Given a field $K$ of characteristic $p>0$ and $f \in K[x]$ an irreducible polynomial, find the separable closure of $K$ in $L=K[x]/(f).$

Question

I have shown that we can find some irreducible, separable polynomial $g \in K[x]$ such that $f(x)=g(x^{p^n})$ for some $n \in \mathbb{Z}_{\geq 0}.$ What I have so far towards the problem in the title is the following:

Let $\alpha \in \overline{K}$ be a zero of $f \in K[x]$, so $L=K[x]/(f)=K(\alpha)$. By the first paragraph, $f(x)=g(x^{p^n})$ so $\alpha^{p^n}$ is a zero of the separable polynomial $g \in K[x]$. Therefore, $K(\alpha^{p^n})$ is a separable extension of $K,$ and $K(\alpha^{p^n}) \subset K(\alpha) = L$ so $K(\alpha^{p^n})$ is contained in $K_s,$ the separable closure of $K$ in $L$. I'm not sure what else I can say, or which directions to look in, so any hints would be useful.

Answer 1

Hint: Working with what you have already shown, an important observation is the following: for any field, $K$, of characteristic $p>0$ and any element, $\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \al$, that is algebraic over $K$, we have $(K(\al)/K)_s=(K(\al^p)/K)_s$ (here $(E/K)_s$ means the separable closure of $K$ in $E$, i.e. the set/ field of all elements of $E$ that are separable over $K$). You should try proving this on your own, but an explanation is given below if needed.

Full explanation:

The reason the above statement works is due to a criterion for separability: that an element, $\al$, algebraic over some field, $K$ with characteristic $p>0$, is separable over $K$ iff $K(\al)=K(\al^p)$ (this even holds if $\al$ is not assumed to be algebraic over $K$). On the one hand it is obvious that $(K(\al)/K)_s\geq (K(\al^p)/K)_s$ (as $K(\al)\geq K(\al^p)$).
On the other hand, for any $\be\in K(\al)-K(\al^p)$, we have $\be^p\in K(\al^p)$ (you can see this by writing $\be=\sum_i k_i\al^i$ for some $k_i$s in $K$ and raising both sides to the $p$th power) and $\be\not\in K(\al^p)$, so $K(\be)\neq K(\be^p)$, i.e. $\be$ is inseparable over $K$. This means no element of $K(\al)-K(\al^p)$ is separable over $K$, so any element in $(K(\al)/K)_s$ must also lie in $(K(\al^p)/K)_s$, hence $(K(\al)/K)_s\leq (K(\al^p)/K)_s$. It follows that $(K(\al)/K)_s=(K(\al^p)/K)_s$.
It now follows pretty easily that the separable closure you are looking for is just $K(\al^{p^n})$, as $\al^{p^n}$ is separable over $K$, although in the same vein you could also say that the separable closure is $K(\al^{p^{n+1}})$ or $K(\al^{p^{n+2}})$, etc.