Lemma 97.8.1. Let $S$ be a locally Noetherian scheme. Assume
$\mathcal{X}$ is an algebraic stack,
$U$ is a scheme locally of finite type over $S$, and
$(\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ is a smooth surjective morphism.
Then, for any $\mathcal{F} = \mathcal{F}_{\mathcal{X}, k, x_0}$ as in Section 97.3 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
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 71.16.2 we can find a finite extension $l/k$ and a point $\mathop{\mathrm{Spec}}(l) \to U_{x_0}$ over $k$. We have
by Lemma 97.7.1 and the fact that $\mathcal{X}$ satisfies (RS). Thus we see that
by Formal Deformation Theory, Lemmas 89.29.3 and 89.29.4 (these are applicable by Lemmas 97.5.2 and 97.6.1 and Remark 97.6.2). 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 97.3.3 and Formal Deformation Theory, Lemma 89.26.2 we see that
Note that $\mathcal{R}$ is an algebraic stack, see Algebraic Stacks, Lemma 93.14.2. Also, $\mathcal{R}$ is representable by an algebraic space $R$ smooth over $U$ (via either projection, see Algebraic Stacks, Lemma 93.16.2). Hence, choose an scheme $U'$ and a surjective étale 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 (97.3.1.1)
is smooth by Lemma 97.3.2. The induced map on tangent spaces
is $l$-linear (by Formal Deformation Theory, Lemma 89.12.4) and surjective (as smooth maps of predeformation categories induce surjective maps on tangent spaces by Formal Deformation Theory, Lemma 89.8.8). 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
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. $\square$
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)
There are also: