Lemma 29.41.8. A universally closed morphism of schemes is quasi-compact.

Due to Bjorn Poonen.

**Proof.**
Let $f : X \to S$ be a morphism. Assume that $f$ is not quasi-compact. Our goal is to show that $f$ is not universally closed. By Schemes, Lemma 26.19.2 there exists an affine open $V \subset S$ such that $f^{-1}(V)$ is not quasi-compact. To achieve our goal it suffices to show that $f^{-1}(V) \to V$ is not universally closed, hence we may assume that $S = \mathop{\mathrm{Spec}}(A)$ for some ring $A$.

Write $X = \bigcup _{i \in I} X_ i$ where the $X_ i$ are affine open subschemes of $X$. Let $T = \mathop{\mathrm{Spec}}(A[y_ i ; i \in I])$. Let $T_ i = D(y_ i) \subset T$. Let $Z$ be the closed set $(X \times _ S T) - \bigcup _{i \in I} (X_ i \times _ S T_ i)$. It suffices to prove that the image $f_ T(Z)$ of $Z$ under $f_ T : X \times _ S T \to T$ is not closed.

There exists a point $s \in S$ such that there is no neighborhood $U$ of $s$ in $S$ such that $X_ U$ is quasi-compact. Otherwise we could cover $S$ with finitely many such $U$ and Schemes, Lemma 26.19.2 would imply $f$ quasi-compact. Fix such an $s \in S$.

First we check that $f_ T(Z_ s) \ne T_ s$. Let $t \in T$ be the point lying over $s$ with $\kappa (t) = \kappa (s)$ such that $y_ i = 1$ in $\kappa (t)$ for all $i$. Then $t \in T_ i$ for all $i$, and the fiber of $Z_ s \to T_ s$ above $t$ is isomorphic to $(X - \bigcup _{i \in I} X_ i)_ s$, which is empty. Thus $t \in T_ s - f_ T(Z_ s)$.

Assume $f_ T(Z)$ is closed in $T$. Then there exists an element $g \in A[y_ i; i \in I]$ 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 $\kappa (s)$. Hence this coefficient is invertible on some neighborhood $U$ of $s$. Let $J$ be the finite set of $j \in I$ such that $y_ j$ appears in $g$. Since $X_ U$ is not quasi-compact, we may choose a point $x \in X - \bigcup _{j \in J} X_ j$ lying above some $u \in U$. Since $g$ has a coefficient that is invertible on $U$, we can find a point $t' \in T$ lying above $u$ such that $t' \not\in V(g)$ and $t' \in V(y_ i)$ for all $i \notin J$. This is true because $V(y_ i; i \in I, i \not\in J) = \mathop{\mathrm{Spec}}(A[t_ j; j\in J])$ and the set of points of this scheme lying over $u$ is bijective with $\mathop{\mathrm{Spec}}(\kappa (u)[t_ j; j \in J])$. In other words $t' \notin T_ i$ for each $i \notin J$. By Schemes, Lemma 26.17.5 we can find a point $z$ of $X \times _ S T$ 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. $\square$

## 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.

## Comments (0)