$k[x,y]/(f^p-a)$ is a reduced ring

Question

Let $k$ be a an imperfect field of characteristic $p>0$ and let $a\in k$ be an element without a $p$'th root. Let $0\neq f=\alpha x+\beta y$. Then the ring $k[x,y]/(f^p-a)$ is reduced.

Showing this is equivalent to showing that whenever $f^p-a$ divides $g^n$ then $f^p-a$ already divides $g$. Intuitively I think the reason for this is that in an algebraic closure of $k$ we have $(f-a^{1/p})^p$ divides $g^n$. So $f-a^{1/p}$ divides $g$ (because it is a prime element). So in $k$ we must have that $f^p-a$ divides $g$. How can I make this rigorous? Especially the last step?

Answer 1

In fact, it is an integral domain: the polynomial $f^p-a$ is irreducible in $k[x,y]$. In order to show this we use the Eisenstein's criterion. Note that $f^p-a=\alpha^px^p+\beta^py^p-a$. Multiplying by $\alpha^{-p}$ doesn't change anything, so we may instead consider the polynomial $x^p+b^py^p-c$, $c\notin k^p$. Now note that the polynomial $b^py^p-c$ is irreducible in $k[y]$ (see here), and we are done.