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Let $X$ be a smooth projective curve of genus $g$ over an algebraically closed field and consider the intersection pairing on the surface $X \times X$. I remember hearing that $\Delta^2 = 2-2g$: how does one prove this? I understand that the self-intersection is defined by intersecting $\Delta$ with a general divisor linearly equivalent to $\Delta$, but I'm in the dark about how to compute such things.

Also, what are $(X \times \{ * \})^2$ and $(\{ * \}\times X)^2$? My intuition suggests that these are both $0$, but again I don't know how to compute.

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I'm not an algebraist, but over $\mathbb{C}$, $\Delta^2=2-2g=\chi(X)$ follows since the normal bundle of $\Delta$ in $X\times X$ is isomorphic to the tangent bundle of $X$. The second follow since $X\times\{*_1\}$ and $X\times \{*_2\}$ are disjoint and homologous for any $*_1\neq *_2\in X$. – Adam Mar 25 '12 at 20:17

1) Use the adjunction formula. For the surface $S=X\times X$ it reads:
$$ \Delta^2+\Delta \cdot K_S=2g-2 $$ Now, denoting by $p_i$ the two projections projections $S\to X$, we have $K_S=p_1^*K_X+ p_2^*K_X$
and plugging into the adjunction formula yields
$$ 2g-2= \Delta^2+ \Delta \cdot (p_1^*K_X)+ \Delta \cdot (p_1^*K_X)= \Delta^2+2g-2+2g-2 $$ from which follows the required formula $\Delta^2=2-2g \:$.

2) If $P, Q$ are any points of $X$, the fibers $p_1^*(P)$ and $p_1^*(Q)$ are algebraically equivalent so that $ (\lbrace P\rbrace\times X)^2 =(p_1^*(P))^2 = p_1^*(P)\cdot p_1^*(Q) =0$ since algebraic equivalence implies numerical equivalence.

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