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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 $*_1\neq *_2\in X$.2012-03-25

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