Let $f: X \rightarrow Y$ be a finite étale covering of degree 1 of varieties over some field $k$, so
$$
[K(Y):K(X)] = 1.
$$
If $k=\mathbb{C}$ and the varieties are smooth, one can apply complex analytic methods to very quickly show that $f$ is an isomorphism.
I would however prefer an algebraic proof of this fact. Does anyone know such a proof?
Second question: Is $f$ still necessarily an isomorphism if we drop the assumption on $k= \mathbb{C}$ and/or smoothness? Can we also weaken the assumption of the schemes being varieties? It would be nice to know the most general situation where this possibly works.
Thanks!
Best Answer
Suppose that $f:X \to S$ is a finite flat morphism, of some degree $d$. Then $f_*\mathcal O_X$ is a locally free sheaf of $\mathcal O_S$-algebras, of rank $d$. We can recover $X$ from $f_*\mathcal O_X$ (via the relative Spec construction; this is discussed in an exercise in Harshorne in one of the early sections of Chapter II).
If $d = 1$, then $f_*\mathcal O_X$ is an invertible sheaf of $\mathcal O_S$-algebras, and so $f_*\mathcal O_X = \mathcal O_S$. Thus, taking relative Specs, we have $X = S$.
More concretely, if Spec $A$ is open affine in $S$, then its preimage in $X$ is Spec $B$, where $B$ is locally free of rank $d$ over $A$. If $d = 1$, then this forces $A = B$, hence Spec $B \to $ Spec $A$ is an isomorphism, and so $X \to S$ is an isomorphism (since being an isomorphism can be checked locally on the target, which is what we just did).