16
$\begingroup$

Maybe this is trivial but lets give it a try anyways..

Obviously there is a forgetful functor from schemes to topological space.. but is it surjective on objects? i.e. I ask whether any topological space is a result of using the forgetful functor on a certain scheme? Certainly this is not true if we consider ONLY affine schemes (which are spectral spaces i.e. Kolmogorov, Compact, compactness preserved upon finite intersection of open compacts, nonempty irreducibles contain a generic point).. but ... hmm wait.. maybe I should require the topological space have the property that each irreducible component must have at least a generic point?

Improve this question
$\endgroup$

6 Answers 6

Reset to default
15
$\begingroup$

Both Manny Reyes and wikipedia are slightly off on their statement of Hochster's characterization of spectral spaces (i.e., topological spaces homeomorphic to the spectrum of some commutative ring).

Both are missing the condition that the quasi-compact opens form a base for the topology. A perhaps more transparent characterization of spectral spaces is that they are precisely the inverse limits of finite $T_0$ spaces (see paragraph 2 on the first page of Hochster's paper for both of these). In particular, from the latter characterization it is easy to see that a compact Hausdorff space is spectral iff it is totally disconnected ($\iff$ zero-dimensional $\iff$ Boolean). Thus most compact sober spaces are not spectral, e.g. the closed unit interval is not a spectral space.

Note also that the paper ends with various characterizations of the topological space of a scheme (Proposition 16): these are precisely the spaces homeomorphic to an open subset of a spectral space.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ I think that Hochster uses the older language of preschemes. In modern terms, he proved that the underlying spaces of separated schemes are exactly the open subsets of spectral spaces (his "locally spectral and quasispectral" spaces) but that the underlying spaces of all schemes are more generally the spaces which are locally spectral. $\endgroup$
    – Andrew Dudzik
    Commented Mar 26, 2012 at 18:59
8
$\begingroup$

M. Hochster classified the prime spectra of commutative rings as spectral spaces---quasi-compact T0 spaces whose quasi-compact open subsets are closed under finite intersection and each of whose nonempty irreducible closed subspaces has a unique generic point (see his paper). He additionally classifies the underlying topological spaces of schemes as open subspaces of spectral spaces (Prop. 16 of his paper).

In particular, as you indicate, any irreducible closed subspace of such a space has a unique generic point. (I believe this may even be an exercise in Hartshorne, I'll look for a reference when I get a chance...) So in particular, the two-point space with the indiscrete topology can't be the underlying topological space of a scheme. I'm sure there's a T0 example. I'll edit this if I can think of an example.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ The Hartshorne ref is Exercise 3.17 in Chapter II and it is on Zariski spaces which are the noetherian topological spaces with unique generic points for their irreducible non-empty closed subsets. The space underlying any noetherian scheme is a Zariski space which is part (a). $\endgroup$
    – Greg Stevenson
    Commented Oct 27, 2009 at 6:07
3
$\begingroup$

This doesn't strictly answer the question but is an interesting example: Spec Qbar \otimes_Q Qbar is isomorphic to the Galois group G_Q. Here G_Q has the profinite topology and Spec Qbar \otimes_Q Qbar has the Zariski topology. This is quite a surprise!

Its easy to see this too: Spec Qbar \otimes_Q Qbar is the inverse limit of Spec K \otimes Qbar = Gal(K/Q) (ranging over Galois number fields K).

Improve this answer
$\endgroup$
1
  • $\begingroup$ This is the first step in the simplicial resolution of Spec Q as the sheaf quotient of Spec Qbar by G_Q. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 27, 2009 at 6:00
2
$\begingroup$

Your idea of requiring that every irreducible closed subset have a generic point (or, equivalently, that you add a generic point to every irreducible closed subset) is called a 'sober space' (and the process of adding a point if it isn't there is called 'soberification'). Wikipedia claims that Hochster proved every compact sober space is Spec of a ring.

Improve this answer
$\endgroup$
1
$\begingroup$

@Scott: Unfortunately, I am a bit new here.. so I am not able to insert comments until i gain 50 points, so please bear with this for a while Im about to get those points.

@Peter Clark: I dont think Hochster missed quasicompact opens forms a base for the topology, and I am sure that any spectral space is the spectrum of a commutative ring. In fact, the part with quasicompact opens forming a base is in the definition of spectral spaces, see second paragraph of the paper of Hochster. Wikipedia is the one that missed the part of the definition of spectral space (sober spaces are not enough.. original definition by Hochster never even mentioned sober spaces).

So I'd like to modify things a bit: Is a T0, sober space with quasicompact opens being a base an open subspace of a spectral space (the only thing I removed here is the condition of being compact)? I fear that if you take the alexandroff compactification of a such a space (has all property of spectral but is not compact) then finite intersection of open compacts may not necessarily mean open compact.. I havent yet given this a serious thought. But if this can work with alexandroff compactification, then I have a characterization of spaces that are a result of schemes.

Im doing a new edit.. I think I have proven the following:

A topological space the result of a scheme (by the forgetful functor) iff It is T0, compact opens forms a basis (NOTE: by compact I mean quasicompact and not quasicompact and hausdorff), finite interesections of compact open is compact open, and every irreducible component has a generic point.

=> This side is easy.. since any scheme has this property (last proposition of hochster proves this also.. since any open subspace of a spectral space has this property)

<= if you have such a topological space then take the alexandroff compactification of it. This compactification turns out to be a spectral space (i.e. all the properties of the topological space AND compactness property). The space is an open subspace of its one point compactification which is spectral, thus an open subspace of a spectral space is the topological space that is a result of a scheme (last proposition of Hochster).

Improve this answer
$\endgroup$
1
  • $\begingroup$ "@Peter Clark: I dont think Hochster missed quasicompact opens forms a base for the topology." Right. (I never said that he did.) $\endgroup$
    – Pete L. Clark
    Commented Jul 31, 2010 at 15:29
0
$\begingroup$

Just another quickie here.. Prop. 16 of Hochsters paper (thanks to Manny for pointing it out) indeed gives us the kind of characterization I need. So for zero dimensional Hausdorff spaces, the answer is immediately YES (just take the Alexandroff compactification of such space and you get a spectral space, and the underlying original Hausdorff zero dimensional space is an open subspace of it)...

Naturally one can try to compactify a T0, sober spaces and ask if it is then a spectral space. Im not yet sure on that, I may post something after my doctor visit.. there might be some problem regarding preservation of compactness upon finite intersection of open compacts (I call them finite intersection property for open compacts)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .