**Proof.**
Assume (1), i.e., let $\varphi : U \to X$ be an étale morphism from a scheme towards $X$, and let $u, u'$ be points of $U$ lying over $x$ such that the fibre of $|R| \to |U| \times _{|X|} |U|$ over $(u, u')$ is a finite set. In this proof we think of a point $u = \mathop{\mathrm{Spec}}(\kappa (u))$ as a scheme. Note that $u \to U$, $u' \to U$ are monomorphisms (see Schemes, Lemma 26.23.7), hence $u \times _ X u' \to R = U \times _ X U$ is a monomorphism. In this language the assumption really means that $u \times _ X u'$ is a scheme whose underlying topological space has finitely many points. Let $\psi : W \to X$ be an étale morphism from a scheme towards $X$. Let $w, w' \in W$ be points of $W$ mapping to $x$. We have to show that $w \times _ X w'$ is a scheme whose underlying topological space has finitely many points. Consider the fibre product diagram

\[ \xymatrix{ W \times _ X U \ar[r]_ p \ar[d]_ q & U \ar[d]^\varphi \\ W \ar[r]^\psi & X } \]

As $x$ is the image of $u$ and $u'$ we may pick points $\tilde w, \tilde w'$ in $W \times _ X U$ with $q(\tilde w) = w$, $q(\tilde w') = w'$, $u = p(\tilde w)$ and $u' = p(\tilde w')$, see Properties of Spaces, Lemma 66.4.3. As $p$, $q$ are étale the field extensions $\kappa (w) \subset \kappa (\tilde w) \supset \kappa (u)$ and $\kappa (w') \subset \kappa (\tilde w') \supset \kappa (u')$ are finite separable, see Remark 68.4.1. Then we get a commutative diagram

\[ \xymatrix{ w \times _ X w' \ar[d] & \tilde w \times _ X \tilde w' \ar[l] \ar[d] \ar[r] & u \times _ X u' \ar[d] \\ w \times _ X w' & \tilde w \times _ S \tilde w' \ar[l] \ar[r] & u \times _ S u' } \]

where the squares are fibre product squares. The lower horizontal morphisms are étale and quasi-compact, as any scheme of the form $\mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k')$ is affine, and by our observations about the field extensions above. Thus we see that the top horizontal arrows are étale and quasi-compact and hence have finite fibres. We have seen above that $|u \times _ X u'|$ is finite, so we conclude that $|w \times _ X w'|$ is finite. In other words, (2) holds.

Assume (2). Let $U \to X$ be an étale morphism from a scheme $U$ such that $x$ is in the image of $|U| \to |X|$. Let $u \in U$ be a point mapping to $x$. Then we have seen in the previous paragraph that $u = \mathop{\mathrm{Spec}}(\kappa (u)) \to X$ has the property that $u \times _ X u$ has a finite underlying topological space. On the other hand, the projection maps $u \times _ X u \to u$ are the composition

\[ u \times _ X u \longrightarrow u \times _ X U \longrightarrow u \times _ X X = u, \]

i.e., the composition of a monomorphism (the base change of the monomorphism $u \to U$) by an étale morphism (the base change of the étale morphism $U \to X$). Hence $u \times _ X U$ is a disjoint union of spectra of fields finite separable over $\kappa (u)$ (see Remark 68.4.1). Since $u \times _ X u$ is finite the image of it in $u \times _ X U$ is a finite disjoint union of spectra of fields finite separable over $\kappa (u)$. By Schemes, Lemma 26.23.11 we conclude that $u \times _ X u$ is a finite disjoint union of spectra of fields finite separable over $\kappa (u)$. In other words, we see that $u \times _ X u \to u$ is quasi-compact and étale. This means that (3) holds.

Let us prove that (3) implies (4). Let $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism from the spectrum of a field into $X$, in the equivalence class of $x$ such that the two projections $t, s : R = \mathop{\mathrm{Spec}}(k) \times _ X \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ are quasi-compact and étale. This means in particular that $R$ is an étale equivalence relation on $\mathop{\mathrm{Spec}}(k)$. By Spaces, Theorem 65.10.5 we know that the quotient sheaf $X' = \mathop{\mathrm{Spec}}(k)/R$ is an algebraic space. By Groupoids, Lemma 39.20.6 the map $X' \to X$ is a monomorphism. Since $s, t$ are quasi-compact, we see that $R$ is quasi-compact and hence Properties of Spaces, Lemma 66.15.3 applies to $X'$, and we see that $X' = \mathop{\mathrm{Spec}}(k')$ for some field $k'$. Hence we get a factorization

\[ \mathop{\mathrm{Spec}}(k) \longrightarrow \mathop{\mathrm{Spec}}(k') \longrightarrow X \]

which shows that $\mathop{\mathrm{Spec}}(k') \to X$ is a monomorphism mapping to $x \in |X|$. In other words (4) holds.

Finally, we prove that (4) implies (1). Let $\mathop{\mathrm{Spec}}(k) \to X$ be a monomorphism with $k$ a field in the equivalence class of $x$. Let $U \to X$ be a surjective étale morphism from a scheme $U$ to $X$. Let $u \in U$ be a point over $x$. Since $\mathop{\mathrm{Spec}}(k) \times _ X u$ is nonempty, and since $\mathop{\mathrm{Spec}}(k) \times _ X u \to u$ is a monomorphism we conclude that $\mathop{\mathrm{Spec}}(k) \times _ X u = u$ (see Schemes, Lemma 26.23.11). Hence $u \to U \to X$ factors through $\mathop{\mathrm{Spec}}(k) \to X$, here is a picture

\[ \xymatrix{ u \ar[r] \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & X } \]

Since the right vertical arrow is étale this implies that $\kappa (u)/k$ is a finite separable extension. Hence we conclude that

\[ u \times _ X u = u \times _{\mathop{\mathrm{Spec}}(k)} u \]

is a finite scheme, and we win by the discussion of the meaning of property (1) in the first paragraph of this proof.
$\square$

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