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.

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    $\begingroup$ 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. $\endgroup$ Nov 30 '10 at 12:49
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    $\begingroup$ Dear auniket, If $X$ is a line then Pic$(X)$ is trivial, so the relation that David Speyer is referring to is trivially satisfied. $\endgroup$
    – Emerton
    Nov 30 '10 at 14:27
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    $\begingroup$ 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. $\endgroup$ Nov 30 '10 at 14:41
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    $\begingroup$ 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 $V\cap X = U$? $\endgroup$ Nov 30 '10 at 17:06
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    $\begingroup$ 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) $\endgroup$ 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.


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?

  • $\begingroup$ Maybe these should be posted as questions, rather than answers? BTW, welcome, Juan! $\endgroup$
    – 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?


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