Lemma 107.11.1. There exist an open substack $\mathcal{C}\! \mathit{urves}^{grc, 1} \subset \mathcal{C}\! \mathit{urves}$ 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 $\mathcal{C}\! \mathit{urves}^{grc, 1}$,

2. the geometric fibres of the morphism $X \to S$ are reduced, connected, and have dimension $1$,

2. given a scheme $X$ 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 $\mathcal{C}\! \mathit{urves}^{grc, 1}$,

2. $X$ is geometrically reduced, geometrically connected, and has dimension $1$.

Proof. By Lemmas 107.10.1, 107.10.2, 107.8.1, and 107.8.2 it is clear that we have

$\mathcal{C}\! \mathit{urves}^{grc, 1} \subset \mathcal{C}\! \mathit{urves}^{geomred} \cap \mathcal{C}\! \mathit{urves}^{CM, 1}$

if it exists. Let $f : X \to S$ be a family of curves such that $f$ is Cohen-Macaulay, has geometrically reduced fibres, and has relative dimension $1$. By More on Morphisms of Spaces, Lemma 74.36.9 in the Stein factorization

$X \to T \to S$

the morphism $T \to S$ is étale. This implies that there is an open and closed subscheme $S' \subset S$ such that $X \times _ S S' \to S'$ has geometrically connected fibres (in the decomposition of Morphisms, Lemma 29.47.5 for the finite locally free morphism $T \to S$ this corresponds to $S_1$). Formation of this open commutes with arbitrary base change because the number of connected components of geometric fibres is invariant under base change (it is also true that the Stein factorization commutes with base change in our particular case but we don't need this to conclude). Thus we get the open substack with the desired properties by the method discussed in Section 107.6. $\square$

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