TITLE: self-intersections in a product of two curves
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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.