# If $K$ is a field extension of $F$ and if $\alpha\in K$ is not separable over $F$, show that $\alpha^{p^m}$ is separable over $F$ for some

If $K$ is a field extension of $F$ and if $\alpha\in K$ is not separable over $F$, show that $\alpha^{p^m}$ is separable over $F$ for some $m\geq 0$, where $p =$char$(F)$.

I know that $x^p-\alpha^p=(x-\alpha)^p$ and that $x^{p^m}-\alpha^{p^m}=(x-\alpha)(x^{p^m-1}+x^{p^m-2}\alpha+...+x\alpha^{p^m-2}+\alpha^{p^m-1})$, but I do not know how to use this and the fact that $\alpha$ is not separable over P, could anyone help me please? Thank you very much.

I assume you want $$K$$ algebraic over $$F$$.
Consider the minimum polynomial $$f$$ of $$\alpha$$. Write it as $$f(x)=a_0+a_1x+a_2x^2+\cdots +a_dx^d$$. If $$a_j\ne0$$ for some $$j$$ with $$p\nmid j$$ then $$f'$$ is nonzero and $$\alpha$$ is separable. Otherwise $$f(x)=a_0+a_p x^p+a_{2p}x^{2p}+\cdots=g(x^p)$$ where $$g$$ is now the minimum polynomial of $$\alpha^p$$. If $$g'$$ is nonzero, then $$\alpha^p$$ is separable, otherwise $$f(x)=h(x^{p^2})$$ etc. This process will eventually conclude...
• Why does $a_d=0$? – J. Doe May 9 at 13:06