Let $\alpha$ be algebraic over $K$.
Prove: $$f_K ^{\alpha} =\prod_{\sigma\in X(K(\alpha)/K)} (X-\sigma(\alpha))^i$$ Where $i$ is the inseparable degree of $K(\alpha)$ over $K$.
Also, $X(K(\alpha)/K)= Hom_K(L, \Omega)$ where $\Omega$ is an algebraically closed field that contains $K$.
I do know that in the separable case, we have $$f_K ^{\alpha} =\prod_{\sigma\in X(K(\alpha)/K)} (X-\sigma(\alpha))$$ Any hints?
If you already proved it for $K(\alpha)/K$ separable then take the least $m$ such that $K(\alpha^{p^m})/K$ is separable, $K(\alpha)/K(\alpha^{p^m})$ is purely inseparable, $p^m=[K(\alpha):K(\alpha^{p^m})]$ is the inseparable degree, $\alpha^{p^m}$ is a root of $\prod_\sigma (X-\sigma(\alpha^{p^m}))$ so $\alpha$ is a root of $\prod_\sigma (X-\sigma(\alpha))^{p^m} \in K[x]$ a polynomial of degree $p^m\, [K(\alpha^{p^m}):K]=[K(\alpha):K]$, thus the $K$-minimal polynomial of $\alpha$.