Lemma 83.5.15. In the situation of Definition 83.5.13. The following are equivalent:
The morphism $j$ is a set-theoretic equivalence relation.
The morphism $j$ is universally injective and $j(|R|) \subset |U \times _ B U|$ contains the image of $|j'|$ for any of the morphisms $j'$ as in Equation (83.5.5.1).
For every algebraically closed field $k$ over $B$ of sufficiently large cardinality the map $j : R(k) \to U(k) \times U(k)$ is injective and its image is an equivalence relation.
If $j$ is decent, or locally separated, or quasi-separated these are also equivalent to
For every algebraically closed field $k$ over $B$ the map $j : R(k) \to U(k) \times U(k)$ is injective and its image is an equivalence relation.
Proof.
The implications (1) $\Rightarrow $ (2) and (2) $\Rightarrow $ (3) follow from Lemma 83.5.14 and the definitions. The same lemma shows that (3) implies $j$ is a set-theoretic pre-equivalence relation. But of course condition (3) also implies that $j$ is universally injective, see Morphisms of Spaces, Lemma 67.19.2, so that $j$ is indeed a set-theoretic equivalence relation. At this point we know that (1), (2), (3) are all equivalent.
Condition (4) implies (3) without any further hypotheses on $j$. Assume $j$ is decent, or locally separated, or quasi-separated and the equivalent conditions (1), (2), (3) hold. By More on Morphisms of Spaces, Lemma 76.3.4 we see that $j$ is radicial. Let $k$ be any algebraically closed field over $B$. Let $\overline{u}, \overline{u}' \in U(k)$ with $\overline{u} \sim _ R \overline{u}'$. We see that $R \times _{U \times _ B U, (\overline{u}, \overline{u}')} \mathop{\mathrm{Spec}}(k)$ is nonempty. Hence, as $j$ is radicial, its reduction is the spectrum of a field purely inseparable over $k$. As $k = \overline{k}$ we see that it is the spectrum of $k$. Whence a point $\overline{r} \in R(k)$ with $t(\overline{r}) = \overline{u}$ and $s(\overline{r}) = \overline{u}'$ as desired.
$\square$
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