Let $f: X \to S$ be an arithmetic surface, where $S=\operatorname{Spec } O_K$ for a number field $K$. It is well known that if we want to introduce a reasonable intersection theory on $X$ we have to deal with Arakelov geometry. In this question I want to stick to the usual theory of schemes, so no fibres at infinity are involved.

I'd like to know if there exist a unique pairing:

$$(,):\operatorname{Div} X\times\operatorname{Div} X\to \operatorname{Div} S$$

Satisfying the following properties:

1) it is bilinear and symmetric

2) It descends to the so called Deligne pairing:

$$\operatorname{Pic} X\times\operatorname{Pic} X\to \operatorname{Pic}S$$

3) If $D,E$ are two smooth divisors on $X$ meeting transversally, then:

$$(D,E)=\sum_{x\in D\cap E} [k(x): k(f(x))] f(x)$$

My idea is that one can follow the classical proof of the existence of an intersection pairing for algenbraic surfaces (see Hartshorne chap. 5 for example) and just modify things for arithmetic surfaces. What do you think?

**Remark:** Keep in mind that I'm not asking about the existence of an intersection pairing, but just about this "cup product" at the level of divisors. For example, if we composed such a map with the degree of divisors on $S$, this wouldn't give an intersection pairing because the degree is not $0$ on principal divisors.

notdivisor classes). Then he uses a moving lemma (basically the trick of writing each divisor class as a difference of very ample divisor classes) to prove that this gives a well-defined pairing on divisorclasses. $\endgroup$ – Jason Starr Jul 27 '18 at 15:46