98.18 Strong Rim-Schlessinger
In the rest of this chapter the following strictly stronger version of the Rim-Schlessinger conditions will play an important role.
Definition 98.18.1. Let $S$ be a scheme. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. We say $\mathcal{X}$ satisfies condition (RS*) if given a fibre product diagram
\[ \xymatrix{ B' \ar[r] & B \\ A' = A \times _ B B' \ar[u] \ar[r] & A \ar[u] } \]
of $S$-algebras, with $B' \to B$ surjective with square zero kernel, the functor of fibre categories
\[ \mathcal{X}_{\mathop{\mathrm{Spec}}(A')} \longrightarrow \mathcal{X}_{\mathop{\mathrm{Spec}}(A)} \times _{\mathcal{X}_{\mathop{\mathrm{Spec}}(B)}} \mathcal{X}_{\mathop{\mathrm{Spec}}(B')} \]
is an equivalence of categories.
We make some observations: with $A \to B \leftarrow B'$ as in Definition 98.18.1
we have $\mathop{\mathrm{Spec}}(A') = \mathop{\mathrm{Spec}}(A) \amalg _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(B')$ in the category of schemes, see More on Morphisms, Lemma 37.14.3, and
if $\mathcal{X}$ is an algebraic stack, then $\mathcal{X}$ satisfies (RS*) by Lemma 98.18.2.
If $S$ is locally Noetherian, then
if $A$, $B$, $B'$ are of finite type over $S$ and $B$ is finite over $A$, then $A'$ is of finite type over $S$1, and
if $\mathcal{X}$ satisfies (RS*), then $\mathcal{X}$ satisfies (RS) because (RS) covers exactly those cases of (RS*) where $A$, $B$, $B'$ are Artinian local.
Lemma 98.18.2. Let $\mathcal{X}$ be an algebraic stack over a base $S$. Then $\mathcal{X}$ satisfies (RS*).
Proof.
This is implied by Lemma 98.4.1, see remarks following Definition 98.18.1.
$\square$
Lemma 98.18.3. 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.
The proof is exactly the same as the proof of Lemma 98.5.3.
$\square$
Comments (2)
Comment #5509 by Martin Olsson on
Comment #5705 by Johan on