Lemma 107.16.1. There exist an open substacks

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

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 $\mathcal{C}\! \mathit{urves}^{smooth}$, resp. $\mathcal{C}\! \mathit{urves}^{smooth, 1}$,

2. $f$ is smooth, resp. smooth of relative dimension $1$,

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 $\mathcal{C}\! \mathit{urves}^{smooth}$, resp. $\mathcal{C}\! \mathit{urves}^{smooth, 1}$,

2. $X$ is smooth over $k$, resp. $X$ is smooth over $k$ and $X$ is equidimensional of dimension $1$.

Proof. To prove the statements regarding $\mathcal{C}\! \mathit{urves}^{smooth}$ it suffices to show that given a family of curves $f : X \to S$, there is an open subscheme $S' \subset S$ such that $S' \times _ S X \to S'$ is smooth and such that the formation of this open commutes with base change. We know that there is a maximal open $U \subset X$ such that $U \to S$ is smooth and that formation of $U$ commutes with arbitrary base change, see Morphisms of Spaces, Lemma 65.37.9. If $T = |X| \setminus |U|$ then $f(T)$ is closed in $S$ as $f$ is proper. Setting $S' = S \setminus f(T)$ we obtain the desired open.

Let $f : X \to S$ be a family of curves with $f$ smooth. Then the fibres $X_ s$ are smooth over $\kappa (s)$ and hence Cohen-Macaulay (for example you can see this using Algebra, Lemmas 10.136.5 and 10.134.3). Thus we see that we may set

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

and the desired equivalences follow from what we've already shown for $\mathcal{C}\! \mathit{urves}^{smooth}$ and Lemma 107.8.2. $\square$

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