Lemma 55.14.5. Let $R$ be a discrete valuation ring with fraction field $K$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. The following are equivalent

1. there exists a proper model of $C$ which is at-worst-nodal of relative dimension $1$ over $R$,

2. there exists a minimal model of $C$ which is at-worst-nodal of relative dimension $1$ over $R$, and

3. any minimal model of $C$ is at-worst-nodal of relative dimension $1$ over $R$.

Proof. To make sense out of this statement, recall that a minimal model is defined as a regular proper model without exceptional curves of the first kind (Definition 55.8.4), that minimal models exist (Proposition 55.8.6), and that minimal models are unique if the genus of $C$ is $> 0$ (Lemma 55.10.1). Keeping this in mind the implications (2) $\Rightarrow$ (1) and (3) $\Rightarrow$ (2) are clear.

Assume (1). Let $X$ be a proper model of $C$ which is at-worst-nodal of relative dimension $1$ over $R$. Applying Lemma 55.14.3 we see that we may assume $X$ is regular as well. Let

$X = X_ m \to X_{m - 1} \to \ldots \to X_1 \to X_0$

be as in Lemma 55.8.5. By Lemma 55.14.4 and induction this implies $X_0$ is at-worst-nodal of relative dimension $1$ over $R$.

To finish the proof we have to show that (2) implies (3). This is clear if the genus of $C$ is $> 0$, since then the minimal model is unique (see discussion above). On the other hand, if the minimal model is not unique, then the morphism $X \to \mathop{\mathrm{Spec}}(R)$ is smooth for any minimal model as its special fibre will be isomorphic to $\mathbf{P}^1_ k$ by Lemma 55.12.4. $\square$

There are also:

• 2 comment(s) on Section 55.14: Semistable reduction

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).