2011
DOI: 10.1016/j.jpaa.2010.12.007
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On the classification of non-normal cubic hypersurfaces

Abstract: a b s t r a c tIn this article we study the classification of non-normal cubic hypersurfaces over an algebraically closed field K of arbitrary characteristic. Let X ⊂ P r K be an irreducible nonnormal cubic hypersurface. If r ≥ 5, then X is necessarily a cone (Remark 2.3). In view of this fact it suffices to classify irreducible non-normal cubic hypersurfaces X ⊂ P r K for r ≤ 4.We prove that there are precisely five non-normal cubic equations (resp. six non-normal cubic equations) when char K ̸ = 2, 3 (resp. … Show more

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Cited by 13 publications
(7 citation statements)
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“…Essentially, these varieties are completely classified: see [Fuj82] for the case of normal varieties, and [Rei94, BS07] for the non-normal case. For surfaces of degree 3 and 4 we refer to [LPS11,LPS12]. For the roots of this classification, originated from work of Schläfli and Cayley, see for instance [Abh60,BW79].…”
Section: Reduction To Surfacesmentioning
confidence: 99%
“…Essentially, these varieties are completely classified: see [Fuj82] for the case of normal varieties, and [Rei94, BS07] for the non-normal case. For surfaces of degree 3 and 4 we refer to [LPS11,LPS12]. For the roots of this classification, originated from work of Schläfli and Cayley, see for instance [Abh60,BW79].…”
Section: Reduction To Surfacesmentioning
confidence: 99%
“…Non-normal cubic hypersurfaces over algebraically closed fields are classified in [3]. The same argument also gives the following classification over Q.…”
Section: Classification Of Non-normal Hypercubics Over the Rationalsmentioning
confidence: 82%
“…(cf. [3], Theorem 3.1) Let W ⊂ P n Q be a geometrically integral and geometrically non-normal hypersurface given by a homogeneous cubic polynomial F ∈ Q[t 0 , . .…”
Section: Classification Of Non-normal Hypercubics Over the Rationalsmentioning
confidence: 99%
“…By [14,Theorem 3.1], we can choose coordinates {x, y, z} of P 2 k such that C is defined by x 3 + y 2 z = 0. Then p 1 = [0 : 0 : 1] and p 2 = [0 : 1 : 0].…”
Section: Proposition 41mentioning
confidence: 99%