1
$\begingroup$

Let $X$ be an affine, complex surface with isolated singularities and $\pi:\widetilde{X} \to X$ be a resolution of singularities (not necessarily minimal) i.e., $\widetilde{X}$ is non-singular and $\pi^{-1}(X_{\mathrm{sm}}) \cong X_{\mathrm{sm}}$, where $X_{\mathrm{sm}}$ is the regular locus of $X$ and the isomorphism is simply the restriction of $\pi$. Denote by $E$ the exception divisor associated to the morphism $\pi$ (set-theoretically isomorphic to $\widetilde{X} \backslash \pi^{-1}(X_{\mathrm{sm}})$).

Let $A$ be a local artinian ring, $f_A: C_A \to \mathrm{Spec}(A)$ a smooth family of irreducible, affine curves and an $A$-morphism $g_A: C_A \to X \times_{\mathbb{C}} \mathrm{Spec}(A)$. Denote by $C_o$ the special fiber of $C_A$ (under the morphism $f_A)$ and $g_o:C_o \to X$ the restriction of $g_A$ to the special fiber. Suppose that $C_o$ does not contract to a point on $X$. Since $C_o$ is a curve, the universal property of blow-up implies that the morphism $g_o$ lifts to $\widetilde{X}$ i.e., there exists a morphsim $h_o:C_o \to \widetilde{X}$ such that $g_o=\pi \circ h_o$.

Is it then true that the morphism $g_A$ also lifts to $\widetilde{X} \times \mbox{Spec}(A)$ i.e., there exists a morphism $h_A: C_A \to \widetilde{X} \times \mbox{Spec}(A)$ such that $g_A=(\pi \times \mathrm{id}) \circ h_A$? If not true in general, is there any known condition under which this holds true? Any hint/reference will be most useful.

$\endgroup$
1
  • 1
    $\begingroup$ That usually is not true. For an example, consider the blowing up of the affine plane $\mathbb{A}^2_k$ at the origin, and consider $A$ equal to the first order neighborhood (i.e., $\mathcal{O}/\mathfrak{m}^2$) for the family of affine lines in the afine plane at a point parameterizing an affine line $L$ containing the origin. A first-order deformation of $L$ lifts to the blowing up if and only if the first-order deformation "vanishes" at the origin in $L$. $\endgroup$ Commented Mar 26, 2019 at 22:39

1 Answer 1

3
$\begingroup$

This is an answer that uses the language of Artin rings as in Michael Schlessinger's papers, Michael Artin's papers, etc. The statements are relative to a fixed square-zero extension of Artinian local rings, $$0 \to M \hookrightarrow A'\twoheadrightarrow A \to 0.$$ Denote the residue field $A/\mathfrak{m} = A'/\mathfrak{m}'$ by $A_0$. For every scheme $T'$ over $\text{Spec}\ A'$, denote $\text{Spec}\ A \times_{\text{Spec}\ A'} T'$ by $T$, and denote $\text{Spec}\ A_0\times_{\text{Spec}\ A'} T'$ by $T_0$.

Let $X'$ be an $A'$-scheme. Let $\mathcal{I}$ be a quasi-coherent ideal sheaf on $X'$. Denote the blowing up of $X'$ along $\mathcal{I}$ by $$\nu:\widetilde{X}'\to X'.$$ Denote the inverse image ideal sheaf of $\mathcal{I}$ by $\mathcal{O}_{\widetilde{X}'}(-\underline{E}')$. This is an invertible sheaf on $\widetilde{X}'$. Recall the universal property of the blowing up: the pair $$(\nu:\widetilde{X}'\to X', \nu^*\mathcal{I}\twoheadrightarrow \mathcal{O}_{\widetilde{X}'}(-\underline{E}')),$$ of an $X'$-scheme and an invertible quotient of the pullback of $\mathcal{I}$ is universal among such pairs for which each induced homomorphism, $$ \nu^*\text{Sym}^d_{\mathcal{O}_{X'}}(\mathcal{I}) \twoheadrightarrow \mathcal{O}_{\widetilde{X}'}(-d\underline{E}), $$ factors through the quotient, $$\nu^*\text{Sym}^d_{\mathcal{O}_{X'}}(\mathcal{I}) \to \nu^*(\mathcal{I}^d).$$ For more on this, see the answer to the following MathOverflow question: Which functor does the blowing up represent?.

Definition. A morphism to $\widetilde{X}'$ is $E'$-flat if the pullback of the following injective sheaf homomorphism is still injective, $$\mathcal{O}_{\widetilde{X}'}(-\underline{E}') \hookrightarrow \mathcal{O}_{\widetilde{X}'}.$$

Let $Y'\to \text{Spec}\ A'$ be a flat, finitely presented morphism. Let $$f':Y'\to X',$$ be a morphism of $A'$-schemes. Define $Z'\hookrightarrow Y'$ to be the closed subscheme defined by the inverse image ideal sheaf of $\mathcal{I}$. Denote by $\mathcal{T}$ the kernel of the induced morphism, $$M\otimes_{A_0} \mathcal{O}_{Z_0} \twoheadrightarrow M\cdot \mathcal{O}_{Z'}.$$

Proposition. For every $E$-flat $X$-morphism, $$e:Y\to \widetilde{X},$$ there exists an $X'$-morphism $e':Y'\to \widetilde{X}'$ extending $e$ if and only if the closed subscheme $Z'$ of $Y'$ defined by the inverse image ideal sheaf of $\mathcal{I}$ is $A'$-flat, and in this case $e'$ is also $E'$-flat. Moreover, this holds if and only if $\mathcal{T}\to M\otimes_{A_0}\mathcal{O}_{Z_0}$ is the zero homomorphism.

Proof. By the local flatness criterion, every extension is $E'$-flat, and thus also $Z'$ is $A'$-flat. Conversely, if $Z'$ is $A'$-flat, then the ideal sheaf of $Z'$ is $A'$-flat. Since the restriction of this ideal sheaf to $Y$ is an invertible $\mathcal{O}_Y$-module, the $A'$-flat ideal sheaf of $Z'$ is an invertible $\mathcal{O}_{Y'}$-module. This invertible quotient of the pullback of $\mathcal{I}$ satisfies the universal property of the blowing up, and thus gives an extension. QED

Among all $A_0$-module quotients $$M \twoheadrightarrow N,$$ such that the following composition is the zero homomorphism, $$\mathcal{T} \hookrightarrow M\otimes_{A_0} \mathcal{O}_{Z_0} \twoheadrightarrow N\otimes_{A_0} \mathcal{O}_{Z_0},$$ there exists an initial such quotient. Denote this initial $A_0$-module quotient by $$q:M\twoheadrightarrow M_e.$$

Corollary. The induced pushout of $A'$, $$A'_e := (A'\oplus M_e)/\Delta(M) = A'/\text{Ker}(q),$$ is the initial quotient of $A'$ such that $e$ extends to an $X'$-morphism on $Y_e:=\text{Spec}\ A'_e \times_{\text{Spec}\ A}. Y'.$

For more on the "obstruction" to the extension of $e$ given by this element in $\text{Hom}(\mathcal{T},\mathcal{O}_{Z_0})\otimes_{A_0} M$, please confer Section 2 of the following.

MR2007396 (2004i:14002)
Olsson, Martin; Starr, Jason
Quot functors for Deligne-Mumford stacks.
Special issue in honor of Steven L. Kleiman.
Comm. Algebra 31 (2003), no. 8, 4069–4096.
https://math.berkeley.edu/~molsson/quot2a.pdf

$\endgroup$

You must log in to answer this question.

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