Timeline for Non-algebraizable Formal Scheme?
Current License: CC BY-SA 4.0
31 events
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| Jul 4, 2022 at 5:43 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added DOI
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| Jul 3, 2022 at 21:01 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
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| Nov 9, 2010 at 10:40 | answer | added | ACL | timeline score: 5 | |
| Nov 9, 2010 at 10:32 | answer | added | ACL | timeline score: 11 | |
| Nov 9, 2010 at 5:39 | vote | accept | jlk | ||
| Nov 9, 2010 at 5:38 | history | edited | jlk | CC BY-SA 2.5 |
Added links to more examples.
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| Nov 7, 2010 at 20:31 | comment | added | BCnrd | Dear Bhargav: Your final comment about the affine case is a good point. I am not sure of the best "fix" to the question, but one which I have in mind is to give examples of proper smooth formal $\mathbf{Z}_p$-schemes which are not formal completions of proper smooth $\mathbf{Z}_p$-schemes. Since the $\mathbf{Z}_p$-points of a $p$-adic open disk cannot be covered by countably many proper analytic sets (by Baire), the various deformation space examples (for abelian variety or K3 surface) with known existence of ample line bundle when given "abstractly" as proper over dvr do the job. | |
| Nov 7, 2010 at 20:16 | comment | added | Bhargav | Brian, thanks for bringing this up. Jesse actually emailed me about this last night. I think my example gives an example of a formal X-scheme that's not algebraic as a formal X-scheme, though it is algebraic as an abstract scheme as you point out, so not an example sought after. In fact, unless I'm mistaken, any (noetherian) affine formal scheme is the completion of a (noetherian) affine scheme more-or-less by definition, so there can't really be a simple example. | |
| Nov 7, 2010 at 19:39 | comment | added | BCnrd | Dear Bhargav & AByer: Bhargav's construction has projection to the first factor identifying his formal scheme with a suitable completion of the affine line, so it is identified with the $m_R$-adic completion of the affine line (and so doesn't give an example of the sort requested in the question). Meanwhile, AByer's construction seems to want to be the zero locus of $\prod_{n \ge 1} (y - x^n)$ in the $x$-adic completion of $k[x,y]$, but this makes no sense since the terms in the product don't tend $x$-adically to 1. So I am confused; what is meant by "union"? | |
| Nov 7, 2010 at 15:43 | comment | added | Arend Bayer | Here is another elementary example that should work, similar to Bhargav's: Take the infinite union of the subschemes $V(y-x^n)$ for $n \ge 1$ in $A^2$. In any infinitesimal neighborhood $x^N = 0$ of the $y$-axis this becomes a finite union, i.e. an algebraic scheme, and thus we can define a formal scheme as their limit. | |
| Nov 7, 2010 at 13:29 | history | edited | BCnrd | CC BY-SA 2.5 |
fixed spelling of "algebraizable" in title; Post Made Community Wiki
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| Nov 7, 2010 at 3:07 | comment | added | jlk | @Emerton: Thanks! When I posed the question, I was not expecting the standard example to be so interesting! | |
| Nov 7, 2010 at 1:57 | comment | added | Emerton | Dear jlk, In regard to your question "As an aside, ..." (which I hadn't noticed before now), I see that you have answered it in the most recent addition to your question. When one is in a context in which "algebraic" is more general than "projective", so that algebraicity can't be tested by deforming an ample line bundle, I'm not sure if there are other general principles one can apply to test for (non)-algebraicity. (None are coming to mind, but maybe someone else will have something to suggest. In fact, perhaps you could ask this as a separate question ... .) | |
| Nov 7, 2010 at 1:33 | history | edited | jlk | CC BY-SA 2.5 |
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| Nov 6, 2010 at 23:59 | history | edited | jlk | CC BY-SA 2.5 |
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| Nov 6, 2010 at 22:26 | history | edited | jlk | CC BY-SA 2.5 |
Added link
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| Nov 6, 2010 at 21:42 | comment | added | jlk | @FP: Thank you! I had not realized that result ("every algebraic $K3$ surface is projective") was still true in a relative setting. | |
| Nov 6, 2010 at 21:09 | comment | added | Francesco Polizzi | @jlk if $\mathcal{X}$ were algebraizable, it would be projective $over$ $\textrm{Spec}(\bar{A})$ (since every algebraic $K3$ surface is projective). It follows that there exists a line bundle $\mathcal{L}$ which is $f$-ample etc etc... | |
| Nov 6, 2010 at 21:07 | history | edited | jlk | CC BY-SA 2.5 |
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| Nov 6, 2010 at 20:40 | history | edited | jlk | CC BY-SA 2.5 |
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| Nov 6, 2010 at 20:28 | comment | added | jlk | @FP: Sorry! I was using $X$ to denote both the algebrization and the original K3 surface. Why is an algebrization automatically projective? | |
| Nov 6, 2010 at 20:27 | history | edited | jlk | CC BY-SA 2.5 |
Fixed confusion about algebrization versus fiber
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| Nov 6, 2010 at 20:25 | comment | added | jlk | @Emerton: As an aside, I see how "deforming an ample line bundle imposes a number of conditions". Is there a nice heuristic for why the $20$-th dimension can't arise from deformations to a non-projective algebraic surface? A non-singular K3-surface can only be deformed to a non-singular surface, and it is a theorem that a proper, non-singular algebraic surface is automatically projective. But this is non-trivial theorem about surfaces, and thus not a great heuristic. | |
| Nov 6, 2010 at 20:17 | comment | added | Francesco Polizzi | @jlk $X$ is a projective variety, so it surely has an ample line bundle $L$ on it. Sernesi's proof, roughly speaking, shows that it is not possible to extend $L$ to a line bundle $\mathcal{L}$ on $\mathcal{X}$. This means that the deformations of $X$ which are projective form a proper subspace in the space of all deformations or, in other words, that the general deformation of $X$ is not algebraic, see the comments of Emerton above. | |
| Nov 6, 2010 at 20:14 | comment | added | Bhargav | The CY examples are obviously great and important, but here's a stupider one. Let (R,m) be a DVR, and let f(t) be a convergent power series over R for the m-adic topology. Then f defines maps A^1 -> A^1 over R/m^i which are compatible. Hence, their graphs glue to give a formal closed subscheme of A^1 x A^1 over R, and this formal subscheme is not algebraic. In fewer words, Chow's lemma fails horribly and easily for non-proper maps. | |
| Nov 6, 2010 at 20:12 | history | edited | jlk | CC BY-SA 2.5 |
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| Nov 6, 2010 at 20:09 | comment | added | Emerton | ... the dimension down to g(g+1)/2 (the dimension of the moduli space of g-dimensional abelian varieties). | |
| Nov 6, 2010 at 20:03 | comment | added | Emerton | ... deformation theory, then deforming the ample line bundle on $X$ (so as to guarantee algebraicity) adds a condition, and so cuts the deformation space down to be 19 dim'l. I learned the details of this from a paper of Deligne and Illusie on K3s in the Lecture Notes volume on Algebraic Surfaces (Eight hundred and something). The case of abelian varieties is similar: the deformation space of a $g$-dimensional a.v. has dimension $g^2$ (the same as the dimension of the space of complex tori of dimension $g$), but deforming an ample line bundle imposes a number of conditions, and cuts ... | |
| Nov 6, 2010 at 19:59 | comment | added | Emerton | Dear jlk, As a commentary on the example of Francesco Polizzi below: the general yoga, when looks at formal deformations, is that picture in formal geometry should be the same as the picture in complex analytic geometry: so the complex analytic K3s form a 20-dim'l space (of which the Specf$(\overline{A})$ in Francesco's answer is a formal neighbourhood around the point corresponding to his initially chosen $K3$ surface $X$), while the algebraic K3s lie in a collection of 19-dim'l subfamilies (so the algebraizable locus in Francesco's $\mathcal X$ is codimension 1; if one looks at the ... | |
| Nov 6, 2010 at 19:51 | answer | added | Francesco Polizzi | timeline score: 16 | |
| Nov 6, 2010 at 19:29 | history | asked | jlk | CC BY-SA 2.5 |