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Let $F$ and $G$ be polynomial in $K[x,y,z], K$ be an algebraically closed field. Let $F$ be homogene0us of degree $n$ and $G$ homogene0us of degree $m.$ Then, $F$ and $G$ define two curves in the projective plane $\mathbb{P}^2_K$. Let $F$ and $G$ be coprime (i.e. they do not have common irreducible factors).

Question: Is it true that $F$ and $G$ intersect in a finite number of point? Why?

Question 2: If $P\in F\cap G$ why $I(P,F\cap G)=I(P,F_*\cap G_*)$ where $I(P,F\cap G)$ is the intersection number and $F_*$ is the dehomogeneous polynomial obtained by $F$?

I find this fact while studying Bezout's theorem. But I can't see it. Can you also give me some title of good books as introduction of algebraic geometry?

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Minato
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  • Do you know Projective Dimension Theorem? It can be derived from there. – Krish Sep 05 '17 at 09:20
  • No, i only know that this is true in the affine plane with $F$ and $G$ polynomial in $K[x,y]$ not necessary homogeneus but still coprime. – Minato Sep 05 '17 at 09:24
  • Can you show that there are only finitely many irreducible components in the intersection? From there, you have to leverage the fact that $F$ and $G$ are coprime into showing that no irreducible component can have dimension greater than $0$. – Arthur Sep 05 '17 at 09:28
  • Here are two relevant posts for your first question: 1, 2 – Viktor Vaughn Sep 05 '17 at 12:01

2 Answers2

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If $F$ and $G$ are irreducible polynomials then you can apply bezout to obtain that their intersection consists of $mn$ points (considering multeplicity). Now suppose they have more than one irreducible factor, then consider separately each of them (they are finetely many) and apply bezout to each couple of irreducible factors of both curves (you can do that because they are coprime). In this way you obtain finetely many points of intersections for finitely many couples, hence they are finite.

SC30
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This is based on OP's comment.

Let $C:=\mathcal{Z}(F)$ and $D:=\mathcal{Z}(G)$. We can choose a line $L \subset \mathbb{P}^2_K$ such that both $L \cap C$ and $L\cap D$ are finite sets (why?). Now choose coordinates of $\mathbb{P}^2_K$ in such a way that $L=\mathcal{Z}(z)$. Then $U:=\mathbb{P}^2_K - L$ is affine and $$C \cap D=\bigg((C\cap U)\cap (D\cap U)\bigg) \cup X$$ where $X$ is a finite set contained in $(C\cap L) \cup (D \cap L).$ Note that, now $C\cap U$ and $D \cap U$ are affine plane curves.

Krish
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