MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Clearly the question in the title has a positive answer for analytic (or smooth, or continuous ...) curves, but what about the algebraic category? More specifically, given an irreducible polynomial curve $X \subseteq \mathbb{CP}^2$ and points $P_1, \ldots, P_k$ on $X$, when can we find another curve $Y$ (defined by a polynomial) such that the $Y$ intersects $X$ only at $P_1, \ldots, P_k$?

I find the question to be nontrivial even for $k = 1$. Here are some observations for $k = 1$ case:

  1. If $P$ is a point on $X$ with multiplicity $\deg X - 1$, then a tangent of $X$ through $P$ intersects $X$ only at $P$ (by Bezout's theorem).

  2. If $X$ is a rational curve and $X \setminus \{P\} \cong \mathbb{C}$, then there is a curve $Y$ such that $X \cap Y = \{P\}$.

  3. Let $X$ be a non-singular cubic. Give it a group structure such that the origin is an inflection point. Then for all $P \in X$, there exists $Y$ such that $Y \cap X = \{P\}$ iff $P$ is a torsion point in the group.

If $X$ (of degree $d$) is non-singular at $P$, then the most direct approach for finding a $Y$ of degree $e$ intersecting $X$ only at $P$ seemed to blow it up $de$ times and look for the conditions under which $Y$ goes through each of the points on $X$ in the $i$-th infinitesimal neighborhood of $P$, $0 \leq i \leq de - 1$. But the conditions on the coefficients of the polynomial defining $Y$ did not appear very tractable.

Edit: I would like to make a correction to observation 3. This is what I know about a non-singular cubic curve $X$: If $P$ is an inflection point, then there is a curve $Y$ such that $Y \cap X = P$ (take $Y$ to be the tangent of $X$ at $P$). If $P$ is a non-torsion point (for the group structure on $X$ for which the origin is an inflection point), then there is no such $Y$. I don't know what happens for torsion points.

share|cite|improve this question
An equivalent question is whether there exists a relation between the points $P_1$, ..., $P_k$ in Pic(X). But this reframing doesn't suggest an algorithm to me. – David Speyer Nov 30 '10 at 12:49
Dear auniket, If $X$ is a line then Pic$(X)$ is trivial, so the relation that David Speyer is referring to is trivially satisfied. – Emerton Nov 30 '10 at 14:27
Minor correction to my comment: the question is whether there is a relation between $P_1$, $P_2$, ..., $P_k$ and $\mathcal{O}(1)$. If there is a degree $d$ curve which meets $P_i$ to order $a_i$ (so $\sum a_i = dk$) then $\sum a_i [P_i] = d [ \mathcal{O}(1)]$. And, as Emerton says, if $X$ has genus $0$ (a line or a conic) then $Pic(X) = \mathbb{Z}$ and the question is trivial. – David Speyer Nov 30 '10 at 14:41
I believe an (amusing to me) equivalent formulation is: For what open affine sets $U \subseteq X$, is there an open affine subset $V \subseteq \mathbb{CP}^2$ such that $V\cap X = U$? – Karl Schwede Nov 30 '10 at 17:06
Sufficiency is more subtle but, for smooth plane curves, it is true. Consider the line bundle $\mathcal{O}(d)$ restricted to the curve $X$. The hypothesis that $\sum a_i [P_i] = d [\mathcal{O}(1)]$ means that $\mathcal{O}(1)|_X$ has a section $f$ which vanishes to order $a_i$ at $P_i$. We want to show that this section arises from a degree $d$ homogenous polynomial. In other words, we want to show that it is in the image of $H^0(\mathbb{P}^2, \mathcal{O}(d)) \to H^0(X, \mathcal{O}(d)|_X)$. (continued) – David Speyer Nov 30 '10 at 18:21

Here are some considerations on the case $X$ smooth.

Let $d$ be the degree of $X$ and let $L$ be the restriction to $X$ of $O_{P^2}(1)$.
If $k=1$ then the condition is precisely that the line bundle $L(-dP)$ is a torsion point of $Pic^0(X)$. In fact let $m$ be such that $mL(-dP)$ is trivial. Since the map $H^0(P^2,{\cal O}_{P^2}(m))\to H^0(X, mL)$ is onto, there exists a curve $Y$ of degree $m$ that intersects $X$ precisely at $P$ with multiplicity $md$. So the condition is satisfied for at most countably many points $P\in X$, unless $X$ is rational. One can argue in a similar (more complicated) way for $k>1$.

I don't know if the remark that follows is useful. If $X$ is smooth of genus $g$, $P\in X$ is fixed and $k=g+1$, then one can consider the image of the map $X^g\to Pic^0(X)$ defined by mapping $(P_1,...,P_g)$ to $(g+1)L(-d(P+P_1+...+P_g))$. This map is surjective, so the above argument implies that, given $P$, one can find $g$ points such that there exists a curve $Y$ that intersects $X$ only at $P, P_1,\dots P_{g}$. Since the subvarieties {P_i=P} of $X^g$ and the weak diagonal map to proper subvarieties of $Pic^0(X)$ and the torsion points are dense in $Pic^0(X)$, one can find $P,P_1,...P_g$ distinct. More generally, if $k>g$ one can assign $k-g$ points and find a curve $Y$ that meets $X$ at those points and at precisely $g$ additional points.

share|cite|improve this answer

I don't even see an answer to the following simpler problem: let Z be a finite (reduced) set of points in the projective plane. Is Z a set-theoretic complete intersection?

By the way, another interesting question is to let the curve X be in projective n-space, take an arbitrary set of points on X, and ask if it is set-theoretically cut out by a hypersurface. Assuming the answer to be no, can such sets of points be characterized somehow?

share|cite|improve this answer
Maybe these should be posted as questions, rather than answers? BTW, welcome, Juan! – quim Dec 1 '10 at 11:15

This is actually an answer (a bit belated!) to one of the questions Juan posed in his answer. I did not want to put it as just a comment, since I thought the question was good, and the answer is cute enough to be interesting to some people.

The answer to Juan's first question is affirmative: "Every finite set of points in the projective plane is a set-theoretic complete intersection." Here are the arguments: we may assume number of points is $d + 2$, $d \geq 0$. Choose $\mathbb{C}^2 \subseteq \mathbb{P}^2$ and coordinates $(x,y)$ on $\mathbb{C}^2$ such that

  1. one of the points is on the intersection of $y$-axis and the line at infinity,

  2. The others are in $\mathbb{C}^2$,

  3. The finite points have mutually distinct $x$-coordinates.

Now by Lagrange interpolation we can find a polynomial curve $C$ with equation of the form $y = f(x)$ which passes through each of the finite points. We may (and will) assume $\deg(f) \geq 2$, so that $C$ passes also through the other point which is at infinity. Let $a_1, \ldots, a_{d+1}$ be the $x$-coordinates of the finite points. Let $g_1 := z\prod_{i=1}^{d+1}(x - a_1z)$ and $g_2(x,y,z)$ be the homogenization of $y - f(x)$ (with respect to $z$). Then $g_1 = 0$ and $g_2 = 0$ intersects precisely at the given points - answering Juan's question.

Note that $g_2$ is irreducible. Replacing $g_1$ by an element of the form $g_1^p + \lambda g_2^q$ for suitable $\lambda$, $p$ and $q$, we may ensure that $g_1$ is also irreducible. Is it possible to ensure that the curves are non-singular?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.