2009
DOI: 10.1016/j.jpaa.2008.05.001
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Polar syzygies in characteristic zero: The monomial case

Abstract: a b s t r a c tGiven a set of forms f = {f 1 , . . . , f m } ⊂ R = k[x 1 , . . . , x n ], where k is a field of characteristic zero, we focus on the first syzygy module Z of the transposed Jacobian module D(f), whose elements are called differential syzygies of f. There is a distinct submodule P ⊂ Z coming from the polynomial relations of f through its transposed Jacobian matrix, the elements of which are called polar syzygies of f. We say that f is polarizable if equality P = Z holds.This paper is concerned w… Show more

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Cited by 10 publications
(18 citation statements)
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References 16 publications
(37 reference statements)
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“…One can now determine the first row of the Betti diagram using the formulation of (2) given in [11, Proposition 2.1] when j = i + 2:…”
Section: An Example Where the Whole Betti Diagram Is Determinedmentioning
confidence: 99%
See 1 more Smart Citation
“…One can now determine the first row of the Betti diagram using the formulation of (2) given in [11, Proposition 2.1] when j = i + 2:…”
Section: An Example Where the Whole Betti Diagram Is Determinedmentioning
confidence: 99%
“…As a general aim, one would like to establish a correspondence between algebraic properties of the ideal I (or the ring R/I, or the subalgebra K[f ] ⊂ R) and the graph theoretical data of G(I) (or of any other graph associated to I such as G(I) c ). Two illustrations of this correspondence are the combinatorial characterizations of normality and polarizability of the subalgebra K[f ] ⊂ R given in [12] and [2], respectively. In this note, we shall focus on properties of I related to its minimal graded free resolution (1) 0…”
Section: Introductionmentioning
confidence: 99%
“…We close with some general questions. In [1] the answer to the first of these questions is affirmative when f are monomials, hence so is the answer to the second question under the same assumption.…”
Section: A Class Of Polar Mapsmentioning
confidence: 97%
“…Then the presentation ideal of k[f ] is a codimension 3 almost complete intersection in which the quadrics generate a maximal regular sequence, while the fourth generator lives in degree 3. It can be shown that f is polarizable and that D is a free module generated by the differentials of the three quadrics (see [1,Example 5.21]).…”
Section: Complete Intersectionsmentioning
confidence: 99%
“…Thus, the initial step could be vacuous or, alternatively, would start from n = s 1 + 2, while in the inductive step one would then assume that n > s 1 + 2. Taking up the second alternative, the initial step has F = {x 2 1 , x 1 x 2 , . .…”
Section: The Degree Of the Inverse Mapmentioning
confidence: 99%