The following lemma will help us understand families of semistable curves.
Lemma 109.21.1. Let $f : X \to S$ be a prestable family of curves of genus $g \geq 1$. Let $s \in S$ be a point of the base scheme. Let $m \geq 2$. The following are equivalent
$X_ s$ does not have a rational tail (Algebraic Curves, Example 53.22.1), and
$f^*f_*\omega _{X/S}^{\otimes m} \to \omega _{X/S}^{\otimes m}$, is surjective over $f^{-1}(U)$ for some $s \in U \subset S$ open.
Proof. Assume (2). Using the material in Section 109.19 we conclude that $\omega _{X_ s}^{\otimes m}$ is globally generated. However, if $C \subset X_ s$ is a rational tail, then $\deg (\omega _{X_ s}|_ C) < 0$ by Algebraic Curves, Lemma 53.22.2 hence $H^0(C, \omega _{X_ s}|_ C) = 0$ by Varieties, Lemma 33.44.12 which contradicts the fact that it is globally generated. This proves (1).
Assume (1). First assume that $g \geq 2$. Assumption (1) implies $\omega _{X_ s}^{\otimes m}$ is globally generated, see Algebraic Curves, Lemma 53.22.6. Moreover, we have
by duality, see Algebraic Curves, Lemma 53.4.2. Since $\omega _{X_ s}^{\otimes m}$ is globally generated we find that the restriction to each irreducible component has nonegative degree. Hence the restriction of $\omega _{X_ s}^{\otimes 1 - m}$ to each irreducible component has nonpositive degree. Since $\deg (\omega _{X_ s}^{\otimes 1 - m}) = (1 - m)(2g - 2) < 0$ by Riemann-Roch (Algebraic Curves, Lemma 53.5.2) we conclude that the $H^0$ is zero by Varieties, Lemma 33.44.13. By cohomology and base change we conclude that
is a perfect complex whose formation commutes with arbitrary base change (Derived Categories of Spaces, Lemma 75.25.4). The vanishing proved above tells us that $E \otimes ^\mathbf {L} \kappa (s)$ is equal to $H^0(X_ s, \omega _{X_ s}^{\otimes m})$ placed in degree $0$. After shrinking $S$ we find $E = f_*\omega _{X/S}^{\otimes m}$ is a locally free $\mathcal{O}_ S$-module placed in degree $0$ (and its formation commutes with arbitrary base change as we've already said), see Derived Categories of Spaces, Lemma 75.26.5. The map $f^*f_*\omega _{X/S}^{\otimes m} \to \omega _{X/S}^{\otimes m}$ is surjective after restricting to $X_ s$. Thus it is surjective in an open neighbourhood of $X_ s$. Since $f$ is proper, this open neighbourhood contains $f^{-1}(U)$ for some open neighbourhood $U$ of $s$ in $S$.
Assume (1) and $g = 1$. By Algebraic Curves, Lemma 53.22.6 the assumption (1) means that $\omega _{X_ s}$ is isomorphic to $\mathcal{O}_{X_ s}$. If we can show that after shrinking $S$ the invertible sheaf $\omega _{X/S}$ because trivial, then we are done. We may assume $S$ is affine. After shrinking $S$ further, we can write
sitting in degrees $0$ and $1$ compatibly with further base change, see Lemma 109.9.3. By duality this means that
sitting in degrees $0$ and $1$1. In particular we obtain an isomorphism $\mathcal{O}_ S \to f_*\omega _{X/S}$ which is compatible with base change since formation of $Rf_*\omega _{X/S}$ is compatible with base change (see reference given above). By adjointness, we get a global section $\sigma \in \Gamma (X, \omega _{X/S})$. The restriction of this section to the fibre $X_ s$ is nonzero (a basis element in fact) and as $\omega _{X_ s}$ is trivial on the fibres, this section is nonwhere zero on $X_ s$. Thus it nowhere zero in an open neighbourhood of $X_ s$. Since $f$ is proper, this open neighbourhood contains $f^{-1}(U)$ for some open neighbourhood $U$ of $s$ in $S$. $\square$
Motivated by Lemma 109.21.1 we make the following definition.
Definition 109.21.2. Let $f : X \to S$ be a family of curves. We say $f$ is a semistable family of curves if
$X \to S$ is a prestable family of curves, and
$X_ s$ has genus $\geq 1$ and does not have a rational tail for all $s \in S$.
In particular, a prestable family of curves of genus $0$ is never semistable. Let $X$ be a proper scheme over a field $k$ with $\dim (X) \leq 1$. Then $X \to \mathop{\mathrm{Spec}}(k)$ is a family of curves and hence we can ask whether or not it is semistable. Unwinding the definitions we see the following are equivalent
$X$ is semistable,
$X$ is prestable, has genus $\geq 1$, and does not have a rational tail,
$X_{\overline{k}}$ is connected, is smooth over $\overline{k}$ apart from a finite number of nodes, has genus $\geq 1$, and has no irreducible component isomorphic to $\mathbf{P}^1_{\overline{k}}$ which meets the rest of $X_{\overline{k}}$ in only one point.
To see the equivalence of (2) and (3) use that $X$ has no rational tails if and only if $X_{\overline{k}}$ has no rational tails by Algebraic Curves, Lemma 53.22.6. This shows that our definition agrees with most definitions one finds in the literature.
Lemma 109.21.3. There exist an open substack $\mathcal{C}\! \mathit{urves}^{semistable} \subset \mathcal{C}\! \mathit{urves}$ such that
given a family of curves $f : X \to S$ the following are equivalent
the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}$,
$X \to S$ is a semistable family of curves,
given $X$ a scheme proper over a field $k$ with $\dim (X) \leq 1$ the following are equivalent
the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}$,
the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, the genus of $X$ is $\geq 1$, and $X$ has no rational tails,
the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, and $\omega _{X_ s}^{\otimes m}$ is globally generated for $m \geq 2$.
Proof. The equivalence of (2)(b) and (2)(c) is Algebraic Curves, Lemma 53.22.6. In the rest of the proof we will work with (2)(b) in accordance with Definition 109.21.2.
By the discussion in Section 109.6 it suffices to look at families $f : X \to S$ of prestable curves. By Lemma 109.21.1 we obtain the desired openness of the locus in question. Formation of this open commutes with arbitrary base change, because the (non)existence of rational tails is insensitive to ground field extensions by Algebraic Curves, Lemma 53.22.6. $\square$
Lemma 109.21.4. There is a decomposition into open and closed substacks where each $\mathcal{C}\! \mathit{urves}^{semistable}_ g$ is characterized as follows:
given a family of curves $f : X \to S$ the following are equivalent
the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}_ g$,
$X \to S$ is a semistable family of curves and $R^1f_*\mathcal{O}_ X$ is a locally free $\mathcal{O}_ S$-module of rank $g$,
given $X$ a scheme proper over a field $k$ with $\dim (X) \leq 1$ the following are equivalent
the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{semistable}_ g$,
the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, the genus of $X$ is $g$, and $X$ has no rational tail,
the singularities of $X$ are at-worst-nodal, $\dim (X) = 1$, $k = H^0(X, \mathcal{O}_ X)$, the genus of $X$ is $g$, and $\omega _{X_ s}^{\otimes m}$ is globally generated for $m \geq 2$.
\[ \mathcal{C}\! \mathit{urves}^{semistable} = \coprod \nolimits _{g \geq 1} \mathcal{C}\! \mathit{urves}^{semistable}_ g \]
Lemma 109.21.5. The morphisms $\mathcal{C}\! \mathit{urves}^{semistable} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ and $\mathcal{C}\! \mathit{urves}^{semistable}_ g \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ are smooth.
Proof. Since $\mathcal{C}\! \mathit{urves}^{semistable}$ is an open substack of $\mathcal{C}\! \mathit{urves}^{nodal}$ this follows from Lemma 109.18.2. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)