The Stacks project

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
\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 ( 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 ( mapping to the given object of the right hand side. This proves that ( is essentially surjective. We omit the proof that it is fully faithful. $\square$

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