## 107.20 Prestable curves

The following definition is equivalent to what appears to be the generally accepted notion of a prestable family of curves.

Definition 107.20.1. Let $f : X \to S$ be a family of curves. We say $f$ is a prestable family of curves if

1. $f$ is at-worst-nodal of relative dimension $1$, and

2. $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and this holds after any base change1.

Let $X$ be a proper scheme over a field $k$ with $\dim (X) \leq 1$. Then $X \to \mathop{\mathrm{Spec}}(k)$ is a family of curves and hence we can ask whether or not it is prestable2 in the sense of the definition. Unwinding the definitions we see the following are equivalent

1. $X$ is prestable,

2. the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, and $k = H^0(X, \mathcal{O}_ X)$,

3. $X_{\overline{k}}$ is connected and it is smooth over $\overline{k}$ apart from a finite number of nodes (Algebraic Curves, Definition 53.16.2).

This shows that our definition agrees with most definitions one finds in the literature.

Lemma 107.20.2. There exist an open substack $\mathcal{C}\! \mathit{urves}^{prestable} \subset \mathcal{C}\! \mathit{urves}$ such that

1. given a family of curves $f : X \to S$ the following are equivalent

1. the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{prestable}$,

2. $X \to S$ is a prestable family of curves,

2. given $X$ a scheme proper over a field $k$ with $\dim (X) \leq 1$ the following are equivalent

1. the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{prestable}$,

2. the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, and $k = H^0(X, \mathcal{O}_ X)$.

Proof. Given a family of curves $X \to S$ we see that it is prestable if and only if the classifying morphism factors both through $\mathcal{C}\! \mathit{urves}^{nodal}$ and $\mathcal{C}\! \mathit{urves}^{h0, 1}$. An alternative is to use $\mathcal{C}\! \mathit{urves}^{grc, 1}$ (since a nodal curve is geometrically reduced hence has $H^0$ equal to the ground field if and only if it is connected). In a formula

$\mathcal{C}\! \mathit{urves}^{prestable} = \mathcal{C}\! \mathit{urves}^{nodal} \cap \mathcal{C}\! \mathit{urves}^{h0, 1} = \mathcal{C}\! \mathit{urves}^{nodal} \cap \mathcal{C}\! \mathit{urves}^{grc, 1}$

Thus the lemma follows from Lemmas 107.9.1 and 107.18.1. $\square$

For each genus $g \geq 0$ we have the algebraic stack classifying the prestable curves of genus $g$. In fact, from now on we will say that $X \to S$ is a prestable family of curves of genus $g$ if and only if the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through the open substack $\mathcal{C}\! \mathit{urves}^{prestable}_ g$ of Lemma 107.20.3.

Lemma 107.20.3. There is a decomposition into open and closed substacks

$\mathcal{C}\! \mathit{urves}^{prestable} = \coprod \nolimits _{g \geq 0} \mathcal{C}\! \mathit{urves}^{prestable}_ g$

where each $\mathcal{C}\! \mathit{urves}^{prestable}_ g$ is characterized as follows:

1. given a family of curves $f : X \to S$ the following are equivalent

1. the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{prestable}_ g$,

2. $X \to S$ is a prestable family of curves and $R^1f_*\mathcal{O}_ X$ is a locally free $\mathcal{O}_ S$-module of rank $g$,

2. given $X$ a scheme proper over a field $k$ with $\dim (X) \leq 1$ the following are equivalent

1. the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{prestable}_ g$,

2. the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, and the genus of $X$ is $g$.

Proof. Since we have seen that $\mathcal{C}\! \mathit{urves}^{prestable}$ is contained in $\mathcal{C}\! \mathit{urves}^{h0, 1}$, this follows from Lemmas 107.20.2 and 107.9.4. $\square$

Lemma 107.20.4. The morphisms $\mathcal{C}\! \mathit{urves}^{prestable} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ and $\mathcal{C}\! \mathit{urves}^{prestable}_ g \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ are smooth.

Proof. Since $\mathcal{C}\! \mathit{urves}^{prestable}$ is an open substack of $\mathcal{C}\! \mathit{urves}^{nodal}$ this follows from Lemma 107.18.2. $\square$

 In fact, it suffices to require $f_*\mathcal{O}_ X = \mathcal{O}_ S$ because the Stein factorization of $f$ is étale in this case, see More on Morphisms of Spaces, Lemma 74.36.9. The condition may also be replaced by asking the geometric fibres to be connected, see Lemma 107.11.2.
 We can't use the term “prestable curve” here because curve implies irreducible. See discussion in Algebraic Curves, Section 53.20.

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