Pathologies on the Hilbert scheme of points

TL;DR

This paper demonstrates that the Hilbert scheme of points in higher-dimensional affine space exhibits non-reduced components and characteristic-dependent behavior, confirming a form of Vakil's Murphy's Law using advanced geometric techniques.

Contribution

It proves the non-reducedness and characteristic-dependent components of the Hilbert scheme of points in higher dimensions, extending Vakil's Murphy's Law to this setting.

Findings

Hilbert scheme of points is non-reduced in higher dimensions.

Components of the scheme depend on the characteristic p.

Vakil's Murphy's Law holds for this scheme.

Abstract

We prove that the Hilbert scheme of points on a higher dimensional affine space is non-reduced and has components lying entirely in characteristic p for all primes p. In fact, we show that Vakil's Murphy's Law holds up to retraction for this scheme. Our main tool is a generalized version of the Bialynicki-Birula decomposition.

Tables

Table 1. Table 3.2. Reference for computations of pieces of Hom T ( J , T / J ) ( α , β ) \operatorname{Hom}_{T}(J,T/J)_{(\alpha,\beta)} , characteristic not equal to two. The pieces marked with Cor  3.5 or Lem  3.9 vanish, others do not.

${\mathrm{Hom}}_T{(J,T/J)}_{(\alpha ,\beta )}$		$\beta \le -2$		$\beta =-1$		$\beta =0$		$\beta \ge 1$
$\alpha \le -2$		Cor 3.5		Cor 3.5		Cor 3.5		Cor 3.5
$\alpha =-1$		Cor 3.5		Lem 3.9		Cor 3.8		$\star$
$\alpha =0$		Cor 3.5		Cor 3.8		$\star$		$\star$
$\alpha =1$		Cor 3.5		Cor 3.15		$\star$		$\star$
$\alpha \ge 2$		Cor 3.5		$\star$		$\star$		$\star$

Equations

$${\mathbb{G}}_a^r\times {\mathcal{H}}_{\mathbb{Z}}^{+}$$

$${\dim }_{\mathbb{k}}{\mathrm{Hom}}_T(I_R,T/I_R{)}_{<0}=\dim T.$$

$$J:=I\cdot T+(x_1,\dots ,x_n{)}^{a+1}+(y_1,\dots ,y_n{)}^2+(x_1{y}_1+\dots +x_n{y}_n).$$

$$(Z,z)$$

$$J:=I\cdot T+{\mathfrak{m}}_x^{a+1}+{\mathfrak{m}}_y^{b+1}+(Q).$$

$${\mathrm{Hom}}_T(J,T/J{)}_{(\alpha ,\beta )}:=\left\{\phi :J\to T/J|\phi \left(J_{(\gamma ,\delta )}\right)\subset (T/J{)}_{(\gamma +\alpha ,\delta +\beta )}\text{mbox}forall\gamma ,\delta \right\}.$$

$$s:\frac{T}{J}\to \frac{T}{I\cdot T+(Q)}$$

$$\underset{j}{⨁}T(-d_{1j},-d_{2j})$$

$$⨁T(-d_{1j},-d_{2j})$$

$$(y_1-l_1,y_2-l_2,\dots ,y_n-l_n,f_1,\dots ,f_{d-1},Q)$$

$$0=\phi \left({\mathfrak{m}}_x^{a+1-a^{\prime }}i\right)={\mathfrak{m}}_x^{a+1-a^{\prime }}\phi (i).$$

$${\left(\frac{T}{I\cdot T+(Q)}\right)}_{(a+1+\alpha ,\beta )}\to {\left(\frac{T}{J}\right)}_{(a+1+\alpha ,\beta )}$$

$$\mathrm{Hom}{\left({\mathfrak{p}}_x,\frac{T}{I\cdot T+(Q)}\right)}_{(\alpha ,\beta )}=0\text{mbox}and\mathrm{Hom}{\left({\mathfrak{p}}_y,\frac{T}{I\cdot T+(Q)}\right)}_{(\alpha ,\beta )}=0$$

$${\mathrm{Hom}}_T(T,M)\to {\mathrm{Hom}}_T({\mathfrak{p}}_x,M)$$

$${\phi }^{\prime \prime }:{\mathfrak{m}}_x^{a+1}\to \frac{T}{I\cdot T+(Q)},$$

$${\mathrm{Hom}}_T{\left(\frac{{\mathfrak{m}}_x^{a+1}}{{\mathfrak{m}}_x^{a+1}\cap (Q)},\frac{T}{J}\right)}_{(-1,0)}=0.$$

$$\phi \in {\mathrm{Hom}}_T{\left(\frac{{\mathfrak{m}}_x^{a+1}}{{\mathfrak{m}}_x^{a+1}\cap (Q)},\frac{T}{J}\right)}_{(-1,0)}$$

$$0=\phi (mQ)=\sum y_i\phi (m{x}_i),$$

$$\sum y_i{\phi }^{\prime }(m{x}_i)\equiv Q{n}_m=\sum y_i{x}_i{n}_m\mathrm{mod}IT+{\mathfrak{m}}_x^{a+1},$$

$$x_i\psi (m_1)-x_j\psi (m_2)=x_i{n}_{m_1}-x_j{n}_{m_2}$$

$$=\phi (x_i{m}_1-x_j{m}_2)=\phi (0)=0\mathrm{mod}IT+{\mathfrak{m}}_x^{a+1}.$$

$${\mathrm{Hom}}_T{\left(\frac{{\mathfrak{m}}_y^{b+1}}{{\mathfrak{m}}_y^{b+1}\cap (Q)},\frac{T}{J}\right)}_{(0,-1)}=0.$$

$$(T_{[J]}\mathcal{H}{)}_{(-1,0)}=⟨{\partial }_{x_1},\dots ,{\partial }_{x_n}⟩\text{mbox}and(T_{[J]}\mathcal{H}{)}_{(0,-1)}=⟨{\partial }_{y_1},\dots ,{\partial }_{y_n}⟩$$

$${\mathfrak{m}}^a={\mathfrak{m}}^a\phi (Q)=\phi ({\mathfrak{m}}^{a}Q)\subset \phi ({\mathfrak{m}}_x^{a+1})=0,$$

$$T_{(\alpha ,\beta )}\simeq S^{\alpha }V\otimes S^{\beta }{V}^{\ast }.$$

$$(\ell ┘\phi )(f)=\phi (\ell \cdot f)$$

$$G=\left\{\left(\begin{array}{cc}{I}_n& A\\ 0& I_n\end{array}\right)|A\in {\mathbb{M}}_{n\times n}\right\}.$$

$$\mathfrak{g}\to {\mathrm{Hom}}_T(J,T/J{)}_{(1,-1)}.$$

$${\mathfrak{g}}^{\prime }\to {\mathrm{Hom}}_T(J/Q,T/J{)}_{(1,-1)}.$$

$$0\to S(-1,-b-1)\otimes S^b{V}^{\ast }\to S(0,-b-1)\otimes S^{b+1}{V}^{\ast }\to M,$$

Full text

Institute of Mathematics, Polish Academy of Sciences and Institute of Mathematics, Faculty of Mathematics, Informatics and Mechanics, University of Warsaw. Email: jjelisiejew@mimuw.edu.pl. Partially supported by NCN grant 2017/26/D/ST1/00913.

1. Introduction

Vakil defined singularity type as an equivalence class of pointed schemes under the relation generated by $(X,x)\sim(Y,y)$ if there is a smooth morphism $(X,x)\to(Y,y)$. He defined that Murphy’s Law holds for a moduli space if every singularity type of finite type over $\mathbb{Z}$ appears on this space. In [Vak06] Vakil gave numerous examples when this happens. A notable item missing in his list is the Hilbert scheme of points. In fact little was known about its singularities: for example the following classical questions raised by Fogarty and Hartshorne were open.

Question 1.1** ([Fog68, p. 520], [Ame10, Problem 1.25], [CEVV09]).**

Is $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{n}_{\mathbb{Z}})$ reduced for all $n$? Is $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{n}_{\mathbb{C}})$ reduced for all $n$?

Question 1.2** **([Har10, p. 148], [Ame10, Problem 1.2],

[Lan18]).

Do all finite $\Bbbk$-schemes, for a finite field $\Bbbk$, lift to characteristic zero?

Question 1.1 was completely open, and $\operatorname{Hilb}_{4}(\mathbb{A}^{3}_{\mathbb{C}})$ is the only known reduced but singular Hilbert scheme of points [RA16, BCR17]. It was explicitly asked in [Ame10, Problem 1.6] whether $\operatorname{Hilb}_{8}(\mathbb{A}^{4}_{\mathbb{C}})$ is reduced. There was a bit of progress on Question 1.2 in recent years. Classically, Berthelot and Ogus [BO78, §3] note that the maximal ideal of the algebra $\mathbb{F}_{p}[x_{1},\ldots,x_{6}]/(x_{1}^{p},x_{2}^{p},\ldots,x_{6}^{p},x_{1}x_{2}+x_{3}x_{4}+x_{5}x_{6})$ admits no divided power structure. Bhatt [Bha12, 3.16] mentioned that this algebra does not lift to $W_{2}(\mathbb{F}_{p})=\mathbb{Z}/p^{2}$ and Zdanowicz [Zda18, §3.2] gave a short direct proof of this fact. Langer [Lan18] proved that for $p=2$ this algebra does lift to characteristic zero. He also refined Zdanowicz’s method to give, for every local Artin ring $A$ with $pA\neq 0$ and residue characteristic $p$, a finite $\mathbb{F}_{p}$-scheme nonliftable to $A$. The constructed scheme depends on $A$ and Langer writes “in principle these schemes could be liftable to characteristic [math] over some more ramified rings but we are unable to check whether this really happens.”

In this paper we prove that the answers to Questions 1.1-1.2 above are negative; both pathologies occur as special cases of a Murphy-type Law, which we now describe. We say that Murphy’s Law holds up to retraction for a space $\mathcal{M}$ if for every singularity type $\mathfrak{S}$ there is a representative $(Y,y)$ of $\mathfrak{S}$, an open subscheme $(X,x)$ of $\mathcal{M}$ and a retraction $(X,x)\to(Y,y)$. Here, a retraction $(X,x)\to(Y,y)$ is a morphism of pointed schemes together with a section. The aim of this paper is to prove the following theorem.

Theorem 1.3**.**

Murphy’s Law holds up to retraction for $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{16}_{\mathbb{Z}})$.

On the infinitesimal level, Theorem 1.3 implies that for every singularity type $\mathfrak{S}$, there exists a representative $(Y,y)$ of $\mathfrak{S}$ and a point on $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{16}_{\mathbb{Z}})$ with complete local ring $\hat{\mathcal{O}}_{Y,y}[[t_{1},\ldots,t_{r}]]/I$, for some $r$ and $I$ such that $\hat{\mathcal{O}}_{Y,y}\cap I=0$. In particular, $\hat{\mathcal{O}}_{Y,y}$ is a subring of the complete local ring.

Murphy’s Law up to retraction holds also for the scheme $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{P}^{16}_{\mathbb{Z}})$, as $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{16}_{\mathbb{Z}})$ is its open subscheme. More generally, the forgetful functor from embedded to abstract deformations of a finite scheme is smooth [Art76, p. 4], hence the above pathologies appear also for the abstract deformations of finite schemes and for Hilbert schemes of points on every smooth quasi-projective variety of dimension at least sixteen.

The negative answers to Questions 1.1-1.2 with $n=16$ follow from Theorem 1.3 applied to singularity types $[\operatorname{Spec}(\mathbb{Z}[u]/u^{2}),(u)]$ and $[\operatorname{Spec}(\mathbb{Z}/p),0]$ respectively, see Section 5. More precisely, for Question 1.2 we obtain finite schemes $R$ which do not lift to any ring $A$ with $pA\neq 0$. In Section 5 we give explicit examples of this behavior, for $\Bbbk=\mathbb{Z}/p$, $p=2,3,5$. Using the singularity type $[\operatorname{Spec}(\mathbb{Z}/p^{\nu}),(p)]_{{\nu}=2,3,\ldots}$ we also obtain finite $(\mathbb{Z}/p)$-schemes $R$ that do lift to $W_{\nu}(\mathbb{Z}/p)=\mathbb{Z}/p^{\nu}$ but do not lift to any $A$ with $p^{\nu}A\neq 0$ and so forth.

The main difficulty with analysing finite schemes is their lack of structure: for example, they admit no non-trivial line bundles. Our proof of Theorem 1.3 proceeds by a series of reductions from objects with more structure, such as projective schemes. The main role is played by the generalized Białynicki-Birula decomposition, which we now recall.

Classically [BB73], for a smooth variety $X$ with a $\mathbb{G}_{m}$-action and $X^{\mathbb{G}_{m}}=\coprod_{i=1}^{r}Y_{i}$ with $Y_{i}$ connected, the BB decomposition is a variety $X^{+}=\coprod_{i=1}^{r}X^{+}_{i}$ with a map $\theta_{0}\colon X^{+}\to X$ and a retraction $\pi\colon X^{+}\to X^{\mathbb{G}_{m}}$ which makes $X^{+}_{i}$ a locally trivial affine fiber bundle over $Y_{i}$.

The generalized Białynicki-Birula decomposition [Dri13, Jel19, JS19] extends this construction to $\mathbb{G}_{m}$-schemes $X$, not necessarily smooth, normal or reduced. We apply it to the scheme $\mathcal{H}_{\mathbb{Z}}:=\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{r}_{\mathbb{Z}})$ with the standard $\mathbb{G}_{m}$-action on $\mathbb{A}^{r}_{\mathbb{Z}}$. We obtain a locally finite type $\mathbb{Z}$-scheme $\mathcal{H}^{+}_{\mathbb{Z}}=\operatorname{Hilb}^{+}_{\mathrm{pts}}(\mathbb{A}^{r}_{\mathbb{Z}})$ together with a map $\theta_{0}\colon\mathcal{H}^{+}_{\mathbb{Z}}\to\mathcal{H}_{\mathbb{Z}}$ and a retraction $\pi\colon\mathcal{H}^{+}_{\mathbb{Z}}\to\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}$ with section $i$ ($\mathcal{H}^{+}_{\mathbb{Z}}$ is not in general a bundle over $\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}$). The $\Bbbk$-points in the image of $\theta_{0}$ are exactly the $\Bbbk$-schemes supported at the origin of $\mathbb{A}^{r}_{\Bbbk}$, so $\operatorname{im}(\theta_{0})$ is nowhere dense. To remedy this, we define $\theta\colon\mathbb{G}_{a}^{r}\times\mathcal{H}^{+}_{\mathbb{Z}}\to\mathcal{H}_{\mathbb{Z}}$ which maps $(v,[R])$ to the scheme $\theta_{0}([R])$ translated by $v$. The structure is summarized in the diagram below, see Section 4 for details.

[TABLE]

The maps $\theta_{0}$ and $\theta$ are injective on $\Bbbk$-points for all fields $\Bbbk$ and we identify the points of $\mathcal{H}^{+}_{\mathbb{Z}}$ with their images in $\mathcal{H}_{\mathbb{Z}}$. The tangent space to $[R]\in\mathcal{H}_{\Bbbk}(\Bbbk)$ is equal to $\operatorname{Hom}_{T}(I_{R},T/I_{R})$, where $T=H^{0}(\mathbb{A}^{r},\mathcal{O}_{\mathbb{A}^{r}_{\Bbbk}})$. If $[R]$ is $\mathbb{G}_{m}$-fixed, then this space is graded and in this case $T_{\mathcal{H}^{+}_{\Bbbk},[R]}=\operatorname{Hom}_{T}(I_{R},T/I_{R})_{\geq 0}$. It follows that $d\theta_{[R]}$ is surjective if and only if

[TABLE]

If (1.2) holds, we say that $R$ has trivial negative tangents (TNT for short). If $R$ has TNT, then $\theta$ is an open immersion on its neighbourhood, see [Jel19]. This property is important enough to give it a name: we say that $(X,x)$ locally retracts to $(Y,y)$ if there are open neighbourhoods $x\in U\subset X$ and $y\in V\subset Y$ and a retraction $(U,x)\to(V,y)$. For example, if $R$ has TNT then $(\mathcal{H}_{\mathbb{Z}},[R])$ locally retracts to $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}},[R])$.

Now we discuss the proof of Theorem 1.3. We first present a natural but unsuccessful line of argument and then refine it to obtain a proof. Fix a singularity type $\mathfrak{S}$. Vakil [Vak06] proved that there is a smooth surface $Z\subset\mathbb{P}^{4}$ whose embedded deformations in $\mathbb{P}^{4}$ are of type $\mathfrak{S}$. For $M\gg 0$ the $\mathbb{G}_{m}$-equivariant deformations of the cone $V(I(Z)_{\geq M})\subset\mathbb{A}^{5}$ are also of singularity type $\mathfrak{S}$. Let $\mathfrak{n}$ be the ideal of the origin in $\mathbb{A}^{5}$. Then the $\mathbb{G}_{m}$-equivariant deformations of the zero-dimensional truncation $R_{0}:=V(I(Z)_{\geq M}+\mathfrak{n}^{M+2})$ are of singularity type $\mathfrak{S}$, see [Erm12]. In other words, $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}},[R_{0}])$ has type $\mathfrak{S}$. Thus, $(\mathcal{H}^{+}_{\mathbb{Z}},[R_{0}])$ has singularity type $\mathfrak{S}$ up to retraction. However, there is no reason for $\theta$ to be an open immersion around $[R_{0}]$ and we cannot say anything about the singularity type of $[R_{0}]\in\mathcal{H}_{\mathbb{Z}}$. To obtain a scheme such that that $\theta$ is an open immersion in its neighbourhood, we need a refinement that produces a scheme with TNT.

The refinement is based on the concept of TNT frames. Let $I\subset S=\Bbbk[x_{1},\ldots,x_{n}]$ be a homogeneous ideal and $a\geq 2$. A TNT frame of size $a$ for $I$ is an ideal $J\subset T=S[y_{1},\ldots,y_{n}]=\Bbbk[x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}]$ given by

[TABLE]

Informally, the TNT frame is a bifurcated reduction of $I$ to dimension zero. The quadric $Q=\sum x_{i}y_{i}$ is technically useful for the proof that $\operatorname{Spec}(T/J)$ has TNT because the deformations of $J$ inside $V(Q)$ do not admit any negative tangents under mild assumptions on $I$, see Corollary 3.8. Since $J$ is homogeneous with respect to both $x_{i}$’s and $y_{i}$’s, the stabilizer of $[J]$ contains a two dimensional torus $\mathbb{G}_{x}\times\mathbb{G}_{y}$. Let $\mathbb{G}_{xy}\subset\mathbb{G}_{x}\times\mathbb{G}_{y}$ be the torus acting diagonally.

The concept of TNT frames is geared towards the following result.

Proposition 1.4**.**

Let $\operatorname{char}\Bbbk\neq 2$ and let $I\subset S$ be a homogeneous ideal with $I_{2}=0$. Assume that $\operatorname{depth}(S_{+},S/I)\geq 3$ and $\dim S\geq 3$. Then

1. (1)

$\operatorname{Spec}(T/J)$* has TNT, hence $(\mathcal{H}_{\mathbb{Z}},[J])$ locally retracts to $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}},[J])$,* 2. (2)

$(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}},[J])$* locally retracts to $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{x}\times\mathbb{G}_{y}},[J])$,* 3. (3)

$(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{x}\times\mathbb{G}_{y}},[J])$* locally retracts to $(\operatorname{Hilb}^{\mathbb{G}_{x}}(\mathbb{A}^{n}),[I+(x_{1},\ldots,x_{n})^{a+1}])$.*

For $a\gg 0$ the $\mathbb{G}_{x}$-equivariant deformations of $I+(x_{1},\ldots,x_{n})^{a+1}\subset S$ and $I\subset S$ are canonically isomorphic [Erm12]. For such a number $a$, the composition of the local retractions from Proposition 1.4 gives a local retraction of $(\mathcal{H}_{\mathbb{Z}},[J])$ to $(\operatorname{Hilb}^{\mathbb{G}_{x}}(\mathbb{A}^{n}),[I])$. The analogue of Proposition 1.4 holds also in $\operatorname{char}\Bbbk=2$, for a slightly modified $J$, see Section 3.3.

We return to the proof of Theorem 1.3. Fix a singularity type $\mathfrak{S}$, a surface $Z\subset\mathbb{P}^{4}$ and its truncation $V(I(Z)_{\geq M})\subset\mathbb{A}^{5}$ as above. The ideal $I(Z)_{\geq M}$ does not satisfy the depth assumption of Proposition 1.4, so fix a linear embedding $\mathbb{A}^{5}\hookrightarrow\mathbb{A}^{8}$ and consider the extended ideal $I:=I(Z)_{\geq M}\cdot S$ in $S=H^{0}(\mathbb{A}^{8},\mathcal{O}_{\mathbb{A}^{8}})$. Let $J$ be a TNT frame for $I$ with $a\gg 0$. Proposition 1.4 implies that the (not necessarily equivariant!) deformations of $[J]$ in $\operatorname{Spec}(T)=\mathbb{A}^{16}$ locally retract to the $\mathbb{G}_{x}$-equivariant deformations of $[I]$ in $\mathbb{A}^{8}$. These in turn retract to the equivariant deformations of $[I(Z)_{\geq M}]$ in $\mathbb{A}^{5}$. Thus, $(\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{16}_{\mathbb{Z}}),[J])$ locally retracts to a scheme of singularity type $\mathfrak{S}$ and the proof is concluded.

In the course of the proof we give two side-results. First, in Corollary 3.10 we present a class of ideals, generalizing TNT frames above, which have TNT. By [Jel19, Theorem 1.2], each of those ideals lies on an elementary component of the Hilbert scheme; thus we obtain a new, very large class of elementary components. Second, in Corollary 3.17 we prove that thickenings of maximal linear subspaces of the quadric $V(Q)$ are rigid.

Theorem 1.3 together with related combinatorics, in particular TNT frames which subtly balance prescribed homogeneous deformations and the TNT condition, is the main novelty of this paper. The Białynicki-Birula decomposition of $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{n}_{\Bbbk})$, in the equicharacteristic setting, was introduced in [Jel19]. The ambient dimension $n=16$ in Theorem 1.3 is chosen to make the argument transparent and is probably not optimal. Hazarding a guess, we would say that it can be reduced to $n=6$ or even $n=4$. In any case, the case $n=3$ is special because of the superpotential description [DS09, BBS13]; it would be interesting to know the answers to Questions 1.1-1.2 in this case.

The above results exhibit pathologies of the space of based rank $n$ algebras, see [Poo08, §4]. Hilbert schemes of points also appear prominently in the study of secant and cactus varieties and in algebraic complexity [Lan12]. The non-reducedness of $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{n}_{\mathbb{C}})$ strongly suggests that the equations for these varieties obtained in [BB14] are only set-theoretic, not ideal-theoretic. To make this suggestion rigorous, one would need to know that the Gorenstein locus of $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{n}_{\mathbb{C}})$ is non-reduced, but this remains open. Another interesting open question is whether there are generically non-reduced components of the Hilbert scheme of points.

The outline is as follows: in Section 3 we prove the necessary prerequisite results on the tangent spaces and maps. In Subsections 3.1-3.2 we deal with $\operatorname{char}\Bbbk\neq 2$, while in Subsection 3.3 we present the modified construction for characteristic two. Section 4 contains main ideas of the paper: we discuss Białynicki-Birula decompositions, prove a generalized version of Proposition 1.4 and finally prove Theorem 1.3. In Section 5 we discuss consequences of specific singularity types and give explicit examples of non-reduced points on the Hilbert scheme and components lying in positive characteristic.

Acknowledgements

I am very grateful to Piotr Achinger, Jarosław Buczyński and Maciek Zdanowicz for helpful and inspiring conversations and insightful comments of the early versions of this paper. The paper was prepared during the Simons Semester Varieties: Arithmetic and Transformations which is supported by the grant 346300 for IMPAN from the Simons Foundation and the matching 2015-2019 Polish MNiSW fund.

2. Pointed schemes and smooth equivalence

A pointed scheme $(X,x)$ is a scheme $X$ of finite type over $\mathbb{Z}$ together with a point $x$ of the underlying topological space of $X$. A morphism of pointed schemes $f\colon(X,x)\to(Y,y)$ is a morphism of schemes $f^{\prime}\colon X\to Y$ such that $f^{\prime}(x)=y$. We say that $f$ is smooth if the underlying morphism $f^{\prime}\colon X\to Y$ is smooth. We say that pointed schemes $(X,x)$ and $(Y,y)$ are smoothly equivalent if there exists a pointed scheme $(Z,z)$ and smooth maps

[TABLE]

Lemma 2.1**.**

Being smoothly equivalent is an equivalence relation of pointed schemes.

Proof.

The relation is clearly reflexive and symmetric. To prove transitivity, suppose that the pairs $(X,x)$, $(Y,y)$ and $(Y,y)$, $(T,t)$ are smoothly equivalent. By definition there exist pointed schemes $(Z,z)$ and $(V,v)$ together with smooth maps forming the diagram

[TABLE]

Let $W:=Z\times_{Y}V$ and let $\kappa(-)$ denote the residue field. The algebra $\kappa(z)\otimes_{\kappa(y)}\kappa(v)$ is a tensor product of nonzero algebras over a field, hence it is nonzero. Therefore the scheme $\operatorname{Spec}(\kappa(z)\otimes_{\kappa(y)}\kappa(v))\simeq\operatorname{Spec}(\kappa(z))\times_{\operatorname{Spec}(\kappa(y))}\operatorname{Spec}(\kappa(v))$ is nonempty. Choose a point $p$ of this scheme. The natural maps $\operatorname{Spec}(\kappa(z))\to Z$ and $\operatorname{Spec}(\kappa(v))\to V$ induce a map $\operatorname{Spec}(\kappa(p))\to W$. Let $w\in W$ be the image of $p$. The pointed scheme $(W,w)$ comes with smooth maps to $(X,x)$ and $(T,t)$ and proves that those schemes are smoothly equivalent. ∎

Lemma 2.1 shows in particular that the above definition of smooth equivalence agrees with Vakil’s one given in the introduction. The equivalence classes of smooth equivalence are called singularity types. The singularity type of $(X,x)$ is denoted by $[(X,x)]$ or simply $[X]$ if $X$ has only one point. For example, the singularity type $[\operatorname{Spec}(\mathbb{F}_{p})]$ consists of pointed schemes $(X,x)$ over $\operatorname{Spec}(\mathbb{F}_{p})$ and such that $x\in X$ is smooth. A retraction $(X,x)\to(Y,y)$ is a pair $(f,s)$ where $f\colon(X,x)\to(Y,y)$ and $s\colon(Y,y)\to(X,x)$ are morphisms of pointed schemes such that $f\circ s=\operatorname{id}_{Y}$. For each such retraction, the residue fields of $x$ and $y$ are isomorphic. The retractions we encounter in this article come from diagrams similar to (1.1).

3. Tangent spaces

In this section we prove tangent-map-surjectivity lemmas, such as the TNT condition, needed for the proof of a generalized version of Proposition 1.4. Specifically, the aim is to prove Proposition 3.10, Corollary 3.16 (for characteristic two, respectively Proposition 3.23, Corollary 3.26), which are applied in Section 4. These results follow from the chain of quite technical partial results, obtained using linear algebra and representation theory. We encourage the reader to consult Section 4 for motivation before diving into details.

Throughout, let $\Bbbk$ be field. Let $I\subset S=\Bbbk[\mathbf{x}]$ be a homogeneous ideal. Let $T=S[\mathbf{y}]=\Bbbk[x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}]$. Fix $a\geq 2$, $b\geq 1$ and let $\mathfrak{m}_{x}:=(x_{1},\ldots,x_{n})\subset T$, $\mathfrak{m}_{y}:=(y_{1},\ldots,y_{n})\subset T$, $Q:=\sum x_{i}y_{i}$ and

[TABLE]

The ideal $J$ is $\mathbb{N}^{2}$-graded by $(\deg\mathbf{x},\deg\mathbf{y})$. Throughout this section the word homogeneous for elements of $T$ refers to this bi-grading. For elements of $S$, the word homogeneous refers to the usual grading by the total degree (when viewing $S$ as subring of $T$, these agree). The ideal $J$ is generated by elements of degrees $(*,0)$, $(0,b+1)$, and $(1,1)$. The space $\operatorname{Hom}_{T}(J,T/J)$ is graded by

[TABLE]

This section is devoted to showing that certain homogeneous pieces of $\operatorname{Hom}_{T}(J,T/J)$ vanish or are as small as possible. The following Table 3.2 provides for each such piece a reference to the corresponding result. Stars denote pieces which are not of interest. The table is subtly asymmetric with respect to $\mathbf{x}$ and $\mathbf{y}$.

Now we introduce two tricks which recur in our computation of the homogeneous components of the space (3.1). Consider the canonical epimorphism $p\colon T/(I\cdot T+(Q))\to T/J$ and note that there exists a unique homogeneous linear section of the map $p$:

[TABLE]

and that $s$ is zero in degrees $(*,\geq a+1)$, $(\geq b+1,*)$ and an isomorphism in degrees $(\leq a,\leq b)$. Our first trick is the following lemma.

Lemma 3.1** (lifting homogeneous homomorphisms).**

Let $N$ be a graded $T$-module with a minimal presentation

[TABLE]

and $\psi$ be a homomorphism $\psi\colon N\to T/J$ of degree $(\alpha,\beta)$. Suppose that $d_{1j}+\alpha\leq a$ and $d_{2j}+\beta\leq b$ for all $j$. Then $\psi$ lifts to a homomorphism $N\to T/(I\cdot T+(Q))$ of degree $(\alpha,\beta)$.

Proof.

Let $K:=\frac{J}{IT+(Q)}$. Take a lifting of $\psi$ to a homogeneous chain complex map

[TABLE]

The map $\rho$ maps the generator of $T(-d_{1j},-d_{2j})$ into $K_{d_{1j}+\alpha,d_{2j}+\beta}$. But $d_{1j}+\alpha\leq a$ and $d_{2j}+\beta\leq b$, hence $K_{d_{1j}+\alpha,d_{2j}+\beta}=0$. Therefore $\rho=0$, which completes the claim. ∎

Our second trick is as follows: suppose that $K\subset T$ is an ideal and $M$ is a $T$-module with $\operatorname{depth}(K,M)\geq 2$. Then $\operatorname{Ext}^{i}(T/K,M)=0$ for $i=0,1$, see [Eis95, Proposition 18.4], so $M=\operatorname{Hom}_{T}(T,M)\to\operatorname{Hom}_{T}(K,M)$ is an isomorphism. To make the trick applicable to $K=I\cdot T+(Q)$, we give a lower bound of the depth of ${T/(I\cdot T+(Q))}$, as follows.

Lemma 3.2**.**

Let $d=\operatorname{depth}(S_{+},S/I)$. Suppose that $d\geq 2$ and $f_{1},\ldots,f_{d-1}\in S_{+}$ is a regular sequence on $S/I$ consisting of homogeneous elements. Then $Q$ is a non-zero divisor on both $T/(I\cdot T)$ and $T/(I\cdot T+f_{1}\cdot T)$. Moreover, $f_{1},\ldots,f_{d-1}$ is a regular sequence on $T/(I\cdot T+(Q))$.

Proof.

By base change, we may assume that $\Bbbk$ is algebraically closed.

The quotient module $M=S/(I+(f_{1},\ldots,f_{d-1}))$ has $S_{+}$-depth at least one, hence there exists a quadric $f_{d}=\sum_{i=1}^{n}x_{i}l_{i}\in S_{2}$ which is a regular element for $M$. Thus, the sequence

[TABLE]

is regular for the $T$-module $T/(I\cdot T)$. This sequence consists of elements homogeneous with respect to the total degree, hence each of its permutations is also a regular sequence [Mat86, Theorem 16.3]. In particular, the sequences $(f_{1},Q,f_{2},\ldots,f_{d-1},y_{1}-l_{1},\ldots,y_{n}-l_{n})$ and $(Q,f_{1},\ldots,f_{d-1},y_{1}-l_{1},\ldots,y_{n}-l_{n})$ are regular for $T/(I\cdot T)$. The first one implies that $Q$ is regular on $T/(I\cdot T+(f_{1}))$ and the second that $Q$ is regular on $T/(I\cdot T)$ and that $f_{1},\ldots,f_{d-1}$ is regular on $T/(I\cdot T+(Q))$. ∎

Now we begin direct computations of specific degrees of the tangent space at $J$.

3.1. Negative tangents

In this section we verify that $\operatorname{Spec}(T/J)$ has TNT.

Lemma 3.3** ($I$-ignoring lemma).**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 2$. Let $\alpha$, $\beta$ be integers, with $\alpha\leq-1$. Let $\varphi\in\operatorname{Hom}(J,T/J)_{(\alpha,\beta)}$ be a homomorphism such that $\varphi(\mathfrak{m}_{x}^{a+1})=0$. Then $\varphi(I)=0$.

Proof.

Choose any homogeneous $i\in I$ and let $a^{\prime}$ be its degree. If $a^{\prime}\geq a+1$, then $i\in\mathfrak{m}_{x}^{a+1}$ and so $\varphi(i)=0$. Otherwise, we have $\mathfrak{m}_{x}^{a+1-a^{\prime}}i\subset\mathfrak{m}_{x}^{a+1}$, so

[TABLE]

But $\deg(S_{a+1-a^{\prime}}\varphi(i))=(a+1+\alpha,\beta)$ and $a+1+\alpha<a+1$. Therefore,

[TABLE]

is an isomorphism. Hence, $\varphi(i)$ uniquely lifts to an element $F$ of $T/(I\cdot T+(Q))$ such that $\mathfrak{m}_{x}^{a+1-a^{\prime}}F=0$. By Lemma 3.2, we have $\operatorname{depth}(\mathfrak{m}_{x},T/(I\cdot T+(Q)))>0$, hence $F=0$, so $\varphi(i)=0$. As $i\in I$ was chosen arbitrarily, we have $\varphi(I)=0$. ∎

Lemma 3.4** (depth lemma).**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 3$ and $n\geq 3$. Fix $(\alpha,\beta)$ so that $\alpha<0$ or $\beta<0$. Let $\mathfrak{p}_{x}$, $\mathfrak{p}_{y}$ be homogeneous ideals with radicals $\mathfrak{m}_{x}$, $\mathfrak{m}_{y}$ respectively. Then

[TABLE]

Proof.

Let $M=T/(I\cdot T+(Q))$ viewed as a $T$-module. By Lemma 3.2, $\operatorname{depth}\left(\mathfrak{p}_{x},M\right)\geq 2$. Hence $\operatorname{Ext}^{i}(T/\mathfrak{p}_{x},M)=0$ for $i=0,1$, so

[TABLE]

is an isomorphism. But $\operatorname{Hom}_{T}(T,M)=M$ has no non-zero elements of degree $(\alpha,\beta)$ with $\alpha<0$ or $\beta<0$. The same argument applies with $\mathfrak{p}_{y}$ instead of $\mathfrak{p}_{x}$; the depth assumption is satisfied as the sequence $(y_{1},y_{2})$ is regular. ∎

Corollary 3.5**.**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 3$ and $n\geq 3$. Then $(T_{[J]}\mathcal{H})_{(\alpha,\beta)}=0$ for $\alpha\leq-2$ or $\beta\leq-2$.

Proof.

Consider the case $\alpha\leq-2$. Take an element $\varphi\in\operatorname{Hom}(J,T/J)_{(\alpha,\beta)}$. By degree reasons, $\varphi$ sends $\mathfrak{m}_{y}^{b+1}$ and $Q$ to zero. Consider $\varphi^{\prime}=\varphi_{|\mathfrak{m}_{x}^{a+1}}\colon\mathfrak{m}_{x}^{a+1}\to T/J$. This map sends generators of $\mathfrak{m}_{x}^{a+1}$ to elements of $(T/J)_{(\leq a-1,\leq b-1)}$ and the syzygies of $\mathfrak{m}_{x}^{a+1}$ are linear, of degree $(1,0)$. Hence, by Lemma 3.1, the map $\varphi^{\prime}$ lifts to a map

[TABLE]

of degree $(\alpha,\beta)$ with $\alpha\leq-2$. Such a map is zero by Lemma 3.4 applied to $\mathfrak{p}_{x}=\mathfrak{m}_{x}^{a+1}$. Therefore, $\varphi(\mathfrak{m}_{x}^{a+1})=0$. From Lemma 3.3 it follows that $\varphi(I)=0$, hence $\varphi=0$. The case $\beta\leq-2$ is symmetric, minus the use of Lemma 3.3. ∎

Now we will analyse homomorphisms of degree $(\alpha,\beta)$ with $\alpha+\beta=-1$. These do exist (e.g. the tangents corresponding to the $\mathbb{G}_{a}^{2n}$-action by translations), so the depth considerations as above are not directly applicable.

Lemma 3.6**.**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 2$. Then

[TABLE]

Proof.

Let

[TABLE]

Take the unique lift of $\varphi$ to a linear map $\varphi^{\prime}\colon S_{a+1}\to(S/I)_{a}$. Let $m\in S_{a}$ be a monomial. Then $\deg(mQ)=(a+1,1)$ and

[TABLE]

hence there exists a form $n_{m}\in(S/I)_{a-1}$ such that

[TABLE]

so $\varphi(mx_{i})\equiv x_{i}n_{m}\mod IT+\mathfrak{m}_{x}^{a+1}$. We define a linear map $\psi\colon S_{a}\to(S/I)_{a-1}$ by $\psi(m):=n_{m}$. Suppose that $m_{1},m_{2}\in S_{a}$ are monomials such that $x_{i}m_{1}=x_{j}m_{2}$ for some $i,j$. Then

[TABLE]

But $\deg(x_{i}\psi(m_{1})-x_{j}\psi(m_{2}))=a$, hence $x_{i}\psi(m_{1})-x_{j}\psi(m_{2})\in IT$ and so $\psi$ extends to a $S$-module homomorphism $\psi^{\prime}\colon S_{\geq a}\to S/I$ of degree $-1$. But $\operatorname{depth}(S_{+},S/I)\geq 2$, so $\operatorname{Ext}^{i}(S/S_{\geq a},S/I)=0$ for $i=1,2$, hence $\operatorname{Hom}(S,S/I)\to\operatorname{Hom}(S_{\geq a},S/I)$ is an isomorphism. In particular, there are no homomorphisms of negative degrees, so $\psi^{\prime}=0$. Accordingly, $\psi=0$, hence $\varphi=0$. ∎

Lemma 3.7**.**

Let $n\geq 2$. Then

[TABLE]

Proof.

The proof of Lemma 3.6 can be repeated with $\mathbf{x}$ interchanged with $\mathbf{y}$ and $I=0$. ∎

Let $\partial_{x_{1}},\ldots,\partial_{x_{n}},\partial_{y_{1}},\ldots,\partial_{y_{n}}$ be the derivations with respect to variables of $T$. By Leibniz’s rule, each such derivation $\partial$ induces a $T$-linear map $J\to T/J$. This map is an element of the tangent space $T_{[J]}\mathcal{H}$. By slight abuse, we denote it by $\partial$. Geometrically, these elements arise from the action of $\mathbb{G}_{a}^{2n}$ on $\mathbb{A}^{2n}=\operatorname{Spec}(T)$ by translation.

Corollary 3.8**.**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 2$ and $n\geq 2$. Then

[TABLE]

Proof.

Choose any homomorphism $\varphi\colon J\to T/J$ of degree $(-1,0)$. Then $\varphi(Q)\in\left\langle y_{1},\ldots,y_{n}\right\rangle$. But $\partial_{x_{i}}(Q)=y_{i}$, hence there is a unique linear combination $D$ of $\left\{\partial_{x_{i}}\right\}$ such that $(\varphi-D)(Q)=0$. Replacing $\varphi$ by $\varphi-D$, we may assume $\varphi(Q)=0$. By Lemma 3.6, we have $\varphi(\mathfrak{m}_{x}^{a+1})=0$. By Lemma 3.3, we have $\varphi(I)=0$. Finally, $\varphi(\mathfrak{m}_{y}^{b+1})=0$ by degree reasons, so $\varphi=0$ and the claim follows for $(T_{[J]}\mathcal{H})_{(-1,0)}$. The argument for degree $(0,-1)$ is symmetric. ∎

Out of all tangents of negative degrees, there is a single degree left to consider: $(-1,-1)$.

Lemma 3.9** ($(-1,-1)$ tangents).**

Suppose that $\mathfrak{m}_{x}^{a}\not\subset I$. Then $(T_{[J]}\mathcal{H})_{(-1,-1)}=0$.

Proof.

Let $\varphi\colon J\to T/J$ a homomorphism of degree $(-1,-1)$. Then $\varphi(\mathfrak{m}_{x}^{a+1})=\varphi(\mathfrak{m}_{y}^{b+1})=\varphi(I)=0$ by degree reasons. Moreover, $\varphi(Q)\in\Bbbk$. If $\varphi(Q)\neq 0$, then

[TABLE]

so $\mathfrak{m}^{a}\subset I$, a contradiction. Hence $\varphi(Q)=0$, so $\varphi=0$. ∎

Proposition 3.10** (TNT for $J$).**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 3$ and $n\geq 3$. Then the scheme $\operatorname{Spec}(T/J)$ has TNT.

Proof.

This follows from Corollary 3.5, Corollary 3.8, and Lemma 3.9. ∎

3.2. Degree zero tangents in characteristic $\neq 2$

Proposition 3.10 is the essential part to obtain the local retraction from Point 1 in Proposition 1.4. To obtain the retraction from Point 2, we need to compute the tangents of $[\operatorname{Spec}(T/J)]$ that have degree $(\alpha,-\alpha)$ for $\alpha>0$. All homomorphisms coming from such tangents kill $I$ by degree reasons. Hence, the material in this section is independent of the choice of $I$. To emphasise this further, we introduce the linear space $V=\left\langle x_{1},\ldots,x_{n}\right\rangle$ and identify $y_{1},\ldots,y_{n}$ with the dual space $V^{*}$. Then

[TABLE]

for all $\alpha,\beta$ and, in particular, $Q$ becomes the trace element in $V\otimes V^{*}$, hence is $\operatorname{SL}(V)$-invariant. For $\mathbf{a},\mathbf{b}\in\mathbb{N}^{n}$ we let $\mathbf{x}^{\mathbf{a}}$, $\mathbf{y}^{\mathbf{b}}$ denote the monomials $x_{1}^{\mathbf{a}_{1}}\ldots x_{n}^{\mathbf{a}_{n}}$, $y_{1}^{\mathbf{b}_{1}}\ldots y_{n}^{\mathbf{b}_{n}}$ respectively. We will denote the space $(\operatorname{S}^{\alpha}V^{*})^{*}$ by $\operatorname{\Gamma}^{\alpha}V$ and denote by $\mathbf{x}^{[\mathbf{a}]}\in\operatorname{\Gamma}^{\alpha}V$ the functional which is dual to $\mathbf{y}^{\mathbf{a}}$. There is an contraction action $(-)\,\lrcorner\,(-)\colon V^{*}\otimes\operatorname{\Gamma}^{\alpha}V\to\operatorname{\Gamma}^{\alpha-1}V$ given by

[TABLE]

for all $\ell\in V^{*}$, $\varphi\in\operatorname{\Gamma}^{\alpha}V$ and $f\in\operatorname{S}^{\alpha-1}V^{*}$. For $\operatorname{char}\Bbbk>\alpha$ there is a unique $\operatorname{GL}(V)$-equivariant isomorphism $\operatorname{\Gamma}^{\alpha}V\simeq\operatorname{S}^{\alpha}V$ and it sends $\mathbf{x}^{[\mathbf{a}]}$ to $\mathbf{x}^{\mathbf{a}}/\mathbf{a}!$. Under this isomorphism, contraction corresponds to differentiation $\ell\circ\varphi\mapsto\partial_{\ell}(\varphi)$. See [Eis95, A2.4] or [BB14, §3] for details.

Now we introduce the group $G$ giving trivial degree zero tangents; this is the degree-zero counterpart of $\mathbb{G}_{a}^{2n}$. Let $G\subset\operatorname{GL}_{2n,\mathbb{Z}}$ be a subgroup given in the basis $x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}$ by

[TABLE]

The group $G$ is smooth and acts naturally on $\mathcal{H}_{\mathbb{Z}}$ and $\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}}$. The Lie algebra $\mathfrak{g}$ of $G_{\Bbbk}$ maps $y_{i}$’s to combinations of $x_{j}$’s. Hence, the tangent to the orbit map $G_{\Bbbk}\ni g\mapsto g\cdot[J]\in\mathcal{H}_{\Bbbk}^{\mathbb{G}_{m}}$ is

[TABLE]

Let $G^{\prime}$ be the stabilizer of $Q$ in $G$. In the description (3.4), it consists of anti-symmetric matrices $A$. Let $\mathfrak{g}^{\prime}$ be the Lie algebra of $G^{\prime}$. The action of $\mathfrak{g}^{\prime}$ annihilates $Q$ and we obtain a tangent map

[TABLE]

Now we proceed to computations. Throughout this subsection, $\otimes$ denotes $\otimes_{\Bbbk}$. The $S$-module $T_{*,b+1}$ is free. Let $M:=(T/Q)_{*,b+1}$. This is an $S$-module with presentation

[TABLE]

where the twists correspond to the fact that generators of $M$ have $\mathbf{x}$-degree $(0,b+1)$ and its syzygies have degree $(1,b+1)$ with respect to natural bi-grading. The presentation map is just the multiplication by $Q$. Explicitly, it is given by

[TABLE]

The module $M$ plays a key role in the computation of $(1,-1)$ tangents of $[\operatorname{Spec}(T/J)]$. There is a natural injection $M\subset\mathfrak{m}_{y}^{b+1}$, hence we obtain restriction maps

[TABLE]

Lemma 3.11**.**

Suppose $I_{2}=0$. Then all maps in the diagram (3.9) are bijective.

Proof.

Left-side horizontal maps are bijective, because the homomorphisms kill other generators of $J/Q$ by degree reasons. The right-side horizontal maps are bijective since homomorphisms of degree $(1,-1)$ into $T/\mathfrak{m}_{y}^{b+1}$ annihilate $T_{(*,\geq b+2)}$. Finally, $I_{2}=0$ implies that the surjection $T/J\to T/Q$ is bijective in degrees $(\leq 1,\leq 2)$. Then the rightmost downward arrow is bijective by applying Lemma 3.1 to (3.7). ∎

We concentrate on analysing $\operatorname{Hom}_{S}(M,T/Q)_{(1,-1)}$. For brevity, let $K:=\operatorname{Hom}_{S}\left(M,\frac{T}{Q}\right)_{(1,-1)}$. From the presentation (3.7) we obtain an exact sequence

[TABLE]

which simplifies to $0\to K\to\operatorname{\Gamma}^{b+1}V\otimes\left(T/Q\right)_{(1,b)}\to\operatorname{\Gamma}^{b}V\otimes\left(T/Q\right)_{(2,b)}$. Hence, we obtain a commutative diagram with exact columns and bottom row.

[TABLE]

Let us write down $\Phi$ and $\Phi_{0}$ explicitly. The map $\Phi$ comes from applying $\operatorname{Hom}_{S}(-,T)_{(1,-1)}$ to the map (3.8). Therefore, it is given in coordinates by

[TABLE]

where $\,\lrcorner\,$ denotes contraction, as defined in (3.3).

Proposition 3.12**.**

Suppose that $b=1$ and $\operatorname{char}\Bbbk\neq 2$. Then the map

[TABLE]

obtained by composing (3.6) and (3.9) is bijective.

Proof.

For an element of $\mathfrak{g}^{\prime}$, the image of $y_{i}$ is read off the image of $y_{i}^{b+1}=y_{i}^{2}$ in the corresponding homomorphism, so the map (3.12) is injective and it is enough to check that $\dim\operatorname{Hom}_{S}(M,T/Q)_{(1,-1)}=\dim\mathfrak{g}^{\prime}=\dim\Lambda^{2}V$. We do this directly.

Since $b=1$, the map $\Phi\colon\operatorname{\Gamma}^{2}V\otimes V\otimes V^{*}\to V\otimes\operatorname{S}^{2}V\otimes V^{*}$ has source and target of the same dimension. We will prove that it is bijective. It is enough to prove surjectivity. By (3.11), we have $\Phi=\Psi\otimes\operatorname{id}_{V^{*}}$ for the map $\Psi\colon\operatorname{\Gamma}^{2}V\otimes V\to V\otimes S^{2}V$ given by

[TABLE]

It is enough to prove that $\Psi$ is surjective. For pairwise distinct $i$, $j$, $k$, we have

[TABLE]

The same holds for not necessarily distinct $i$, $j$, $k$ under the convention that $x_{i}\cdot x_{i}=2x_{i}^{[2]}$ in $\operatorname{\Gamma}^{2}V$. Thus, the map $\Psi$ is surjective, hence the map $\Phi$ is bijective. In particular, $\Phi$ and $\Phi_{0}$ are injective. From the snake lemma applied to (3.10), we have $K\simeq\operatorname{coker}\Phi_{0}$, in particular $\dim K=\dim\Lambda^{2}V$ as claimed. ∎

Remark 3.13**.**

Proposition 3.12 fails for $\operatorname{char}\Bbbk=2$; in particular $\Psi$ and $\Phi$ are not injective.

The restriction to $b=1$, while sufficient for our purposes, is not very satisfactory. For $b>1$ we have the following result in large enough characteristics. We will not use it in the proof of Theorem 1.3, so we only sketch a proof.

Proposition 3.14**.**

Let $b\geq 1$ be arbitrary and $\operatorname{char}\Bbbk=0$. Then the map $\mathfrak{g}^{\prime}\to\operatorname{Hom}_{S}(M,T/Q)_{(1,-1)}$ is bijective.

Sketch of proof.

By Proposition 3.12 we may assume $b>1$ (we only need this for notational reasons in Schur functors). The maps in Diagram (3.10) split into maps between simple $\operatorname{GL}(V)$-modules, which are indexed by Young diagrams [FH91, §6].

The cokernel of $\Phi$ is isomorphic to $\operatorname{\mathbb{S}}_{b,2}V\otimes\operatorname{S}^{b}V^{*}$. Similarly, the cokernel of $\operatorname{id}\otimes(-\cdot Q)$ is isomorphic to $\operatorname{\Gamma}^{b+1}V\otimes\operatorname{\mathbb{S}}_{b+1,b,\ldots,b}V$. Hence, the cokernel of $\overline{\Phi}$ is obtained from $\operatorname{\Gamma}^{b}V\otimes V^{\otimes 2}\otimes\operatorname{S}^{b}V^{*}\simeq\operatorname{S}^{b}V\otimes V^{\otimes 2}\otimes\operatorname{S}^{b}V^{*}$ by applying two partial symmetrizations (corresponding to dividing by images of $\operatorname{id}\otimes(-\cdot Q)$ and $\Phi$), which together imply that $\operatorname{coker}\overline{\Phi}=\operatorname{\mathbb{S}}_{2b,b+2,b,\ldots,b}V$. Now, a dimension count on the bottom row shows that $\dim K=\dim\Lambda^{2}V$ as claimed. ∎

Corollary 3.15**.**

Suppose that $I_{2}=0$ and $\operatorname{char}\Bbbk\neq 2$. Suppose further that either $b=1$ or $\operatorname{char}\Bbbk=0$. Then the map (3.5) is bijective.

Proof.

Let

[TABLE]

Take an element $\varphi\in\operatorname{Hom}(J,T/J)_{(1,-1)}$. By degree reasons, $\varphi(\mathfrak{m}_{x}^{a+1})=\varphi(I)=0$ and $\varphi(Q)\in T_{2,0}=S_{2}$. There is a unique element $D\in\mathfrak{lt}$ such that $D(Q)=\varphi(Q)$. Replacing $\varphi$ by $\varphi-D$, we can assume $\varphi(Q)=0$. Therefore, $\varphi$ comes from an element of

[TABLE]

By Lemma 3.11 and Proposition 3.12 or Proposition 3.14, there exists a unique element of $\mathfrak{g}^{\prime}$ mapping to $\varphi$, which concludes the proof. ∎

Corollary 3.16**.**

Let $n\geq 3$ and $\operatorname{char}\Bbbk\neq 2$. Suppose that $\operatorname{depth}(S_{+},S/I)\geq 3$ and $I_{2}=0$. Suppose further that either $b=1$ or $\operatorname{char}\Bbbk=0$. The map

[TABLE]

is bijective.

Proof.

This follows from Corollary 3.15 and Corollary 3.5. ∎

The above computations of degree-zero tangents can be translated to a geometric statement, which is of some independent interest at least as a motivation.

Corollary 3.17**.**

Let $\operatorname{char}\Bbbk\neq 2$ and $n\geq 3$. Assume either $b=1$ or $\operatorname{char}\Bbbk=0$. Let

[TABLE]

be a thickening of a projective subspace on a quadric. Then $L_{\mathrm{y}}^{b}$ is a rigid scheme: we have $T^{1}_{L_{\mathrm{y}}^{b}}=0$, where $T^{1}$ is the Schlessinger’s functor [Har10, §3].

Proof.

Let $B=T/((Q)+(y_{1},\ldots,y_{n})^{b+1})$ be the coordinate ring of $L_{\mathrm{y}}^{b}$. Since $n\geq 3$, we have $\operatorname{depth}(\mathfrak{m}_{y},B)\geq 2$, so $T^{1}_{L_{\mathrm{y}}^{b}}\simeq(T^{1}_{B})_{0}$ naturally, see e.g. [Har10, Theorem 5.4]. Corollary 3.15 proves that $(T^{1}_{B})_{0}=0$, directly from the construction of $T^{1}_{B}$. ∎

3.3. Tangents in characteristic $2$

Assumption 3.18**.**

In this subsection $\Bbbk$ is a field of characteristic two.

In this case, Proposition 3.12 fails and we need to replace $\mathfrak{m}_{y}^{2}$ in the definition of $J$ by another ideal. There are many possible replacements; in any case the symmetry has to be broken. We choose $\mathfrak{p}\subset\Bbbk[\mathbf{y}]$ defined by

[TABLE]

mainly for the (relatively) simple computations. Let $I\subset S=\Bbbk[x_{1},\ldots,x_{n}]$ be a homogeneous ideal and $a\geq 2$. A tweaked frame of size $a$ for $I$ is an ideal $\mathcal{J}\subset T=\Bbbk[\mathbf{x},\mathbf{y}]$ defined by

[TABLE]

As in Subsection 3.1, we calculate some graded pieces of $T_{[\mathcal{J}]}\mathcal{H}=\operatorname{Hom}_{T}(\mathcal{J},T/\mathcal{J})$. Most of the arguments will directly pass to this setup. There is some additional work needed in degrees $(*,-1)$. Anyway, for clarity we provide statements and sketches of proofs of all steps.

An important additional piece is the following easy result about the syzygies of $(y_{1}^{2},y_{2}^{2},\ldots,y_{n}^{2},Q)$.

Lemma 3.19**.**

Let $\alpha\in\{0,1\}$ and $n\geq 3$. Suppose that $F_{1},\ldots,F_{n}\in T_{\alpha,1}$ are forms satisfying $\sum_{i=1}^{n}x_{i}^{2}F_{i}\in(Q)$. Then $F_{i}\in(Q)$ for $i=1,\ldots,n$. In particular, if $\alpha=0$, then $F_{i}=0$ for all $i$.

Proof.

Let $\sum x_{i}^{2}F_{i}=Q\cdot G$, then $\deg(G)=(\alpha+1,0)$. Differentiate with respect to $y_{j}$ to obtain

[TABLE]

In follows that $G\in(x_{1}^{2},\ldots,x_{n}^{2})+(x_{j})$. Intersecting over all $x_{j}$, we get $G=\sum_{i=1}^{n}\lambda_{i}x_{i}^{2}$ for some $\lambda_{i}\in\Bbbk$. The forms $\frac{\partial F_{i}}{\partial y_{j}}$ have degree $(\alpha,0)\leq(1,0)$, hence $\frac{\partial F_{i}}{\partial y_{j}}=\lambda_{i}x_{j}$. Consequently, we have $F_{i}=\sum_{j=1}^{n}\frac{\partial F_{i}}{\partial y_{j}}y_{j}=\lambda_{i}\sum_{j}x_{j}y_{j}\in(Q)$. ∎

Lemma 3.20**.**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 3$ and $n\geq 3$. Then $(T_{[\mathcal{J}]}\mathcal{H})_{(\alpha,\beta)}=0$ for $\alpha\leq-2$ or $\beta\leq-2$.

Proof.

The non-Koszul syzygies of $\mathfrak{p}$ are linear, hence the proof of Corollary 3.5 applies without changes. ∎

Lemma 3.21**.**

Suppose that $n\geq 3$ and $I_{2}=0$. Then

[TABLE]

Proof.

Take a homomorphism $\varphi$ as in the left-hand side of (3.15) and its unique lift of $\varphi$ to a linear map $\varphi^{\prime}\colon T_{0,2}\to T_{0,1}$. First, from $Q^{2}=\sum_{i=1}^{n}x_{i}^{2}y_{i}^{2}$ we get

[TABLE]

so $\varphi^{\prime}(y_{i}^{2})=0$ for $i=1,\ldots,n$, by Lemma 3.19. Next, $y_{1}Q=\sum_{i=1}^{n}x_{i}(y_{1}y_{i})$ so we have $0=\sum_{i=1}^{n}x_{i}\varphi(y_{1}y_{i})$ and

[TABLE]

for some $\lambda\in\Bbbk$. Comparing coefficients near $x$’s, we get $\varphi^{\prime}(y_{1}y_{i})=\lambda y_{i}$. In particular $0=\varphi^{\prime}(y_{1}^{2})=\lambda y_{1}$, hence $\lambda=0$, thus $\varphi^{\prime}=0$. ∎

Corollary 3.22**.**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 2$, $I_{2}=0$ and $n\geq 3$. Then

[TABLE]

Proof.

Homomorphisms of degree $(-1,0)$ kill $\mathfrak{p}$ and this case reduces to the one considered in Corollary 3.8. Consider a homomorphism $\varphi\in\operatorname{Hom}_{T}(\mathcal{J},T/\mathcal{J})_{(0,-1)}$. Then $\varphi(Q)\in\left\langle x_{1},\ldots,x_{n}\right\rangle$. But $\partial_{y_{i}}(Q)=x_{i}$, hence there is a unique linear combination $D$ of $\left\{\partial_{y_{i}}\right\}$ such that $(\varphi-D)(Q)=0$. Replacing $\varphi$ by $\varphi-D$, we may assume $\varphi(Q)=0$. By Lemma 3.21 we have $\varphi(\mathfrak{p})=0$. Moreover, $\varphi(I+\mathfrak{m}_{x}^{a+1})=0$ for degree reasons. This shows that $\varphi=0$. ∎

The following Proposition 3.23 summarizes the above discussion. We stress once more, that we assume $\operatorname{char}\Bbbk=2$, see Assumption 3.18.

Proposition 3.23**.**

Suppose that $\operatorname{depth}(S_{+},S/I)\geq 3$ and $n\geq 3$. Then the scheme $\operatorname{Spec}(T/\mathcal{J})$ has TNT.

Proof.

This follows from Lemma 3.20, Corollary 3.22, and a direct analogue of Lemma 3.9. ∎

Now we analyse degree zero tangents of $\mathcal{J}$. In (3.4) we defined the group $G\subset\operatorname{GL}_{2n,\mathbb{Z}}$ by

[TABLE]

We also introduce its Lie algebra $\mathfrak{g}$ and the tangent map

[TABLE]

As before, we will prove that it is bijective.

Lemma 3.24** (special liftings).**

Let $n\geq 4$. Let $\varphi\colon\mathcal{J}\to T/\mathcal{J}$ be a homomorphism of degree $(1,-1)$. Then $\varphi(y_{i}^{2})=0$ for all $i$. Moreover, there exists a special lifting of $\varphi_{|\mathfrak{p}}$, that is, a degree $(1,-1)$ linear map

[TABLE]

such that

1. (1)

$\varphi^{\prime}(g)\mod(Q)=\varphi(g)$* for every generator of $\mathfrak{p}$,* 2. (2)

$\varphi^{\prime}(y_{i}^{2})=0$* and $\varphi^{\prime}(y_{1}y_{i})\in(y_{1},y_{i})$ for all $i$.*

Proof.

Take a homomorphism $\varphi\colon\mathcal{J}/Q\to T/\mathcal{J}$ of degree $(1,-1)$ and any lift to a linear map $\varphi^{\prime}\colon T_{0,2}\to T_{1,1}$. The syzygy $Q\cdot Q=\sum x_{i}^{2}y_{i}^{2}$ implies $\sum_{i=1}^{n}x_{i}^{2}\varphi^{\prime}(y_{i}^{2})\in(Q)$. By Lemma 3.19 we have $\varphi^{\prime}(y_{i}^{2})=0$ for all $i$, after possible changing the lifting.

The syzygy $y_{i}\cdot(y_{1}y_{i})=y_{1}\cdot y_{i}^{2}$ implies that $y_{i}\varphi^{\prime}(y_{1}y_{i})\in\mathfrak{p}+(Q)$. Take $p\in\mathfrak{p}$ and $l\in T_{0,1}$ such that

[TABLE]

Let $l=\sum_{j=1}^{n}l_{j}y_{j}$ for $l_{j}\in\Bbbk$. Replacing $p$ with $p+l_{1}y_{1}Q$, we may assume $l_{1}=0$. Pick $j\neq i$ and any $j^{\prime}\neq 1,i,j$ (we use $n\geq 4$). The monomial $x_{j^{\prime}}y_{j^{\prime}}y_{j}$ appears in $lQ$ with coefficient $l_{j}$, but does not appear elsewhere in (3.17), so $l_{j}=0$. Then $l=l_{i}y_{i}$. Then $p\in(y_{i})\cap\mathfrak{p}=y_{i}(y_{1},y_{i})$, so $p=y_{i}p^{\prime}$ for some $p^{\prime}\in(y_{1},y_{i})$. Clearly, $\varphi^{\prime}(y_{1}y_{i})-p^{\prime}\in(Q)\subset\mathcal{J}$, hence we may replace $\varphi^{\prime}(y_{1}y_{i})$ by $p^{\prime}$. The lifting thus obtained satisfies the conditions. ∎

Proposition 3.25**.**

Let $n\geq 4$ and $I_{2}=0$. The map (3.16) is bijective.

Proof.

Take a lift $\varphi^{\prime}$ as in Lemma 3.24 and the unique lift $\hat{Q}\in T_{2,0}$ of $\varphi(Q)$. Recall from Lemma 3.24 that $\varphi^{\prime}(y_{i}^{2})=0$ and $\varphi(y_{i}^{2})=0$ for all $i$. Write

[TABLE]

where $L_{1}=M_{1}=0$. Since $y_{1}Q=\sum_{i=1}^{n}x_{i}(y_{1}y_{i})$, we have $y_{1}\varphi(Q)=\sum_{i=1}^{n}x_{i}\varphi(y_{1}y_{i})$. Since $I_{2}=0$, we have

[TABLE]

for some $\mu\in T_{1,0}$. Putting (3.18) into (3.19), we get

[TABLE]

Comparing coefficients of $y_{i}$ in (3.20), for $i>1$, we obtain $L_{2}=L_{3}=\ldots=L_{n}=-\mu$. Moreover, $\varphi^{\prime}(y_{1}\cdot y_{1})=\varphi^{\prime}(y_{1}^{2})=0$ by the first paragraph, so $L_{1}=0$ and $M_{1}=0$. From the equation (3.20), we compute $\hat{Q}=\sum_{i\geq 2}x_{i}M_{i}+\mu x_{1}$.

Define $D\in\mathfrak{g}$ by setting $D(y_{1})=\mu$ and $D(y_{i})=M_{i}$ for $i\geq 2$. Then $D(y_{i}^{2})=0=\varphi^{\prime}(y_{i}^{2})$ and moreover $\varphi^{\prime}(y_{1}y_{i})=\mu y_{i}+y_{1}M_{i}=D(y_{1}y_{i})$ for all $i\geq 2$. Therefore $\varphi$ is the image of $D\in\mathfrak{g}$. ∎

Corollary 3.26**.**

Let $n\geq 4$ and $I_{2}=0$. The map

[TABLE]

is bijective.

Proof.

This follows from Proposition 3.25 and Lemma 3.20. ∎

4. Białynicki-Birula decompositions and retractions

In this section we formally define Białynicki-Birula decompositions and apply the results from the previous section to obtain the local retractions and prove Proposition 1.4 and Theorem 1.3. In total, these proofs apply three BB decompositions; we consider three different linear $\mathbb{G}_{m}$-actions, which correspond to the three retractions from Proposition 1.4.

Let $\mathbb{G}_{m}:=\operatorname{Spec}(\mathbb{Z}[t^{\pm 1}])$ and $\overline{\mathbb{G}}_{m}:=\operatorname{Spec}(\mathbb{Z}[t^{-1}])\simeq\mathbb{A}^{1}_{\mathbb{Z}}$ be its compactification at infinity. Every $\mathbb{G}_{m}$-action on $\mathbb{A}^{r}_{\mathbb{Z}}$ induces a $\mathbb{G}_{m}$-action on $\mathcal{H}_{\mathbb{Z}}$. The Białynicki-Birula decomposition of the $\mathbb{G}_{m}$-scheme $\mathcal{H}_{\mathbb{Z}}$ is a functor from $\mathbb{Z}$-schemes to sets given by

[TABLE]

This functor is represented by a scheme $\mathcal{H}^{+}_{\mathbb{Z}}$, whose connected components are quasi-projective over $\mathbb{Z}$. This scheme comes with naturals maps:

• •

Forgetting about the limit by restricting $\varphi$ to $\varphi_{|1\times B}\colon B\to\mathcal{H}_{\mathbb{Z}}$. This induces a map $\theta_{0}\colon\mathcal{H}^{+}_{\mathbb{Z}}\to\mathcal{H}_{\mathbb{Z}}$.

• •

Restricting to the limit by restricting $\varphi$ to $\varphi_{|\infty\times B}\colon B\to\mathcal{H}_{\mathbb{Z}}$. The family $\varphi_{|\infty\times B}$ is equivariant, hence the image lies in $\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}$. We obtain a map

[TABLE]

• •

Embedding of fixed points. The trivial $\mathbb{G}_{m}$-action on $\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}$ extends to $\overline{\mathbb{G}}_{m}\times\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}\to\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}$ and hence induces a map $i\colon\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}\to\mathcal{H}^{+}_{\mathbb{Z}}$. We have $\pi\circ i=\operatorname{id}$ and $\theta_{0}\circ i\colon\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}\to\mathcal{H}_{\mathbb{Z}}$ is the embedding of fixed points. In particular, $\pi$ is a retraction.

The existence of Białynicki-Birula decompositions for Hilbert schemes of points and totally divergent $\mathbb{G}_{m}$-action is proven in [Jel19, Proposition 3.1]; in that paper $\Bbbk$ denotes a field, but the proof holds equally well for $\Bbbk=\mathbb{Z}$: indeed the proof of [Jel19, Proposition 3.1] goes through without changes for $\Bbbk=\mathbb{Z}$ and its main nontrivial ingredient is the existence of the multigraded Hilbert scheme which holds over any commutative ring $\Bbbk$ [HS04]. Alternatively and more explicitly, one can take the standard affine $\mathbb{G}_{m}$-stable covering $\{U_{\lambda}\}_{\lambda}$ of $\mathcal{H}^{+}_{\mathbb{Z}}$, see [MS05, §18.1] and then glue $\mathcal{H}^{+}_{\mathbb{Z}}$ from $U_{\lambda}^{+}$, as in [JS19, Proposition 5.3]; neither of these two steps depends on the base ring.

Remark 4.1**.**

The image of $\theta_{0}$ is frequently nowhere dense. This happens in particular for the dilation action $\mathbb{G}_{m}\times\mathbb{A}^{r}\to\mathbb{A}^{r}$, given by $t\cdot(x_{1},\ldots,x_{r})=(tx_{1},\ldots,tx_{r})$, see [Jel19, Proposition 3.2]. Hence, we cannot in general hope to prove that $\theta_{0}$ is an open immersion. To remedy this, we choose a smooth algebraic group $\mathbb{Z}$-scheme $G$ acting on $\mathbb{A}^{r}$ and extend the map $\theta_{0}$ to

[TABLE]

which maps $(g,x)$ to $g\cdot\theta_{0}(x)$. Below, $G$ will be either $\mathbb{G}_{a}^{r}$ acting by translation or the unipotent group $G$ defined in (3.4) and recalled below.

Recall the group $G$ of linear transformations given in the basis $x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}$ by

[TABLE]

and its Lie algebra $\mathfrak{g}$.

We would now like to apply Białynicki-Birula decompositions to prove Proposition 1.4 for frames (for $\operatorname{char}\Bbbk\neq 2$) and tweaked frames (for $\operatorname{char}\Bbbk=2$). To avoid dichotomy in proofs and for clarity, we abstract the necessary properties into a standalone definition.

Definition 4.2** (Frame-like ideals).**

Let $I\subset S$ be a homogeneous ideal and $J\subset T=\Bbbk[\mathbf{x},\mathbf{y}]$ be a $\mathbb{N}^{2}$-homogeneous ideal of the form

[TABLE]

where $\mathfrak{p}\subset(y_{1},\ldots,y_{n})^{2}$. We say that $J$ is frame-like if the following conditions hold

1. (a)

the scheme $\operatorname{Spec}(T/J)$ has TNT, 2. (b)

the map $\mathfrak{g}\to\bigoplus_{\alpha\geq 1}\operatorname{Hom}_{T}(J,T/J)_{(\alpha,-\alpha)}$ is bijective, 3. (c)

there exists a $b$ such that $\mathfrak{p}\supset T_{2,b+1}$ and $I_{b}=0$.

Lemma 4.3**.**

Let $\operatorname{char}\Bbbk\neq 2$ and $n=\dim S\geq 3$. Let $I\subset S$ be an ideal with $\operatorname{depth}(S_{+},S/I)\geq 3$, $I_{2}=0$. Let $J$ be a frame of size $a$ for $I$. Then $J$ is frame-like (for all $b\geq 1$).

Proof.

This follows from Proposition 3.10 and Corollary 3.16. ∎

Lemma 4.4**.**

Let $\operatorname{char}\Bbbk=2$ and $n=\dim S\geq 4$. Let $I\subset S$ be an ideal with $\operatorname{depth}(S_{+},S/I)\geq 3$, $I_{n}=0$. Let $\mathcal{J}$ be a tweaked frame of size $a$ for $I$. Then $\mathcal{J}$ is frame-like (for all $b\geq n$).

Proof.

This follows from Proposition 3.23, Corollary 3.26 and from $(y_{1}^{2},\ldots,y_{n}^{2})\supset\mathfrak{m}_{y}^{n+1}$. ∎

Let us fix a frame-like ideal $J$. The ideal $J$ is $\mathbb{N}^{2}$-graded by $(\deg\mathbf{x},\deg\mathbf{y})$ and hence its stabilizer contains a two-dimensional torus $\mathbb{G}_{m}^{2}$. We consider three of its one-dimensional sub-tori: $\mathbb{G}_{x},\mathbb{G}_{y},\mathbb{G}_{xy}$. They act on $T$ by respectively

[TABLE]

We identify $\Bbbk$-points of $\mathcal{H}_{\mathbb{Z}}$ with finite subschemes of affine space and with their ideals, with the conversion that $I\subset S$ and $J\subset T$, so for example $(\mathcal{H}_{\mathbb{Z}},[J])$ is the pointed scheme $(\operatorname{Hilb}(\mathbb{A}^{2n}_{\mathbb{Z}}),[J])$. Finally, we consider the following Diagram (4.1) of Hilbert schemes, where subscripts indicate the points of interest (see below for explanations).

[TABLE]

The scheme $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}},[J])^{+,\mathbb{G}_{y}}$ is the Białynicki-Birula decomposition of $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}},[J])$ with respect to the natural $\mathbb{G}_{y}$-action. The scheme $\mathcal{H}{\mathrm{flag}}_{\mathbb{Z}}$ is the flag Hilbert scheme, which parameterizes deformations of pairs of finite subschemes $R_{1}\subset R_{2}$ in affine space (the flag Hilbert scheme is constructed as a closed subscheme of $\mathcal{H}_{\mathbb{Z}}\times\mathcal{H}_{\mathbb{Z}}$).

The map $\theta_{xy}$ was denoted by $\theta$ in the introduction. It is the forgetful map $\mathbb{G}_{a}^{2n}\times\mathcal{H}^{+}_{\mathbb{Z}}$ to $\mathbb{G}_{a}^{2n}\times\mathcal{H}_{\mathbb{Z}}$ composed with the translation action $\mathbb{G}_{a}^{2n}\times\mathcal{H}_{\mathbb{Z}}\to\mathcal{H}_{\mathbb{Z}}$. Similarly, the map $\theta_{x}$ is the forgetful map followed by the $G$-action on $\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}}$.

The proof of Proposition 1.4 is a journey on Diagram (4.1), from its upper-right corner to the lower-left one. Specifically, each of the three parts of the proposition asserts the existence of a local retraction for one “hook” on this diagram. First two retractions will be obtained from the BB decompositions corresponding to $(\theta_{xy},\pi_{xy},i_{xy})$ and $(\theta_{x},\pi_{x},i_{x})$ respectively. The last one is easily deduced from $(\operatorname{pr}_{1},\operatorname{pr}_{2})$. The conditions (a), (b) of the definition of frame-like ideal imply that $d\theta_{xy}$ and $d\theta_{x}$ are bijective in relevant points.

Proposition 4.5**.**

There exists an open $\mathbb{G}_{xy}$-stable neighbourhood $[J]\in U_{xy}\subset\mathcal{H}^{+}_{\mathbb{Z}}$ such that $(\theta_{xy})_{|\mathbb{G}_{a}^{2n}\times U_{xy}}\colon\mathbb{G}_{a}^{2n}\times U_{xy}\to\mathcal{H}_{\mathbb{Z}}$ is an open immersion.

Proof.

The map $\theta_{xy}$ at $(0,[J])$ induces an injection on obstruction spaces [Jel19, Thm 4.2] and a bijection on tangent spaces by Condition 4.2(a). Hence it is étale at this point. The torus $\mathbb{G}_{xy}$ normalizes $\mathbb{G}_{a}^{2n}$ in the automorphism scheme of $\mathcal{H}_{\mathbb{Z}}$, hence we obtain a semidirect subgroup $H:=\mathbb{G}_{xy}\ltimes\mathbb{G}_{a}^{2n}\subset\operatorname{Aut}(\mathcal{H}_{\mathbb{Z}})$. This subgroup acts on $\mathbb{G}_{a}^{2n}\times\mathcal{H}^{+}_{\mathbb{Z}}$ by $(t,v_{1})\cdot(v_{2},x):=(t(v_{1}+v_{2})t^{-1},tx)$ and $\theta_{xy}$ is $H$-equivariant. Hence, the étale locus of $\theta_{xy}$ has the form $\mathbb{G}_{a}^{2n}\times U$ for some $\mathbb{G}_{m}$-stable open $U\subset(\mathcal{H}_{\mathbb{Z}},[J])^{+,\mathbb{G}_{xy}}$. Now, $\theta_{xy}$ is universally injective, hence $(\theta_{xy})_{|U}$ is an open immersion, see [sta17, Tag 025G]. Take $U_{xy}:=U$. ∎

To repeat the argument above for $\theta_{x}$ we need to check universal injectivity for the $G$-action.

Lemma 4.6**.**

*Let $[J]\in(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}})^{+,\mathbb{G}_{y}}$ be a point with limit point $[J_{0}]\in\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{x}\times\mathbb{G}_{y}}$. Assume that $\mathfrak{g}\to\operatorname{Hom}_{T}(J_{0},T/J_{0})$ is injective. Then the stabilizer of $[J]$ in $G$ is trivial. *

Proof.

Suppose that there exists a point of the stabilizer of $[J]$ and identify it with a matrix $A$ as in (3.4). For every $t\in\overline{\mathbb{G}}_{m}$ the element of $G$ corresponding to the matrix $t^{-1}A$ stabilizes $t\cdot[J]$. Therefore, the tangent vector $A\in\mathfrak{g}$ maps to zero in $\operatorname{Hom}_{T}(J_{0},T/J_{0})$. Since the map is injective, $A=0$. ∎

Proposition 4.7**.**

There exists an open $\mathbb{G}_{y}$-stable neighbourhood $[J]\in U_{y}\subset\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}}$ such that $(\theta_{x})_{|G\times U_{y}}\colon G\times U_{y}\to\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}}$ is an open immersion.

Proof.

Let $\mathfrak{g}$ be the Lie algebra of $G$. The tangent map $d(\theta_{x})$ at $[J]$ is

[TABLE]

so it is bijective by Condition 4.2(b). As in the proof of Proposition 4.5, we find a $\mathbb{G}_{y}$-stable $U_{y}$ such that $(\theta_{x})_{|G\times U_{y}}$ is étale. Then by Lemma 4.6 the map $(\theta_{x})_{|G\times U_{y}}$ is universally injective, hence an open immersion [sta17, Tag 025G]. ∎

Proposition 4.8**.**

The map $\operatorname{pr}_{2}$ is an isomorphism on the connected component of $[IT+\mathfrak{m}_{x}^{a+1}\subset J]\in\mathcal{H}{\mathrm{flag}}_{\mathbb{Z}}^{\mathbb{G}_{x}\times\mathbb{G}_{y}}$.

Proof.

This is an easy consequence of the bi-homogeneity of $J$. The map $(T/(IT+\mathfrak{m}_{x}^{a+1}))_{(*,0)}\to(T/J)_{(*,0)}$ is bijective. Consider a base ring $A$ and an equivariant deformation $T_{A}/\widetilde{J}$ of $T/J$ over $A$. Let $g_{1},\ldots,g_{r}$ be homogeneous generators of $I$ and $\widetilde{g}_{1},\ldots,\widetilde{g}_{r}$ be their unique lifts in $\widetilde{J}$. Let $\widetilde{I}:=(\widetilde{g}_{1},\ldots,\widetilde{g}_{r})+\mathfrak{m}_{x}^{a+1}$, then $(T_{A}/\widetilde{I})_{(*,0)}\to(T_{A}/\widetilde{J})_{(*,0)}$ is bijective as well. Take any syzygy $s\in T^{r}$ between $g_{1},\ldots,g_{r}$. Since $T_{A}/\widetilde{J}$ is $A$-flat, there is a syzygy $\widetilde{s}\in T_{A}^{r}$ lifting $s$. This means that $\widetilde{s}(\widetilde{g}_{1},\ldots,\widetilde{g}_{r})\in\widetilde{J}$. We have $\deg(s)=\deg(\widetilde{s})=(*,0)$, so $\widetilde{s}(\widetilde{g}_{1},\ldots,\widetilde{g}_{r})\in\widetilde{I}$. Hence, $[\widetilde{I}\subset\widetilde{J}]$ is the required deformation of $[IT+\mathfrak{m}_{x}^{a+1}\subset J]$. Uniqueness is evident. ∎

Proposition 4.9**.**

The projection $\operatorname{pr}_{1}$ is a retraction of the connected component of $[IT+\mathfrak{m}_{x}^{a+1}\subset J]\in\mathcal{H}{\mathrm{flag}}_{\mathbb{Z}}^{\mathbb{G}_{x}\times\mathbb{G}_{y}}$ to the connected component of $[I+\mathfrak{m}_{x}^{a+1}]\in\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{x}}$.

Proof.

The element $Q$ is a non-zero-divisor on $T/I\cdot T$ by Lemma 3.2. Moreover, $T_{2,b}\subset\mathfrak{p}$ by Condition 4.2(c), thus every of $I+(x_{1},\ldots,x_{n})^{a+1}\subset S$ induces a deformation of $J$ by keeping $Q$ and $\mathfrak{p}$ fixed. This gives the required section of $\operatorname{pr}_{1}$. ∎

Now, we prove the following abstract version of Proposition 1.4 from the introduction.

Proposition 4.10**.**

Let as before, $J$ be a frame-like ideal

1. (1)

the scheme $(\mathcal{H}_{\mathbb{Z}},[J])$ locally retracts to $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}},[J])$, 2. (2)

the scheme $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{xy}},[J])$ locally retracts to $(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{x}\times\mathbb{G}_{y}},[J])$, 3. (3)

$(\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{x}\times\mathbb{G}_{y}},[J])$* locally retracts to $(\operatorname{Hilb}^{\mathbb{G}_{x}}(\mathbb{A}^{n}),[I+(x_{1},\ldots,x_{n})^{a+1}])$.*

Proof.

The existence of the local retractions (1), (2), (3) is proven respectively in Proposition 4.5, Proposition 4.7, Propositions 4.8-4.9. ∎

Proof of Theorem 1.3.

Fix a singularity type $\mathfrak{S}$ and let $n=8$. By [Vak06, Proposition 4.4] there exists a field $\Bbbk$ and a smooth general type surface $Z$ over $\Bbbk$ and an embedding $Z\hookrightarrow\mathbb{P}^{4}_{\mathbb{Z}}$ such that $[Z]\in\operatorname{Hilb}(\mathbb{P}^{4}_{\mathbb{Z}})$ has singularity type $\mathfrak{S}$. Let $S_{0}:=\bigoplus_{i}H^{0}(\mathbb{P}^{4},\mathcal{O}(i))=\Bbbk[x_{1},\ldots,x_{5}]$. Let $M\gg 0$, in particular $M>n$, and let $I_{0}:=I(Z)_{\geq M}\subset S_{0}$, so that $[I_{0}]\in\operatorname{Hilb}^{\mathbb{G}_{m}}(\mathbb{A}^{5})$ has singularity type $\mathfrak{S}$, see [HS04, Lemma 4.1]. Let $S=S_{0}[x_{6},x_{7},x_{8}]$ be a polynomial ring over $S_{0}$ and let $I=I_{0}\cdot S$. Let

[TABLE]

Fix an action of $\mathbb{G}_{\mathrm{res}}=\mathbb{G}_{m}$ on $S$ acting with weight one on coordinates $x_{6},x_{7},x_{8}$ and fixing $S_{0}$. Since $I$ is generated by elements of $S_{0}$, we have $(\operatorname{Tan}_{I})=(\operatorname{Tan}_{I})_{\geq 0}$ with respect to the grading induced by $\mathbb{G}_{\mathrm{res}}$. Let $\mathcal{H}=\operatorname{Hilb}^{\mathbb{G}_{m}}(\mathbb{A}^{8}_{\mathbb{Z}})$ and let $\theta\colon\mathcal{H}^{+}\to\mathcal{H}$ be its Białynicki-Birula decomposition with respect to the $\mathbb{G}_{\mathrm{res}}$-action. Since $\operatorname{Tan}_{I}$ is non-negatively graded, the map $\theta$ is an open immersion near $[I]$. Hence, on an neighbourhood $U$ of $[I]\in\mathcal{H}$ there is a retraction

[TABLE]

But a neighbourhood of $[I]\in\mathcal{H}^{\mathbb{G}_{\mathrm{res}}}$ is canonically isomorphic to a neighbourhood of $[I_{0}]\in\operatorname{Hilb}^{\mathbb{G}_{m}}(\mathbb{A}^{5})$. Hence, $(\operatorname{Hilb}^{\mathbb{G}_{m}}(\mathbb{A}^{5}),[I_{0}])$ is a local retract of $(\operatorname{Hilb}^{\mathbb{G}_{m}}(\mathbb{A}^{8}),[I])$. We note that

1. (1)

$I_{n}=0$, 2. (2)

$\operatorname{depth}(S_{+},S/I)\geq 3$, because $x_{6},x_{7},x_{8}$ form a regular sequence.

We fix $a\geq\operatorname{reg}(I)+1$, where $I$ is the regularity of ideal $I$. Then by [Erm12, Proposition 3.1] the component of $H$ containing $[I+\mathfrak{m}_{x}^{a+1}]$ is isomorphic to the component of $H$ containing $[I]$ (here the choice of large enough $a$ is crucial).

If $\operatorname{char}\Bbbk\neq 2$, then let $J$ be a TNT frame for $I$ of size $a$. If $\operatorname{char}\Bbbk=2$, then let $J$ be a tweaked frame for $I$. By Lemma 4.3 or Lemma 4.4, the ideal $J$ is $n$-frame-like.

By Proposition 4.10 we obtain a retraction from an open neighbourhood of $[J]$ in $(\mathcal{H}_{\mathbb{Z}},[J])$ to a neighbourhood of $[I+\mathfrak{m}_{x}^{a+1}]$ in $H:=\operatorname{Hilb}^{\mathbb{G}_{m}}\left(\mathbb{A}^{8}\right)$, which is isomorphic to a neighbourhood of $[I]$. Composing with (4.3), we obtain the desired retraction. ∎

Remark 4.11**.**

Erman [Erm12] proved that $\coprod_{n}\operatorname{Hilb}^{\mathbb{G}_{m}}(\mathbb{A}^{n})$ satisfies Murphy’s Law by a different reduction from $Z$ to $I(Z)$ using a sufficiently positive embedding of $Z$. His method is not applicable here, since the obtained $I(Z)$ is generated by quadrics and we require $I_{2}=0$.

5. Corollaries of Theorem 1.3 and

examples

Corollary 5.1** (Answer to Question 1.1).**

There are non-reduced points on the schemes $\mathcal{H}_{\mathbb{Z}}$ and on $\mathcal{H}_{\mathbb{K}}$, where $\mathbb{K}$ is any field.

Proof.

For every pointed scheme $(Y,y)$ in the singularity type of $[\operatorname{Spec}(\mathbb{Z}[u]/u^{2}),V(u)]$ the ring $\mathcal{O}_{Y,y}$ is $\mathbb{Z}$-flat but not reduced, and hence contains an element $v$ such that $\mathbb{Z}[v]/v^{2}\subset\mathcal{O}_{Y,y}$. As in Theorem 1.3, suppose that $(Y,y)$ is such a scheme with a retraction $(X,x)\to(Y,y)$ from an open subscheme $X$ of $\mathcal{H}_{\mathbb{Z}}$. Then the pullback $\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$ has a section, hence is an injective homomorphism. In particular, $\mathcal{O}_{X,x}$ is non-reduced as well. This proves the claim for $\mathcal{H}_{\mathbb{Z}}$. The injective homomorphisms

[TABLE]

stay injective under $(-)\otimes_{\mathbb{Z}}\mathbb{Q}$, hence the claim follows for $\mathcal{H}_{\mathbb{Q}}$. To prove the claim for $\mathcal{H}_{\mathbb{F}_{p}}$, we argue as above for the singularity $[\operatorname{Spec}(\mathbb{F}_{p}[u]/(u^{2}))]$. The claim for arbitrary field $\mathbb{K}$ now follows from base change. ∎

Corollary 5.2** (Answer to Question 1.2).**

The scheme $\mathcal{H}_{\mathbb{Z}}\to\operatorname{Spec}(\mathbb{Z})$ has components lying entirely in the fiber over $\operatorname{Spec}(\mathbb{Z}/p)$ for all primes $p$.

Proof.

For every pointed scheme $(Y,y)$ in the singularity type of $[\operatorname{Spec}(\mathbb{Z}/p)]$ we have $p\mathcal{O}_{Y,y}=0$. As in Theorem 1.3, suppose that $(Y,y)$ is such a scheme with a retraction $(X,x)\to(Y,y)$ from an open subscheme $X$ of $\mathcal{H}_{\mathbb{Z}}$. Then the pullback map $\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$ implies that $p\mathcal{O}_{X,x}=0$, hence each component containing $x$ lies entirely in characteristic $p$. ∎

We can extend Corollary 5.2 by considering higher infinitesimal neighbourhoods.

Corollary 5.3**.**

For every prime $p$ and every ${\nu}\geq 0$ there exists a finite field $\Bbbk$ and a finite irreducible $\mathbb{F}_{p}$-scheme $R$ with residue field $\Bbbk$ which lifts to $\mathbb{Z}/p^{\nu}$ but not to any ring $A$ with $p^{{\nu}}A\neq 0$. In particular $R$ does not lift to $\mathbb{Z}/p^{{\nu}+1}$.

Proof.

Using Theorem 1.3, choose a member $(Y,y)$ in the singularity type of $\mathfrak{S}=[\operatorname{Spec}(\mathbb{Z}/p^{\nu})]$ such that there exists a retraction $(X,x)\to(Y,y)$ from an open subscheme $X$ of $\mathcal{H}_{\mathbb{Z}}$. Let $\Bbbk:=\kappa(x)=\kappa(y)$ and $[R]:=x$. Since $(Y,y)$ in the type of $\mathfrak{S}$ in the smooth equivalence relation, we have $p^{\nu}\mathcal{O}_{Y,y}=0$ and a morphism $\varphi\colon\operatorname{Spec}(\mathbb{Z}/p^{\nu})\to Y$. Composing $\varphi$ with the section of $X\to Y$, we get a lifting of $R$ to $\operatorname{Spec}(\mathbb{Z}/p^{\nu})$. The scheme $R$ embeds into $\mathbb{A}^{16}_{\Bbbk}$, hence its lifting over $A$ would embed into $\mathbb{A}^{16}_{A}$ and give a morphism $\operatorname{Spec}(A)\to\mathcal{H}_{\mathbb{Z}}$, which (after perhaps localizing $A$) restricts to $\operatorname{Spec}(A)\to X$. Hence we obtain $\operatorname{Spec}(A)\to Y$ and so $p^{{\nu}}A=0$. ∎

So far our arguments built upon Vakil’s construction, which in turn depends on Mnëv-Sturmfels universality for incidence schemes [LV13] and on results about abelian covers [Vak06, §4]. The Mnëv-Sturmfels construction requires $\mathbb{P}^{2}_{\Bbbk}$ to have enough $\Bbbk$-points, hence usually it does not work over $\Bbbk=\mathbb{F}_{p}$ (this is the reason why in Corollary 5.3 we do not obtain algebras with residue field $\mathbb{F}_{p}$). The theory of abelian covers, while in principle constructive, is not very prone to become explicit either.

In this final part we explicitly construct appropriate points of the Hilbert scheme by hand, bypassing Vakil’s work, for several singularity types. First, we note that one can obtain explicit examples of non-reduced points on $\mathcal{H}_{\mathbb{Z}}$ by taking a TNT frame for the truncation of the cone over a curve from Mumford’s famous example [Har10, §13] or the examples of Martin-Deschamps and Perrin [MDP96]. We give an explicit example by framing a degree $3$, genus $-2$ reducible curve from [MDP96, Prop 0.6].

Example 5.4**.**

Let $\Bbbk$ be of characteristic zero. Let $S=\Bbbk[x_{1},\ldots,x_{7},y_{1},\ldots,y_{7}]$ be a polynomial ring and $K=(x_{1}^{2},\ x_{1}x_{2},\ x_{2}^{2}(x_{3}+x_{4}),\ x_{1}x_{4}^{3}+x_{2}(x_{3}+x_{4})x_{3}^{2})$. The regularity of $K$ is four. Let $I=K\cap(x_{1},\ldots,x_{4})^{4}$ and let $J$ be a TNT frame for $I$ with $a=5$. Then $[J]\in\mathcal{H}_{\Bbbk}$ is non-reduced.

Below, we give explicit components of $\operatorname{Hilb}_{\mathrm{pts}}(\mathbb{A}^{6}_{\mathbb{Z}})$ lying in characteristic $p$ for small $p$; in fact we give $\mathbb{F}_{p}$-points of these components. The proof is obtained by replacing the construction of Theorem 1.3 by some explicit computations. Let $\Bbbk=\mathbb{F}_{p}$ and let $R\subset\mathbb{A}^{n}_{\Bbbk}$ be a finite scheme given by a homogeneous ideal. The examples below employ the following line of argument:

1. (1)

check that $\dim(\operatorname{GL}_{n}\cdot[R])=\dim_{\Bbbk}T_{\mathcal{H}_{\Bbbk}^{\mathbb{G}_{m}},[R]}$, 2. (2)

conclude that $[R]\in\mathcal{H}_{\Bbbk}^{\mathbb{G}_{m}}$ is smooth, 3. (3)

verify that $R$ does not lift to $W_{2}(\Bbbk)$. 4. (4)

use 1-3 to conclude that the component of $\mathcal{H}_{\mathbb{Z}}^{\mathbb{G}_{m}}$ containing $[R]$ lies entirely over $\Bbbk$. This is a known argument, see e.g. [Eke04, Lemma 5.7]. If this holds, also a neighbourhood of $[R]\in\mathcal{H}^{+}_{\mathbb{Z}}$ lies over $\Bbbk$. 5. (5)

check that $R$ has TNT and conclude that $\theta\colon\mathcal{H}^{+}_{\mathbb{Z}}\to\mathcal{H}_{\mathbb{Z}}$ is an open immersion in a neighbourhood of $[R]$.

The heart of all examples is the observation that the ideal

[TABLE]

satisfies Properties (1)-(3). It remains to reduce $K$ to dimension zero so as not to lose these properties and additionally gain TNT. We present one such reduction below.

Example 5.5**.**

Let $q=p^{e}$ be a prime power, let $\Bbbk=\mathbb{F}_{p}$ as before and consider the ideal $I^{\prime}=(x_{1}x_{2}+x_{3}x_{4}+x_{5}x_{6})+(x_{2}^{q},x_{4}^{q},x_{6}^{q})\subset S=\Bbbk[x_{1},\ldots,x_{6}]$. Let $I$ be its saturation and let $J=I+(x_{1},x_{3},x_{5})^{q+1}$. Below all unjustified claims are checked with Macaulay2 for $q=3,4,5$. Hence, we obtain examples in characteristics $\leq 5$. First, the stabilizer of $J$ is $10$-dimensional, given by

[TABLE]

where $\lambda\in\Bbbk^{*}$ and $D\in\operatorname{GL}_{3}$. Hence the $\operatorname{GL}_{6}$-orbit is $26$-dimensional.

To prove that $J$ does not lift to $W_{2}(\Bbbk)$, we argue similarly as in [Zda18, Proposition 6.1.1]. Let $\widetilde{S}:=W_{2}(\Bbbk)[X_{1},\ldots,X_{6}]$. Suppose that $R$ lifts to $W_{2}(\Bbbk)$. Then there exists an ideal $\widetilde{J}\subset\widetilde{S}$ such that $\widetilde{S}/\widetilde{J}$ is an embedded deformation of $S/J$ over $W_{2}(\Bbbk)$. In particular, the syzygy

[TABLE]

lifts to a syzygy between generators of $\widetilde{J}$, which means that

[TABLE]

We have $p\widetilde{S}\simeq S$ as $\widetilde{S}$-modules and $p\widetilde{J}\simeq J$ in this isomorphism. Equation (5.2) translates into

[TABLE]

But $(x_{1}x_{2})^{2}(x_{3}x_{4}+x_{5}x_{6})\equiv(x_{1}x_{2})^{3}\equiv 0\mod J$, hence (5.3) is equivalent to $2x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}\in J$, which is false. Hence, we obtain a contradiction (the argument for $q=4$ should be ramified here). Finally, we check directly that for $\operatorname{Tan}:=\operatorname{Hom}_{S}(J,S/J)$ we have $\dim_{\Bbbk}(\operatorname{Tan})_{0}=26$, which verifies Property 1 and that $\dim_{\Bbbk}(\operatorname{Tan})_{<0}=6$, which verifies Property 5.