## 108.16 Smooth curves

In this section we study open substacks of $\mathcal{C}\! \mathit{urves}$ parametrizing smooth “curves”.

Lemma 108.16.1. 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 66.37.9. 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 10.137.5 and 10.135.3). 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 108.8.2. $\square$

Lemma 108.16.2. The morphism $\mathcal{C}\! \mathit{urves}^{smooth} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is smooth.

Proof. Follows immediately from the observation that $\mathcal{C}\! \mathit{urves}^{smooth} \subset \mathcal{C}\! \mathit{urves}^{lci+}$ and Lemma 108.15.2. $\square$

Lemma 108.16.3. There exist an open substack $\mathcal{C}\! \mathit{urves}^{smooth, h0} \subset \mathcal{C}\! \mathit{urves}$ 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 $\mathcal{C}\! \mathit{urves}^{smooth}$,

2. $f_*\mathcal{O}_ X = \mathcal{O}_ S$, this holds after any base change, and $f$ is 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 $\mathcal{C}\! \mathit{urves}^{smooth, h0}$,

2. $X$ is smooth, $\dim (X) = 1$, and $k = H^0(X, \mathcal{O}_ X)$,

3. $X$ is smooth, $\dim (X) = 1$, and $X$ is geometrically connected,

4. $X$ is smooth, $\dim (X) = 1$, and $X$ is geometrically integral, and

5. $X_{\overline{k}}$ is a smooth curve.

Proof. If we set

$\mathcal{C}\! \mathit{urves}^{smooth, h0} = \mathcal{C}\! \mathit{urves}^{smooth} \cap \mathcal{C}\! \mathit{urves}^{h0, 1}$

then we see that (1) holds by Lemmas 108.9.1 and 108.16.1. In fact, this also gives the equivalence of (2)(a) and (2)(b). To finish the proof we have to show that (2)(b) is equivalent to each of (2)(c), (2)(d), and (2)(e).

A smooth scheme over a field is geometrically normal (Varieties, Lemma 33.25.4), smoothness is preserved under base change (Morphisms, Lemma 29.34.5), and being smooth is fpqc local on the target (Descent, Lemma 35.23.27). Keeping this in mind, the equivalence of (2)(b), (2)(c), 2(d), and (2)(e) follows from Varieties, Lemma 33.10.7. $\square$

Definition 108.16.4. We denote $\mathcal{M}$ and we name it the moduli stack of smooth proper curves the algebraic stack $\mathcal{C}\! \mathit{urves}^{smooth, h0}$ parametrizing families of curves introduced in Lemma 108.16.3. For $g \geq 0$ we denote $\mathcal{M}_ g$ and we name it the moduli stack of smooth proper curves of genus $g$ the algebraic stack introduced in Lemma 108.16.5.

Here is the obligatory lemma.

Lemma 108.16.5. There is a decomposition into open and closed substacks

$\mathcal{M} = \coprod \nolimits _{g \geq 0} \mathcal{M}_ g$

where each $\mathcal{M}_ g$ is characterized as follows:

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 $\mathcal{M}_ g$,

2. $X \to S$ is smooth, $f_*\mathcal{O}_ X = \mathcal{O}_ S$, this holds after any base change, and $R^1f_*\mathcal{O}_ X$ is a locally free $\mathcal{O}_ S$-module of rank $g$,

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 $\mathcal{M}_ g$,

2. $X$ is smooth, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, and $X$ has genus $g$,

3. $X$ is smooth, $\dim (X) = 1$, $X$ is geometrically connected, and $X$ has genus $g$,

4. $X$ is smooth, $\dim (X) = 1$, $X$ is geometrically integral, and $X$ has genus $g$, and

5. $X_{\overline{k}}$ is a smooth curve of genus $g$.

Lemma 108.16.6. The morphisms $\mathcal{M} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ and $\mathcal{M}_ g \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ are smooth.

Proof. Since $\mathcal{M}$ is an open substack of $\mathcal{C}\! \mathit{urves}^{lci+}$ this follows from Lemma 108.15.2. $\square$

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