## 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. $\square$

[1] 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.
[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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