Etale Covers and the Etale Fundamental Group — Examples

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.

A finite étale algebra over $\mathbb{C}$ is a finite product of copies of $\mathbb{C}$ by [F1] and the finite-étale geometric-fibre assertion in [F2]. Its fibre functor is therefore the usual finite-set functor on disjoint unions of the basepoint. A natural automorphism of this functor fixes the singleton fibre of the identity cover, and by naturality for all maps from that singleton it fixes every point of every finite fibre. Hence ${\pi }_1^{\mathrm{e}\mathrm{t}}(\mathrm{Spec}\mathbb{C})=1$.

The algebra $\mathbb{C}=\mathbb{R}[T]\mathrm{/}(T^2+1)$ is free of rank two over $\mathbb{R}$, and $2T$ is invertible in it, so it is finite étale by [F2]. Its spectrum is connected. Its two geometric points over the chosen complex basepoint correspond to the embeddings sending $T$ to $i$ and to $-i$. Complex conjugation interchanges them; it is the unique nonidentity deck transformation, since an automorphism is determined by its action on the image of $T$. Thus the cover is Galois of order two. By [F2], ${\pi }_1^{\mathrm{e}\mathrm{t}}(\mathrm{Spec}\mathbb{R})$ surjects onto that deck group, and cannot be trivial. This differs from step 1.1 after the stated base change and refutes the claim.