This is a question in Liu's book.

Let $X$ be a quasi-compact scheme. Show that $X$ contains a closed point.

Well I'm unable to do this question, so any help would be appreciated. This question also makes me curious to know about the meaning/use of closed points of a scheme in general - by that I mean a scheme which is not an algebraic variety/local scheme over a field, which has a geometric meaning. Thanks!


Kevin Lin's answer regarding the meaning of closed points is quite reasonable, especialy in the case when the scheme in question underlies a classical variety. I want to add some additional remarks and examples for thinking about more general schemes.

Here are some tautological remarks: recall that a point $x$ in a scheme $X$ is called a specialization of $y$ if $x$ lies in the Zariski closure of $y$ (and $y$ is called a generalization of $x$). So tautologically, a closed points is one that cannot be specialized any further (just as a generic point cannot be generalized any further). What does specialization really mean: ring theoretically, it means taking the image under a homomorphism; so if $\mathfrak p$ and $\mathfrak q$ are prime ideals of a ring $A$, then $\mathfrak q$ is a specialization of $\mathfrak p$ in $\text{Spec} A$ if and only if $\mathfrak q$ contains $\mathfrak p$, i.e. if $A/\mathfrak p$ surjects onto $A/\mathfrak q$. It is perhaps best to think of an example: say $A$ is $\mathbb C[x,y]$, $\mathfrak p$ is the prime ideal gen'd by $(x-1)$ and $\mathfrak q$ is the prime (actually maximal ideal) gen'd by $(x-1,y)$. Then in $A/\mathfrak p$, we have "specialized" the value of $x$ to equal $1$ (because we have declared $x-1 = 0$) but $y$ is still a free variable. When we pass to the further quotient $A/\mathfrak q$, we have specialized both $x$ and $y$: $x$ is specialized to $1$ and $y$ is specialized to $0$. At this point, we can't specialize any more; technically, this is because $\mathfrak q$ is a maximal ideal of $A$, so a closed point of $\text{Spec} A$; intuitively, it is because both $x$ and $y$ have now both been "specialized" to actual numbers, and so we can't specialize any further.

But suppose now we set $B = \mathbb Z[x,y]$, and take $\mathfrak p$ and $\mathfrak q$ to be the same, i.e. gen'd by $x-1$ and by $(x-1,y)$ respectively. Then $\mathfrak q$ is not maximal; there is more capacity for specialization. How is this? Well, $x$ and $y$ are now taking values in $\mathbb Z$ (rather than the field $\mathbb C$) and so we can also reduce both $x$ and $y$ modulo some prime, say $5$; this gives a prime ideal $\mathfrak r = (x-1,y,5)$ in $B$ containing $\mathfrak q$. Now $\mathfrak r$ is maximal, and so we are done specializing.

So if you have a scheme that is finite type over $\mathbb Z$, the closed points will correspond to "actual points", in Kevin's terminology, but defined over finite fields. The points of the scheme whose coordinates are integers, say, will not be closed. One has the choice of thinking them of them as "actual points" which nevertheless can be specialized further by reducing modulo primes, or as subvarieties rather than "actual points", by identifying them with their Zariski closures (for a picture of this, see the drawing of Mumford that Kevin links to).


Zorn's lemma implies there is a minimal nonempty closed set $F \subset X$ with no proper closed subsets (because the closed sets have the finite intersection property in view of quasi-compactness). It is sufficient to find a closed point in $F$. Now $F$ is in itself a scheme, and it has an open subset $U=\mathrm{Spec}\, A$ for $A$ a ring. This must be all of $F$ by minimality. A maximal ideal in $A$ gives a closed point in $F$, hence in $X$.

There are people here who could give a much better answer to your other question, so I'll leave it.

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    $\begingroup$ So that's why we needed compactness, I missed that. Thanks! $\endgroup$ – Soarer Jul 28 '10 at 12:17
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    $\begingroup$ Just to be precise: F is not a scheme, or better, it has different scheme structures which are not determined by F alone. Anyway, what matters for the argument is that it has one, so one can choose for instance the reduced scheme structure. $\endgroup$ – Andrea Ferretti Aug 7 '10 at 0:05
  • $\begingroup$ Yeah, good point--I wasn't being very precise there. $\endgroup$ – Akhil Mathew Aug 7 '10 at 0:54
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    $\begingroup$ (Manually posting a comment by a deleted user) Hey akhil: If you find a closed point in F, doesnt that mean by minimality that the point {x} IS F? $\endgroup$ – Zev Chonoles Jun 8 '12 at 15:25
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    $\begingroup$ (Manually posting a comment by Matt E) Yes, but a priori one doesn't know that F is a single point. $\endgroup$ – Zev Chonoles Jun 8 '12 at 15:26

Closed points should be thought of as being "actual points", whereas non-closed points can correspond to all sorts of different things: subvarieties, "fat" or "fuzzy" points, generic points, etc. You might be interested in reading this blog post about Mumford's drawing of $\operatorname{Spec} \mathbb{Z}[x]$.

One possible way to justify the claim that closed points are the "actual points" is the fact that if we have, for instance, a smooth variety over $\mathbb{C}$, then its analytification will be a complex manifold. The closed points of the former will then correspond exactly to the points of the latter.

  • $\begingroup$ Thanks for the answer. Just want a clarification: when you say fat points, are you talking about points with different residue fields? $\endgroup$ – Soarer Aug 7 '10 at 13:40

Sorry for posting again on this old thread but I'd like to know what's wrong, if something, with the following, because I think it should works.

We have an open affine covering $X = \bigcup_{i} U_{i}$ with $U_{i} = \mbox{Spec}(A_{i})$. Let $\xi_{1}$ be a closed point in $U_{1}$ (just take a maximal ideal). If $\xi_{1}$ is still closed in $X$ we are done. Otherwise, take $\xi_{2} \neq \xi_{1}$ with $\xi_{2} \in \overline{\left\{\xi_{1}\right\}}$. Hence $\xi_{2} \in U_{i}$ for some $i \neq 1$, say $i = 2$. Then repeat the reasoning above again. The process must stop since there are only finitely many $U_{i}$ and so $X$ has a closed point.

  • $\begingroup$ And which $\xi_i$ should be your closed point? Please check the details before giving an answer. $\endgroup$ – Martin Brandenburg Dec 21 '10 at 16:11
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    $\begingroup$ I don't understand the preceding comment, and this looks okay to me. $\endgroup$ – Matt E Jan 24 '11 at 20:29

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