Non-algebraizable Formal Scheme?

What is an example of a formal scheme that is not algebraizable?

Recall that, if $X$ is a locally noetherian scheme and $Z$ is a closed subset (of the underlying topological space), then one can form the formal completion of $X$ along $Z$ which is sometimes denoted $X_{/Z}$. This is a formal scheme whose underlying topological space is $Z$.

What is a formal scheme that is not of this form?

Update: Emerton and Francesco Polizzi suggested several examples that arise in the study of deformations of varieties with trivial canonical bundle. It'd be nice to see some more elementary, explicit examples as well.

Update 2: In comments, Francesco Polizzi mentioned that further examples can be found in [Hironaka-Matsumura, "Formal functions and formal embeddings" J. math. soc. Japan 20, Theorem 5.3.3 ] and [Hartshorne, Ample subvarieties of algebraic varieties, p. 205].

This is too long to fit into comments:

@FP: Thanks! I'm not sure I quite follow the argument for non-algebraizability in the book. Sernesi states that, if $X \to \text{Spec}(\bar{A})$ is an algebrization, then $X$ would admit a non-trivial line bundle "since $X$ is of finite type over an integral scheme." Furthermore, he states that this line bundle can be chosen to "correspond to a Cartier divisor whose support does not contain $X_{s}$ [the special fiber] and has nonempty intersection with $X_{s}$." (note: The notation $X$, $X_s$ is different in the text.)

It is not clear to me why such a line bundle exists: $\mathbb{A}^n$ is a finite type scheme over an integral scheme that has no non-trivial line bundles.

I understand how this shows that there is no algebraization by a $\bar{A}$-projective scheme, but why is there no algebraization by an arbitrary scheme?

I was a little nervous about the argument (Raynaud has an example of a family of Abelian varieties over a nodal curve with non-projective total space), but my concern was needless.

Here is one argument. Let $X_0/\mathbb{C}$ be an algebraic $K3$-surface. We assume algebraizability and derive a contradiction. The statement about existence of non-algebraic deformations is very strong: In fact, there exists a 1st order deformation $f_1 \colon X_1 \to \text{Spec}(\mathbb{C}[t]/(t^2))$ with the property that the restriction of any line bundles $L_1$ on $X_1$ to $X_0$ is numerically trivial. We use this deformation to derive a contradiction.

By definition, there exists a morphism $f_1 \colon \text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[x_1, \dots, x_{20}]])$ with property that the versal deformation restricts to $X_1$. Now factor this morphism as $\text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[t]]) \to \text{Spec}(\mathbb{C}[[x_1, \dots, x_{20}]])$ (by lifting the images of $x_1, \dots, x_{20}$ under $f_1^{*}$).

If $X_{t} \to \text{Spec}(\mathbb{C}[[t]])$ is the restriction of the versal deformation, then the generic fiber is an algebraic $K3$-surface, hence admits an ample line bundle. The total space $X_{t}$ of the family is regular, so it is possible to extend this line bundle to a line bundle $L_{t}$ on $X_{t}$. But then the restriction of $L_{t}$ to the special fiber is not numerically trivial (by flatness); however, no such line bundle can even lift to 1st order. Contradiction.

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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 ... – Emerton Nov 6 '10 at 19:59
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. – Bhargav Nov 6 '10 at 20:14
@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... – Francesco Polizzi Nov 6 '10 at 21:09
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 ... .) – Emerton Nov 7 '10 at 1:57
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"? – BCnrd Nov 7 '10 at 19:39

I think the following should work.

Let $X$ be a smooth, complex, projective $K3$ surface, and let $\bar{A}$ be the base of the formal semi-universal deformation of $X$. It is well-known that

$\bar{A}=\mathbb{C}[[X_1, \ldots, X_{20}]]$.

Let $\mathcal{X} \to \textrm{Specf}(\bar{A})$ be the corresponding formal scheme. Then $\mathcal{X}$ is not algebraizable. Roughly speaking, the reason is that the general deformation of $X$ is a $K3$ surface which is $not$ algebraic.

For a complete proof, see [Sernesi, Deformations of algebraic schemes, Example 2.5.12].

EDIT. As it is also remarked in Sernesi's book, this example shows that a smooth, complex, projective variety $X$ need not have an algebraic formally versal deformation, even if the functor $\textrm{Def}_X$ is prorepresentable and unobstructed.

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Yes, this is the first example that came to my mind! – Emerton Nov 6 '10 at 19:52
I guess so. The universal deformation of an abelian variety of dimension $> 1$ would be another, I guess. – Emerton Nov 6 '10 at 19:56
The question asked for non-algebraization as abstract (locally noetherian) scheme, not equipped with auxiliary structure (such as map to a specific affine scheme). By replacing $\mathbf{C}$ with $\mathbf{Q}$, perhaps an approximation argument can prove that if the above example admits algebraization as an abstract scheme then it also does as a proper flat scheme over the indicated base (hence a contradiction, as explained above). – BCnrd Nov 7 '10 at 13:54
Dear Francesco: Do you see what such an approximation argument might be? I thought for a little bit and didn't see how to make it work. I have vague recollection that Artin constructed examples of non-algebraizable formal singularities (perhaps over any alg. closed field of char. 0?), but I don't remember anything more about that. – BCnrd Nov 7 '10 at 19:41
Dear Brian and jlk, I do not see how to make an approximation argument work either. About Artin's examples, I looked for them but I could not find the exact reference. However, during my search I met further examples of non-algebraizable formal schemes, maybe (if you do not know them already) you could find them interesting. The references are [Hironaka-Matsumura, "Formal functions and formal embeddings" J. math. soc. Japan 20, Theorem 5.3.3 ] and [Hartshorne, Ample subvarieties of algebraic varieties, p. 205]. Regards, Francesco – Francesco Polizzi Nov 8 '10 at 9:04

Bhargav's example is really an example of a non-algebraic formal subscheme of the affine plane. Such examples are ubiquitous in foliation theory : a differential equation and, more generally, a foliation on a (smooth) algebraic variety has local leaves which are smooth formal schemes. This follows from the formal Frobenius theorem (in positive characteristic, the foliation needs to have p-curvature zero). Sometimes, these leaves are the formal completions of an algebraic subvariety, but often not. However, these leaves are isomorphic, at formal schemes, to the formal completion at the origin of an affine space; from the intrinsic point of view, they thus are algebraizable.

The theorems of Hironaka, Matsumura, Hartshorne to which Francesco Polizzi refers are in the same spirit, but concern formal subschemes along an algebraic subvariety. They don't apply to formal subschemes based at a point.

Actually, Arakelov geometry allows to establish analogs of these theorems and algebraize some formal subschemes based at a point (eg leaves of a foliation). See papers of Bost (Pub. Math IHES, vol. 93, 2001), and of Bost and myself (Manin Festschrift, 2010).

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@ACL: Interesting. Thank you for the response. – jlk Nov 9 '10 at 19:05

I find it more or less illusory to ask for non-algebraizable formal schemes which would not fit into the scope of deformation theory. Indeed, a formal scheme $\hat X$ over $C[[t]]$, say, is nothing but a family of schemes $(X_n)$, where $X_n$ is a scheme over $C[t]/(t^{n+1})$ together with isomorphisms of $X_n$ with $X_{n+1}\otimes C[t]/(t^{n+1})$.

On the other hand, I wonder whether classical examples of non-algebraic analytic spaces, or algebraic spaces, could be constructed in the category of formal schemes, but I have no precise answer to give.

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I was discussing this with jlk and Bhargav yesterday, and Bhargav pointed out to be that there are interesting formal schemes which are not deformations in this sense. Indeed, if $X$ is a curve over $k$, $L$ is a line bundle on $X$ of positive degree and $\hat{X}$ is the formal neighborhood of $X$ in the total space of $L$, then $\hat{X}$ is a formal scheme which does not have any map to $\mathrm{Spec} k[[t]]$ for which the preimage of the closed point is $X$. – David Speyer Nov 9 '10 at 14:09