The one-step Shafarevich gap in embedding dimension five

Abstract

Let $k$ be an algebraically closed field of characteristic zero and let $S=k[x_1,\dots ,x_5]$ with maximal ideal $\mathfrak{m}=(x_1,\dots ,x_5)$. For a codimension-$r$ subspace $Q\subset S_2$, set $I_Q=(Q)+{\mathfrak{m}}^3$. Then $S/I_Q$ has Hilbert function $(1,5,r)$. We prove that the translated one-step locus defined by these ideals is contained in the smoothable component for every $r\in 6,7,\dots ,15$. We introduce a finite field differential rank certificate proving dominance, for $6\le r\le 14$, of the Erman--Velasco map $\mathrm{GL}\ast 5\times ({\mathbb{A}}^5{)}^r⇢\mathrm{Gr}(r,{\mathrm{Sym}}^2{k}^5)$, $(g,a^{(1)},\dots ,a^{(r)})\mapsto g\cdot ⟨q(a^{(1)}),\dots ,q(a^{(r)})⟩$, where $q(a)=\sum \ast {i=1}^5{a}_i{y}_i^2-{\left({\sum }_{i=1}^5{a}_i{y}_i\right)}^2$. The endpoint $r=15$ is handled separately by a flat degeneration of $21$ general reduced points to the fat point defined by ${\mathfrak{m}}^3$. Combined with the known small cases and with the known elementary components for $r=3$ and $r=5$, this gives the complete one-step classification in embedding dimension five: the one-step loci with Hilbert function $(1,5,r)$ are smoothable for all $r\ne 3,5$, and the cases $r=3,5$ are precisely the generically reduced elementary component cases. In this sense the one-step Shafarevich gap in embedding dimension five is completely resolved.

9 pages; updated to a more general one-step statement; ancillary M2 verification script