0
$\begingroup$

Let $X\to S=\operatorname{Spec}(O_K)$ be an arithmetic surface. We denote with $X_s$ the fiber over $s\in S$ and let $\operatorname{Div}_s(X)$ be the set of divisors on $X$ with support contained in $X_s$. If $D$ and $E$ are two divisors with no common components, One can define a bilinear and symmetric map (which is NOT an intersection pairing): $$D.E:=\sum_{s\in S}\sum_{x\in X_s}i_x(D,E)[k(x):k(s)][s]$$ Here $i_x(D,E)$ is the local intersection number at $x$ calculated in the classical way as length of a module.

Now look at the following remark from Liu's book (page 388):

Remark 1.31 Let $D \in \operatorname{Div}(X)$. We can write $D = D_1 + V_1 + V_2$, where $D_1$ is a combination of horizontal prime divisors, $V_1 \in \operatorname{Div}_s(X)$, and $V_2$ is vertical, with support in $\pi^{-1}(S \setminus \{s\})$. We then have $D \cdot X_s = D_1 \cdot X_s$. The divisor $D$ induces, in a natural way, a divisor $D_\eta$ on $X_\eta$, and we have $D_\eta = (D_1)_\eta$. It follows from Proposition 1.30 that $$ D \cdot X_s = \deg_{K(S)} (D_1)_\eta. $$

(Original image here.)

I understand why elements of $V_2$ don't contribute to $D.E$, but what about the elements of $V_1$? Two irreducible curves in $X_s$ might have intersection points! Why is it enough to consider just $D_1$? What if we try to “intersect” two elements of $X_s$? In this case $D_1 = 0$.

$\endgroup$
  • $\begingroup$ Have a look at Proposition 1.21.(a) (page 384). $\endgroup$ – Ariyan Javanpeykar Aug 19 '18 at 10:58

Your Answer

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

Browse other questions tagged or ask your own question.