I'm reading the classical paper of Arakelov "Intersection Theory of Divisors on an Arithmetic Surface". At the very beginning he uses the notion of model of a curve.

In specific we have a number field $K$ with ring of integers $R$, $X$ is a smooth complete algebraic curve over $K$. Here Arakelov writes: "Let $V$ be any smooth and complete model of $X$ over $R$."

What does it mean? I didn't find the definition nor googling it, neither looking on the books "Algebraic Geometry" of Hartshorne and "Introduction to Intersection Theory in Algebraic Geometry" of Fulton.

  • 5
    $\begingroup$ See Liu's "Algebraic Geometry and Arithmetic Curves", Chapter 10. $\endgroup$
    – KCd
    Oct 14, 2011 at 13:37
  • $\begingroup$ Thank you very much, that´s perfect! $\endgroup$ Oct 14, 2011 at 14:13

1 Answer 1


The real answer has been given by KCd in the comments to the question. I answer my own question to have an explicit answer written under the question and to mark the question as answered (because in fact it is).

According to Definition $1.1$ of Liu´s Algebraic Geometry and Arithmetic Curves, Chapter 10 paragraph 10.1.1:

Let $S$ be a Dedekind scheme of dimension $1$, with function field $K$. Let $C$ be a normal, connected, projective curve over $K$. We call a normal fibered surface $\mathcal{C} \rightarrow S$ together with an isomorphism $f:\mathcal{C}_\eta \simeq C$ a model of $C$ over $S$. We will say that and a regular model of $C$ if $\mathcal{C}$ is regular. More generally, we will say that a model $(\mathcal{C},f)$ verifies a property $(P)$ if $\mathcal{C} \rightarrow S$ verifies $(P)$. The property $(P)$ can, for example, be the fact of being smooth, minimal regular, or regular with normal crossings, etc. A morphism $\mathcal{C} \rightarrow \mathcal{C}^´$ of two models of $C$ is a morphism of $S$-schemes that is compatible with the isomorphisms $\mathcal{C}_\eta \simeq C$, $\mathcal{C}^´_\eta \simeq C$.

  • $\begingroup$ I think there is a typo here. You meant $\mathcal{C}_{\eta}$ in the second line. Otherwise I think this does not make sense. $\endgroup$ May 11, 2017 at 6:39
  • $\begingroup$ @Bombyxmori old post, but I think it's correct as written: the curve over $S$ is denoted with a calligraphic $\mathcal C$ (which happens to look very similar to the original curve $C$), and $\mathcal C_{\eta}$ is its generic fiber. $\endgroup$ Sep 30, 2022 at 3:10

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