# Intersection of an affine cubic and quartic: at least nine points?

Hello, I have posed myself the following problem: suppose that two affine algebraic with no common components curves be given. To fix ideas, suppose that we have a cubic $C$ and a quartic $D$. More precisely, let $C=\{(u,v)\in C^2\colon P(u,v)=0\}$ and $D=\{(u,v)\in C^2\colon Q(u,v)=0\}$, with $P$ and $Q$ polynomials and $deg(P)=3$, $deg(Q)=4$. Could we say that $C\cap D\cap \mathbb{C}^2$ has at least 9 points?

I have also worked out an answer, (which should be: yes). All points are to be counted with multiplicity. Let $\tilde C$ and $\tilde D$ be the projective extensions of $C$ and $D$: $\tilde C$ has three points at infinity and $\tilde D$ has four, so $\tilde C \cap \tilde D$ has at most three points at infinity. Now we can apply Bézout’s theorem: we have that $\tilde C$ and $\tilde D$ intersect in exactly twelve points. This implies that $C\cap D\cap \mathbb{C}^2$ has at least $12-3=9$ points, QED. It seems correct to me, but still something sounds wrong. Many thanks for any answers or comments.

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The answer is no: consider the irreducible smooth curves $C: u^3-v=0$ and $D: u^4-uv+1=0$. You can find yourself where your arguments fail. – Qing Liu Feb 25 '11 at 23:59
Think about the case when $C \cup D$ consists of seven parallel lines! – Georges Elencwajg Feb 26 '11 at 0:24

## 2 Answers

It is possible for all 12 intersection points to be at infinity (remember that they are counted with multiplicity). For instance you might have seven parallel lines, as Georges said.

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Your proof is correct if and only $C$ and $D$ intersect transversally on the line at infinity. In other words all the intersection points at infinity have multiplicity $1$ as you assume in your proof. The correct statement is that you get $12-m$ intersection points where $m$ is the number of intersection points at infinity counted with multiplicity. This $m$ could be anything between $0$ and $12$.

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