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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
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Think about the case when $C \cup D$ consists of seven parallel lines! –  Georges Elencwajg Feb 26 '11 at 0:24

2 Answers 2

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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