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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?

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  • $\begingroup$ I think you meant to say: a morphism is étale if and only if it is flat and unramified. $\endgroup$
    – Clark Barwick
    Commented Jan 28, 2010 at 1:59
  • $\begingroup$ @CB: Yes, I sure did -- thanks. I will delete the above comment and reproduce the first sentence below. $\endgroup$
    – Pete L. Clark
    Commented Jan 28, 2010 at 2:02
  • $\begingroup$ What do you mean by an "infinite degree algebraic cover"? $\endgroup$
    – Pete L. Clark
    Commented Jan 28, 2010 at 2:02
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    $\begingroup$ @HH: Please modify the question accordingly. But anyway, you don't seem to be taking my point: the category of etale covers of a curve is far from being closed under passage to inverse limits. I am asking for a nontrivial example of an etale covering of a curve which is not just a finite covering. Is your example an inverse limit of finite coverings? $\endgroup$
    – Pete L. Clark
    Commented Jan 28, 2010 at 5:32
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    $\begingroup$ Yes, exactly. Unramifiedness is not preserved by inverse limits. And by "not preserved" I don't just mean "not always preserved" but "hardly ever preserved except in rather trivial cases". I believe the following is the basic idea (I hope I'm not being misled by the analogy to the topological case): the fiber of an inverse limit of finite maps is going to be quasi-compact, but the fiberwise criterion for unramifiedness shows that unramified + quasi-compact fibers implies finite fibers. $\endgroup$
    – Pete L. Clark
    Commented Jan 28, 2010 at 6:30

3 Answers 3

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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

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  • $\begingroup$ Thanks! And welcome to mathoverflow. Pete and Scott were also very helpful, but unfortunately I can pick only one answer. $\endgroup$
    – H. Hasson
    Commented Jan 29, 2010 at 4:35
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(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.

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  • $\begingroup$ I was going to give a similar example, where the base curve is a Neron n-gon. (Here you are taking n = 1, which is arguably nicer, since it makes the base irreducible.) But I wasn't sure whether this would be discounted because of the reducibility of the covering scheme. As a matter of terminology, though, the etale fundamental group of a scheme is defined as an inverse limit over finite coverings, so it should still be $\hat{\mathbb{Z}}$. $\endgroup$
    – Pete L. Clark
    Commented Jan 28, 2010 at 6:42
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    $\begingroup$ By the way, there's still something interesting to resolve here. No one has given an example of an "infinite degree algebraic covering" to which all the usual nice adjectives apply: i.e., for which the base is a smooth, geometrically integral projective variety and the covering is geometrically integral and locally of finite type. If you replace "locally of finite type" by "finite type", it's easy to see that there is no such thing. But is there an example of the above? $\endgroup$
    – Pete L. Clark
    Commented Jan 28, 2010 at 6:49
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    $\begingroup$ Pete, I'm not sure who says that that's 'the' definition of the etale fundamental group, but I would say that it's not the best one. I'd want the definition that would classify total spaces which are etale locally trivial. That way you'd get the usual relation between maps from pi_1 and the etale cohomology group H^1. Otherwise, you wouldn't. $\endgroup$
    – JBorger
    Commented Jan 28, 2010 at 11:25
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    $\begingroup$ Dear Pete, What Jim means is that H^1(X,G) should equal continuous homs. from pi_1 to G for any G (not just finite G). For the nodal curve X, H^1(X,Z) is non-zero, so there there should be a hom. from pi_1 to Z. The fact the dual graph is making this non-zero contribution to the H^1 is not coincidental. It is related to the fact that this part of the H^1 is of motivic weight 0. (See the discussion at the secret blogging seminar for more on this.) $\endgroup$
    – Emerton
    Commented Jan 29, 2010 at 4:01
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    $\begingroup$ pi_1 can be defined as a limit of discrete groups, by the usual automorphisms-of-fiber-functor technique, but we simply don't restrict covers to be finite. $\endgroup$
    – S. Carnahan ♦
    Commented Jan 30, 2010 at 18:18
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Perhaps http://math.harvard.edu/~kwickelg/papers/VW.pdf helps you.

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  • $\begingroup$ I've read this paper. Their definition of a cover is pro-finite etale (whereas I'm asking whether any "cover" is pro-finite etale). Warning 2.5.(a) says that not every integral, flat, formally unramified cover is profinite etale -- but the example he gives is in char=p. How about over C? Is the only reason we only ever look at profinite covers Riemann existence? Or is there a way that all covers are profinite? $\endgroup$
    – H. Hasson
    Commented Jan 27, 2010 at 20:34
  • $\begingroup$ To your question, the difference between "unramified" and "formally unramified" is probably essential. $\endgroup$
    – Lars
    Commented Jan 27, 2010 at 20:56

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