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?
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$\begingroup$ I changed the title ad removed the tags "deformation theory" and "hilbert schemes", which in my opinion were superfluous. $\endgroup$– Francesco PolizziDec 11, 2012 at 16:07
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2 Answers
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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 $L$.
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 $L$.
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 $L$.
EDIT 1. 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 $L$.
EDIT 2. 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 Edit 1, 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 $C=C \cup \ldots \cup C_m$.
My original answer only dealt with the case $m=d$ and $d_1=d_2= \ldots =d_m=1$.
answered Dec 11, 2012 at 16:00
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2$\begingroup$ Even simpler: any two different planes in the family meet along a line. All d points must belong to it. $\endgroup$– quimDec 11, 2012 at 18:27
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1$\begingroup$ Actually, there is another possibility: that the line L is a component of C, then there are no restrictions on the rest: C is composed of a line through the d points plus a curve of degree $d-1$. The argument "fails" in that the projection is not birational on C in that case. $\endgroup$– quimDec 12, 2012 at 10:47
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$\begingroup$ Of course you are right. If the line $L$ through the $d$ points is component of the curve, then it is contracted by any projection with center on it. I will edit the answer, thank you for pointing this out. $\endgroup$ Dec 12, 2012 at 12:05
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2$\begingroup$ Hi Francesco, the projection from a generic $p \in L$ may not be birational! The correct answer seems to be that either $L$ is a component of $C$, or there are hyperplanes $H_1, \ldots, H_m$ containing $L$ and curves $C_j \subseteq H_j$ of degree $d_j$ such that $1 \leq m \leq n$, $d_1 + \cdots d_m = d$ and $C = C_1 \cup \cdots \cup C_m$. $\endgroup$– pinakiDec 12, 2012 at 20:26
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$\begingroup$ Hi auniket, you are right! In fact, it seems that I overlooked many possibilities :-). I edited the answer, thank you. $\endgroup$ Dec 13, 2012 at 8:47
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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 $L$.
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 $C$.
Actually, this is projecting from $L$ rather than projecting from a point of $L$.
