The Stacks project

Lemma 96.8.1. Let $S$ be a locally Noetherian scheme. Assume

  1. $\mathcal{X}$ is an algebraic stack,

  2. $U$ is a scheme locally of finite type over $S$, and

  3. $(\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ is a smooth surjective morphism.

Then, for any $\mathcal{F} = \mathcal{F}_{\mathcal{X}, k, x_0}$ as in Section 96.3 the tangent space $T\mathcal{F}$ and infinitesimal automorphism space $\text{Inf}(\mathcal{F})$ have finite dimension over $k$

Proof. Let us write $\mathcal{U} = (\mathit{Sch}/U)_{fppf}$. By our definition of algebraic stacks the $1$-morphism $\mathcal{U} \to \mathcal{X}$ is representable by algebraic spaces. Hence in particular the 2-fibre product

\[ \mathcal{U}_{x_0} = (\mathit{Sch}/\mathop{\mathrm{Spec}}(k))_{fppf} \times _\mathcal {X} \mathcal{U} \]

is representable by an algebraic space $U_{x_0}$ over $\mathop{\mathrm{Spec}}(k)$. Then $U_{x_0} \to \mathop{\mathrm{Spec}}(k)$ is smooth and surjective (in particular $U_{x_0}$ is nonempty). By Spaces over Fields, Lemma 70.16.2 we can find a finite extension $l \supset k$ and a point $\mathop{\mathrm{Spec}}(l) \to U_{x_0}$ over $k$. We have

\[ (\mathcal{F}_{\mathcal{X}, k , x_0})_{l/k} = \mathcal{F}_{\mathcal{X}, l, x_{l, 0}} \]

by Lemma 96.7.1 and the fact that $\mathcal{X}$ satisfies (RS). Thus we see that

\[ T\mathcal{F} \otimes _ k l \cong T\mathcal{F}_{\mathcal{X}, l, x_{l, 0}} \quad \text{and}\quad \text{Inf}(\mathcal{F}) \otimes _ k l \cong \text{Inf}(\mathcal{F}_{\mathcal{X}, l, x_{l, 0}}) \]

by Formal Deformation Theory, Lemmas 88.29.3 and 88.29.4 (these are applicable by Lemmas 96.5.2 and 96.6.1 and Remark 96.6.2). Hence it suffices to prove that $T\mathcal{F}_{\mathcal{X}, l, x_{l, 0}}$ and $\text{Inf}(\mathcal{F}_{\mathcal{X}, l, x_{l, 0}})$ have finite dimension over $l$. Note that $x_{l, 0}$ comes from a point $u_0$ of $\mathcal{U}$ over $l$.

We interrupt the flow of the argument to show that the lemma for infinitesimal automorphisms follows from the lemma for tangent spaces. Namely, let $\mathcal{R} = \mathcal{U} \times _\mathcal {X} \mathcal{U}$. Let $r_0$ be the $l$-valued point $(u_0, u_0, \text{id}_{x_0})$ of $\mathcal{R}$. Combining Lemma 96.3.3 and Formal Deformation Theory, Lemma 88.26.2 we see that

\[ \text{Inf}(\mathcal{F}_{\mathcal{X}, l, x_{l, 0}}) \subset T\mathcal{F}_{\mathcal{R}, l, r_0} \]

Note that $\mathcal{R}$ is an algebraic stack, see Algebraic Stacks, Lemma 92.14.2. Also, $\mathcal{R}$ is representable by an algebraic space $R$ smooth over $U$ (via either projection, see Algebraic Stacks, Lemma 92.16.2). Hence, choose an scheme $U'$ and a surjective ├ętale morphism $U' \to R$ we see that $U'$ is smooth over $U$, hence locally of finite type over $S$. As $(\mathit{Sch}/U')_{fppf} \to \mathcal{R}$ is surjective and smooth, we have reduced the question to the case of tangent spaces.

The functor (

\[ \mathcal{F}_{\mathcal{U}, l, u_0} \longrightarrow \mathcal{F}_{\mathcal{X}, l, x_{l, 0}} \]

is smooth by Lemma 96.3.2. The induced map on tangent spaces

\[ T\mathcal{F}_{\mathcal{U}, l, u_0} \longrightarrow T\mathcal{F}_{\mathcal{X}, l, x_{l, 0}} \]

is $l$-linear (by Formal Deformation Theory, Lemma 88.12.4) and surjective (as smooth maps of predeformation categories induce surjective maps on tangent spaces by Formal Deformation Theory, Lemma 88.8.8). Hence it suffices to prove that the tangent space of the deformation space associated to the representable algebraic stack $\mathcal{U}$ at the point $u_0$ is finite dimensional. Let $\mathop{\mathrm{Spec}}(R) \subset U$ be an affine open such that $u_0 : \mathop{\mathrm{Spec}}(l) \to U$ factors through $\mathop{\mathrm{Spec}}(R)$ and such that $\mathop{\mathrm{Spec}}(R) \to S$ factors through $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$. Let $\mathfrak m_ R \subset R$ be the kernel of the $\Lambda $-algebra map $\varphi _0 : R \to l$ corresponding to $u_0$. Note that $R$, being of finite type over the Noetherian ring $\Lambda $, is a Noetherian ring. Hence $\mathfrak m_ R = (f_1, \ldots , f_ n)$ is a finitely generated ideal. We have

\[ T\mathcal{F}_{\mathcal{U}, l, u_0} = \{ \varphi : R \to l[\epsilon ] \mid \varphi \text{ is a } \Lambda \text{-algebra map and } \varphi \bmod \epsilon = \varphi _0\} \]

An element of the right hand side is determined by its values on $f_1, \ldots , f_ n$ hence the dimension is at most $n$ and we win. Some details omitted. $\square$

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