109.7 Curves with finite reduced automorphism groups
Let $X$ be a proper scheme over a field $k$ of dimension $\leq 1$, i.e., an object of $\mathcal{C}\! \mathit{urves}$ over $k$. By Lemma 109.5.1 the automorphism group algebraic space $\mathit{Aut}(X)$ is finite type and separated over $k$. In particular, $\mathit{Aut}(X)$ is a group scheme, see More on Groupoids in Spaces, Lemma 79.10.2. If the characteristic of $k$ is zero, then $\mathit{Aut}(X)$ is reduced and even smooth over $k$ (Groupoids, Lemma 39.8.2). However, in general $\mathit{Aut}(X)$ is not reduced, even if $X$ is geometrically reduced.
Example 109.7.1 (Non-reduced automorphism group). Let $k$ be an algebraically closed field of characteristic $2$. Set $Y = Z = \mathbf{P}^1_ k$. Choose three pairwise distinct $k$-valued points $a, b, c$ in $\mathbf{A}^1_ k$. Thinking of $\mathbf{A}^1_ k \subset \mathbf{P}^1_ k = Y = Z$ as an open subschemes, we get a closed immersion
\[ T = \mathop{\mathrm{Spec}}(k[t]/(t - a)^2) \amalg \mathop{\mathrm{Spec}}(k[t]/(t - b)^2) \amalg \mathop{\mathrm{Spec}}(k[t]/(t - c)^2) \longrightarrow \mathbf{P}^1_ k \]
Let $X$ be the pushout in the diagram
\[ \xymatrix{ T \ar[r] \ar[d] & Y \ar[d] \\ Z \ar[r] & X } \]
Let $U \subset X$ be the affine open part which is the image of $\mathbf{A}^1_ k \amalg \mathbf{A}^1_ k$. Then we have an equalizer diagram
\[ \xymatrix{ \mathcal{O}_ X(U) \ar[r] & k[t] \times k[t] \ar@<1ex>[r] \ar@<-1ex>[r] & k[t]/(t - a)^2 \times k[t]/(t - b)^2 \times k[t]/(t - c)^2 } \]
Over the dual numbers $A = k[\epsilon ]$ we have a nontrivial automorphism of this equalizer diagram sending $t$ to $t + \epsilon $. We leave it to the reader to see that this automorphism extends to an automorphism of $X$ over $A$. On the other hand, the reader easily shows that the automorphism group of $X$ over $k$ is finite. Thus $\mathit{Aut}(X)$ must be non-reduced.
Let $X$ be a proper scheme over a field $k$ of dimension $\leq 1$, i.e., an object of $\mathcal{C}\! \mathit{urves}$ over $k$. If $\mathit{Aut}(X)$ is geometrically reduced, then it need not be the case that it has dimension $0$, even if $X$ is smooth and geometrically connected.
Example 109.7.2 (Smooth positive dimensional automorphism group). Let $k$ be an algebraically closed field. If $X$ is a smooth genus $0$, resp. $1$ curve, then the automorphism group has dimension $3$, resp. $1$. Namely, in the genus $0$ case we have $X \cong \mathbf{P}^1_ k$ by Algebraic Curves, Proposition 53.10.4. Since
\[ \mathit{Aut}(\mathbf{P}^1_ k) = \text{PGL}_{2, k} \]
as functors we see that the dimension is $3$. On the other hand, if the genus of $X$ is $1$, then we see that the map $X = \underline{\mathrm{Hilb}}^1_{X/k} \to \underline{\mathrm{Pic}}^1_{X/k}$ is an isomorphism, see Picard Schemes of Curves, Lemma 44.6.7 and Algebraic Curves, Theorem 53.2.6. Thus $X$ has the structure of an abelian variety (since $\underline{\mathrm{Pic}}^1_{X/k} \cong \underline{\mathrm{Pic}}^0_{X/k}$). In particular the (co)tangent bundle of $X$ are trivial (Groupoids, Lemma 39.6.3). We conclude that $\dim _ k H^0(X, T_ X) = 1$ hence $\dim \mathit{Aut}(X) \leq 1$. On the other hand, the translations (viewing $X$ as a group scheme) provide a $1$-dimensional piece of $\text{Aut}(X)$ and we conlude its dimension is indeed $1$.
It turns out that there is an open substack of $\mathcal{C}\! \mathit{urves}$ parametrizing curves whose automorphism group is geometrically reduced and finite. Here is a precise statement.
Lemma 109.7.3. There exist an open substack $\mathcal{C}\! \mathit{urves}^{DM} \subset \mathcal{C}\! \mathit{urves}$ with the following properties
$\mathcal{C}\! \mathit{urves}^{DM} \subset \mathcal{C}\! \mathit{urves}$ is the maximal open substack which is DM,
given a family of curves $X \to S$ the following are equivalent
the classifying morphism $S \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{DM}$,
the group algebraic space $\mathit{Aut}_ S(X)$ is unramified over $S$,
given $X$ a proper scheme over a field $k$ of dimension $\leq 1$ the following are equivalent
the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}$ factors through $\mathcal{C}\! \mathit{urves}^{DM}$,
$\mathit{Aut}(X)$ is geometrically reduced over $k$ and has dimension $0$,
$\mathit{Aut}(X) \to \mathop{\mathrm{Spec}}(k)$ is unramified.
Proof.
The existence of an open substack with property (1) is Morphisms of Stacks, Lemma 101.22.1. The points of this open substack are characterized by (3)(c) by Morphisms of Stacks, Lemma 101.22.2. The equivalence of (3)(b) and (3)(c) is the statement that an algebraic space $G$ which is locally of finite type, geometrically reduced, and of dimension $0$ over a field $k$, is unramified over $k$. First, $G$ is a scheme by Spaces over Fields, Lemma 72.9.1. Then we can take an affine open in $G$ and observe that it will be proper over $k$ and apply Varieties, Lemma 33.9.3. Minor details omitted.
Part (2) is true because (3) holds. Namely, the morphism $\mathit{Aut}_ S(X) \to S$ is locally of finite type. Thus we can check whether $\mathit{Aut}_ S(X) \to S$ is unramified at all points of $\mathit{Aut}_ S(X)$ by checking on fibres at points of the scheme $S$, see Morphisms of Spaces, Lemma 67.38.10. But after base change to a point of $S$ we fall back into the equivalence of (3)(a) and (3)(c).
$\square$
Lemma 109.7.4. Let $X$ be a proper scheme over a field $k$ of dimension $\leq 1$. Then properties (3)(a), (b), (c) are also equivalent to $\text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X) = 0$.
Proof.
In the discussion above we have seen that $G = \mathit{Aut}(X)$ is a group scheme over $\mathop{\mathrm{Spec}}(k)$ which is finite type and separated; this uses Lemma 109.5.1 and More on Groupoids in Spaces, Lemma 79.10.2. Then $G$ is unramified over $k$ if and only if $\Omega _{G/k} = 0$ (Morphisms, Lemma 29.35.2). By Groupoids, Lemma 39.6.3 the vanishing holds if $T_{G/k, e} = 0$, where $T_{G/k, e}$ is the tangent space to $G$ at the identity element $e \in G(k)$, see Varieties, Definition 33.16.3 and the formula in Varieties, Lemma 33.16.4. Since $\kappa (e) = k$ the tangent space is defined in terms of morphisms $\alpha : \mathop{\mathrm{Spec}}(k[\epsilon ]) \to G = \mathit{Aut}(X)$ whose restriction to $\mathop{\mathrm{Spec}}(k)$ is $e$. It follows that it suffices to show any automorphism
\[ \alpha : X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ]) \longrightarrow X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ]) \]
over $\mathop{\mathrm{Spec}}(k[\epsilon ])$ whose restriction to $\mathop{\mathrm{Spec}}(k)$ is $\text{id}_ X$. Such automorphisms are called infinitesimal automorphisms.
The infinitesimal automorphisms of $X$ correspond $1$-to-$1$ with derivations of $\mathcal{O}_ X$ over $k$. This follows from More on Morphisms, Lemmas 37.9.1 and 37.9.2 (we only need the first one as we don't care about the reverse direction; also, please look at More on Morphisms, Remark 37.9.7 for an elucidation). For a different argument proving this equality we refer the reader to Deformation Problems, Lemma 93.9.3.
$\square$
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