Vakil 5.1 E

Show that in a quasi-compact scheme every point has a closed point in its closure

Solution: Let $X$ be a quasi-compact scheme so that it has a finite cover by open affines $U_i$.

Let $z \in X$, and $\bar z$ its closure. Consider the (finite) sub-collection of $\{U_i\}_{i=1}^N$ that intersect $\bar z$. Because $U_1$ is just the spec of a ring, we can pick a closed point $z_1 \in \bar z \cap U_1$. If $z_1$ is also closed in all other $U_i$ that contain it, we're done, but if it is not closed in some $U_i$ then it is not closed in $X$. However, in that case, we can pick another $z_i \in \bar {z_1} \cap U_i$. Certainly, $z_i$ does not lie in $U_1$, because if it did $z_1$ wouldn't have been a closed point in $U_1$ in the first place.

Now we proceed in the obvious way until we get to a point $z_n$ that is closed in all the open affines that contain it. Notice that this procedure terminates because once we move from $z_i$ to $z_{i+1}$ we can no longer have $z_{i+1}$ contained in any open affine where some $z_{j\le i}$ was closed, because if we did then $z_{i+1} \in \bar{z_j} \cap U_j = \{z_j\}$.

  • $\begingroup$ It seems to me that the very last sentence is unnecessary: once you have $z_n$ closed in all the affine opens in your chosen cover, $z_n$ is closed in $X$. $\endgroup$
    – RghtHndSd
    Jan 22, 2014 at 15:02
  • $\begingroup$ @rghthndsd I think it was just a bad choice to start the sentence with "however". I wanted to show that the process terminates, i.e. that we will eventually get to a point that is closed in all affine opens that contain it. $\endgroup$
    – Rodrigo
    Jan 22, 2014 at 23:00
  • 1
    $\begingroup$ Nice proof! I did not thought about this correctly. $\endgroup$ Jan 6, 2015 at 23:28

1 Answer 1


Maybe an easier answer? Notice it is enough to show that every closed subset $Z$ of $X$ has a closed point. Observe a point $p \in Z$ is closed in $Z$ if and only if it is closed in $X$ so it suffices to show that $Z$ has a closed point. But $Z$ is also a quasicompact scheme so we reduce to the case of showing that a quasicompact sheme $X$ has a closed point. For this, say $X = U_1 \cup \dots \cup U_n$ is an irredundant decomposition of $X$ as a union of open affines. We can then pick a point $p \in U_1$ that is closed in $U_1$ and such that $p \notin U_j$ for $j \neq 1$. Because $p \in (U_2 \cup \dots \cup U_n)^c$ the closure is also in $(U_2 \cup \dots \cup U_n)^c$. It is then easy to check that the closure of $p$ in $X$ is $p$.

  • 1
    $\begingroup$ Why is it possible to pick such a $p$? $\endgroup$
    – user501184
    Feb 4, 2019 at 16:10
  • 4
    $\begingroup$ Observe that $U_1 \cap (U_2 \cup \dots \cup U_n)^c$ is closed in $U_1$ and non-empty. Recall that $U_1$ is affine. The ability to pick such a $p$ follows because any non-empty closed subset of an affine scheme has a closed point (i.e. every proper ideal is contained in a maximal ideal). $\endgroup$
    – equin
    Feb 6, 2019 at 1:36
  • 2
    $\begingroup$ How is the quasi-compactness useful? I don't see why the argument wouldn't work if there were an infinite number of $U_i$ $\endgroup$
    – Evariste
    Dec 20, 2019 at 8:34
  • $\begingroup$ That's a really great question. The only place where I can think that one would need this is in picking an irredundant decomposition. Of course we can make sense of what it means for $X = \cup_{i \in I} U_i$ to be irredundant: $U_i \not \subseteq \cup_{j \neq i} U_j$ for all $i$. But it is not clear to me that such an irredundant decomposition exists. With a countable union it is probably okay. $\endgroup$
    – equin
    Jan 1, 2020 at 17:47
  • 2
    $\begingroup$ And proving $p$ closed in $X$ does not rely on the quasicompactness of $X$: because $\displaystyle (\bigcup_{j \in I, j \neq i}U_j)^c$ is closed containing $p$, the closure $\overline{\{p\}}$ is also in $\displaystyle (\bigcup_{j \in I, j \neq i}U_j)^c \subset U_1$. Hence $\overline{\{p\}} = U_1 \cap \overline{\{p\}} = \{p\}$, i.e, the closure of $p$ in $X$ is $p$. Here even the index set $I$ is not finite, the argument still holds as long as $\{U_j\}_{j \in I}$ covers $X$ and $p$ does not contained in any $U_j$ for $j \neq i$. $\endgroup$
    – onRiv
    Nov 20, 2022 at 8:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .