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 $z_{i+1} \in \bar{z_j} \cap U_j = \{z_j\}$.

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  • 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 $X$. – RghtHndSd Jan 22 '14 at 15:02
  • @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. – Rodrigo Jan 22 '14 at 23:00
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    Nice proof! I did not thought about this correctly. – Bombyx mori Jan 06 '15 at 23:28

1 Answers1


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 $p$.

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    Why is it possible to pick such a $p$? –  Feb 04 '19 at 16:10
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    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). – equin Feb 06 '19 at 01:36
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    How is the quasi-compactness useful? I don't see why the argument wouldn't work if there were an infinite number of $U_i$ – Evariste Dec 20 '19 at 08:34
  • 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. – equin Jan 01 '20 at 17:47
  • Quasi-compactness is definitely necessary and I think @equin is on the right track. See Liu's AG book, exercises 3.3.26 and 3.3.27 for an example of a scheme without a closed point! – Nathan Lowry Jun 29 '21 at 15:42