## 97.5 The Rim-Schlessinger condition

The motivation for the following definition comes from Lemma 97.4.1 and Formal Deformation Theory, Definition 89.16.1 and Lemma 89.16.4.

Definition 97.5.1. Let $S$ be a locally Noetherian scheme. Let $\mathcal{Z}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. We say $\mathcal{Z}$ satisfies condition (RS) if for every pushout

$\xymatrix{ X \ar[r] \ar[d] & X' \ar[d] \\ Y \ar[r] & Y' = Y \amalg _ X X' }$

in the category of schemes over $S$ where

1. $X$, $X'$, $Y$, $Y'$ are spectra of local Artinian rings,

2. $X$, $X'$, $Y$, $Y'$ are of finite type over $S$, and

3. $X \to X'$ (and hence $Y \to Y'$) is a closed immersion

the functor of fibre categories

$\mathcal{Z}_{Y'} \longrightarrow \mathcal{Z}_ Y \times _{\mathcal{Z}_ X} \mathcal{Z}_{X'}$

is an equivalence of categories.

If $A$ is an Artinian local ring with residue field $k$, then any morphism $\mathop{\mathrm{Spec}}(A) \to S$ is affine and of finite type if and only if the induced morphism $\mathop{\mathrm{Spec}}(k) \to S$ is of finite type, see Morphisms, Lemmas 29.11.13 and 29.16.2.

Lemma 97.5.2. Let $\mathcal{X}$ be an algebraic stack over a locally Noetherian base $S$. Then $\mathcal{X}$ satisfies (RS).

Proof. Immediate from the definitions and Lemma 97.4.1. $\square$

Lemma 97.5.3. Let $S$ be a scheme. Let $p : \mathcal{X} \to \mathcal{Y}$ and $q : \mathcal{Z} \to \mathcal{Y}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $\mathcal{X}$, $\mathcal{Y}$, and $\mathcal{Z}$ satisfy (RS), then so does $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$.

Proof. This is formal. Let

$\xymatrix{ X \ar[r] \ar[d] & X' \ar[d] \\ Y \ar[r] & Y' = Y \amalg _ X X' }$

be a diagram as in Definition 97.5.1. We have to show that

$(\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_{Y'} \longrightarrow (\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_ Y \times _{(\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_ X} (\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_{X'}$

is an equivalence. Using the definition of the $2$-fibre product this becomes

97.5.3.1
$$\label{artin-equation-RS-fibre-product} \mathcal{X}_{Y'} \times _{\mathcal{Y}_{Y'}} \mathcal{Z}_{Y'} \longrightarrow (\mathcal{X}_ Y \times _{\mathcal{Y}_ Y} \mathcal{Z}_ Y) \times _{(\mathcal{X}_ X \times _{\mathcal{Y}_ X} \mathcal{Z}_ X)} (\mathcal{X}_{X'} \times _{\mathcal{Y}_{X'}} \mathcal{Z}_{X'}).$$

We are given that each of the functors

$\mathcal{X}_{Y'} \to \mathcal{X}_ Y \times _{\mathcal{Y}_ Y} \mathcal{Z}_ Y, \quad \mathcal{Y}_{Y'} \to \mathcal{X}_ X \times _{\mathcal{Y}_ X} \mathcal{Z}_ X, \quad \mathcal{Z}_{Y'} \to \mathcal{X}_{X'} \times _{\mathcal{Y}_{X'}} \mathcal{Z}_{X'}$

are equivalences. An object of the right hand side of (97.5.3.1) is a system

$((x_ Y, z_ Y, \phi _ Y), (x_{X'}, z_{X'}, \phi _{X'}), (\alpha , \beta )).$

Then $(x_ Y, x_{Y'}, \alpha )$ is isomorphic to the image of an object $x_{Y'}$ in $\mathcal{X}_{Y'}$ and $(z_ Y, z_{Y'}, \beta )$ is isomorphic to the image of an object $z_{Y'}$ of $\mathcal{Z}_{Y'}$. The pair of morphisms $(\phi _ Y, \phi _{X'})$ corresponds to a morphism $\psi$ between the images of $x_{Y'}$ and $z_{Y'}$ in $\mathcal{Y}_{Y'}$. Then $(x_{Y'}, z_{Y'}, \psi )$ is an object of the left hand side of (97.5.3.1) mapping to the given object of the right hand side. This proves that (97.5.3.1) is essentially surjective. We omit the proof that it is fully faithful. $\square$

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