$g(x) = x^p-x-\alpha \in k[x]$. Let $y$ be a root of $g(x)$ Show that $k(y)$ is separable normal over $k$.

Question

Let $char(k)=p$ and $g(x) = x^p-x-\alpha \in k[x]$. Let $y$ be a root of $g(x)$ Show that $k(y)$ is separable normal over $k$.

I have done many problems of this kind but this polynomial seems tricky.

Answer 1

Consider $k(y)$. Notice, that for all $j \in \mathbb{F}_{p}$, $g(y+j)=(y+j)^{p}-(y+j)-\alpha=y^{p}+j^{p}-y-j-\alpha=j^{p}-j=0$. Now, since $\mathbb{F}_{p}\subseteq k \subset k(y)$, $k(y)$ is the minimal $k$ extension containing all the roots of $g(x)$, hence $k(y)$ is the splitting field of $g$ over $k$. Notice that $g'(x)=-1$, so $g$ is seperable. $k(y)/k$ is hence a seperable Galois extension.