In general, the étale topology does not form a topology in the strict sense. However, is there any subcategory of $Sch$ where we can realize the étale topology as an honest topology on some scheme?
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asked Feb 21, 2010 at 21:34
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3$\begingroup$ Sure. If a scheme is discrete and all its residue fields are separably closed, it's the same as the Zariski topology. There's probably nothing beyond this, though. $\endgroup$– JBorgerCommented Feb 21, 2010 at 21:40
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2$\begingroup$ OK, I'll interpret the question as asking: for which schemes is the etale topology essentially the same as the Zariski topology (i.e. every etale covering is refined by a Zariski covering). Since etale morphisms are open, we are asking that every etale morphism onto a Zariski open has a section (at least Zariski-locally). The Spec of a strictly Henselian discrete valuation ring whose field of fractions is separably closed has this property. Beyond that, I don't know. $\endgroup$– JS MilneCommented Feb 22, 2010 at 1:05
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3$\begingroup$ No discrete valuation rings have separably closed fraction fields. (Let l be a prime not equal to the characteristic, and let t be a uniformizer. Then x^l-t is separable and has no roots.) But I take your point -- the integral closure of a discrete valuation ring in a separable closure of its fraction field is surely another example where the two topologies coincide. $\endgroup$– JBorgerCommented Feb 22, 2010 at 3:38
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1$\begingroup$ The Novikov ring with complex coefficients (used in Floer homology) also has a ring of integers with this property. $\endgroup$– S. Carnahan ♦Commented Feb 22, 2010 at 5:39
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1$\begingroup$ I think that for any finite topological space, you can glue similar such objects (and higher dimensional analogues) together to get a (possibly highly nonseparated) scheme homeomorphic to that space. $\endgroup$– S. Carnahan ♦Commented Feb 22, 2010 at 6:01
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1 Answer
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I think Senor Borger speaks truth.
Maybe I am talking non-sense, but maybe this could be a sketch of some way to proceed (maybe not....):
Key feature of a site which we cannot have in a classical topology is that there may be several distinct "inclusion" morphisms from a smaller open into a bigger one.
Hence, we get a "fail" as soon as we encounter a scheme admitting an étale open with several inclusion maps, for example the inclusion of some open into the whole scheme.
Suppose we have a scheme admitting a non-trivial étale covering (i.e. one with non-trivial étale fundamental group), this is also an étale open, but by the action of the étale fundamental group it has several non-equal inclusions into the whole scheme. So we have an effect which is impossible in a classical, "honest" as you say, topology.
In the case our scheme indeed has trivial étale fundamental group, pick a Zariski open which has non-trivial étale fundamental group. Basically, take some finite covering and remove the ramification divisor, the complement is Zariski open. Now compose this Zariski open with the finite covering (which has become étale as the ramification has gone to nowhere land), same problem occurs, so again we get a "fail". Hence, to get honest topology, we need to have trivial étale fund. group for the scheme and all its Zariski opens, well.... that sounds pretty close to being just some closed points again.
