55.15 Semistable reduction in genus zero

In this section we prove the semistable reduction theorem (Theorem 55.18.1) for genus zero curves.

Let $R$ be a discrete valuation ring with fraction field $K$. Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$. If the genus of $C$ is $0$, then $C$ is isomorphic to a conic, see Algebraic Curves, Lemma 53.10.3. Thus there exists a finite separable extension $K'/K$ of degree at most $2$ such that $C(K') \not= \emptyset$, see Algebraic Curves, Lemma 53.9.4. Let $R' \subset K'$ be the integral closure of $R$, see discussion in More on Algebra, Remark 15.111.6. We will show that $C_{K'}$ has semistable reduction over $R'_{\mathfrak m}$ for each maximal ideal $\mathfrak m$ of $R'$ (of course in the current case there are at most two such ideals). After replacing $R$ by $R'_{\mathfrak m}$ and $C$ by $C_{K'}$ we reduce to the case discussed in the next paragraph.

In this paragraph $R$ is a discrete valuation ring with fraction field $K$, $C$ is a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_ C) = K$, of genus $0$, and $C$ has a $K$-rational point. In this case $C \cong \mathbf{P}^1_ K$ by Algebraic Curves, Proposition 53.10.4. Thus we can use $\mathbf{P}^1_ R$ as a model and we see that $C$ has both good and semistable reduction.

Example 55.15.1. Let $R = \mathbf{R}[[\pi ]]$ and consider the scheme

$X = V(T_1^2 + T_2^2 - \pi T_0^2) \subset \mathbf{P}^2_ R$

The base change of $X$ to $\mathbf{C}[[\pi ]]$ is isomorphic to the scheme defined in Example 55.10.3 because we have the factorization $T_1^2 + T_2^2 = (T_1 + iT_2)(T_1 - iT_2)$ over $\mathbf{C}$. Thus $X$ is regular and its special fibre is irreducible yet singular, hence $X$ is the unique minimal model of its generic fibre (use Lemma 55.12.4). It follows that an extension is needed even in genus $0$.

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