A finite normal extension is separable over its purely inseparable fixed field

The fixed field $E=M^G$ satisfies $F\subseteq E\subseteq M$ by [L4], and $[M:E]=\mathrm{\mid }G\mathrm{\mid }$: both bounds $[M:E]\ge \mathrm{\mid }G\mathrm{\mid }$ and $[M:E]\le \mathrm{\mid }G\mathrm{\mid }$ hold by [L5], applied to the finite group $G$ of automorphisms of $M$. In particular $M\mathrm{/}E$ is a finite extension of degree $\mathrm{\mid }G\mathrm{\mid }$.

The extension $M\mathrm{/}E$ is finite Galois. It is finite by step 2.1. For $\alpha \in M$ let ${\alpha }^G:=\{\sigma (\alpha ):\sigma \in G\}$ be its finite $G$-orbit and put $f:={\prod }_{\beta \in {\alpha }^G}(T-\beta )\in M[T]$, a monic polynomial of degree $\mathrm{\mid }{\alpha }^G\mathrm{\mid }$ with $f(\alpha )=0$ whose roots in $M$ are the distinct elements of the orbit. Every $\sigma \in G$ permutes ${\alpha }^G$, hence fixes the coefficients of $f$, which are the elementary symmetric functions of the orbit; those coefficients therefore lie in $E=M^G$, that is $f\in E[T]$. It follows that the minimal polynomial $m$ of $\alpha$ over $E$, whose existence and divisibility property are given by [L2], divides $f$ in $E[T]$; being a divisor of a polynomial that is a product of distinct linear factors, $m$ itself is a product of distinct linear factors over $M$. Thus the minimal polynomial over $E$ of every $\alpha \in M$ splits over $M$ with distinct roots, so $M\mathrm{/}E$ is normal by [L6] and separable by [L7]. Therefore $M\mathrm{/}E$ is finite Galois by [L7].

The extension $E\mathrm{/}F$ is purely inseparable, and $E=F$ in characteristic zero. Since $M\mathrm{/}F$ is finite it is algebraic by [L8], and by [L1] and [L9] $M$ is a splitting field over $F$ of the product $p_1\cdots p_n$ of the minimal polynomials of a finite generating list of $M\mathrm{/}F$. Let $a\in E$ with minimal polynomial $q$ over $F$, and let $b\in M$ be any root of $q$. The assignment $a\mapsto b$ defines an isomorphism $F(a)\to F(b)$ of $F$-extensions: the $F$-algebra map $F[T]\to M$ with $T\mapsto b$ kills $q$ and so factors through $F[T]\mathrm{/}(q)\cong F(a)$ by [L13], and it is injective because $F(a)$ is a field. Moreover $M$ is a splitting field of $p_1\cdots p_n$ over both $F(a)$ and $F(b)$, since all roots of the $p_j$ lie in $M$ and generate it over $F$, hence also over each of these intermediate fields. By [L10] applied over the base field $F(a)$ the isomorphism $a\mapsto b$ extends to an isomorphism $M\to M$, which is an $F$-automorphism because it fixes $F$; so $b=\sigma (a)=a$ for this $\sigma \in G$, since $a$ lies in the fixed field $E$. Hence $q$ has exactly one distinct root in $M$, and by normality of $M\mathrm{/}F$ it has all its roots in $M$ by [L6]. If $F$ has characteristic $p>0$, write $q(T)=g(T^{p^e})$ as in [L11] with $g$ irreducible and separable; the distinct roots of $q$ in $M$ correspond bijectively to the roots of $g$ in $M$, because $z\mapsto z^{p^e}$ is injective by [L12], so $g$ has exactly one root and $\mathrm{deg}g=1$; writing $g(T)=T-\beta$ with $\beta \in F$ gives $q(T)=T^{p^e}-\beta$ and hence $a^{p^e}=\beta \in F$. Thus every element of $E$ has a $p$-power in $F$, so $E\mathrm{/}F$ is purely inseparable by [L12]. If instead $F$ has characteristic zero, then the irreducible $q$ is separable by [L11], so $q$ has $\mathrm{deg}q$ distinct roots in $M$ by [L7]; having exactly one root forces $\mathrm{deg}q=1$ and $a\in F$. Hence $E=F$ in characteristic zero. This proves all three clauses of the statement.