my reasoning: Let $E/F$ be normal and let $G=\text{Aut}(E/F)$. Then $E^G/F$ is purely inseparable, and $E/E^G$ is separable. We have that $\text{Aut}(E/E^G)=\text{Aut}(E/F)$. Because $E/F$ is normal, we have that $E/E^G$ is normal and hence $E/E^G$ is Galois, and therefore a normal subgroup $H\triangleleft\text{Aut}(E/E^G)=\text{Aut}(E/F)$ corresponds to a normal extension $E^H/E^G$. It is known that if $C\subseteq B\subseteq A$ is a tower of field extensions and $A/C$ is normal and $B/C$ is purely inseparable, then $A/B$ is normal. Thus $E^H/F$ is normal.
Conversely, given a normal subextension $L/F$ of $E/F$ that is the fixed field $L=E^H$ of some subgroup $H\subseteq\text{Aut}(E/F)=\text{Aut}(E/E^G)$, then $L$ in fact contains $E^G$, and $L/F$ normal implies $L/E^G$ normal, hence the subgroup of $G$ that fixes $L$, namely $H$, is normal in $G$.
Please take note of this question: it is not an obvious one!
However, constructing a counterexample is evading me. Here is a try:Let $\mathbb{F}_p$ be a finite field where $3\nmid p-1$ (so that there are no cube roots of unity), let $F=\mathbb{F}_p(T)$, let $f=x^{3p}-T\in F[x]$, let $E$ be the splitting field of $f$ over $F$. Then $E=F(\sqrt[3p]{T},\sqrt[3]{1})$. This has as a subfield $M=F(\sqrt[3]{T})$, which is separable, but not normal, over $F$. EDIT: Nevermind, this doesn't work. $M$ isn't the fixed field of a normal subgroup of $\text{Aut}(E/F)$.