This comes from Vakil 20.2 C, self-study.
We have to show the intersection form of $X = \mathbb P^1 \times \mathbb P^1$ is given by $l \cdot l = m \cdot m = 0$ and $l \cdot m = 1$, where $l$ is the curve $\mathbb P^1 \times \{0\}$ and $m$ is the curve $\{0\} \times \mathbb P^1$.
By exercise 20.2 A, we see $l \cdot l = h^0(l, \mathcal O_l)$. I suppose if we could show this was a negative degree line bundle, we would be done by 19.2.3, but I do not see why that should be the case.
We also have just proven Adjunction: given a curve $C$ in a smooth projective surface $X$, if $K_X$ is the canonical divisor, then
$$C\cdot (K_X + C) = 2g_C – 2$$
where $g_C$ is the genus of $C$. Right after that (and right before this problem), we showed Riemann-Roch for Surfaces:
If $D$ is a Weil divisor on $X$, where $X$ is as above, then
$$\chi(X, \mathcal O_X(D)) = \frac{D \cdot (D – K_X)}{2} + \chi(X, \mathcal O_X)$$
How do I finish?
Best Answer
Your formula is wrong, we don't have $l \cdot l = h^0(l, \mathcal O_l)$. Indeed in this case $l \cdot l = 0$ and $h^0(l, \mathcal O_l) = 1$.
Actually you don't need so much machinery. It's easy to see that $\mathbb P^1 \times \{ p\}$ is linearly equivalent to $\mathbb P^1 \times \{ q\}$ for any $p,q \in \mathbb P^1$. By basic properties of intersection product, you obtain $l \cdot l = 0$.
If you want to use the adjunction formula, then as Mohan suggests you can try to compute the canonical class. To do this, you can notice that $\Omega_X \cong p_1^* \Omega_{X_1} \oplus p_2^* \Omega_{X_2}$ if $X = X_1 \times X_2$ and $p_i : X_i \to X$ is the projection.