Lemma 96.20.1. Let $S$ be a locally Noetherian scheme. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ having (RS*). Let $x$ be an object of $\mathcal{X}$ over an affine scheme $U$ of finite type over $S$. Let $u_ n \in U$, $n \geq 1$ be finite type points such that (a) there are no specializations $u_ n \leadsto u_ m$ for $n \not= m$, and (b) $x$ is not versal at $u_ n$ for all $n$. Then there exist morphisms

$x \to x_1 \to x_2 \to \ldots \quad \text{in }\mathcal{X}\text{ lying over }\quad U \to U_1 \to U_2 \to \ldots$

over $S$ such that

1. for each $n$ the morphism $U \to U_ n$ is a first order thickening,

2. for each $n$ we have a short exact sequence

$0 \to \kappa (u_ n) \to \mathcal{O}_{U_ n} \to \mathcal{O}_{U_{n - 1}} \to 0$

with $U_0 = U$ for $n = 1$,

3. for each $n$ there does not exist a pair $(W, \alpha )$ consisting of an open neighbourhood $W \subset U_ n$ of $u_ n$ and a morphism $\alpha : x_ n|_ W \to x$ such that the composition

$x|_{U \cap W} \xrightarrow {\text{restriction of }x \to x_ n} x_ n|_ W \xrightarrow {\alpha } x$

is the canonical morphism $x|_{U \cap W} \to x$.

Proof. Since there are no specializations among the points $u_ n$ (and in particular the $u_ n$ are pairwise distinct), for every $n$ we can find an open $U' \subset U$ such that $u_ n \in U'$ and $u_ i \not\in U'$ for $i = 1, \ldots , n - 1$. By Lemma 96.19.1 for each $n \geq 1$ we can find

$x \to y_ n \quad \text{in }\mathcal{X}\text{ lying over}\quad U \to T_ n$

such that

1. the morphism $U \to T_ n$ is a first order thickening,

2. we have a short exact sequence

$0 \to \kappa (u_ n) \to \mathcal{O}_{T_ n} \to \mathcal{O}_ U \to 0$
3. there does not exist a pair $(W, \alpha )$ consisting of an open neighbourhood $W \subset T_ n$ of $u_ n$ and a morphism $\beta : y_ n|_ W \to x$ such that the composition

$x|_{U \cap W} \xrightarrow {\text{restriction of }x \to y_ n} y_ n|_ W \xrightarrow {\beta } x$

is the canonical morphism $x|_{U \cap W} \to x$.

Thus we can define inductively

$U_1 = T_1, \quad U_{n + 1} = U_ n \amalg _ U T_{n + 1}$

Setting $x_1 = y_1$ and using (RS*) we find inductively $x_{n + 1}$ over $U_{n + 1}$ restricting to $x_ n$ over $U_ n$ and $y_{n + 1}$ over $T_{n + 1}$. Property (1) for $U \to U_ n$ follows from the construction of the pushout in More on Morphisms, Lemma 37.14.3. Property (2) for $U_ n$ similarly follows from property (2) for $T_ n$ by the construction of the pushout. After shrinking to an open neighbourhood $U'$ of $u_ n$ as discussed above, property (3) for $(U_ n, x_ n)$ follows from property (3) for $(T_ n, y_ n)$ simply because the corresponding open subschemes of $T_ n$ and $U_ n$ are isomorphic. Some details omitted. $\square$

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