I vaguely recall the following fact that I'd like to use in my research. It should be easy to see that this holds (if it does) but I can't seem to prove it.

Let $p:X\longrightarrow S$ be a (regular) arithmetic surface over a Dedekind scheme $S$.

Let $P:S\longrightarrow X$ be a section and let $\omega$ be a non-zero rational section of $\omega_{X/S}$. Let $K_{X} = \mathrm{div}(\omega) $ be the canonical divisor defined by $\omega$. (A better notation for $K_X$ would be $K_{X/S}$, maybe.)

By definition, the intersection number $(K_X,P)$ is defined as $$\sum_{s } i_s (K_{X}, P) \log( \mathrm{card}( k (s)) ), $$ where the sum runs over the closed points of $S$ and $$i_s(K_{X},P) = \sum_{x} i_x(K_{X}, P) [k(x):k(s)],$$ where the sum runs over the closed points of $X_s$ and $i_x$ denotes the intersection number at $x$.

Now, I wonder if the following equality is trivial to see.

Write $\omega = df$ for some rational function $f\in K(X)$. (We assume this to be possible.) Do we have that

$$(K_{X}, P) = \sum_{s} \mathrm{ord}_s(P^\ast\omega) \log(\mathrm{card}(k(s)))?$$

To see this, it suffices to prove the following equality: $$\mathrm{ord}_s(P^\ast \omega) =i_s (K_{X}, P) .$$ Does this hold?

  • $\begingroup$ $ord_s(P^*\omega)$ doesn't necessarily make sense. Plus, $P^*\omega$ may not be what you think it is. Consider the case $S = \Spec \mathbb{Z}$ and, I don't know, $X = \mathbb{P}^1, P=(0:1)$. $\endgroup$ – Felipe Voloch Jul 23 '11 at 16:18
  • $\begingroup$ Thanks for your comment. I edited the question and hopefully $\mathrm{ord}_s(P^\ast \omega)$ makes sense now. I really want $P^\ast \omega$ to be a nonzero rational section of $P^\ast \omega_{X/S}$. Maybe there are smoothness conditions that need to be fulfilled? $\endgroup$ – Enchanted Jul 23 '11 at 17:14

This is correct if $P(S)$ is not contained in the support of $\mathrm{div}(\omega)$. It comes essentially from the definition of $i_x(K_X, P)$. You don't need $\omega$ to be an exact differential from. However the intersection number depends on the choice of $\omega$ (as well as the Weil divisor $K_X$). You can check this by yourself by multiplying $\omega$ by a non-zero constant in $K(S)$ and see the effect on the total intersection number. If $P(S)$ is contained in the support of $K_X$, then you can't define $i_x(K_X, P)$.

EDIT. Let me add some more details. Denote by $K(X)$ the field of rational functions on $X$, viewed as a constant sheaf on $X$. Then $\omega\in \omega_{X/S}\otimes K(X)$. Hence $\omega\cdot\omega_{X/S}^{\vee}$ is a subsheaf of $K(X)$, hence equal (not only isomorphic) to $O_X(-D)$ for some Cartier divisor $D$ on $X$. We have $$ \omega_{X/S}=\omega\cdot O_X(D). $$ A straightforward local computation shows that $\mathrm{div}(\omega)=D$ as Cartier divisors. Let us identify $P$ with $P(S)$. Let $I\subset O_X$ be the ideal sheaf defining $P$ in $X$. As $P$ is not contained in the support of $D$, $D|_P$ is a well defined Cartier divisor on $P$. Namely, if a local equation of $D$ at some point $x\in P$ is given by $f_x\in K(X)$, then we can write $f_x=a/b$ with $a, b\notin I_x$ (here we use the fact that $O_{X,x}$ is a UFD). Then a local equation of $D$ restricted to $P$ is $\bar{a}/\bar{b}$ where $\bar{c}$ means the image of $c$ in $O_{X,x}/I$.

The above equality restricted to $P$ reads $$P^{*}(\omega_{X/S})=P^{*}(\omega) \cdot O_P(D|_P).$$ So $P^{*}(\omega)$ is a rational section of

$P^{*}\omega_{X/S}$ and its divisor on $S$ is $D|_P$.

To get an intersection number independent of the choice of a rational section $\omega$, you have to use Arakelov intersection theory.

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    $\begingroup$ Under the condition on $P(S)$, $P^*\omega$ can be viewed as a rational section of $P^*\omega_{X/S}$. Then its order in the DVR $O_{S,s}$ makes sense. If you want, $K_{X}$ is a Cartier divisor, and the divisor of $P^*\omega$ is equal to $P^*K_{X}$. Probably Felip was concerned with the case when $P(S)$ is contained in the support of $K_X$. $\endgroup$ – Qing Liu Jul 23 '11 at 19:08
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    $\begingroup$ Isn't $P^*\omega =0$ if e.g. $S=Spec \mathbb{Z}$? After all $\Omega^1_{Spec \mathbb{Z}} = 0$. You can pull back the divisor allright, but not the differential. $\endgroup$ – Felipe Voloch Jul 23 '11 at 19:08
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    $\begingroup$ @Felipe: $P^*\omega_{X/S}\ne \Omega^1_{P/S}$. The first is an invertible sheaf while the second is equal to $0$. $\endgroup$ – Qing Liu Jul 23 '11 at 21:53
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    $\begingroup$ $1^*dx = d1 = 0$, Enchanted was pulling back a differential, not a line bundle. $\endgroup$ – Felipe Voloch Jul 24 '11 at 0:18
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    $\begingroup$ I think he/she is trying to give a meaning to $P^*\omega$. For a general coherent sheaf $L$ and a regular section $s$, $P^*s$ is a regular section of $P^*L$. In the case $L=\Omega_{X/S}$, a commun meaning for $P^*s$ is its image by the canonical map $P^*\Omega_{X/S}\to \Omega_{P/S}$. The two notions differ for $L=\Omega_{X/S}$ and this is sometimes confusing. In the current context, his/her $\omega$ is a rational differential form, so it is more reasonable to view $P^*\omega$ as a rational section of $P^*\omega_{X/S}$. Then the desired equality holds under the condition on the support, ... $\endgroup$ – Qing Liu Jul 24 '11 at 9:08

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