The Stacks project

Lemma 71.5.1. Let $S$ be a scheme. Let $X$, $Y$ be integral algebraic spaces over $S$ Let $x \in |X|$ and $y \in |Y|$ be the generic points. Let $f : X \to Y$ be locally of finite type. Assume $f$ is dominant (Morphisms of Spaces, Definition 66.18.1). The following are equivalent:

  1. the transcendence degree of $x/y$ is $0$,

  2. the extension $\kappa (x)/\kappa (y)$ (see proof) is finite,

  3. there exist nonempty affine opens $U \subset X$ and $V \subset Y$ such that $f(U) \subset V$ and $f|_ U : U \to V$ is finite,

  4. $f$ is quasi-finite at $x$, and

  5. $x$ is the only point of $|X|$ mapping to $y$.

If $f$ is separated or if $f$ is quasi-compact, then these are also equivalent to

  1. there exists a nonempty affine open $V \subset Y$ such that $f^{-1}(V) \to V$ is finite.

Proof. By elementary topology, we see that $f(x) = y$ as $f$ is dominant. Let $Y' \subset Y$ be the schematic locus of $Y$ and let $X' \subset f^{-1}(Y')$ be the schematic locus of $f^{-1}(Y')$. By the discussion above, using Decent Spaces, Proposition 67.12.4 and Theorem 67.10.2, we see that $x \in |X'|$ and $y \in |Y'|$. Then $f|_{X'} : X' \to Y'$ is a morphism of integral schemes which is locally of finite type. Thus we see that (1), (2), (3) are equivalent by Morphisms, Lemma 29.51.7.

Condition (4) implies condition (1) by Morphisms of Spaces, Lemma 66.33.3 applied to $X \to Y \to Y$. On the other hand, condition (3) implies condition (4) as a finite morphism is quasi-finite and as $x \in U$ because $x$ is the generic point. Thus (1) – (4) are equivalent.

Assume the equivalent conditions (1) – (4). Suppose that $x' \mapsto y$. Then $x \leadsto x'$ is a specialization in the fibre of $|X| \to |Y|$ over $y$. If $x' \not= x$, then $f$ is not quasi-finite at $x$ by Decent Spaces, Lemma 67.18.9. Hence $x = x'$ and (5) holds. Conversely, if (5) holds, then (5) holds for the morphism of schemes $X' \to Y'$ (see above) and we can use Morphisms, Lemma 29.51.7 to see that (1) holds.

Observe that (6) implies the equivalent conditions (1) – (5) without any further assumptions on $f$. To finish the proof we have to show the equivalent conditions (1) – (5) imply (6). This follows from Decent Spaces, Lemma 67.21.4. $\square$

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