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Suppose that $E$ and $F$ are two complex compact Riemann surfaces with genus greater or equal than $2$.
Set $$S=E \times F$$ the surface composed by the cartesian product of thees curves. What can i say about the intersection number $E.F$ and the self intersection $E^2$ or $F^2$?

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  • $\begingroup$ Draw a picture of what this surface looks like. Where are $E$ and $F$ in your picture? How many points of intersection do they have? $\endgroup$ – user64687 Apr 14 '15 at 11:19
  • $\begingroup$ Inside $E\times F$ the curves $E$ and $F$ meet just once, transversally. $\endgroup$ – AdLibitum Apr 14 '15 at 11:20
  • $\begingroup$ i think that $E^2=0$ and $E.F=1$ but i would like to write a formal proof $\endgroup$ – dario Apr 14 '15 at 11:20
  • $\begingroup$ how can i show this? $\endgroup$ – dario Apr 14 '15 at 12:21
  • $\begingroup$ @dario Compute $E^2$ by writing it as $p^{-1}(y) \cdot p^{-1}(y)$ with $y$ some point in $F$ and $p$ the projection $E\times F\to F$. $\endgroup$ – Ariyan Javanpeykar Apr 14 '15 at 12:57
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You can calculate E.F by taking curves representing the classes [E] and [F] which intersect transversely and counting their points of intersection.

In particular, pick points $e \in E$ and $f \in F$ and take $E \times \{f\}$ and $\{e\} \times F$. Clearly these only meet at the single point $(e, f)$, and moreover it's easy to calculate the tangent spaces and see that the intersection is transverse. So $E.F = 1$.

For $E.E$, take points $f, f' \in F$ and consider the curves $E \times \{f\}$ and $E \times \{f'\}$. These don't meet at all, so they surely meet transversely. Therefore $E^2 = 0$.

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