The Stacks project

Lemma 67.4.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent:

  1. for every affine scheme $U$, any étale morphism $\varphi : U \to X$ setting $R = U \times _ X U$ the fibres of both

    \[ |U| \longrightarrow |X| \quad \text{and}\quad |R| \longrightarrow |X| \]

    over $x$ are finite,

  2. there exist schemes $U_ i$ and étale morphisms $U_ i \to X$ such that $\coprod U_ i \to X$ is surjective and for each $i$, setting $R_ i = U_ i \times _ X U_ i$ the fibres of both

    \[ |U_ i| \longrightarrow |X| \quad \text{and}\quad |R_ i| \longrightarrow |X| \]

    over $x$ are finite,

  3. there exists a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ with $k$ a field in the equivalence class of $x$, and for any affine scheme $U$ and étale morphism $U \to X$ the fibre product $\mathop{\mathrm{Spec}}(k) \times _ X U$ is a finite scheme over $k$,

  4. there exists a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ with $k$ a field in the equivalence class of $x$, and

  5. there exists a quasi-compact morphism $\mathop{\mathrm{Spec}}(k) \to X$ with $k$ a field in the equivalence class of $x$.

Proof. The equivalence of (1) and (3) follows on applying Lemma 67.4.4 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 67.4.2 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 67.4.3 (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 67.4.1) 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 66.8.8.

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 67.4.1). Since the image of $|F| \to |U|$ is the fibre of $|U| \to |X|$ over $x$ (Properties of Spaces, Lemma 65.4.3), 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 65.4.3 we conclude that (1) holds. $\square$


Comments (2)

Comment #7747 by Laurent Moret-Bailly on

I suggest to add the condition (6) Every morphism with a field in the class of is quasi-compact. Clearly this implies (5). Proof that (4) implies (6): if is a quasi-compact monomorphism, then factors as (consider the fibre product and observe that every monomorphism from a nonempty scheme to is an isomorphism), so is quasi-compact by composition. (Ideally, the property that factors through , without quasi-compactness, could be stated earlier; I have not found it.)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03JV. Beware of the difference between the letter 'O' and the digit '0'.