Separability of $k(\alpha)$ over $k$

Question

Let $f(X)=X^p-X-a$ and $k$ be a field of characteristic $p$, then why is $k(\alpha)$ is separable over $k$ ?, where $\alpha$ is one of the roots of $f$

The roots are of the form $\alpha+i$ for $i=0,\dots,p-1$, so all of them lie in $k(\alpha)$ how to show that there are no elements $\beta$ in $k(\alpha)$ s.t. their minimal polynomial is not of the form $g(X^p)$

Answer 1

An algebraic extension $L/k$ is separable iff it is generated by separable elements (theorem 4.4 in Lang, Algebra).

Here $L=k(\alpha)$ is generated by $\alpha$, which is separable since its conjugates are distinct. Therefore $k(\alpha)/k$ is separable.

Answer 2

Note that all field extensions can be written as a tower $E\subseteq F_{sep}\subseteq F$ where $F_{sep}$ is the field containing $E$ and all separable elements of $F/E$. Now as the minimal polynomial of $\alpha$ is separable, we know that $F_{sep}/E$ is non-trivial. But then since the degree is multiplicative, it must have degree $p$, hence $F/F_{sep}$ is trivial, i.e. all elements of $k(\alpha)$ in your case are separable.