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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!

Joachim
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1 Answers1

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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).

Matt E
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  • And so should also argue that why the degree of $f$ coincides with the rank of the pushforward of the structure sheaf. – Ehsan M. Kermani Mar 26 '13 at 22:51
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    @EhsanM.Kermani: Dear Ehsan, For a general finite flat morphism, one definition of the degree is that is the rank of the pushforward of the structure sheaf. In the context of the question, and using its notation, you can check this rank at the generic point of $Y$, where it is equal to $[K(X):K(Y)]$, since $K(X)$ is the stalk of the pushforward at the generic point of $Y$. Thus indeed in this case the two definitions of degree (via the rank of the pushforward or via the degree of extension of functions fields) coincide. Regards, – Matt E Mar 26 '13 at 23:03
  • @MattE, nice, thanks! Earlier i was having trouble with the degree, see http://math.stackexchange.com/questions/340565/definition-of-degree-of-finite-morphism-plus-context/340596#340596 . I then understood that the degree is in general defined only for dominant morphisms, as the degree of the field extension as you wrote in the comment. So do i understand correctly that if we assume finiteness and flatness, the degree is also the rank of $B$ as an $A$-module? Could you explain why/give a reference to this? Also, finiteness and flatness here are absolutely necessary, right? – Joachim Mar 27 '13 at 13:10
  • Lastly, are these two things the only necessary assumptions, i.e. does this proof work for arbitrary schemes? – Joachim Mar 27 '13 at 13:12
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    Ah, after some thought i solved it already, and (for whoever will be reading this somewhere in the future) i found a reference in stacks project definition 24.46.1. – Joachim Mar 27 '13 at 14:24