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My question is: Is it possible to write any scheme as a (1-categorical) colimit of a diagram of affines? If no, what are some examples?

I ask this question because I have read that one can write any derived scheme as a colimit (in the $\infty$-categorical sense) over a diagram of affine derived schemes. So I am curious what is the analogous statement is in the classical setting.

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    $\begingroup$ Cool. Can I just take the stupid coequalizer of $ \coprod_{i,j} U_{ij} \rightrightarrows \coprod_{i} U_i $? (where $ X = \bigcup U_i$) Or is it something more intricate, that I am missing? $\endgroup$ – Anette Sep 17 '16 at 2:01
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    $\begingroup$ Yes, the Cech nerve will work (as long as $U_{ij}$ are affines, so you want quasi-separatedness I suppose). It actually will work for the derived setting too (the full Cech nerve with the colimit being the homotopy colimit). If the intersections are not affine you will have to compose with the colimit now covering these intersections by affines. $\endgroup$ – user40276 Sep 17 '16 at 2:18
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    $\begingroup$ There's a cleaner using colimits in LRS. Consider your scheme as a presheaf in rings. Now take the left Kan extension of the functor Spec going to LRS (locally ringed spaces) along the inclusion of affine schemes as representable presheaves. This gives you a characterization of schemes as colimits of affine schemes. But the colimit is being taken on LRS. I don't know if this particular colimit will be preserved (in general, they're not preserved at all). $\endgroup$ – user40276 Sep 17 '16 at 2:48
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    $\begingroup$ @user40276: left kan extensions along the Yoneda embedding into presheaves always preserve colimits, as they have a right adjoint. In this case the right adjoint is given by sending a LRS to its functor of points. As such a functor of points is always a sheaf, we even have a adjunction between Zariski sheaves and LRS. $\endgroup$ – David Carchedi Sep 17 '16 at 8:04
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    $\begingroup$ @user40276 Fully faithful functors always reflect limits and colimits, which is what we need here. $\endgroup$ – Marc Hoyois Sep 17 '16 at 16:34
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Yes, this is just a basic fact in category theory, if interpreted correctly. For $C$ any category, and $F$ any preheaf on $C,$ $F$ is the colimit in presheaves of the diagram $C/F \to C \stackrel{y}{\hookrightarrow} Psh(C),$ which sends a morphism $f:y(C) \to F,$ to $y(C),$ where $y$ is the Yoneda embedding. This follows immediately from the Yoneda lemma. Now, if $F$ is a sheaf for some Grothendiek topology, then since the sheafification functor $a$ is a left adjoint, it preserves all colimits, so we also have that $F$ is the colimit of the diagram $C/F \to C \stackrel{y}{\hookrightarrow} Sh(C).$ Note that this diagram consists entirely of representables (provided the Grothendieck topology is subcanonical, i.e. each representable is a sheaf). Now, take $C$ to the the category of affine schemes, and let the Grothendieck topology be the Zariski topology. The functor of points of any scheme, in particular, is a sheaf. The result now follows.

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  • $\begingroup$ The category of functorial schemes has the desired property more or less by definition, as you have described. A little bit more has to be checked when one works within the category of geometric schemes. In fact, then the statement here is one step towards the equivalence of categories between functorial and geometric schemes. $\endgroup$ – HeinrichD Sep 17 '16 at 7:17
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    $\begingroup$ Well, since every scheme can be covered by affine schemes, there is an equivalence of categories between sheaves on the Zariski site of (geometric) schemes and sheaves on the Zariski site of affine schemes (by the comparison lemma for sites), and both sites are subcanonical. So hence one has the geometric schemes embed fully faithfully into sheaves on affine schemes. And then one can use the above argument. $\endgroup$ – David Carchedi Sep 17 '16 at 7:25
  • $\begingroup$ I think that your argument is circular, at least when we work in the category of geometric schemes (locally ringed spaces etc.). $\endgroup$ – HeinrichD Sep 17 '16 at 7:48
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    $\begingroup$ I fail to see how the argument is circular. The Zariski site on geometric schemes being subcanonical just means that every scheme is the colimit of the Cech nerve of any Zariski cover, which follows almost immediately from the fact a topological space is a colimit of the Cech nerve of any open cover. Note that even if the Zariski cover is by affines, the Cech nerve need not consist of affines, so this is a weaker statement. $\endgroup$ – David Carchedi Sep 17 '16 at 7:59
  • $\begingroup$ @HeinrichD There's an adjunction between LRS and Zariski sheaves given by the nerve and realization approach ncatlab.org/nlab/show/nerve+and+realization . $\endgroup$ – user40276 Sep 17 '16 at 15:37
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Yes. Let's assume your scheme $X$ has affine diagonal. This implies that, if $U_i$ is an affine open cover, then $U_{ij}$ is also an affine open cover. Then $X$ is the co-equalizer you wrote.

However, if $X$ does not have affine diagonal, it's more annoying. The problem is that the $U_{ij}$ will not be affine and you need to cover them further. So, start with an open affine cover $U_i$, then pick an affine open cover $V_k^{ij}$ of each intersection $U_{ij}$. Then pick an affine open cover of each triple intersection $V^{ij}_k \cap V^{ab}_h$... so on and so forth. You get a big Cech-type diagram and its colimit will still be $X$. (the buzzword here is "hypercover")

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    $\begingroup$ You don't actually need to look at triple intersections; the diagram with just the $V^{ij}_k$ and the $U_i$ already has colimit $X$. $\endgroup$ – Eric Wofsey Sep 17 '16 at 2:25
  • $\begingroup$ You beat me to it Eric! :) (But thank you Yosemite Sam for pointing me towards hypercovers, it was something anyway that I needed to look at) $\endgroup$ – Anette Sep 17 '16 at 2:49
  • $\begingroup$ my bad, I got carried away $\endgroup$ – Yosemite Sam Sep 17 '16 at 3:49
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    $\begingroup$ In fact, if you want a hypercover by affines, there is no need to refine your intersections after the first stage. Once you've chosen the $U_{i}$s and the $V^{ij}_k$s, you can form a 1-coskeletal hypercover and all the pieces will be affine, because the $U_i$s have affine diagonal. This is useful because coskeletal hypercovers are colimits in the ∞-category of Zariski sheaves, but arbitrary hypercovers need not be. $\endgroup$ – Marc Hoyois Sep 17 '16 at 16:52
  • $\begingroup$ @MarcHoyois Just out of curiosity. Is it possible to use the same argument for derived schemes if the intersection of affines are not affines (that is using the Cech cover together with this $V_{ij}^k$)? $\endgroup$ – user40276 Sep 17 '16 at 17:10
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Actually, every scheme is the canonical colimit of all affine schemes mapping into it: $$X = \underset{\substack{U \to X\\U \text{ affine}}}{\mathrm{colim}} U$$ In order to avoid size issues, we can in fact restrict ourselves to open affine subschemes $U \hookrightarrow X$ here. (So the diagram scheme doesn't really depend on the choice of an open affine cover of $X$. But the proof, that $X$ is the colimit, clearly requires such a choice.)

The statement above is equivalent to the statement that the functor of points $\mathsf{Sch} \to [\mathsf{CRing},\mathsf{Set}]$ is fully faithful. In functorial algebraic geometry, one defines $\mathsf{Sch}$ as a certain full subcategory of $[\mathsf{CRing},\mathsf{Set}]$, so in that setting there would be nothing to prove.

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    $\begingroup$ Hi Henrich. This is just a redrafting of what I explained in my answer. The colimit formula is the same. $\endgroup$ – David Carchedi Sep 17 '16 at 7:28
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    $\begingroup$ Yes, I know. But I assume that someone who is asking such a basic question on schemes, or in fact most readers interested in this question, are not familiar yet with Grothendieck topologies, subcanonical sites, Cech nerves, etc. All this is not necessary to understand and verify this colimit. This is why I wanted to make it more direct. (It's like explaining freshmen group theory by developing category theory first and then specialize to groupoids with one object.) $\endgroup$ – HeinrichD Sep 17 '16 at 12:52
  • $\begingroup$ @DavidCarchedi And your answer is a redrafting of my comment about point wise Kan extensions :) . Ok, just kidding. Now, we have a problem with both of these answer if this colimit is not the same as the one taken in Sch which I think is what the OP was looking for. $\endgroup$ – user40276 Sep 17 '16 at 15:47
  • $\begingroup$ I was being silly as noted my Marc in the comments above. So this is in fact a colimit in Sch. $\endgroup$ – user40276 Sep 17 '16 at 17:03

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