Semistable reduction for genus zero

Trivial case

models.tex

@@ -5771,7 +5771,7 @@ \section{Semistable reduction}
 The following are equivalent
 \begin{enumerate}
 \item there exists a proper smooth model for $C$,
-\item there exists a minimal model for $C$ is smooth over $R$,
+\item there exists a minimal model for $C$ which is smooth over $R$,
 \item any minimal model is smooth over $R$.
 \end{enumerate}
 \end{lemma}
@@ -5780,8 +5780,14 @@ \section{Semistable reduction}
 If $X$ is a smooth proper model, then the special fibre is
 connected (Lemma \ref{lemma-regular-model-connected})
 and smooth, hence irreducible. This immediately implies that
-it is minimal. The rest of the proof follows exactly as in
-the proof of Lemma \ref{lemma-semistable}.
+it is minimal. Thus (1) implies (2). 
+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 (Lemma \ref{lemma-minimal-model-unique}).
+On the other hand, if the minimal model is not unique, then
+the morphism $X \to \Spec(R)$ is smooth for any minimal model
+as its special fibre will be isomorphic to $\mathbf{P}^1_k$
+by Lemma \ref{lemma-nonuniqueness}.
 \end{proof}
 
 \begin{definition}
@@ -5799,6 +5805,89 @@ \section{Semistable reduction}
 
 
 
+\section{Semistable reduction in genus zero}
+\label{section-semistable-reduction-genus-zero}
+
+\noindent
+In this section we prove the semistable reduction theorem
+(Theorem \ref{theorem-semistable-reduction})
+for genus zero curves.
+
+\medskip\noindent
+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 Curves, Lemma \ref{curves-lemma-genus-zero}.
+Thus there exists a finite separable extension $K'/K$ of
+degree at most $2$ such that $C(K') \not = \emptyset$, see
+Curves, Lemma \ref{curves-lemma-smooth-plane-curve-point-over-separable}.
+Let $R' \subset K'$ be the integral closure of $R$, see
+discussion in More on Algebra, Remark
+\ref{more-algebra-remark-finite-separable-extension}.
+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.
+
+\medskip\noindent
+In this paragraph $R$ is a discrete valuation ring with fraction field $K$,
+$C$ is a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_C) = K$,
+and $C$ has a $K$-rational point. In this case $C \cong \mathbf{P}^1_K$
+by Curves, Proposition \ref{curves-proposition-projective-line}.
+Thus we can use $\mathbf{P}^1_R$ as a model and we see that
+$C$ has both good and semistable reduction.
+
+\begin{example}
+\label{example-extension-necessary-genus-zero}
+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 \ref{example-nonunique-in-genus-zero}
+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 \ref{lemma-nonuniqueness}).
+It follows that an extension is needed even in genus $0$.
+\end{example}
+
+
+
+
+
+
+
+
+
+\section{Semistable reduction for curves}
+\label{section-semistable-reduction-theorem}
+
+\noindent
+In this section we finish the proof of the theorem.
+
+\begin{theorem}
+\label{theorem-semistable-reduction}
+Let $R$ be a discrete valuation ring with fraction field $K$.
+Let $C$ be a smooth projective curve over $K$ with $H^0(C, \mathcal{O}_C) = K$.
+Then there exists an extension of discrete valuation rings
+$R \subset R'$ which induces a finite separable extension of
+fraction fields $K \subset K'$ such that $C_{K'}$ has
+semistable reduction.
+\end{theorem}
+
+\begin{proof}
+For the case of genus zero, see
+Section \ref{section-semistable-reduction-genus-zero}.
+For higher genus: coming soon.
+\end{proof}
+
+
+
+
+