Lemma 109.21.3 (0E70)—The Stacks project

Lemma 109.21.3. There exist an open substack $\mathcal{C}\! \mathit{urves}^{semistable} \subset \mathcal{C}\! \mathit{urves}$ such that

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 $\mathcal{C}\! \mathit{urves}^{semistable}$,
2. $X \to S$ is a semistable family of curves,
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 $\mathcal{C}\! \mathit{urves}^{semistable}$,
2. the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, the genus of $X$ is $\geq 1$, and $X$ has no rational tails,
3. the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, and $\omega _{X_ s}^{\otimes m}$ is globally generated for $m \geq 2$.

Proof. The equivalence of (2)(b) and (2)(c) is Algebraic Curves, Lemma 53.22.6. In the rest of the proof we will work with (2)(b) in accordance with Definition 109.21.2.

By the discussion in Section 109.6 it suffices to look at families $f : X \to S$ of prestable curves. By Lemma 109.21.1 we obtain the desired openness of the locus in question. Formation of this open commutes with arbitrary base change, because the (non)existence of rational tails is insensitive to ground field extensions by Algebraic Curves, Lemma 53.22.6. $\square$

The Stacks project

Comments (0)

Post a comment