2011
DOI: 10.1016/j.jalgebra.2011.09.019
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The implicit equation of a multigraded hypersurface

Abstract: In this article we analyze the implicitization problem of the image of a rational map φ : X P n , with X a toric variety of dimension n − 1 defined by its Cox ring R. Let I := (f 0 , . . . , f n ) be n + 1 homogeneous elements of R. We blow-up the base locus of φ, V (I), and we approximate the Rees algebra Rees R (I) of this blow-up by the symmetric algebra Sym R (I). We provide under suitable assumptions, resolutions Z • for Sym R (I) graded by the divisor group of X , Cl(X ), such that the determinant of a g… Show more

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Cited by 23 publications
(53 citation statements)
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“…Taking (ν 1 , ν 2 ) = (2a − 1, b − 1) one has that in STEP 1 a basis of syzygies (h Equality (3.2) also follows from Theorem 5.5 in [Bot11b].…”
Section: 4mentioning
confidence: 97%
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“…Taking (ν 1 , ν 2 ) = (2a − 1, b − 1) one has that in STEP 1 a basis of syzygies (h Equality (3.2) also follows from Theorem 5.5 in [Bot11b].…”
Section: 4mentioning
confidence: 97%
“…Goldman et al studied the cases of planar and space curves [JG09,HWJG10,JWG10]. A generalization of the linear syzygy method when the support of the input polynomials is a square (that is, bihomogeneous of degree (d, d)) was proposed by Busé and Dohm [BD07], and for any polygon by Botbol, Dickenstein and Dohm [BDD09], and Botbol [Bot09,Bot11a,Bot11b]. This method is particularly adapted when the polynomials defining the parametrization are sparse, which is often the case.…”
Section: Introductionmentioning
confidence: 99%
“…Following Theorem 5.1, ν = (5, 0). The ideal I U has two syzygies in degree (2, 0) and three syzygies in bidegree (1,1). The syzygies in degree (2, 0) span a free R-module and determine a basis for the syzygies of bidegree (5, 0) by multiplying by {s 3 , s 2 t, st 2 , t 3 } .…”
Section: Applications To Implicitization Of Tensor Product Surfacesmentioning
confidence: 99%
“…Proof. From the proof of Lemma 7.3 in Botbol [1], we know that Z ν is acyclic and that (Z 3 ) ν = 0. Hence Z ν has the form…”
Section: 2mentioning
confidence: 99%
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