$d$ points on a curve which are in the base locus of a pencil of planes - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:34:11Z http://mathoverflow.net/feeds/question/116094 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116094/d-points-on-a-curve-which-are-in-the-base-locus-of-a-pencil-of-planes $d$ points on a curve which are in the base locus of a pencil of planes Naga Venkata 2012-12-11T15:15:14Z 2012-12-15T15:22:05Z <p>Let $C$ be a reduced curve in $\mathbb{P}^3$ of degree $d$. Does there exist $d$ points on $C$ such that there exists a $1-$dimensional family of hyperplanes in $\mathbb{P}^3$ passing through these points?</p> http://mathoverflow.net/questions/116094/d-points-on-a-curve-which-are-in-the-base-locus-of-a-pencil-of-planes/116099#116099 Answer by Francesco Polizzi for $d$ points on a curve which are in the base locus of a pencil of planes Francesco Polizzi 2012-12-11T16:00:54Z 2012-12-13T14:13:39Z <p>Assume that the curve $C$ is not degenerate and $d \geq 3$. Then, in order to fulfill your assumptions, $C$ must be reducible in $d$ lines, and furthermore each of these lines meets another fixed line $L$. </p> <p>In fact, suppose that there are $d$ points on $C$ such that there exists a $1$-dimensional family of planes through these points. Since the base locus of a pencil of planes is a line, this means that the $d$ points belong to a line $L$. </p> <p>Now take a general point on $p \in L$ and project $C$ from $p$ to $\mathbb{P}^2$. The projection is birational, hence we obtain a plane a curve $C'$ of degree $d$ with a point $p'$ of multiplicity $d$. Then $C'$ is necessarily the union of $d$ lines through $p'$. It follows that the curve $C$ is the union of $d$ lines in $\mathbb{P}^3$, and each of these lines is incident to $L$. </p> <p><strong>EDIT 1.</strong> As quid points out there is actually another possibility, namely when $C=L \cup D$, with $D$ any curve of degree $d-1$. In this case the projection from $p \in L$ is not birational, since it contracts $L$.</p> <p><strong>EDIT 2.</strong> As Auniked points out, one cannot assume that the projection is birational. In fact, the complete answer seems to be the following. Either $L$ is a component of $C$ and we are in the situation described in <strong>Edit 1</strong>, or there exist planes $H_1, \ldots, H_m$ containing $L$ and curves $C_i \subset H_i$ of degree $d_i$ such that $d_1+ \ldots + d_m=d$ and $C=C \cup \ldots \cup C_m$.</p> <p>My original answer only dealt with the case $m=d$ and $d_1=d_2= \ldots =d_m=1$.</p> http://mathoverflow.net/questions/116094/d-points-on-a-curve-which-are-in-the-base-locus-of-a-pencil-of-planes/116446#116446 Answer by quim for $d$ points on a curve which are in the base locus of a pencil of planes quim 2012-12-15T13:22:11Z 2012-12-15T15:22:05Z <p>A slightly different way to prove Francesco+auniket's result is as follows. First, given two different planes in the family, all $d$ points must lie in their intersection, which is a line $L$. </p> <p>Second, every plane intersecting $C$ properly does so in $d$ points (counting multiplicities). So for every plane $H$ through $L$ intersecting $C$ properly, $H\cap C \subset L$. If there is any such plane, then general planes through $L$ intersect $C$ properly. Let $H_1, \dots, H_m$ be the (therefore finitely many) planes through $L$ which meet $C$ nonproperly. Then $C \subset \bigcup H_i$, each $H_i$ contains a component $C_i$ of $C$ of degree $d_i$ which goes through $d_i$ of the $d$ points. If there is no such plane on the other hand, then $L$ is a component of $C$.</p> <p>Actually, this is projecting from $L$ rather than projecting from a point of $L$.</p>