The Stacks project

Proof. Since a stable family of curves is also prestable, it is immediate that (2) implies (1). Conversely, given a prestable family of curves over $R$ with generic fibre $C$, we can contract it to a stable family of curves by Lemma 109.23.4. Since the generic fibre already is stable, it does not get changed by this procedure and the proof is complete. $\square$

Proof. Let $Y$ be the minimal model for $C$. Recall that $Y$ exists, is unique, and is at-worst-nodal of relative dimension $1$ over $R$, see Semistable Reduction, Proposition 55.8.6 and Lemmas 55.10.1 and 55.14.5 (applies because we have $X$). There is a contraction morphism

such that $Z$ is a stable family of curves of genus $g$ over $R$ (Lemma 109.23.4). We claim there is a unique isomorphism of models $X \to Z$. By symmetry the same is true for $X'$ and this will finish the proof.

By Semistable Reduction, Lemma 55.14.3 there exists a sequence

such that $X_{i + 1} \to X_ i$ is the blowing up of a closed point $x_ i$ where $X_ i$ is singular, $X_ i \to \mathop{\mathrm{Spec}}(R)$ is at-worst-nodal of relative dimension $1$, and $X_ m$ is regular. By Semistable Reduction, Lemma 55.8.5 there is a sequence

of proper regular models of $C$, such that each morphism is a contraction of an exceptional curve of the first kind¹. By Semistable Reduction, Lemma 55.14.4 each $Y_ i$ is at-worst-nodal of relative dimension $1$ over $R$. To prove the claim it suffices to show that there is an isomorphism $X \to Z$ compatible with the morphisms $X_ m \to X$ and $X_ m = Y_ n \to Y \to Z$. Let $s \in \mathop{\mathrm{Spec}}(R)$ be the closed point. By either Lemma 109.23.2 or Lemma 109.23.4 we reduce to proving that the morphisms $X_{m, s} \to X_ s$ and $X_{m, s} \to Z_ s$ are both equal to the canonical morphism of Algebraic Curves, Lemma 53.24.2.

For a morphism $c : U \to V$ of schemes over $\kappa (s)$ we say $c$ has property (*) if $\dim (U_ v) \leq 1$ for $v \in V$, $\mathcal{O}_ V = c_*\mathcal{O}_ U$, and $R^1c_*\mathcal{O}_ U = 0$. This property is stable under composition. Since both $X_ s$ and $Z_ s$ are stable genus $g$ curves over $\kappa (s)$, it suffices to show that each of the morphisms $Y_ s \to Z_ s$, $X_{i + 1, s} \to X_{i, s}$, and $Y_{i + 1, s} \to Y_{i, s}$, satisfy property (*), see Algebraic Curves, Lemma 53.24.2.

Property (*) holds for $Y_ s \to Z_ s$ by construction.

The morphisms $c : X_{i + 1, s} \to X_{i, s}$ are constructed and studied in the proof of Semistable Reduction, Lemma 55.14.3. It suffices to check (*) étale locally on $X_{i, s}$. Hence it suffices to check (*) for the base change of the morphism “$X_1 \to X_0$” in Semistable Reduction, Example 55.14.1 to $R/\pi R$. We leave the explicit calculation to the reader.

The morphism $c : Y_{i + 1, s} \to Y_{i, s}$ is the restriction of the blow down of an exceptional curve $E \subset Y_{i + 1}$ of the first kind, i.e., $b : Y_{i + 1} \to Y_ i$ is a contraction of $E$, i.e., $b$ is a blowing up of a regular point on the surface $Y_ i$ (Resolution of Surfaces, Section 54.16). Then $\mathcal{O}_{Y_ i} = b_*\mathcal{O}_{Y_{i + 1}}$ and $R^1b_*\mathcal{O}_{Y_{i + 1}} = 0$, see for example Resolution of Surfaces, Lemma 54.3.4. We conclude that $\mathcal{O}_{Y_{i, s}} = c_*\mathcal{O}_{Y_{i + 1, s}}$ and $R^1c_*\mathcal{O}_{Y_{i + 1, s}} = 0$ by More on Morphisms, Lemmas 37.72.1, 37.72.2, and 37.72.4 (only gives surjectivity of $\mathcal{O}_{Y_{i, s}} \to c_*\mathcal{O}_{Y_{i + 1, s}}$ but injectivity follows easily from the fact that $Y_{i, s}$ is reduced and $c$ changes things only over one closed point). This finishes the proof. $\square$

Proof. Part (1) is an immediate consequence of Lemma 109.24.1 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 109.24.1 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 70.18.4 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 70.18.1 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. $\square$