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Related papers: The Jacobian ideal of a hyperplane arrangement

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Let $\mathcal{A}$ denote a central hyperplane arrangement of rank $n$ in affine space $\mathbb{K}^n$ over an infinite field $\mathbb{K}$ and let $l_1,\ldots, l_m\in R:= \mathbb K[x_1,\ldots,x_n]$ denote the linear forms defining the…

Commutative Algebra · Mathematics 2021-01-11 Ricardo Burity , Aron Simis , Stefan Tohaneanu

In this article we describe two new characterizations of freeness for hyperplane arrangements via the study of the generic initial ideal and of the sectional matrix of the Jacobian ideal of arrangements.

Algebraic Geometry · Mathematics 2018-01-31 Anna Maria Bigatti , Elisa Palezzato , Michele Torielli

Let f_1,...,f_r be homogeneous polynomials in K[x_1,...,x_n], K a field. Put F=y_1f_1+...+y_rf_r in K[x,y] and let I be the ideal of K[x,y] generated by the partials of F relative to the x_i and y_j. The Jacobian ring of F is the quotient…

Algebraic Geometry · Mathematics 2007-05-23 Alan Adolphson , Steven Sperber

Let $ V $ a vector space of dimension $n$. A $V$ family $ \{H_1, \ldots, H_p \} $ of vectorial hyperplanes being distinct two by two defines an arrangement $ {\cal A}_p = {\cal A} ( H_1, \ldots ,H_p ) $ of $ V $. For $ i \in \{ 1, \ldots, p…

Algebraic Geometry · Mathematics 2016-10-12 Philippe Maisonobe

We study the classes of free and plus-one generated hyperplane arrangements. Specifically, we describe how to compute the associated prime ideals of the Jacobian ideal of such an arrangement from its lattice of intersection. Moreover, we…

Combinatorics · Mathematics 2020-07-20 Elisa Palezzato , Michele Torielli

We compute the multiplier ideals of hyperplane arrangements via the interpretation of these ideals in terms of spaces of arcs, due to Ein, Lazarsfeld and the author.

Algebraic Geometry · Mathematics 2007-05-23 Mircea Mustata

A hyperplane arrangement $\cA$ is said to be free if the corresponding Jacobian ideal $J_\cA$ is Cohen-Macaulay. If $\cA$ is free then $J_\cA$ is unmixed (i.e. equidimensional). Freeness is an important property, yet its presence is not…

Commutative Algebra · Mathematics 2025-08-19 Juan Migliore , Uwe Nagel

Freeness is an important property of a hypersurface arrangement, although its presence is not well understood. A hypersurface arrangement in $\PP^n$ is free if $S/J$ is Cohen-Macaulay (CM), where $S = K[x_0,\ldots,x_n]$ and $J$ is the…

Algebraic Geometry · Mathematics 2024-07-12 Juan Migliore , Uwe Nagel

We study the Hadamard product of the linear forms defining a hyperplane arrangement with those of its dual, which we view as generating an ideal in a certain polynomial ring. We use this ideal, which we call the ideal of pairs, to study…

Combinatorics · Mathematics 2022-02-08 Avi Steiner , Graham Denham

We investigate which homogeneous polynomials are determined by their Jacobian ideals, and extend and complete previous results due to J. Carlson and Ph. Griffiths, K. Ueda and M. Yoshinaga, and A. Dimca and E. Sernesi.

Algebraic Geometry · Mathematics 2014-04-22 Zhenjian Wang

We consider polynomial maps, which we call degree $d$-linear maps, that satisfy the Jacobian condition. We prove that certain infinite families of elements, which appear in the coefficients of the formal inverse of such maps, are in the…

Commutative Algebra · Mathematics 2021-11-09 Mario DeFranco

Steinberg showed that when a finite reflection group acts on a real or complex vector space of finite dimension, the Jacobian determinant of a set of basic invariants factors into linear forms which define the reflecting hyperplanes. This…

Representation Theory · Mathematics 2007-05-23 Julia Hartmann , Anne V. Shepler

Let $k$ be a field of characteristic $p>0$ and $R$ be a subalgebra of $k[X]=k[x_1,...,x_n]$. Let $J(R)$ be the ideal in $k[X]$ defined by $J(R)\Omega_{k[X]/k}^n=k[X]\Omega_{R/k}^n$. It is shown that if it is a principal ideal then $J(R)^q$…

Commutative Algebra · Mathematics 2011-06-28 A. V. Gavrilov

We prove that the ideal generated by the maximal minors of the higher-order Jacobian matrix of a weighted homogeneous polynomial is also weighted homogeneous. As an application, we give a partial answer to a conjecture concerning the…

Algebraic Geometry · Mathematics 2025-07-28 Wágner Badilla-Céspedes , Abel Castorena , Daniel Duarte

We establish a series of results showing that the Jacobian ideal is contained in the test ideal. We first prove a new result in characteristic $p$ for complete rings over a field $K$. Then we prove some results showing that Jacobian ideals…

Commutative Algebra · Mathematics 2022-09-20 Zhan Jiang

The Jacobian ideal provides the set of infinitesimally trivial deformations for a homogeneous polynomial, or for the corresponding complex projective hypersurface. In this article, we investigate whether the associated linear deformation is…

Algebraic Geometry · Mathematics 2016-12-22 Zhenjian Wang

Let $ V $ a vector space of dimension $n$. A family $ \{H_1, \ldots, H_p \} $ of vectorial hyperplans $V$ defines an arrangement $ {\cal A} $ of $ V $. For $ i \in \{ 1, \ldots, p \} $, let $ l_i $ be a linear form on $V$ with $H_i$ as…

Algebraic Geometry · Mathematics 2016-10-12 Philippe Maisonobe

The Jacobian group of a graph is a finite abelian group through which we can study the graph in an algebraic way. When the graph is a finite abelian covering of another graph, the Jacobian group is equipped with the action of the Galois…

Combinatorics · Mathematics 2023-03-02 Takenori Kataoka

A topological hyperplane is a subspace of R^n (or a homeomorph of it) that is topologically equivalent to an ordinary straight hyperplane. An arrangement of topological hyperplanes in R^n is a finite set H such that k topological…

Combinatorics · Mathematics 2010-01-24 David Forge , Thomas Zaslavsky

We show that every convergent power series with monomial extended Jacobian ideal is right equivalent to a Thom-Sebastiani polynomial. This solves a problem posed by Hauser and Schicho. On the combinatorial side, we introduce a notion of…

Algebraic Geometry · Mathematics 2022-07-08 Raul Epure , Mathias Schulze
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