96.5 The Rim-Schlessinger condition
The motivation for the following definition comes from Lemma 96.4.1 and Formal Deformation Theory, Definition 88.16.1 and Lemma 88.16.4.
Definition 96.5.1. 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
$X$, $X'$, $Y$, $Y'$ are spectra of local Artinian rings,
$X$, $X'$, $Y$, $Y'$ are of finite type over $S$, and
$X \to X'$ (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 96.5.2. 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 96.4.1.
$\square$
Lemma 96.5.3. 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 96.5.1. 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
96.5.3.1
\begin{equation} \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'}). \end{equation}
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 (96.5.3.1) 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 (96.5.3.1) mapping to the given object of the right hand side. This proves that (96.5.3.1) is essentially surjective. We omit the proof that it is fully faithful.
$\square$
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