The Stacks project

Lemma 66.9.7. Let $S$ be a scheme. A universally closed morphism of algebraic spaces over $S$ is quasi-compact.

Proof. This proof is a repeat of the proof in the case of schemes, see Morphisms, Lemma 29.41.8. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $f$ is not quasi-compact. Our goal is to show that $f$ is not universally closed. By Lemma 66.8.8 there exists an affine scheme $Z$ and a morphism $Z \to Y$ such that $Z \times _ Y X \to Z$ is not quasi-compact. To achieve our goal it suffices to show that $Z \times _ Y X \to Z$ is not universally closed, hence we may assume that $Y = \mathop{\mathrm{Spec}}(B)$ for some ring $B$.

Write $X = \bigcup _{i \in I} X_ i$ where the $X_ i$ are quasi-compact open subspaces of $X$. For example, choose a surjective étale morphism $U \to X$ where $U$ is a scheme, choose an affine open covering $U = \bigcup U_ i$ and let $X_ i \subset X$ be the image of $U_ i$. We will use later that the morphisms $X_ i \to Y$ are quasi-compact, see Lemma 66.8.9. Let $T = \mathop{\mathrm{Spec}}(B[a_ i ; i \in I])$. Let $T_ i = D(a_ i) \subset T$. Let $Z \subset T \times _ Y X$ be the reduced closed subspace whose underlying closed set of points is $|T \times _ Y Z| \setminus \bigcup _{i \in I} |T_ i \times _ Y X_ i|$, see Properties of Spaces, Lemma 65.12.3. (Note that $T_ i \times _ Y X_ i$ is an open subspace of $T \times _ Y X$ as $T_ i \to T$ and $X_ i \to X$ are open immersions, see Spaces, Lemmas 64.12.3 and 64.12.2.) Here is a diagram

\[ \xymatrix{ Z \ar[r] \ar[rd] & T \times _ Y X \ar[d]^{f_ T} \ar[r]_ q & X \ar[d]^ f \\ & T \ar[r]^ p & Y } \]

It suffices to prove that the image $f_ T(|Z|)$ is not closed in $|T|$.

We claim there exists a point $y \in Y$ such that there is no affine open neighborhood $V$ of $y$ in $Y$ such that $X_ V$ is quasi-compact. If not then we can cover $Y$ with finitely many such $V$ and for each $V$ the morphism $Y_ V \to V$ is quasi-compact by Lemma 66.8.9 and then Lemma 66.8.8 implies $f$ quasi-compact, a contradiction. Fix a $y \in Y$ as in the claim.

Let $t \in T$ be the point lying over $y$ with $\kappa (t) = \kappa (y)$ such that $a_ i = 1$ in $\kappa (t)$ for all $i$. Suppose $z \in |Z|$ with $f_ T(z) = t$. Then $q(t) \in X_ i$ for some $i$. Hence $f_ T(z) \not\in T_ i$ by construction of $Z$, which contradicts the fact that $t \in T_ i$ by construction. Hence we see that $t \in |T| \setminus f_ T(|Z|)$.

Assume $f_ T(|Z|)$ is closed in $|T|$. Then there exists an element $g \in B[a_ i; i \in I]$ with $f_ T(|Z|) \subset V(g)$ but $t \not\in V(g)$. Hence the image of $g$ in $\kappa (t)$ is nonzero. In particular some coefficient of $g$ has nonzero image in $\kappa (y)$. Hence this coefficient is invertible on some affine open neighborhood $V$ of $y$. Let $J$ be the finite set of $j \in I$ such that the variable $a_ j$ appears in $g$. Since $X_ V$ is not quasi-compact and each $X_{i, V}$ is quasi-compact, we may choose a point $x \in |X_ V| \setminus \bigcup _{j \in J} |X_{j, V}|$. In other words, $x \in |X| \setminus \bigcup _{j \in J} |X_ j|$ and $x$ lies above some $v \in V$. Since $g$ has a coefficient that is invertible on $V$, we can find a point $t' \in T$ lying above $v$ such that $t' \not\in V(g)$ and $t' \in V(a_ i)$ for all $i \notin J$. This is true because $V(a_ i; i \in I \setminus J) = \mathop{\mathrm{Spec}}(B[a_ j; j\in J])$ and the set of points of this scheme lying over $v$ is bijective with $\mathop{\mathrm{Spec}}(\kappa (v)[a_ j; j \in J])$ and $g$ restricts to a nonzero element of this polynomial ring by construction. In other words $t' \not\in T_ i$ for each $i \not\in J$. By Properties of Spaces, Lemma 65.4.3 we can find a point $z$ of $X \times _ Y T$ mapping to $x \in X$ and to $t' \in T$. Since $x \not\in |X_ j|$ for $j \in J$ and $t' \not\in T_ i$ for $i \in I \setminus J$ we see that $z \in |Z|$. On the other hand $f_ T(z) = t' \not\in V(g)$ which contradicts $f_ T(Z) \subset V(g)$. Thus the assumption “$f_ T(|Z|)$ closed” is wrong and we conclude indeed that $f_ T$ is not closed as desired. $\square$

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