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I´m asking for a proof of a fact used by Arakelov in his paper: Intersection Theory of Divisors on an Arithmetic Surface (page 1169 row 16). He gives no references or explanations for this fact.

The setup is the following, $X$ is a smooth complete curve over a number field $K$ with ring of integers $\Lambda$. $V$ is a regular complete model of $X$ over $Spec(\Lambda)$, i.e. a regular fibered arithmetic surface with a proper morphism $f:V \rightarrow Spec(\Lambda)$ such that the fiber over the generic point is the original curve $X$. $D$ is an irreducible horizontal divisor on $V$, i.e. a prime divisor in Hartshorne´s notations such that $f(D)=Spec(\Lambda)$.

The claim is the following. There exists a finite field extension $L/K$ with ring of integers $\Gamma$, and a $Spec(\Lambda)$-morphism $\epsilon: Spec(\Gamma) \rightarrow V$ such that $\epsilon(Spec(\Gamma))=D$.

And this is my work. My idea was to prove some finiteness condition on the map $f|_D$ in order to use the fact, stated for example in Matsumura´s Commutative Ring Theory, that an extension of rings $A \subset B$ is integral if and only if every element of $B$ is contained in a finitely generated $A$-module.

Proceeding in that direction I proved that $f|_D$ is of finite type (clear because $f$ is proper), it is generically finite (because by irreducibility of $D$ its preimage on the generic point is a single closed point of $X$) and it is dominant (because $D$ is horizontal).

Using Exercise 3.7. of Hartshorne we get that there exists an open dense subset $W\subseteq Spec(\Lambda)$ such that $f|_D:f|_D^{-1}(W) \rightarrow W$ is finite, moreover the function field $K(D)$ of $D$ is a finite field extension of $K$ (so my guess is $L=K(D)$, also because I don´t see any other possible candidate).

Unfortunately I have no idea on how to proceed further. There any suggestions? (or complete proves... :) ) Thank you in advance!

Edit 1: According to the excellent answer of Matt there is only one point that still bother me. In order to use the Valuative Criterion we need $\Gamma$ to be a Valuation Ring, but to be the ring of integers of a number field is clearly not enough. For example $\mathbb{Z}$ is not a Valuation Ring. So, how can we prove that $\Gamma$ is a Valuation Ring?

Edit 2: This is the answer to the question in Edit 1. The proof was sketched by Matt E in a comment to his own answer, I have just expanded it with details (and maybe errors!). I write it here because it could be useful and because I hope that, if my checkings are wrong, someone could point it out!

We have a map $\varphi: Spec(L) \rightarrow V$ and we want extend it to a map $Spec(\Gamma) \rightarrow V$. Since $f$ is proper it is of finite type and the point $P$ has a neighborhood of the form $Spec(A)$ such that $A$ is a finitely generated $\Lambda$-algebra. Moreover by integrality of $V$ the fraction field of $A$ is (at least) a subfield of $L$. In specific there is an extension of algebras $A \rightarrow \Lambda[l_1,...,l_n]$ where $l_1,...,l_n \in L$. Define $\lambda$ to be the common denominator of the $l_i$, $\lambda \in \Gamma$. Then the map $A \rightarrow L$ factorizes through $Spec(\Gamma[\frac{1}{\lambda}])$ and we have a natural extension of $\varphi$ to $\hat{\varphi}:Spec(\Gamma[\frac{1}{\lambda}]) \rightarrow Spec(A) \subset V$.

Now let $p_1,...,p_m$ the primes of the factorization of $\lambda$ in $\Gamma$. We have natural extensions of $\varphi$ to $\varphi_i:Spec(\Gamma_{p_i}) \rightarrow V$ because of the Valuative Criterion for Properness (Hartshorne, Theorem II.4.7). And because the localization of a Dedekind Domain to a prime ideal is a Discrete Valuation Ring.

Now we want to glue together all those morpisms. It is clear that $Spec(\Gamma)$ is covered by the images of $\hat{\varphi}$ and $\varphi_i$, and the intersection of the image of any two such morphisms is $(0)=Spec(L)$. But they are constructed as extension of $\varphi:Spec(L) \rightarrow V$ so they agree on their intersection. By the proof of Theorem II.3.3 of Hartshorne they give rise to a well defined map $\epsilon: Spec(\Gamma) \rightarrow V$ and we are done.

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up vote 4 down vote accepted

As you write, $D$ meets $X$ in a single closed point. If we call this point Spec $L$, then the composite Spec $L \to V \to $ Spec $\Lambda$ extends to a map Spec $\Gamma \to $ Spec $\Lambda$ (because the corresponding map of rings $\Lambda \to L$ factors through $\Gamma$), and now by properness of $X$ over Spec $\Lambda$, the map Spec $L \to V$ extends to a map $\epsilon:$ Spec $\Gamma \to X$. (Use the valuative criterion.)

Now Spec $\Gamma$ is the closure of its generic point, and the image of its generic point under $\epsilon$ equals the generic point of $D$. Hence the image of $\epsilon$ is contained in $D$ (since $D$ is closed).

By irreducibility of $D$, the image of $\epsilon$ equals $D$, and we are done.

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Thank you! I checked details and indeed everything works! I have only a last question: Do we know that $L$ is finite over $K$? We know that by the machinery I used, but your solution is that elegant that I think there should be an "easy" way to prove that too! –  Giovanni De Gaetano Dec 7 '11 at 13:13
    
I´m sorry but I there´s something which is not really clear. My problem is that, for using Valuative Criterion, I need $\Gamma$ to be a Valuation Ring. But it doesn´t seem to be true ($\mathbb{Z}$ is not a Valuation Ring for example). Am I missing something? There´s a way to avoid that gap? –  Giovanni De Gaetano Dec 7 '11 at 14:23
    
@student: The map Spec $L \to X$ extends to Spec $\Lambda[1/\lambda] \to X$ for some $\lambda \in \Lambda \setminus \{0\}$ automatically (just because it involves only finitely many denominators). Now you just have to extend it locally at each prime $\wp$ dividing $\lambda$, and $\Lambda$ localized at $\wp$ is a discrete valuation ring. (This is a very standard argument, so if you don't understand it straight away, you should think about it. Perhaps a more familiar case is the fact that e.g. $\mathbb P^n(L) = \mathbb P^n(\Lambda)$.) Regards, –  Matt E Dec 7 '11 at 16:24
    
@student: Dear Student, We know that $L$ is finite over $K$ because it is the residue field of a closed point of the curve $X$ over $K$. Regards, –  Matt E Dec 7 '11 at 16:25
    
Thank you for your help and patience! I´m currently checking the details and I´ll accept your answer again as soon as it will be clear in my mind! May I assume that when you write $X$ in your comment you are meaning $V$ and when you write $\Lambda$ you are meaning $\Gamma$? –  Giovanni De Gaetano Dec 7 '11 at 16:55
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