The Stacks project

Lemma 109.14.1. There exist an open substack $\mathcal{C}\! \mathit{urves}^{+} \subset \mathcal{C}\! \mathit{urves}$ such that

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

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

    2. the singular locus of $X \to S$ endowed with any/some closed subspace structure is finite over $S$.

  2. given $X$ a proper scheme over a field $k$ of dimension $\leq 1$ the following are equivalent

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

    2. $X \to \mathop{\mathrm{Spec}}(k)$ is smooth except at finitely many points.

Proof. To prove the lemma it suffices to show that given a family of curves $f : X \to S$, there is an open subscheme $S' \subset S$ such that the fibre of $S' \times _ S X \to S'$ have property (2). (Formation of the open will automatically commute with base change.) By definition the locus $T \subset |X|$ of points where $X \to S$ is not smooth is closed. Let $Z \subset X$ be the closed subspace given by the reduced induced algebraic space structure on $T$ (Properties of Spaces, Definition 66.12.5). Now if $s \in S$ is a point where $Z_ s$ is finite, then there is an open neighbourhood $U_ s \subset S$ of $s$ such that $Z \cap f^{-1}(U_ s) \to U_ s$ is finite, see More on Morphisms of Spaces, Lemma 76.35.2. This proves the lemma. $\square$

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