# Two points of view on constructible sets

This question is aimed at understanding the relationship between two different definitions of the constructible sets in a Noetherian scheme, both of which I encountered in Atiyah-MacDonald's Introduction to Commutative Algebra (henceforth AM). It follows up a question I asked before, which was beautifully answered by user hot_queen.

The setup:

Let $X$ be a set and $\{U_\lambda\}_{\lambda\in\Lambda}\subset 2^X$ a family of subsets of $X$ that is closed under finite intersection, so it serves as a base for a topology $\mathscr{T}$.

Let $\mathscr{F}$ be the smallest family of subsets of $X$ that contains $\mathscr{T}$ and is closed under complementation and finite intersection. By exercise 20 of ch. 7 in AM, $\mathscr{F}$ is equivalently the family of finite unions of locally closed sets. AM defines this as the constructible sets in exercise 21 of the same chapter.

Meanwhile, let $\mathscr{G}$ be the coarsest topology in which every $U_\lambda$ is clopen. In exercises 27-28 of chapter 3 of AM, it is shown that if $X$ is the Spec of a ring $B$ and the $U_\lambda$'s are the standard basic opens of the Zariski topology, then this is precisely the topology in which the images in $X$ of the Specs of all $B$-algebras are taken as the closed sets. AM defines this as the constructible topology in exercise 27. AM notes that the closed sets are exactly the images of Specs, so I infer that it is the closed sets of this topology (i.e. the image in $2^X$ of the family $\mathscr{G}$ under complementation of each member; call it $\mathscr{G}^c$) that AM means to refer to as constructible.

$\mathscr{F}$ is not equal to $\mathscr{G}$ or $\mathscr{G}^c$ in the generality in which I've defined them. In fact $\mathscr{G}$ depends on the base $\{U_\lambda\}$ chosen for $\mathscr{T}$ whereas $\mathscr{F}$ only depends on $\mathscr{T}$. See hot_queen's answer to my question linked above for beautiful simple examples of the differences. (I framed that question in terms of $\mathscr{F}$ vs. $\mathscr{G}$, since I forgot the context in AM ch. 3 of the def. of $\mathscr{G}$, but the examples show the same for $\mathscr{G}^c$ because $\mathscr{F}$ is invariant under complementation.)

I assume, however, since AM is using the same word for them, that if $X$ is the Spec of a noetherian ring, $\mathscr{T}$ is the Zariski topology, and $\{U_\lambda\}$ are the standard basic sets $X_f$ of this topology (i.e. the Specs of the localizations of the underlying ring at single elements), then $\mathscr{F}$ coincides with $\mathscr{G}^c$. So:

My questions:

1) Is this true? (I.e. that $\mathscr{F}=\mathscr{G}^c$ if $X$ is the Spec of a Noetherian ring, $\mathscr{T}$ is the Zariski topology, and the $\{U_\lambda\}$'s are the standard Zariski basis opens?)

2) If yes, which if any of these assumptions can be loosened? Does it remain true if we use a different base for the Zariski topology (e.g. the whole topology)? If yes, does it matter what base? If it doesn't, can the statement be loosened to if we just assume $(X,\mathscr{T})$ is a noetherian space (rather than the Spec of a noetherian ring)? And what are the proofs and/or counterexamples?

No, $\mathscr{F}$ and $\mathscr{G}^c$ are usually very different. For instance, suppose $X$ is irreducible, Noetherian, and $1$-dimensional, so that it has a generic point $g\in X$ and a nonempty set $U\subseteq X$ is Zariski-open iff it contains $g$ and is cofinite. Then $\mathscr{F}$ consists of the sets that either are cofinite and contain $g$ or are finite and do not contain $g$. But $\mathscr{G}$ contains every singleton other than $\{g\}$, and thus contains all subsets of $X\setminus\{g\}$ since it is closed under arbitrary unions. So $\mathscr{G}^c$ contains all subsets of $X$ containing $g$ (in fact, it consists of exactly the sets that either contain $g$ or are finite).

What is true is that $\mathscr{F}=\mathscr{G}\cap\mathscr{G}^c$. That is, the constructible sets are the clopen sets in the constructible topology. In fact, this is true for Spec of any ring as long as you change the definition of $\mathscr{F}$ slightly: $\mathscr{F}$ should be the smallest collection of sets containing the basic Zariski-open sets $U_\lambda$ and closed under finite Boolean operations (in the non-Noetherian case, this is different from your $\mathscr{F}$ since not every Zariski-open set can be obtained as a finite union of basic open sets).

With this definition, let us prove that $\mathscr{F}=\mathscr{G}\cap\mathscr{G}^c$ always holds. Clearly $\mathscr{F}\subseteq\mathscr{G}\cap\mathscr{G}^c$. To prove the reverse inclusion, we will first show that the $\mathscr{G}$-topology is compact. To show this, suppose you have a collection of sets, each of which is either basic open or closed with respect to the Zariski topology, and these sets have the finite intersection property; we wish to show their intersection is nonempty. Let $\{U_i\}$ be the basic open sets in your family and $\{C_j\}$ be the closed sets; we may assume each of these collections is closed under finite intersections. Note that any basic open set is compact in the Zariski topology (since it is Spec of a localization of the ring), and so $U_i\cap \bigcap C_j$ is still nonempty for each $i$. Thus writing $C=\bigcap C_j$, the family $\{U_i\}\cup\{C\}$ still has the finite intersection property. But now $C$ is a closed subset of $X$ and thus is Spec of some ring $A$, and the sets $U_i\cap C$ are the sets of prime ideals of $A$ that do not contain $f_i$, for some elements $f_i\in A$. The finite intersection property of these sets says that for any finite collection of the $f_i$, the multiplicatively closed set they generate does not contain $0$. It follows that the multiplicatively closed set that all the $f_i$ generate does not contain $0$. Thus localizing at this multiplicatively closed set, we get a nonzero ring, and any prime ideal in this ring gives a point of $C\cap\bigcap U_i$.

Now suppose $C\subseteq X$ is clopen for the $\mathscr{G}$-topology. Since $C$ is open, we can write it as a union of sets from $\mathscr{F}$. Since the $\mathscr{G}$-topology is compact and $C$ is closed, $C$ is compact, so actually only finitely many of our sets from $\mathscr{F}$ are needed to cover $C$. Since $\mathscr{F}$ is closed under finite unions, $C\in\mathscr{F}$.

A similar argument can be shown to work more generally whenever you have a topological space $X$ which is sober and such that the compact open subsets of $X$ are closed under finite intersections and form a basis for the topology of $X$ (in particular, this applies to any Noetherian sober space, since any subset of a Noetherian space is compact). Letting $\mathscr{F}$ denote the collection of sets generated by the compact open subsets of $X$ under finite Boolean operations and $\mathscr{G}$ denote the topology generated by $\mathscr{F}$, we can then prove that the clopen sets for the topology $\mathscr{G}$ are exactly the elements of $\mathscr{F}$. (Sobriety of $X$ is used in place of the ring-theoretic argument I gave above that $C\cap\bigcap U_i$ is nonempty.)

In fact, however, this result is not actually more general, because the spaces $X$ satisfying the hypotheses above (called spectral spaces) are exactly the spaces that are homeomorphic to Spec of a ring. This is a fairly difficult theorem of Hochster; see this paper for a proof and much more on the general theory of spectral spaces (the topology $\mathscr{G}$ is what Hochster calls the patch topology).