Lemma 107.8.2. There exist an open substack $\mathcal{C}\! \mathit{urves}^{CM, 1} \subset \mathcal{C}\! \mathit{urves}$ 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 $\mathcal{C}\! \mathit{urves}^{CM, 1}$,

2. the morphism $X \to S$ is Cohen-Macaulay and has relative dimension $1$ (Morphisms of Spaces, Definition 65.33.2),

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 $\mathcal{C}\! \mathit{urves}^{CM, 1}$,

2. $X$ is Cohen-Macaulay and $X$ is equidimensional of dimension $1$.

Proof. By Lemma 107.8.1 it is clear that we have $\mathcal{C}\! \mathit{urves}^{CM, 1} \subset \mathcal{C}\! \mathit{urves}^{CM}$ if it exists. Let $f : X \to S$ be a family of curves such that $f$ is a Cohen-Macaulay morphism. By More on Morphisms of Spaces, Lemma 74.26.8 we have a decomposition

$X = X_0 \amalg X_1$

by open and closed subspaces such that $X_0 \to S$ has relative dimension $0$ and $X_1 \to S$ has relative dimension $1$. Since $f$ is proper the subset

$S' = S \setminus f(|X_0|)$

of $S$ is open and $X \times _ S S' \to S'$ is Cohen-Macaulay and has relative dimension $1$. Moreover, formation of $S'$ commutes with arbitrary base change because this is true for the decomposition above (as relative dimension behaves well with respect to base change, see Morphisms of Spaces, Lemma 65.34.3). Thus we get the open substack with the desired properties by the method discussed in Section 107.6. $\square$

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