Closed points of a scheme correspond to maximal ideals in the affines?

Let $X$ be a scheme. If $x\in X$ is a closed point, then it corresponds to a maximal ideal in $\mathcal{O}_X(U)$ for some affine open subset $U\subseteq X$.

If I take an arbitrary open affine $\mbox{Spec}~A$ in $X$ and a maximal ideal $x$ in $A$, then $\{x\}$ is closed in $\mbox{Spec}~A$, but may not be closed in $X$. It might even be the case that $\overline{\{x\}}$ contains a point belonging to an affine that has empty intersection with $\mbox{Spec}~A$.

Or not? Maybe I'm missing some easy topological property here.

Is there an example for such a situation? Or is it actually the case that $\{x\}$ will also be closed in $X$? If not, under what hypotheses on $X$ are all maximal ideals in the rings of affine opens also closed points in $X$?

If $X=\mathrm{Spec}(A)$ is the spectrum of a discrete valuation ring, then the generic point $\eta$ is an affine open, namely $\mathrm{Spec}(k(\eta))$. Of course $\{\eta\}$ is closed in itself, but not in $\mathrm{Spec}(A)$. So this shows that a closed point in an affine open need not be closed in the ambient scheme.

This does, however, hold for schemes locally of finite type over a field $k$. Namely, if $X$ is such a $k$-scheme and $x\in X$ is closed in some affine open $\mathrm{Spec}(A)$, then $k(x)=A/\mathfrak{p}_x$ is an extension field of $k$ which is of finite type over $k$. So it is finite over $k$ by Zariski's lemma (also known sometimes as the Nullstellensatz). Now if $U=\mathrm{Spec}(B)$ is any affine open containing $x$, then $x$ corresponds to a prime $\mathfrak{p}_x^\prime$ of $B$, and $B/\mathfrak{p}_x^\prime$ is contained in $k(x)$ ($k(x)$ is the fraction field of this domain in fact). Since $k(x)/k$ is finite, we see that $B/\mathfrak{p}_x^\prime$ is a domain finite over a field, and so is itself a field, i.e., $\mathfrak{p}_x^\prime$ is a closed point of $\mathrm{Spec}(B)$. Since the affine opens cover $X$, it follows that $x$ is a closed point of $X$.

Perhaps there are other situations where closed points of affine opens are closed. I learned this particular fact from Qing Liu's book on algebraic geometry. If we're lucky maybe he will see this problem and add some further insight :)

• Even much worse situation can happen: there are schemes without closed points ! (While locally, any affine open subset $U$ has a closed point of $U$). An example is given in the book you refered to (Exercise 3.3.27).
– user18119
Nov 22, 2012 at 22:14
• It's been 13 years but since you are still around, I'm wondering whether writing $k(x)=A/p_x$ is correct. Correct me but $x$ will correspond to $p_x$ (let's write just $p$ for simplicity) and then $k(x)$ is $A_p/pA_p$. Therefore the quotient $A/p$ will be contained in $k(x)$ but they don't need to be equal. Also Liu writes $k\subseteq A/p\subseteq k(x)$. 21 hours ago
• Yes, it is corrrect. For a ring $A$ and a prime ideal $\mathfrak{p}$, $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is the field of fractions of $A/\mathfrak{p}$. This can be seen in various ways, including literally just playing with fractions. In the event that $\mathfrak{p}$ is maximal, this implies that $A/\mathfrak{p}=A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$. 2 hours ago

$$\def\Spec{\operatorname{Spec}}$$A counter-example(the same as the Keenan Kidwell's) Let $$A$$ be a discrete valuation ring. Let $$m$$ be its unique maximal ideal. Let $$f$$ be a generator of $$m$$. Then $$A_f$$ is the field of fractions of $$A$$. Hence $$x = {0}$$ is a closed point of $$D(f) = \Spec(A_f)$$. But it is not closed in $$\Spec(A)$$.

Proposition Let $$X$$ be a Jacobson scheme. Let $$x$$ be a closed point of an open subset $$U$$ of $$X$$. Then $$x$$ is a closed point of $$X$$.

Proof: Let $$\overline{{\{x\}}}$$ be the closure of $$\{x\}$$ in $$X$$. Let $$y \in \overline{{\{x\}}}$$. It suffices to prove that $$x = y$$. Let $$V = \Spec(B)$$ be an affine open neighborhood of $$y$$. Then $$x \in V \cap U$$. There exists an affine open subset $$D(f) = \Spec(B_f)$$ such that $$x \in D(f) \subset V \cap U$$, where $$f \in B$$. Since $$x$$ is closed in $$U$$, $$x$$ is closed in $$D(f)$$. Hence $$x$$ is a maximal ideal of $$B_f$$. Since $$B_f$$ is finitely generated over $$B$$ and $$B$$ is a Jacobson ring, $$x$$ is a maximal ideal of $$B$$. Hence $$x$$ is closed in $$V$$. Hence $$\overline{{\{x\}}}\cap V = \{x\}$$. Hence $$x = y$$ as desired. QED

• The converse is true: if any closed point of any affine open subset of $X$ is closed in $X$, then $X$ is a Jacobson scheme (EGA, IV$_3$, § 10.3-10,4). This means that for all affine open subsets $U$ of $X$, $O_X(U)$ is a Jacobson ring.
– user18119
Nov 22, 2012 at 22:09
• @QiL I did not know that. Thanks for the input. Nov 22, 2012 at 22:34
• If one defines a Jacobson scheme to be a scheme whose underlying topological space is Jacobson (this is one among the possible equivalent definitions, see 01P4), then your proposition is a actually a property of Jacobson spaces, see 005W (there the $0$ subscript denotes "subset of closed points"). Mar 15, 2023 at 12:41