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Bezout thereom that says that two irreducible algebraic curves $C$ and $D$ in $\mathbb{P}^2_\mathbb{C}$ intersect at $nm$ points (counted with multiplicity), where $n$ and $m$ are the degrees of $C$ and $D$, respectively. So in this case we know that $C\cdot D=nm$ is finite.

The question is, if I change $\mathbb{P}^2_\mathbb{C}$ by any other non-singular projective algebraic surface $X$, may I find two irreducible algebraic curves $C,D \subset X$ such that $C \cdot D=\infty$? If yes, what can we say about $X$, $C$ and $D$?

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    $\begingroup$ On any smooth projective surface, $C\cdot D$ is defined in such a way that it's an integer. So no, $C\cdot D$ won't be $\infty$. Of course $C\cap D$ might be an infinite set, even in the plane. $\endgroup$ Commented Nov 19, 2021 at 16:58
  • $\begingroup$ @DonuArapura As far as I know $C \cdot D \geq \#(C \cap D)$ (since it is the sume of the intersection multiplicities at points of $C \cap D$), so I do not understand how $ \# (C \cap D)$ can be inifnite but not $C\cdot D$. On the other hand, when you say that "$C \cap D$ might be infinite, even in the plane", I guess you make reference to the projective plane, but here we have $\# (C \cap D) \leq nm$. You're answer doesn't convince me, but thank you anyway. $\endgroup$
    – BMS
    Commented Nov 19, 2021 at 17:15
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    $\begingroup$ I think Donu means that $C\cap C$ will be infinite. However, $C\cdot C$ wont be. $\endgroup$ Commented Nov 19, 2021 at 17:19
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    $\begingroup$ The integer $C\cdot D$ is never infinite. However, $C\cdot D \ge \#(C\cap D)$ only when $C$ and $D$ have no component in common. So your first sentence is not correct, as Donu Arapura says. $\endgroup$
    – Ben McKay
    Commented Nov 19, 2021 at 17:33
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    $\begingroup$ @BMS Other people have caught my meaning. I second Daniel suggestion of reading that section of Hartshorne. Let me also observe the following which might help you. Lemma: if $C$ and $D$ are distinct irreducible curves on a variety $X$, then $C\cap D$ is finite. Proof: The intersection is a proper Zariski closed subset of $C$. $\endgroup$ Commented Nov 19, 2021 at 19:06

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This question is probably more suitable for MSE. However, since it is still open and it has (currently) three upvotes, let me shortly answer it. Of course, my answer is just an extended version of the comments above, and everything I will say can be found in the classical literature on the subject, such as Beauville, Hartshorne, Barth-Hulek-Peters-Van de Ven, Griffiths-Harris, etc.

The intersection pairing $(C, D)$ on the Picard group $\mathrm{Pic}(X) \simeq H^1(X, \, \mathcal{O}_X^*)$ of a a smooth, complex projective surface $X$ is induced by the intersection pairing in $H^2(X, \, \mathbb{Z})$, in the sense that $$(C, \, D)=c_1(C) \cup c_1(D) \in H^4(X, \, \mathbb{Z}) \simeq \mathbb{Z},$$ where $c_1 \colon \operatorname{Pic}(X) \to H^2(X, \, \mathbb{Z})$ is the first Chern class map, seen as the connecting homomorphism arising from the exponential sequence $$0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^* \to 0.$$

In particular, the intersection number $(C, \, D)$ is always a (possibly negative) integer and it cannot be infinite.

When $C$, $D$ are holomorphic curves with no component in common then, by the Poincaré duality $H^4(X, \, \mathbb{Z}) \simeq H_0(X, \, \mathbb{Z})$, we are really counting the intersection points of $C$ and $D$ (with multiplicity); in fact, the natural orientation given by the complex structure tells us that each intersection point $p \in C \cap D$ must be counted with positive sign.

It follows that, if $(C, \, D)$ is negative, then the two curves $C$, $D$ must have a component in common; this means in particular that the set $C \cap D$ is infinite (but $(C, \, D)$ is not!).

As a typical example, consider the exceptional divisor $E$ of a smooth blow-up: then $E^2=-1$, even if $E \cap E =E$ is an infinite set. In fact, in general, the self-intersection number $(C, \, C)$ can be interpreted as the degree of the normal bundle $N_{C/X} \simeq \mathcal{O}_C(C)$ of $C$ in $X$.

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