Kummer covers of the multiplicative group

The upstairs algebra is $k[u,u^{-1}][T]\mathrm{/}(T^n-u)$: $T$ is automatically invertible because $T^n=u$, so this quotient is $k[t,t^{-1}]$. Division by the monic polynomial shows that $1,T,\dots ,T^{n-1}$ is a free basis over the downstairs ring. The derivative $n{T}^{n-1}$ is a unit, so the relative differentials vanish. By [F1] the map is finite étale of rank $n$. Its source is integral and nonempty, hence connected.

The fibre at $u=1$ consists of the $n$ distinct roots of $T^n-1$ in $k$. Each $\zeta \in {\mu }_n(k)$ gives an automorphism $T\mapsto \zeta T$ over the base, and these act simply transitively on that fibre. By [F2] all automorphisms are determined by one fibre point, so these are the entire deck group. The explicit reconstruction in [F2] gives a continuous surjection from the fundamental group to its opposite deck group; this is the same group because ${\mu }_n(k)$ is abelian.

If $p\mid n$, write $n=p^{a}m$ with $a\ge 1$ and $p\nmid m$. In characteristic $p$, $T^n-1=(T^m-1{)}^{p^a}$. Thus the finite fibre over $1$ has nonzero nilpotents and is not geometrically regular of dimension zero. It cannot be a fibre of an étale morphism by [F1]. This verifies the characteristic restriction and the degree interpretation.