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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