Is every flat unramified cover of quasi-projective curves profinite? - MathOverflow

Is every flat unramified cover of quasi-projective curves profinite?

2010-01-27T20:15:59Z

When I first learned about the etale fundamental group, there was a mythical theorem going around that in the algebraic case all we need to look at is the finite covers, because the infinite degree algebraic covers are inverse limits of the finite ones (obviously unlike the topological case). But I've never seen a convincing source for this theorem.

It seems reasonable that the statement would be: "every flat unramified map of a connected scheme onto a quasi-projective curve is an inverse limit of finite etale covers". Is this true? Do you have a reference for this?

Answer by norondion for Is every flat unramified cover of quasi-projective curves profinite?

2010-01-27T20:23:28Z

Perhaps http://math.harvard.edu/~kwickelg/papers/VW.pdf helps you.

Answer by S. Carnahan for Is every flat unramified cover of quasi-projective curves profinite?

2010-01-28T06:24:12Z

(More editing for cleanliness)

The statement is false. I learned of this example from "James" at this blog post. If you take a nodal cubic curve (notably quasiprojective), there is a flat, unramified cover by an infinite connected chain of copies of P^1, each glued transversely to its successor at a point. This is not profinite. If I'm not mistaken, the etale fundamental group of the nodal cubic over a separably closed field (with a chosen basepoint) is $\mathbb{Z}$, not its profinite completion.

Edit: Regarding the correct definition of etale fundamental group: In SGA1 Exp 5, Grothendieck (and Mme. Raynaud?) build up axiomatics for the theory of the fundamental group using only profinite sets, and the group is defined following one peculiar claim. In the beginning of Exp 5 Section 7, there is the assertion that for any connected locally noetherian scheme $S$, and any geometric point $a: \ast \to S$, the functor that takes an etale cover $X \to S$ to the set of geometric points over $a$ (with the usual morphisms) lands in the category of finite sets. The example I gave above seems to contradict this, but if you look in Exp 1, you find that all of SGA1 is written under a definition of etale morphisms that assumes that they are finite type (which this example is not). Anyway, one reason why Pete Clark only sees profinite definitions for the etale fundamental group, is that people like to use finite type morphisms, while etale morphisms only have to be locally of finite presentation (according to EGA4, and Wikipedia I guess).

As for the question of infinite degree etale covering maps between locally finite type geometrically integral schemes, I don't think one exists, since (if I'm not mistaken) you automatically get an infinite degree algebraic extension of function fields, which is therefore infinitely generated. I'm having trouble thinking through the details of this, though.

Answer by Kirsten Wickelgren for Is every flat unramified cover of quasi-projective curves profinite?

2010-01-29T03:54:01Z

Any modification of the theorem where the definition of "cover" you give is local on the base and contains inverse limits of finite etale covers (e.g. flat plus unramified as in the original question) will also be false because the property of being an inverse limit of finite etale covers is not local on the base.

To see this, proceed similarly to Scott Carnahan's example, but instead of gluing a chain of $\mathbb{P}^1$'s together, glue together $\mathbb{P}^1$'s "indexed by $\operatorname{Spec} \mathbb{C}[...,x_{-1},x_0,x_1,...]/\langle x_i^2-1\rangle$." Explicitly, let the base curve $B$ be two $\mathbb{P}^1$'s glued together at two distinct points. Over each $\mathbb{P}^1$ consider the affine morphism corresponding to the sheaf of algebras $\mathcal{O}_{\mathbb{P}^1}[...,x_{-1},x_0,x_1,...]/\langle x_i^2-1\rangle$. At one of the points, glue together the two possible $x_i$'s. At the other, glue $x_i$ to $x_{i-1}$. Over each $\mathbb{P}^1$, the resulting morphism is an inverse limit of finite covers, but over all of $B$, it is not. This is written down fully in Warning 2.5b of http://math.harvard.edu/~kwickelg/papers/VW.pdf -- Kirsten Wickelgren