$K/L$ is normal and $L/F$ is purely inseparable. Show that $K/F$ is normal (Proof verification)

Question

Let $F \subset L \subset K$ be fields such that $K/L$ is normal and $L/F$ is purely inseparable. Show that $K/F$ is normal.

Let $\sigma: K \to \overline{K}$ be an $F$-homomorphism. Let $\alpha \in L$ and $p(x) \in F[x]$ such that $p(x) = \min(F,\alpha)$, so $p(\sigma(\alpha))=0$. But $L/F$ is purely inseparable, then $\sigma(\alpha) = \alpha$. Thus, $\sigma$ is a $L$-homomorphism. Since $K/L$ is normal, $\sigma(K) = K$. Therefore, $K/F$ is normal.

I'm just in doubt if I can say that $\alpha$ has a minimal.

---

First, I tried to get an irreducible polynomial over $F$ that has a root in $K$ and show that it splits on $K$. But I could not show, I wish someone could help me with this idea.

Thanks for the advance.

Answer 1

Your proof is correct. Since $\alpha \in L$ is algebraic over $F$, it has a minimal polynomial, there is no issue with that.

Notice that you have also shown that if $L/F$ is purely inseparable, then $L/F$ is normal, which is an interesting result. (Compare with What's an example of a non-normal, purely inseparable field extension?.)