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Let $k$ be some field, algebraically closed and of characteristic $0$, if you like.

Let $U= \mathbb{A}^2_k \setminus \{ (0,0) \}$ be the punctured affine plane over $k$. Write $U$ as the union of $U_1 = \mathrm{Spec} k[x^{\pm 1},y]$ and $U_2 = \mathrm{Spec} k[x,y^{\pm 1}]$. Then deformations $U^\prime$ of $U$ to $k[t]/t^2$ are given in the following way:

We have $U^\prime = U_1^\prime\cup U_2^\prime$, where $U_1^\prime = \mathrm{Spec} (k[t]/t^2) [x^{\pm 1},y]$ and $U_2^\prime = \mathrm{Spec} (k[t]/t^2)[x^{\prime},y^{\prime \pm 1 }]$, and they are glued along an isomorphism of $\mathrm{Spec} (k[t]/t^2) [x^{\pm 1},y^{\pm 1}]$ with $\mathrm{Spec} (k[t]/t^2) [x^{\prime \pm 1},y^{\prime \pm 1}]$ which can be brought in a unique way into the form $$ x^\prime = x + t (\sum_{i,j>0} a_{ij} x^{-i} y^{-j})\ ,\ y^\prime = y + t (\sum_{i,j>0} b_{ij} x^{-i} y^{-j})\ , $$ where both sums are finite. In particular, the simplest case is given by an isomorphism of the form $$ x^\prime = x + atx^{-1}y^{-1}\ ,\ y^\prime = y + btx^{-1}y^{-1}\ , $$ where $a,b\in k$ are parameters.

Moreover, deforming $U$ is unobstructed, so there exists a formal lifting of $U$ to a formal scheme $\mathcal{U}$ over $\mathrm{Spf} k[[t]]$. My question is whether there exist algebraic deformations to $\mathrm{Spec} k[[t]]$. In other words:

Question: Does there exists a flat separated scheme $X$ of finite type over $\mathrm{Spec} k[[t]]$ together with an open subscheme $U^\prime\subset X\otimes_{k[[t]]} k[t]/t^2$ such that $U^\prime$ is a nontrivial deformation of $U$?

Note that up to now, I could not construct a single such example, but I couldn't figure out an obstruction either.

One attempt might be to find an affine scheme $X$. This essentially boils down to choosing a subring $R\subset k[x,y]$ which contains a power of the ideal $(x,y)$, and deforming $R$ to $k[[t]]$. One can check that for any given deformation $U^\prime$ to $k[t]/t^2$, one can find such a ring $R$ that lifts to $k[t]/t^2$, and gives rise to the deformation $U^\prime$ (upon removing the origin). However, typically $R$ will be obstructed, and it is far from clear that one can further find a lift of this $R$ to $k[[t]]$.

[Question edited to account for Angelo's comment. Sorry for a stupid second question!]

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About question 2, doesn't it follow by taking global sections of the structure sheaf that $R = k[x,y]$? –  Angelo Jan 14 '13 at 17:37
    
No -- it is not true in general that $R$ is the set of global sections of $\mathrm{Spec} R$ - $x$. One example (which is of course not a complete intersection) is $R=k[X^2,X^3,XY,Y^2,Y^3]\subset k[X,Y]$. You might also have nilpotents at $x$. –  Peter Scholze Jan 14 '13 at 20:03
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If $R$ is a complete intersection it is Cohen-Macaulay; it follows that $R$ is the set of sections of the structure sheaf of $\mathop{\rm Spec}R \smallsetminus \{x\}$, since $\{x\}$ has codimension 2. Am I missing something? –  Angelo Jan 14 '13 at 20:15
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@Angelo: I realize that your question is rhetorical. Perhaps not everybody realizes this. Of course you are correct (as surely you already know). One relevant reference is Proposition 5.10.14, p. 119 of EGA IV_2 (presumably also "Local cohomology", Grothendieck-Hartshorne, etc.). –  Jason Starr Jan 14 '13 at 20:22
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To Jason: I am never 100% sure I am correct, because I have a tendency to overlook key points; so, the question was not completely rhetorical. To Peter: no reason to be sorry. –  Angelo Jan 14 '13 at 20:51

1 Answer 1

I think I can prove the following

Theorem: Let $R$ be a $k$-algebra of finite type with a closed point $x\in \mathrm{Spec} R$ such that $\mathrm{Spec} R\setminus \{x\}\cong \mathbb{A}^2_k\setminus \{(0,0)\}$. Let $\tilde{R}$ be a $t$-adically complete flat lift of $R$ to $k[[t]]$. Then $\mathrm{Spf} \tilde{R}\setminus \{x\}\cong \mathbb{A}^2_{\mathrm{Spf} k[[t]]}\setminus \{(0,0)\}$.

As mentioned in the question, there do exist nontrivial deformations to $k[[t]]/t^n$ for any $n$; for that reason, I find this result surprising: Lifting to $k[[t]]$ poses a strong rigidity that I was not previously aware of.

In particular, there are no examples to the above question with $X$ affine. The proof below also shows that my intended application of algebraic deformations of the punctured affine plane cannot work, so I consider my question as answered in the negative (although strictly speaking the question posed is still open).

Let me start with something seemingly unrelated.

Proposition: Let $K$ be some field, let $X/K$ be a smooth surface, and let $C\cong \mathbb{P}^1\subset X$ be a smooth rational curve in $X$ with self-intersection $C^2 = 1$. Then there is an open subset $U\subset X$ containing $C$ and an open embedding $U\hookrightarrow \mathbb{P}^2$ (carrying $C$ into a line).

I deduced this (hopefully correctly) from the classification of surfaces (showing first that $X$ has to be rational); probably there is a more direct argument.

Now consider $Y=\mathbb{P}_k^2\setminus \{(0:0:1)\}$, with its line $C\subset Y$ at infinity. Let $R$ and $\tilde{R}$ be as above. Glue $Y$ with $\mathrm{Spec} R$ along $\mathbb{A}^2_k\setminus \{(0,0)\}$ to $\bar{Y}$ over $k$; this is projective, with $C$ an ample divisor. As $H^2(\mathbb{P}_k^2,T_{\mathbb{P}_k^2}(-C))=0$, one checks that any deformation of $R$ can be extended to a deformation of $\bar{Y}$ which also lifts $C$. In particular, $\tilde{R}$ can be extended to a proper flat formal scheme $\tilde{\bar{Y}}$ with a lift $\tilde{C}$ of $C$, over $\mathrm{Spf} k[[t]]$. By formal GAGA, this is algebraizable, to a projective surface $Z$ over $\mathrm{Spec} k[[t]]$, and $D\cong \mathbb{P}^1\subset Z$. The smooth locus of the generic fibre of $Z$, together with the generic fibre of $D$, satisfies the assumptions of the Proposition. It follows that there is a birational map from the smooth locus of the generic fibre of $Z$ to $\mathbb{P}^2_{k((t))}$. It cannot contract anything: If a curve $E$ was contracted, then this curve would meet the generic fibre of $D$ (as $D$ is ample), but in a neighborhood of $D$, the map is well-defined. Moreover, it cannot have critical points, as the target $\mathbb{P}^2_{k((t))}$ contains no exceptional curves. It follows that the smooth locus of the generic fibre of $Z$ is open in $\mathbb{P}^2_{k((t))}$. As its image contains an ample curve, it follows that the codimension of the image is at least $2$.

Now consider the reduction map $H^0(Z^{\mathrm{sm}},\mathcal{O}(nD))\to H^0(Y,\mathcal{O}(nC))$, where $Z^{\mathrm{sm}}\subset Z$ denotes the smooth locus; it induces an injection $H^0(Z^{\mathrm{sm}},\mathcal{O}(nD))/t\to H^0(Y,\mathcal{O}(nC))$. Both have the same dimension as $H^0(\mathbb{P}^2,\mathcal{O}(n))$. It follows that the map has to be surjective. It follows that $\mathrm{Proj} \bigoplus H^0(Z^{\mathrm{sm}},\mathcal{O}(nD))$ is a flat deformation of $\mathbb{P}^2_k$, thus is isomorphic to $\mathbb{P}^2_{k[[t]]}$. We get an open embedding $Z^{\mathrm{sm}}\hookrightarrow \mathbb{P}^2_{k[[t]]}$, which restricts to an isomorphism $\mathrm{Spf} \tilde{R}\setminus \{x\}\cong \mathbb{A}^2_{\mathrm{Spf} k[[t]]}\setminus \{(0,0)\}$, as desired.

Addendum: For completeness, I deduce the Proposition (in case $K$ is algebraically closed -- one can then remove this hypothesis) from the paper "Curves with high self-intersection on algebraic surfaces" of Hartshorne (I will use freely notation from there). We may assume that $X$ is proper. First, $X$ is rational: Fix a point $x\in C$. Because $C$ is very free, there is a $1$-parameter family of rational curves which contain $x$. As these curves intersect with multiplicity $1$, no two of them can have the same tangent vector at $x$. It follows that this family is defined over a rational base, thus $X$ is (uni-, and thus) rational. Now we use Theorem 3.5 in Hartshorne's paper. Case a) means that $C\subset X$ is equivalent to a section of a rational Hirzebruch surface $F_e\to \mathbb{P}^1$. The condition $C^2=1$ will then ensure that $e=1$, and that the curve $C$ does not meet the exceptional locus of $F_1\to \mathbb{P}^2$, giving the result. Case b) cannot occur. Case c) could cause trouble. Looking into the proof, only the case $m=2$ might occur. Looking at how the case $m=2$ comes about in Proposition 3.2, we see that we must have $e+n=1$. The case $e=0$, $n=1$ is impossible because of condition b) in Proposition 3.1, and the other case $e=1$, $n=0$ is impossible because of condition e) in Proposition 3.1.

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