0
$\begingroup$

The question is based on the example given in Intersection Theory under the heading Self-intersection. The example is as follows:

Consider a line $L$ in the projective plane $\mathbb{CP}^{2}$: it has self-intersection number $1$ since all other lines cross it once: one can push $L$ off to $L′$, and $L · L′ = 1$ (for any choice) of $L′$, hence $L · L = 1$.

Given that my current knowledge on algebraic geometry is limited I was wondering if one can calculate this number by algebraically (if it is possible) computing the intersection point of $L$ and $L'$?

$\endgroup$
0
$\begingroup$

Yes. If two algebraic curves $C$ and $D$ lie on a surface $X$, $p \in C\cap D$ and $\{c=0\}$ and $\{d=0\}$ are local equations of $C$ and $D$, respectively, on a neighborhood of $p$, then the intersection number at p is $$ {\rm dim}_{\mathbb{C}}\frac{\mathcal{O}_{X,p}}{<c_p, d_p>} $$ where $c_p$ and $d_p$ are the localizations of $c$ and $d$ at $p$. The sum of these for all $p$ gives $C\cdot D$.

Take a look at the first chapter of Beauville's Complex Algebraic Surfaces.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If they do not share a irreducible component... $\endgroup$ – Alan Muniz Feb 5 '16 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.