separable extensions and Frobenius

Question

Let $L/K$ be an algebraic extension of $char(K)=p>0$ with $\alpha \in L$. Then is it true that $K(\alpha)=K(\alpha^p) \Leftrightarrow \alpha$ is separable over $K$?
This seems to be using something like $K(\alpha)=K(\alpha)^p=K(\alpha^p)$, ie. surjectivity of Frobenius $F$ acting on $K(\alpha)$? But this requires $K$ to be perfect...?

Answer 1

Yes, it is true for any field $K$, and it does not require the Frobenius.

Assume that $K(\alpha)=K(\alpha^p)$. Then, there exists $P\in K[X]$ such that $\alpha=P(\alpha^p)$. Hence $f=P(X^p)-X\in K[X]$ satisfies $f(\alpha)=0$.

Now, $f'=-1$ because $K$ has characteristic $p$ and so $gcd(f,f')=1$. Therefore, $f$ is separable. Any divisor of $f$ is thus separable. In particular, $\mu_{\alpha,K}$ is separable and $\alpha$ is separable over $K$.

Assume now that $\alpha$ is separable over $K$. Then $\alpha$ is separable over $K(\alpha^p)$.Let $g=\mu_{\alpha,K(\alpha^p)}$. Then $g\mid X^p-\alpha^p$ and $g$ is separable. Now $X^p-\alpha^p=(X-\alpha)^p$. Since $g$ is separable, necessarily, $g=X-\alpha$. But $g$ has coefficients in $K(\alpha^p)$. Hence $\alpha\in K(\alpha^p)$ and it follows easily that $K(\alpha)=K(\alpha^p)$