Inspired by this question:

Suppose given an algebraic curve $C \subset \mathbb{A}^2$, and a point $x \in C$. Can you find another (closed) curve $D \subset \mathbb{A}^2$ such that $C \cap D = x$?


1 Answer 1


Let $C$ be a smooth curve which is the zero set of a function $y^2-x^3-ax-b$ in $\mathbb{A}^2$. I.e. $C$ is a Weierstrass elliptic restricted to the x-y plane. Let $x \in C$ be a chosen point. Suppose $D$ is another curve in $\mathbb{A}^2$ such that $D \cap C = {x}$ and the defining equation for $D$ has degree $d$.

Taking the closure $\overline{C}$ and $\overline{D}$ in $\mathbb{P}^2$ we see that $\overline{C} \cap \overline{D}$ gives an effective divisor on $\overline{C}$ which is only supported at $x$ and at $e$, the point at infinity (also the identity in the group law). The divisor $\overline{C} \cap \overline{D}$ is rationally equivalent to $3d \cdot e$. I.e. the equation $m \cdot x + n \cdot e \sim_{rat} 3d \cdot e$ holds where $m$ is the multiplicity of $C \cap D$ and $n = 3d - m$. In particular, the degree zero divisor $m \cdot x - m \cdot e$ is rationally equivalent to zero. So $x-e$ is torsion in the group of Cartier divisors on $\overline{C}$.

On the other hand there are many point $x \in C$ which do not give rise to torsion divisors $x-e$ in the group of Cartier divisors on $\overline{C}$ so such points cannot be described as $C \cap D$.

  • $\begingroup$ The question is about $\mathbb{A}^2$, not $\mathbb{P}^2$. $\endgroup$
    – abx
    Sep 10, 2014 at 5:15
  • $\begingroup$ I edited my answer to try and make it more clear. Does that look ok? $\endgroup$ Sep 10, 2014 at 5:41
  • $\begingroup$ Yes, it is fine. $\endgroup$
    – abx
    Sep 10, 2014 at 7:49

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