**Proof.**
The equivalence of (1) and (3) follows on applying Lemma 68.4.4 to every étale morphism $U \to X$ with $U$ affine. It is clear that (3) implies (2). Assume $U_ i \to X$ and $R_ i$ are as in (2). We conclude from Lemma 68.4.2 that for any affine scheme $U$ and étale morphism $U \to X$ the fibre of $|U| \to |X|$ over $x$ is finite. Say this fibre is $\{ u_1, \ldots , u_ n\} $. Then, as Lemma 68.4.3 (1) applies to $U_ i \to X$ for some $i$ such that $x$ is in the image of $|U_ i| \to |X|$, we see that the fibre of $|R = U \times _ X U| \to |U| \times _{|X|} |U|$ is finite over $(u_ a, u_ b)$, $a, b \in \{ 1, \ldots , n\} $. Hence the fibre of $|R| \to |X|$ over $x$ is finite. In this way we see that (1) holds. At this point we know that (1), (2), and (3) are equivalent.

If (4) holds, then for any affine scheme $U$ and étale morphism $U \to X$ the scheme $\mathop{\mathrm{Spec}}(k) \times _ X U$ is on the one hand étale over $k$ (hence a disjoint union of spectra of finite separable extensions of $k$ by Remark 68.4.1) and on the other hand quasi-compact over $U$ (hence quasi-compact). Thus we see that (3) holds. Conversely, if $U_ i \to X$ is as in (2) and $\mathop{\mathrm{Spec}}(k) \to X$ is a monomorphism as in (3), then

\[ \coprod \mathop{\mathrm{Spec}}(k) \times _ X U_ i \longrightarrow \coprod U_ i \]

is quasi-compact (because over each $U_ i$ we see that $\mathop{\mathrm{Spec}}(k) \times _ X U_ i$ is a finite disjoint union spectra of fields). Thus $\mathop{\mathrm{Spec}}(k) \to X$ is quasi-compact by Morphisms of Spaces, Lemma 67.8.8.

It is immediate that (4) implies (5). Conversely, let $\mathop{\mathrm{Spec}}(k) \to X$ be a quasi-compact morphism in the equivalence class of $x$. Let $U \to X$ be an étale morphism with $U$ affine. Consider the fibre product

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

Then $F \to U$ is quasi-compact, hence $F$ is quasi-compact. On the other hand, $F \to \mathop{\mathrm{Spec}}(k)$ is étale, hence $F$ is a finite disjoint union of spectra of finite separable extensions of $k$ (Remark 68.4.1). Since the image of $|F| \to |U|$ is the fibre of $|U| \to |X|$ over $x$ (Properties of Spaces, Lemma 66.4.3), we conclude that the fibre of $|U| \to |X|$ over $x$ is finite. The scheme $F \times _{\mathop{\mathrm{Spec}}(k)} F$ is also a finite union of spectra of fields because it is also quasi-compact and étale over $\mathop{\mathrm{Spec}}(k)$. There is a monomorphism $F \times _ X F \to F \times _{\mathop{\mathrm{Spec}}(k)} F$, hence $F \times _ X F$ is a finite disjoint union of spectra of fields (Schemes, Lemma 26.23.11). Thus the image of $F \times _ X F \to U \times _ X U = R$ is finite. Since this image is the fibre of $|R| \to |X|$ over $x$ by Properties of Spaces, Lemma 66.4.3 we conclude that (1) holds. At this point we know that (1) – (5) are equivalent.

It is clear that (6) implies (5). Conversly, assume $\mathop{\mathrm{Spec}}(k) \to X$ is as in (4) and let $\mathop{\mathrm{Spec}}(k') \to X$ be another morphism with $k'$ a field in the equivalence class of $x$. By Properties of Spaces, Lemma 66.4.11 we have a factorization $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) \to X$ of the given morphism. This is a composition of quasi-compact morphisms and hence quasi-compact (Morphisms of Spaces, Lemma 67.8.5) as desired.
$\square$

## Comments (2)

Comment #7747 by Laurent Moret-Bailly on

Comment #7992 by Stacks Project on