**Proof.**
Part (1) is an immediate consequence of Lemma 108.24.1 and Semistable Reduction, Theorem 55.18.1.

Proof of (2). Let $L/K$ be the finite separable extension found in part (3) of Semistable Reduction, Theorem 55.18.1. Let $A \subset L$ be the integral closure of $R$. Recall that $A$ is a Dedekind domain finite over $R$ with finitely many maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ n$, see More on Algebra, Remark 15.111.6. Set $S = \mathop{\mathrm{Spec}}(A)$, $S_ i = \mathop{\mathrm{Spec}}(A_{\mathfrak m_ i})$, $U = \mathop{\mathrm{Spec}}(L)$, and $U_ i = S_ i \setminus \{ \mathfrak m_ i\} $. Observe that $U \cong U_ i$ for $i = 1, \ldots , n$. Set $X = C_ L$ viewed as a scheme over the open subscheme $U$ of $S$. By our choice of $L$ and $A$ and Lemma 108.24.1 we have stable families of curves $X_ i \to S_ i$ and isomorphisms $X \times _ U U_ i \cong X_ i \times _{S_ i} U_ i$. By Limits of Spaces, Lemma 69.18.4 we can find a finitely presented morphism $Y \to S$ whose base change to $S_ i$ is isomorphic to $X_ i$ for $i = 1, \ldots , n$. Alternatively, you can use that $S = \bigcup _{i = 1, \ldots , n} S_ i$ is an open covering of $S$ and $S_ i \cap S_ j = U$ for $i \not= j$ and use $n - 1$ applications of Limits of Spaces, Lemma 69.18.1 to get $Y \to S$ whose base change to $S_ i$ is isomorphic to $X_ i$ for $i = 1, \ldots , n$. Clearly $Y \to S$ is the stable family of curves we were looking for.
$\square$

## Comments (2)

Comment #7207 by David Holmes on

Comment #7324 by Johan on