**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 72.16.2 we can find a finite extension $l/k$ and a point $\mathop{\mathrm{Spec}}(l) \to U_{x_0}$ over $k$. We have

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

by Lemma 98.7.1 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 90.29.3 and 90.29.4 (these are applicable by Lemmas 98.5.2 and 98.6.1 and Remark 98.6.2). 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 98.3.3 and Formal Deformation Theory, Lemma 90.26.2 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 94.14.2. Also, $\mathcal{R}$ is representable by an algebraic space $R$ smooth over $U$ (via either projection, see Algebraic Stacks, Lemma 94.16.2). 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 (98.3.1.1)

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

is smooth by Lemma 98.3.2. 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 90.12.4) and surjective (as smooth maps of predeformation categories induce surjective maps on tangent spaces by Formal Deformation Theory, Lemma 90.8.8). 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.
$\square$

## Comments (2)

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