Stanley-Reisner ideals with linear powers

TL;DR

This paper characterizes when squarefree monomial ideals with linear resolutions also have linear powers, revealing that this equivalence depends on the degree and is independent of the base field for specific degrees.

Contribution

It provides a complete combinatorial classification of squarefree monomial ideals with linear resolutions and linear powers, resolving two open questions in commutative algebra.

Findings

The equivalence between linear resolutions and linear powers holds for degrees 0,1,2, n-2, n-1, n.

For degrees 3 to n-3, the property depends on the base field and can fail for powers.

In certain degrees, ideals with linear resolutions have all powers with linear quotients.

Abstract

Let $S = K[x_1, \dots, x_n]$ be the standard graded polynomial ring over a field $K$. In this paper, we address and completely solve two fundamental open questions in Commutative Algebra: (i) For which degrees $d$, does there exist a uniform combinatorial characterization of all squarefree monomial ideals in $S$ having $d$-linear resolutions? (ii) For which degrees $d$, does having a linear resolution coincide with having linear powers for all squarefree monomial ideals of $S$ generated in degree $d$? Let $\mathcal{I}_{n,d}(K)$ denote the class of squarefree monomial ideals of $S$ having a $d$-linear resolution. Our main result establishes the equivalence of the following conditions: (a) Any squarefree monomial ideal $I$ in $S$ generated in degree $d$ has a linear resolution, if and only if, $I$ has linear powers. (b) $\mathcal{I}_{n,d}(K)$ is independent of the base field $K$. (c) $d\in\{0,1,2,n{-}2,n{-}1,n\}$. In each of these degrees, we show that a squarefree monomial ideal has a linear resolution if and only if all of its powers admit linear quotients, and we combinatorially classify such ideals. In contrast, for each degree $3\le d\le n{-}3$, we construct fully-supported squarefree monomial ideals $I$ and $J$ in $S$ generated in degree $d$ such that the linear resolution property of $I$ depends on the choice of the base field, $J$ has a linear resolution and $J^2$ does not have a linear resolution.