Lemma 68.18.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $x \in |X|$ with image $y \in |Y|$. Let $F = f^{-1}(\{ y\} )$ with induced topology from $|X|$. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to Y$ be in the equivalence class defining $y$. Set $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$. Let $\tilde x \in |X_ k|$ map to $x \in |X|$. Consider the following conditions

1. $\dim _ x(F) = 0$,

2. $x$ is isolated in $F$,

3. $x$ is closed in $F$ and if $x' \leadsto x$ in $F$, then $x = x'$,

4. $\dim _{\tilde x}(|X_ k|) = 0$,

5. $\tilde x$ is isolated in $|X_ k|$,

6. $\tilde x$ is closed in $|X_ k|$ and if $\tilde x' \leadsto \tilde x$ in $|X_ k|$, then $\tilde x = \tilde x'$,

7. $\dim _{\tilde x}(X_ k) = 0$,

8. $f$ is quasi-finite at $x$.

Then we have

$\xymatrix{ (0ACF) \ar@{=>}[r]_{f\text{ decent}} & (0ACG) \ar@{<=>}[r] & (0ACH) \ar@{<=>}[r] & (0ACI) \ar@{<=>}[r] & (0ACJ) }$

If $Y$ is decent, then conditions (2) and (3) are equivalent to each other and to conditions (5), (6), (7), and (8). If $Y$ and $X$ are decent, then all conditions are equivalent.

Proof. By Lemma 68.18.7 conditions (5), (6), and (7) are equivalent to each other and to the condition that $X_ k \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at $\tilde x$. Thus by Morphisms of Spaces, Lemma 67.27.2 they are also equivalent to (8). If $f$ is decent, then $X_ k$ is a decent algebraic space and Lemma 68.18.7 shows that (4) implies (5).

If $Y$ is decent, then we can pick a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k') \to Y$ in the equivalence class of $y$. In this case Lemma 68.18.6 tells us that $|X_{k'}| \to F$ is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted. $\square$

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.

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 0ACB. Beware of the difference between the letter 'O' and the digit '0'.