Lemma 41.6.4. Let $S$ be a Noetherian scheme. Let $X \to S$ be a quasi-compact unramified morphism. Let $Y \to S$ be a morphism with $Y$ Noetherian. Then $\mathop{\mathrm{Mor}}\nolimits _ S(Y, X)$ is a finite set.

Proof. Assume first $X \to S$ is separated (which is often the case in practice). Since $Y$ is Noetherian it has finitely many connected components. Thus we may assume $Y$ is connected. Choose a point $y \in Y$ with image $s \in S$. Since $X \to S$ is unramified and quasi-compact then fibre $X_ s$ is finite, say $X_ s = \{ x_1, \ldots , x_ n\}$ and $\kappa (x_ i)/\kappa (s)$ is a finite field extension. See Morphisms, Lemma 29.35.10, 29.20.5, and 29.20.10. For each $i$ there are at most finitely many $\kappa (s)$-algebra maps $\kappa (x_ i) \to \kappa (y)$ (by elementary field theory). Thus $\mathop{\mathrm{Mor}}\nolimits _ S(Y, X)$ is finite by Proposition 41.6.3.

General case. There exists a nonempty open $U \subset S$ such that $X_ U \to U$ is finite (in particular separated), see Morphisms, Lemma 29.51.1 (the lemma applies since we've already seen above that a quasi-compact unramified morphism is quasi-finite and since $X \to S$ is quasi-separated by Morphisms, Lemma 29.15.7). Let $Z \subset S$ be the reduced closed subscheme supported on the complement of $U$. By Noetherian induction, we see that $\mathop{\mathrm{Mor}}\nolimits _ Z(Y_ Z, X_ Z)$ is finite (details omitted). By the result of the first paragraph the set $\mathop{\mathrm{Mor}}\nolimits _ U(Y_ U, X_ U)$ is finite. Thus it suffices to show that

$\mathop{\mathrm{Mor}}\nolimits _ S(Y, X) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _ Z(Y_ Z, X_ Z) \times \mathop{\mathrm{Mor}}\nolimits _ U(Y_ U, X_ U)$

is injective. This follows from the fact that the set of points where two morphisms $a, b : Y \to X$ agree is open in $Y$, due to the fact that $\Delta : X \to X \times _ S X$ is open, see Morphisms, Lemma 29.35.13. $\square$

Comment #6706 by WhatJiaranEatsTonight on

In general, there exists a nonempty open U subset S but not X.

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