Abstract:Let $S = k[x_1, \ldots, x_5]$, let $Q \subset S_2$ be a subspace of codimension $r$, and set $I_Q = (Q) + \mathfrak{m}^3$. Then $S/I_Q$ has Hilbert function $(1,5,r)$. The cases $r = 6, 7$ are exceptional in embedding dimension five: at a general member the tangent space to the Hilbert scheme is larger than the locus, so neither locus is a generically reduced elementary component. We prove instead that both loci are smoothable. The proof uses the Erman-Velasco method of smoothable regularity-two ideals. In degree two, the inverse system space coming from their smoothable family is, up to $\mathrm{GL}_d$, spanned by quadrics $q(a) = \sum_i a_i y_i^2 - (\sum_i a_i y_i)^2$. For $d = 5$ and $e = 6, 7$ we prove, by a differential rank test over $\mathbb{F}_{32003}$, that the associated map $\mathrm{GL}_5 \times (\mathbb{A}^5)^e \dashrightarrow \mathrm{Gr}(e, \mathrm{Sym}^2 k^5)$ is dominant. Hence a general one-step algebra with Hilbert function $(1,5,6)$ or $(1,5,7)$ is smoothable. By closedness, the whole translated one-step locus is contained in the smoothable component. It follows that these loci are not irreducible components of the Hilbert schemes $\mathrm{Hilb}^{12}(\mathbb{A}^5)$ and $\mathrm{Hilb}^{13}(\mathbb{A}^5)$. This settles these two embedding dimension five cases on the smoothable side of the Shafarevich gap problem.
From: Chenyang Zhao [view email]
[v1]
Sun, 21 Jun 2026 07:36:17 UTC (10 KB)
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。