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I am following Ravi Vakil's Math 216: Foundations of Algebraic geometry notes, and there is a remark following an exercise that I don't understand at all, and if anyone could enlighten me then that would be brilliant.

The exercise asks one to show that if $X$ is a quasicompact scheme, then every point has a closed point in its closure, which is clear from the preceding exercise asking to show that $X$ is quasicompact if and only if it can be written as a finite union of affine schemes. These I am fine, as well as the following implication that every nonempty closed subset of $X$ contains a closed point.

However, then the notes then go on to state that this will be used in the following way: If a property $P$ is open (that is, if some point $x$ has $P$, then there exists an open neighbourhood $U$ of $x$ such that all points in $U$ have $P$), then to check that all points of a quasicompact scheme have $P$, then it suffices to check only the closed points.

I do not seem to be able to see how this follows at all. It seems to me that everything in the exercises is regarding closed points being in closures, and to show the remark, I want to show that other points are in (all) open neighbourhood(s) if closed points. These seem relatively distinct to me - is this wrong?

These comments/exercises are on pages 139-140 of the notes.

Peter Mortensen
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Joe Tait
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3 Answers3

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Let $\eta$ be a generic point with closure $Y = \overline{\{ \eta \}}$, and let $P$ be an open property. Then, the set of all points of $X$ with property $P$ is open, so its complement is closed; in particular, if $\eta$ does not have property $P$, then no point of $Y$ has property $P$. (This is just point set topology.)

Now, suppose $X$ is a quasicompact scheme. Then, $Y$ must have a closed point; so if all closed points of $X$ have property $P$, then $\eta$ must have property $P$ as well.

Zhen Lin
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First: Are you sure that your argument about existence of closed points is correct? The reason I ask is that you say that it's clear from writing $X$ is a finite union of open affines, but (while I agree that $X$ is quasi-compact if and only if this is possible) I don't myself see how it follows immediately from this. (The arguments I know, e.g. this one, use the topological property of quasi-compactness in an explicit manner.)

Secondly: Suppose $U$ is an open subset of $X$ that contains all the closed points of $X$. The complement of $U$ is a closed subset of $X$. Can it contain any closed point of $X$? If not, then taking into account the facts you state in your question, can it contain any points at all?

Community
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Matt E
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  • With regards to the first point, I think the following argument is sound. If $X$ is a finite union of affines, we can suppose that any point is in ${\rm Spec}\ A$ for some ring $A$. This will be closed in $X$ since ${\rm Spec} A$ is a union of irreducible closed sets. Then, $x$ will correspond to a prime ideal, which is necessarily contained in a maximal ideal, which correspond to a closed point in the closure of $x$.

    With regards to the second, thank you very much :)

     – Joe Tait May 27 '12 at 13:59
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    @Joe Tait: This doesn't work; a point may be closed in some open subset, but not in the whole space. – Martin Brandenburg May 27 '12 at 14:23
  • @MartinBrandenburg Is it not true in the case that the scheme is a finite union of affine schemes though, given that $\sqcup_i {\rm Spec}\ A_i \cong {\rm Spec}\ \prod_i A_i$? I would have thought ${\rm Spec}\ A_i$ would be a closed subset, since it would "equal" (in some way equivalent to) the closure of the ideal generated by $(1,\ldots,1,0,1,\ldots,1)$ in the product. Is this wrong? – Joe Tait May 27 '12 at 16:08
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    @Joe: Dear Joe, You might want to think about whether or not $\mathbb A^n$ is closed in $\mathbb P^n$ (the latter being a well-known union of $n+1$ affine open subsets). Getting this part of the story straight is much more basic than the issue about closed points in quasi-compact schemes that you are asking about. Regards, – Matt E May 27 '12 at 17:39
  • @MattE Okay, sorry, I was thinking of disjoint unions, I think because of the examples before. Thanks, I will look at that again – Joe Tait May 27 '12 at 22:19
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This is a purely topological statement. Suppose that an open property $P$ holds for all closed points. Let $\eta$ be a non-closed point, and let $x \in \bar{\eta}$ be a closed point in the closure (which exists by the exercise!). Then by assumption $x$ has $P$, and since $P$ is open there exists an open neighborhood $U$ of $x$ so that all points of $U$ have $P$.

It is clear that any open neighborhood $U$ of $x$ must contain $\eta$. If $\eta \notin U$, then $U^c:=X-U$ is a closed subset of $X$ containing $\eta$. But $\eta \in U^c$ implies $\bar{\eta} \subset U^c$ which then means that $x \in \bar{\eta} \subset U^c$, a contradiction since $x \in U$.

Since $\eta \in U$, then $\eta$ has property $P$ also. Hopefully this gives you some intuition about the strangeness of non-closed points.

Parsa
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  • Thank you very much - I started along this line, and then went on some wild tangent that was fruitless. Thanks again :) – Joe Tait May 27 '12 at 13:42