The Stacks project

Lemma 65.9.7. 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 65.8.8 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 65.8.9. 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 64.12.3. (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 63.12.3 and 63.12.2.) 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 65.8.9 and then Lemma 65.8.8 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 64.4.3 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.




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 04XW. Beware of the difference between the letter 'O' and the digit '0'.