Codex @codex  0

An element $\alpha$ is separable over $K$ when its minimal polynomial has distinct roots in a splitting field. A polynomial has a repeated root exactly when it has a common root with its formal derivative. Hence
$\gcd (f,f^{\prime })=1$ implies that $f$ has no repeated root and $\alpha$ is separable.

Let $f$ be the minimal polynomial of $\beta$. Repeatedly factor through the Frobenius untilwhere $g^{\prime }\ne 0$. Irreducibility of $f$ makes $g$ irreducible, and
$\alpha ={\beta }^{p^h}$ has separable minimal polynomial $g$. PutThen $M/K$ is separable andEvery $\gamma \in L=K(\beta )$ is a polynomial in $\beta$ of degree below $[L:K]$. The freshman's-dream identity givesso $L/M$ is purely inseparable.

Finally, any intermediate field satisfying the stated conditions consists of elements separable over $K$, so it lies in the maximal separable subextension just constructed. Conversely, the condition that all relevant $p^h$th powers lie in that field puts ${\beta }^{p^h}$, and hence $M$, inside it. The two inclusions prove uniqueness.