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Suppose I have an allergy to non-Hausdorff spaces but I really want to do, say, arithmetic geometry. I wonder if there is some perverse way I could develop scheme theory that would accomodate for my needs.

I am not sure what is the best way to formalize this question but here is an attempt. Does there exist a functor from schemes to the category of Hausdorff topological spaces/continuous maps that is one of the following

  • full and faithful
  • faithful and admits a left adjoint
  • full and admits a left adjoint
  • faithful and admits a right adjoint
  • full and admits a right adjoint?

P.S. I believe that there is a faithful functor from schemes to sets (take the union of the underlying set and all the stalks and do some power-set acrobatics) and there is a faithful functor from sets to Hausdorff spaces (discrete topology) so at least a faithful functor should exist.

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    $\begingroup$ The constant functor from schemes to Hausdorff spaces that sends everything to the empty space is full and has a right adjoint which is the constant functor that sends everything to Spec $\mathbb{Z}$. Similarly, the constant functor with value the one-point space is full and has as left adjoint the constant functor with value the empty scheme. So I don't think "full + adjoint" is anywhere close to what you want. $\endgroup$
    – Achim Krause
    Commented Apr 10, 2019 at 23:00
  • $\begingroup$ @StepanBanach Probably the condition you want is "conservative" (or possibly even "monadic" depending on how strict you are on what a forgetful functor is). However it is unclear to me why what you are trying to do is a sensible thing. Spectral spaces arise naturally in a vast array of places, and there's really no reason to try to avoid them. $\endgroup$
    – Denis Nardin
    Commented Apr 11, 2019 at 18:27
  • $\begingroup$ @DenisNardin honestly, it was just a stupid argument with a friend (one professor said in an introductory topology course that "normal people don't have a need for non-Hausdorff spaces"). I wanted to figure out if algebraic geometers count as "normal people". $\endgroup$
    – user137767
    Commented Apr 11, 2019 at 18:33
  • $\begingroup$ Tim Campion's answer reminds me that you may want to check out Stone spaces of complete types. If my memory serves me, there is a situation remotely similar to the following: to a polynomial algebra one can associate a Stone space of complete types X (compact & Hausdorff) and the prime spectrum Y (compact and not Hausdorff), and a natural, continuous map X --> Y which is (almost?) a bijection. Oh, I see that Denis already mentioned and poo-pooed this idea. $\endgroup$
    – Somatic Custard
    Commented Apr 11, 2019 at 21:18
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    $\begingroup$ You should have put the motivation for the question in the question so it was clearer where it came from. That professor's statement about "normal people" is simply ignorant. What would the professor think if a student who is used to metric spaces said "normal people don't have a need for topological spaces"? Or you could tell the professor that the word "normal" is really overused in math and we don't need yet another meaning for it. $\endgroup$
    – KConrad
    Commented Apr 12, 2019 at 6:54

2 Answers 2

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It's not clear to me that any of these properties are even attained by a functor from schemes to all topological spaces. For instance, the functor sending a scheme to its underlying space isn't full or faithful, nor does it preserve limits (in particular, it doesn't preserve the terminal object or binary products) -- maybe it preserves colimits? Anyway, that makes me doubt that these questions are the right direction to look for a "Hausdorff theory of schemes".

Maybe a more fruitful place to look for a "Hausdorff theory of schemes" is to consider the boolean algebra generated by the Zariski-open sets (as a subalgebra of the boolean algebra of closed points of a scheme $X$). This is the boolean algebra of constructible sets, which is a totally disconnected Hausdorff topology. It's often considered in model-theoretic approaches to algebraic geometry, as far as I understand.

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    $\begingroup$ What is true is that the category of spectral spaces is equivalent to the category of posets in Stone spaces (i.e. totally disconnected, compact Hausdorff spaces), so in a sense the spectrum is only the datum of the underlying Stone space and of the specialization order. However I don't think this is a fruitful avenue $\endgroup$
    – Denis Nardin
    Commented Apr 11, 2019 at 18:20
  • $\begingroup$ @DenisNardin These things are what I know as Priestley spaces: en.wikipedia.org/wiki/Priestley_space $\endgroup$
    – Robert Furber
    Commented Apr 12, 2019 at 2:14
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The replacement in algebraic geometry of the Hausdorff condition is that of a separated scheme. These behave in many respects like Hausdorff spaces, for the Zariski topology.

But despite the Zarski topology not being Hausdorff in general, the etale topology is a very good replacement and improved (Grothendieck) topology. In many ways it is "as good as" the complex topology of a complex variety, so behaves like a Hausdorff topology which is sufficient for many applications.

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  • $\begingroup$ thank you for your suggestion, Dr. Loughran. The question is really obtuse and stupid and wants a "forgetful functor" from all schemes (including non-separated) to Hausdorff spaces. Your suggestion could probably be a helpful starting point. $\endgroup$
    – user137767
    Commented Apr 11, 2019 at 23:32

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