Lemma 67.18.9. 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 67.18.7 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 66.27.2 they are also equivalent to (8). If $f$ is decent, then $X_ k$ is a decent algebraic space and Lemma 67.18.7 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 67.18.6 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. $\square$

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