On the uniqueness of algebraic curves passing through \(n\)-independent nodes

A set of nodes is called \(n\)-independent if each its node has a fundamental polynomial of degree \(n.\) We proved in a previous paper [H. Hakopian and S. Toroyan, On the minimal number of nodes determining uniquelly algebraic curves, accepted in Proceedings of YSU] that the minimal number of \(n\)...

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Published in:arXiv.org 2015-10
Main Authors: Hakopian, H, Toroyan, S
Format: Article
Language:English
Subjects:
Algebra
Nodes
Polynomials
Online Access:Get full text
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Summary:A set of nodes is called \(n\)-independent if each its node has a fundamental polynomial of degree \(n.\) We proved in a previous paper [H. Hakopian and S. Toroyan, On the minimal number of nodes determining uniquelly algebraic curves, accepted in Proceedings of YSU] that the minimal number of \(n\)-independent nodes determining uniquely the curve of degree \(k\le n\) equals to \({\mathcal K}:=(1/2)(k-1)(2n+4-k)+2.\) Or, more precisely, for any \(n\)-independent set of cardinality \({\mathcal K}\) there is at most one curve of degree \(k\le n\) passing through its nodes, while there are \(n\)-independent node sets of cardinality \({\mathcal K}-1\) through which pass at least two such curves. In this paper we bring a simple characterization of the latter sets. Namely, we prove that if two curves of degree \(k\le n\) pass through the nodes of an \(n\)-independent node set \({\mathcal X}\) of cardinality \({\mathcal K}-1\) then all the nodes of \({\mathcal X}\) but one belong to a (maximal) curve of degree \(k-1.\)
ISSN:2331-8422