The Stacks project

Proof. Choose a scheme $U$ locally of finite type over $S$, a finite type point $u_0$ of $U$, and an object $x$ of $\mathcal{X}$ over $U$ such that $x$ is versal at $u_0$. After shrinking $U$ we may assume $U$ is affine and $U$ maps into an affine open $\mathop{\mathrm{Spec}}(\Lambda )$ of $S$. Let $E \subset U$ be the set of finite type points $u$ such that $x$ is not versal at $u$. By Lemma 98.19.2 if $u \in E$ then $u_0$ is not a specialization of $u$. If openness of versality does not hold, then $u_0$ is in the closure $\overline{E}$ of $E$. By Properties, Lemma 28.5.13 we may choose a countable subset $E' \subset E$ with the same closure as $E$. By Properties, Lemma 28.5.12 we may assume there are no specializations among the points of $E'$. Observe that $E'$ has to be (countably) infinite as $u_0$ isn't the specialization of any point of $E'$ as pointed out above. Thus we can write $E' = \{ u_1, u_2, u_3, \ldots \} $, there are no specializations among the $u_ i$, and $u_0$ is in the closure of $E'$.

Choose $x \to x_1 \to x_2 \to \ldots $ lying over $U \to U_1 \to U_2 \to \ldots $ as in Lemma 98.20.1. Write $U_ n = \mathop{\mathrm{Spec}}(R_ n)$ and $U = \mathop{\mathrm{Spec}}(R_0)$. Set $R = \mathop{\mathrm{lim}}\nolimits R_ n$. Observe that $R \to R_0$ is surjective with kernel an ideal of square zero. By assumption (4) we get $\xi $ over $\mathop{\mathrm{Spec}}(R)$ whose base change to $R_ n$ is $x_ n$. By assumption (3) we get that $\xi $ comes from an object $\xi '$ over $U' = \mathop{\mathrm{Spec}}(R')$ for some finite type $\Lambda $-subalgebra $R' \subset R$. After increasing $R'$ we may and do assume that $R' \to R_0$ is surjective, so that $U \subset U'$ is a first order thickening. Thus we now have

By assumption (1) there is an algebraic space $Z$ over $S$ representing

see Algebraic Stacks, Lemma 94.10.11. By construction of $2$-fibre products, a $T$-valued point of $Z$ corresponds to a triple $(a, a', \alpha )$ consisting of morphisms $a : T \to U$, $a' : T \to U'$ and a morphism $\alpha : a^*x \to (a')^*\xi '$. We obtain a commutative diagram

The morphism $i : U \to Z$ comes the isomorphism $x \to \xi '|_ U$. Let $z_0 = i(u_0) \in Z$. By Lemma 98.12.6 we see that $Z \to U'$ is smooth at $z_0$. After replacing $U$ by an affine open neighbourhood of $u_0$, replacing $U'$ by the corresponding open, and replacing $Z$ by the intersection of the inverse images of these opens by $p$ and $p'$, we reach the situation where $Z \to U'$ is smooth along $i(U)$. Note that this also involves replacing $u_ n$ by a subsequence, namely by those indices such that $u_ n$ is in the open. Moreover, condition (3) of Lemma 98.20.1 is clearly preserved by shrinking $U$ (all of the schemes $U$, $U_ n$, $U'$ have the same underlying topological space). Since $U \to U'$ is a first order thickening of affine schemes, we can choose a morphism $i' : U' \to Z$ such that $p' \circ i' = \text{id}_{U'}$ and whose restriction to $U$ is $i$ (More on Morphisms of Spaces, Lemma 76.19.6). Pulling back the universal morphism $p^*x \to (p')^*\xi '$ by $i'$ we obtain a morphism

lying over $p \circ i' : U' \to U$ such that the composition

is the identity. Recall that we have $x_1 \to \xi '$ lying over the morphism $U_1 \to U'$. Composing we get a morphism $x_1 \to x$ whose existence contradicts condition (3) of Lemma 98.20.1. This contradiction finishes the proof. $\square$