Self intersection of curve on $P^1 \times P^1$

Question

Let $l = \mathbb{P}^1 \times \{0\}$, I want to prove $l \cdot l=0$, where $l \cdot l$ is the self intersection of $l$. I am working out of Ravi Vakil chapter 20, this is exercise 20.2.C. His hint is to use that the intersection number of two curves is $C \cdot D = h^0(C \cap D, \mathscr{O}_{C \cap D})$ where $C \cap D$ is the scheme theoretic intersection. Now, $ l \cap l= l$ and I know that the global sections of $\mathbb{P}^1$ is just the constant functions, which are dimension $1$. So why isn't $l \cdot l=1$?

Answer 1

As Mohan said, you need $C,D$ to have no common components (else we would always have $C^2 = 1$ !), so the idea is to move one copy of $l$.

Let $p_2 : \Bbb P^1 \times \Bbb P^1 \to \Bbb P^1$ be the second projection. We have $l \sim p_2^{-1}(\{\infty\}) := l'$, so $l^2 = l \cdot l' = 0$.

In fact, $\Bbb P^1 \times \Bbb P^1$ is a quadric surface in $\Bbb P^3$ : more generally if $X$ is a surface of degree $d$ and $l \subset X$ is a line then one has $l^2 = 2 - d$.

Edit (more details) : The adjunction formula says if $Y \subset X$ is an hypersurface then we have equality of divisors on $Y$ (maybe up to linear equivalence) $$K_Y = (K_X +Y)_{|Y} = K_X \cdot Y + Y^2$$

First we take $X = \Bbb P^3$ and $Y$ is a surface of degree $d$. We read $K_Y = (-4h + dh)_{|Y} = (d-4)h_Y$ where $h$ is the hyperplane class and $h_Y$ is an hyperplane section of $Y$.

Now we take $X$ to be a surface of degree $d$ and $l \subset X$ a line. For a curve, $K_X = 2g -2$ so here $g=0$ and the formula reads $- 2 = (d-4) + l^2$ i.e $l^2 = 2 - d$.