Let $F\subseteq L\subseteq K$ be fields such that $K/L$ is normal and $L/F$ is purely inseparable. Show that $K/F$ is normal. While studying Patrick Morandi's book "Field and Galois Theory", on page49, I came across the following question:
Let $F\subseteq L\subseteq K$ be fields such that $K/L$ is normal and $L/F$ is purely inseparable. Show that $K/F$ is normal.
What I know about normal extension is an equivalent conditon, which says that the following conditions are equivlent 


*

*$K/F$ is normal extension.

*$M$ is an algebraic closure of $K$ and if $\tau :K\rightarrow M$ is an $F-$homo, then $\tau(K)=K$.

*If $F\subseteq L\subseteq K\subseteq N$ are fields and if $\sigma:L\rightarrow N$ is an $F-$ home, then $\sigma (L)\subseteq K$, and there is a $\tau \in Gal(K/F)$ with $\tau \mid_L=\sigma$.

*For any irreducible $f(x)\in F[x]$, if $f$ has a root in $K$, then $f$ splits over $K$.


But I just don't know how to use the above results. I am not familiar with these, so I hope someone  could provide an answer with details. Thank you very much!
 A: Take the characteristic of our fields to be $p$.
Let $\alpha\in K$. As $K/L$ is normal, then $f(\alpha)=0$
where $f$ is monic with has coefficients in $L$, and $f$ splits into linear
factors in $K$. Then $f(X)=X^m +a_1X^{m-1}+\cdots+a_m$. As $L/F$
is purely inseparable, then for each $i$, $a_i^{p^{m_i}}\in K$ for
some $m_i$. Let $m$ be the maximum of the $m_i$ and consider $g=f^{p^m}$.
Prove that $g$ has coefficients in $F$, $g(\alpha)=0$ and $g$ splits
into linear factors in $K$.
A: I would highly recommend to take another characterisation of a normal field extension, namely the 2. point on your list. The result immediately follows from $\operatorname{Hom}_F(K,\overline K) = \operatorname{Hom}_L(K,\overline K)$, because this equality tells you that normality of $K/F$ and $K/L$ are the same.
The reason for this equality is very easy: Let $\sigma$ fix $F$. If $x \in L$, then $\sigma$ fixes $x^{p^e}$ for some $e$, i.e. $\sigma(x)^{p^e}=x^{p^e}$. By the injectivity of the Frobenius morphism, we get $\sigma(x)=x$, i.e. $\sigma$ also fixes $L$.

In general I would say that the characterisation 2. is way more convenient for theoretical purposes (i.e. building up the theory, proving lemmas/propositions...), while the characterisation 4. is of course the one for practical purposes.
