Timeline for Can the étale topology ever be realized as an "honest" topology?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2010 at 18:41 | comment | added | Harry Gindi | It is a result of Hochster that any compact sober space is homeomorphic to the spectrum of some ring. | |
| Feb 23, 2010 at 15:16 | vote | accept | Harry Gindi | ||
| Feb 22, 2010 at 13:08 | comment | added | JS Milne | OK James, you knew what I meant. | |
| Feb 22, 2010 at 6:01 | comment | added | S. Carnahan♦ | 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. | |
| Feb 22, 2010 at 5:39 | comment | added | S. Carnahan♦ | The Novikov ring with complex coefficients (used in Floer homology) also has a ring of integers with this property. | |
| Feb 22, 2010 at 3:38 | comment | added | JBorger | 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. | |
| Feb 22, 2010 at 1:05 | comment | added | JS Milne | 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. | |
| Feb 22, 2010 at 0:52 | answer | added | dh35jvn | timeline score: 4 | |
| Feb 22, 2010 at 0:30 | comment | added | Harry Gindi | I definitely meant it in the first sense. If you can give some sort of convincing argument (not necessarily a formal proof) as an answer I'd be happy to vote it up. If you'd like to elaborate on what you thought I meant, feel free to add it to an answer or maybe even ask your own question if you don't know the answer. I'd be interested to hear it. | |
| Feb 22, 2010 at 0:16 | comment | added | Kevin H. Lin | I think James is right. In any other case, I think you should have strictly more etale opens than actual opens. But I read the question as asking something different, something like: are there any schemes X for which the etale topology on X "agrees" in some way with the Zariski topology on another scheme Y which is not necessarily X? | |
| Feb 21, 2010 at 21:43 | comment | added | Harry Gindi | I figured it would be true for some sort of trivial case, but I'm wondering if there's any time that it's true in a nontrivial case. | |
| Feb 21, 2010 at 21:40 | comment | added | JBorger | 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. | |
| Feb 21, 2010 at 21:34 | history | asked | Harry Gindi | CC BY-SA 2.5 |