Definition 5.23.1. A topological space $X$ is called *spectral* if it is sober, quasi-compact, the intersection of two quasi-compact opens is quasi-compact, and the collection of quasi-compact opens forms a basis for the topology. A continuous map $f : X \to Y$ of spectral spaces is called *spectral* if the inverse image of a quasi-compact open is quasi-compact.

## 5.23 Spectral spaces

The material in this section is taken from [Hochster] and [Hochster-thesis]. In his thesis Hochster proves (among other things) that the spectral spaces are exactly the topological spaces that occur as the spectrum of a ring.

In other words a continuous map of spectral spaces is spectral if and only if it is quasi-compact (Definition 5.12.1).

Let $X$ be a spectral space. The *constructible topology* on $X$ is the topology which has as a subbase of opens the sets $U$ and $U^ c$ where $U$ is a quasi-compact open of $X$. Note that since $X$ is spectral an open $U \subset X$ is retrocompact if and only if $U$ is quasi-compact. Hence the constructible topology can also be characterized as the coarsest topology such that every constructible subset of $X$ is both open and closed. Since the collection of quasi-compact opens is a basis for the topology on $X$ we see that the constructible topology is stronger than the given topology on $X$.

Lemma 5.23.2. Let $X$ be a spectral space. The constructible topology is Hausdorff and quasi-compact.

**Proof.**
Since the collection of all quasi-compact opens forms a basis for the topology on $X$ and $X$ is sober, it is clear that $X$ is Hausdorff in the constructible topology.

Let $\mathcal{B}$ be the collection of subsets $B \subset X$ with $B$ either quasi-compact open or closed with quasi-compact complement. If $B \in \mathcal{B}$ then $B^ c \in \mathcal{B}$. It suffices to show every covering $X = \bigcup _{i \in I} B_ i$ with $B_ i \in \mathcal{B}$ has a finite refinement, see Lemma 5.12.15. Taking complements we see that we have to show that any family $\{ B_ i\} _{i \in I}$ of elements of $\mathcal{B}$ such that $B_{i_1} \cap \ldots \cap B_{i_ n} \not= \emptyset $ for all $n$ and all $i_1, \ldots , i_ n \in I$ has a common point of intersection. We may and do assume $B_ i \not= B_{i'}$ for $i \not= i'$.

To get a contradiction assume $\{ B_ i\} _{i \in I}$ is a family of elements of $\mathcal{B}$ having the finite intersection property but empty intersection. An application of Zorn's lemma shows that we may assume our family is maximal (details omitted). Let $I' \subset I$ be those indices such that $B_ i$ is closed and set $Z = \bigcap _{i \in I'} B_ i$. This is a closed subset of $X$. If $Z$ is reducible, then we can write $Z = Z' \cup Z''$ as a union of two closed subsets, neither equal to $Z$. This means in particular that we can find a quasi-compact open $U' \subset X$ meeting $Z'$ but not $Z''$. Similarly, we can find a quasi-compact open $U'' \subset X$ meeting $Z''$ but not $Z'$. Set $B' = X \setminus U'$ and $B'' = X \setminus U''$. Note that $Z'' \subset B'$ and $Z' \subset B''$. If there exist a finite number of indices $i_1, \ldots , i_ n \in I$ such that $B' \cap B_{i_1} \cap \ldots \cap B_{i_ n} = \emptyset $ as well as a finite number of indices $j_1, \ldots , j_ m \in I$ such that $B'' \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset $ then we find that $Z \cap B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset $. However, the set $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m}$ is quasi-compact hence we would find a finite number of indices $i'_1, \ldots , i'_ l \in I'$ with $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} \cap B_{i'_1} \cap \ldots \cap B_{i'_ l} = \emptyset $, a contradiction. Thus we see that we may add either $B'$ or $B''$ to the given family contradicting maximality. We conclude that $Z$ is irreducible. However, this leads to a contradiction as well, as now every nonempty (by the same argument as above) open $Z \cap B_ i$ for $i \in I \setminus I'$ contains the unique generic point of $Z$. This contradiction proves the lemma. $\square$

Lemma 5.23.3. Let $f : X \to Y$ be a spectral map of spectral spaces. Then

$f$ is continuous in the constructible topology,

the fibres of $f$ are quasi-compact, and

the image is closed in the constructible topology.

**Proof.**
Let $X'$ and $Y'$ denote $X$ and $Y$ endowed with the constructible topology which are quasi-compact Hausdorff spaces by Lemma 5.23.2. Part (1) says $X' \to Y'$ is continuous and follows immediately from the definitions. Part (3) follows as $f(X')$ is a quasi-compact subset of the Hausdorff space $Y'$, see Lemma 5.12.4. We have a commutative diagram

of continuous maps of topological spaces. Since $Y'$ is Hausdorff we see that the fibres $X'_ y$ are closed in $X'$. As $X'$ is quasi-compact we see that $X'_ y$ is quasi-compact (Lemma 5.12.3). As $X'_ y \to X_ y$ is a surjective continuous map we conclude that $X_ y$ is quasi-compact (Lemma 5.12.7). $\square$

Lemma 5.23.4. Let $X$ be a spectral space. Let $E \subset X$ be closed in the constructible topology (for example constructible or closed). Then $E$ with the induced topology is a spectral space.

**Proof.**
Let $Z \subset E$ be a closed irreducible subset. Let $\eta $ be the generic point of the closure $\overline{Z}$ of $Z$ in $X$. To prove that $E$ is sober, we show that $\eta \in E$. If not, then since $E$ is closed in the constructible topology, there exists a constructible subset $F \subset X$ such that $\eta \in F$ and $F \cap E = \emptyset $. By Lemma 5.15.14 this implies $F \cap \overline{Z}$ contains a nonempty open subset of $\overline{Z}$. But this is impossible as $\overline{Z}$ is the closure of $Z$ and $Z \cap F = \emptyset $.

Since $E$ is closed in the constructible topology, it is quasi-compact in the constructible topology (Lemmas 5.12.3 and 5.23.2). Hence a fortiori it is quasi-compact in the topology coming from $X$. If $U \subset X$ is a quasi-compact open, then $E \cap U$ is closed in the constructible topology, hence quasi-compact (as seen above). It follows that the quasi-compact open subsets of $E$ are the intersections $E \cap U$ with $U$ quasi-compact open in $X$. These form a basis for the topology. Finally, given two $U, U' \subset X$ quasi-compact opens, the intersection $(E \cap U) \cap (E \cap U') = E \cap (U \cap U')$ and $U \cap U'$ is quasi-compact as $X$ is spectral. This finishes the proof. $\square$

Lemma 5.23.5. Let $X$ be a spectral space. Let $E \subset X$ be a subset closed in the constructible topology (for example constructible).

If $x \in \overline{E}$, then $x$ is the specialization of a point of $E$.

If $E$ is stable under specialization, then $E$ is closed.

If $E' \subset X$ is open in the constructible topology (for example constructible) and stable under generalization, then $E'$ is open.

**Proof.**
Proof of (1). Let $x \in \overline{E}$. Let $\{ U_ i\} $ be the set of quasi-compact open neighbourhoods of $x$. A finite intersection of the $U_ i$ is another one. The intersection $U_ i \cap E$ is nonempty for all $i$. Since the subsets $U_ i \cap E$ are closed in the constructible topology we see that $\bigcap (U_ i \cap E)$ is nonempty by Lemma 5.23.2 and Lemma 5.12.6. Since $X$ is a sober space and $\{ U_ i\} $ is a fundamental system of open neighbourhoods of $x$, we see that $\bigcap U_ i$ is the set of generalizations of $x$. Thus $x$ is a specialization of a point of $E$.

Part (2) is immediate from (1).

Proof of (3). Assume $E'$ is as in (3). The complement of $E'$ is closed in the constructible topology (Lemma 5.15.2) and closed under specialization (Lemma 5.19.2). Hence the complement is closed by (2), i.e., $E'$ is open. $\square$

Lemma 5.23.6. Let $X$ be a spectral space. Let $x, y \in X$. Then either there exists a third point specializing to both $x$ and $y$, or there exist disjoint open neighbourhoods containing $x$ and $y$.

**Proof.**
Let $\{ U_ i\} $ be the set of quasi-compact open neighbourhoods of $x$. A finite intersection of the $U_ i$ is another one. Let $\{ V_ j\} $ be the set of quasi-compact open neighbourhoods of $y$. A finite intersection of the $V_ j$ is another one. If $U_ i \cap V_ j$ is empty for some $i, j$ we are done. If not, then the intersection $U_ i \cap V_ j$ is nonempty for all $i$ and $j$. The sets $U_ i \cap V_ j$ are closed in the constructible topology on $X$. By Lemma 5.23.2 we see that $\bigcap (U_ i \cap V_ j)$ is nonempty (Lemma 5.12.6). Since $X$ is a sober space and $\{ U_ i\} $ is a fundamental system of open neighbourhoods of $x$, we see that $\bigcap U_ i$ is the set of generalizations of $x$. Similarly, $\bigcap V_ j$ is the set of generalizations of $y$. Thus any element of $\bigcap (U_ i \cap V_ j)$ specializes to both $x$ and $y$.
$\square$

Lemma 5.23.7. Let $X$ be a spectral space. The following are equivalent:

$X$ is profinite,

$X$ is Hausdorff,

$X$ is totally disconnected,

every quasi-compact open is closed,

there are no nontrivial specializations between points,

every point of $X$ is closed,

every point of $X$ is the generic point of an irreducible component of $X$,

add more here.

**Proof.**
Lemma 5.22.2 shows the implication (1) $\Rightarrow $ (3). Irreducible components are closed, so if $X$ is totally disconnected, then every point is closed. So (3) implies (6). The equivalence of (6) and (5) is immediate, and (6) $\Leftrightarrow $ (7) holds because $X$ is sober. Assume (5). Then all constructible subsets of $X$ are closed (Lemma 5.23.5), in particular all quasi-compact opens are closed. So (5) implies (4). Since $X$ is sober, for any two points there is a quasi-compact open containing exactly one of them, hence (4) implies (2). It remains to prove (2) implies (1). Suppose $X$ is Hausdorff. Every quasi-compact open is also closed (Lemma 5.12.4). This implies $X$ is totally disconnected. Hence it is profinite, by Lemma 5.22.2.
$\square$

Lemma 5.23.8. If $X$ is a spectral space, then $\pi _0(X)$ is a profinite space.

Lemma 5.23.9. The product of two spectral spaces is spectral.

**Proof.**
Let $X$, $Y$ be spectral spaces. Denote $p : X \times Y \to X$ and $q : X \times Y \to Y$ the projections. Let $Z \subset X \times Y$ be a closed irreducible subset. Then $p(Z) \subset X$ is irreducible and $q(Z) \subset Y$ is irreducible. Let $x \in X$ be the generic point of the closure of $p(X)$ and let $y \in Y$ be the generic point of the closure of $q(Y)$. If $(x, y) \not\in Z$, then there exist opens $x \in U \subset X$, $y \in V \subset Y$ such that $Z \cap U \times V = \emptyset $. Hence $Z$ is contained in $(X \setminus U) \times Y \cup X \times (Y \setminus V)$. Since $Z$ is irreducible, we see that either $Z \subset (X \setminus U) \times Y$ or $Z \subset X \times (Y \setminus V)$. In the first case $p(Z) \subset (X \setminus U)$ and in the second case $q(Z) \subset (Y \setminus V)$. Both cases are absurd as $x$ is in the closure of $p(Z)$ and $y$ is in the closure of $q(Z)$. Thus we conclude that $(x, y) \in Z$, which means that $(x, y)$ is the generic point for $Z$.

A basis of the topology of $X \times Y$ are the opens of the form $U \times V$ with $U \subset X$ and $V \subset Y$ quasi-compact open (here we use that $X$ and $Y$ are spectral). Then $U \times V$ is quasi-compact as the product of quasi-compact spaces is quasi-compact. Moreover, any quasi-compact open of $X \times Y$ is a finite union of such quasi-compact rectangles $U \times V$. It follows that the intersection of two such is again quasi-compact (since $X$ and $Y$ are spectral). This concludes the proof. $\square$

Lemma 5.23.10. Let $f : X \to Y$ be a continuous map of topological spaces. If

$X$ and $Y$ are spectral,

$f$ is spectral and bijective, and

generalizations (resp. specializations) lift along $f$.

Then $f$ is a homeomorphism.

**Proof.**
Since $f$ is spectral it defines a continuous map between $X$ and $Y$ in the constructible topology. By Lemmas 5.23.2 and 5.17.8 it follows that $X \to Y$ is a homeomorphism in the constructible topology. Let $U \subset X$ be quasi-compact open. Then $f(U)$ is constructible in $Y$. Let $y \in Y$ specialize to a point in $f(U)$. By the last assumption we see that $f^{-1}(y)$ specializes to a point of $U$. Hence $f^{-1}(y) \in U$. Thus $y \in f(U)$. It follows that $f(U)$ is open, see Lemma 5.23.5. Whence $f$ is a homeomorphism. To prove the lemma in case specializations lift along $f$ one shows instead that $f(Z)$ is closed if $X \setminus Z$ is a quasi-compact open of $X$.
$\square$

Lemma 5.23.11. The inverse limit of a directed inverse system of finite sober topological spaces is a spectral topological space.

**Proof.**
Let $I$ be a directed set. Let $X_ i$ be an inverse system of finite sober spaces over $I$. Let $X = \mathop{\mathrm{lim}}\nolimits X_ i$ which exists by Lemma 5.14.1. As a set $X = \mathop{\mathrm{lim}}\nolimits X_ i$. Denote $p_ i : X \to X_ i$ the projection. Because $I$ is directed we may apply Lemma 5.14.2. A basis for the topology is given by the opens $p_ i^{-1}(U_ i)$ for $U_ i \subset X_ i$ open. Since an open covering of $p_ i^{-1}(U_ i)$ is in particular an open covering in the profinite topology, we conclude that $p_ i^{-1}(U_ i)$ is quasi-compact. Given $U_ i \subset X_ i$ and $U_ j \subset X_ j$, then $p_ i^{-1}(U_ i) \cap p_ j^{-1}(U_ j) = p_ k^{-1}(U_ k)$ for some $k \geq i, j$ and open $U_ k \subset X_ k$. Finally, if $Z \subset X$ is irreducible and closed, then $p_ i(Z) \subset X_ i$ is irreducible and therefore has a unique generic point $\xi _ i$ (because $X_ i$ is a finite sober topological space). Then $\xi = \mathop{\mathrm{lim}}\nolimits \xi _ i$ is a generic point of $Z$ (it is a point of $Z$ as $Z$ is closed). This finishes the proof.
$\square$

Lemma 5.23.12. Let $W$ be the topological space with two points, one closed, the other not. A topological space is spectral if and only if it is homeomorphic to a subspace of a product of copies of $W$ which is closed in the constructible topology.

**Proof.**
Write $W = \{ 0, 1\} $ where $0$ is a specialization of $1$ but not vice versa. Let $I$ be a set. The space $\prod _{i \in I} W$ is spectral by Lemma 5.23.11. Thus we see that a subspace of $\prod _{i \in I} W$ closed in the constructible topology is a spectral space by Lemma 5.23.4.

For the converse, let $X$ be a spectral space. Let $U \subset X$ be a quasi-compact open. Consider the continuous map

which maps every point in $U$ to $1$ and every point in $X \setminus U$ to $0$. Taking the product of these maps we obtain a continuous map

By construction the map $f : X \to Y$ is spectral. By Lemma 5.23.3 the image of $f$ is closed in the constructible topology. If $x', x \in X$ are distinct, then since $X$ is sober either $x'$ is not a specialization of $x$ or conversely. In either case (as the quasi-compact opens form a basis for the topology of $X$) there exists a quasi-compact open $U \subset X$ such that $f_ U(x') \not= f_ U(x)$. Thus $f$ is injective. Let $Y = f(X)$ endowed with the induced topology. Let $y' \leadsto y$ be a specialization in $Y$ and say $f(x') = y'$ and $f(x) = y$. Arguing as above we see that $x' \leadsto x$, since otherwise there is a $U$ such that $x \in U$ and $x' \not\in U$, which would imply $f_ U(x') \not\leadsto f_ U(x)$. We conclude that $f : X \to Y$ is a homeomorphism by Lemma 5.23.10. $\square$

Lemma 5.23.13. A topological space is spectral if and only if it is a directed inverse limit of finite sober topological spaces.

**Proof.**
One direction is given by Lemma 5.23.11. For the converse, assume $X$ is spectral. Then we may assume $X \subset \prod _{i \in I} W$ is a subset closed in the constructible topology where $W = \{ 0, 1\} $ as in Lemma 5.23.12. We can write

as a cofiltered limit. For each $J$, let $X_ J \subset \prod _{j \in J} W$ be the image of $X$. Then we see that $X = \mathop{\mathrm{lim}}\nolimits X_ J$ as sets because $X$ is closed in the product with the constructible topology (detail omitted). A formal argument (omitted) on limits shows that $X = \mathop{\mathrm{lim}}\nolimits X_ J$ as topological spaces. $\square$

Lemma 5.23.14. Let $X$ be a topological space and let $c : X \to X'$ be the universal map from $X$ to a sober topological space, see Lemma 5.8.14.

If $X$ is quasi-compact, so is $X'$.

If $X$ is quasi-compact, has a basis of quasi-compact opens, and the intersection of two quasi-compact opens is quasi-compact, then $X'$ is spectral.

If $X$ is Noetherian, then $X'$ is a Noetherian spectral space.

**Proof.**
Let $U \subset X$ be open and let $U' \subset X'$ be the corresponding open, i.e., the open such that $c^{-1}(U') = U$. Then $U$ is quasi-compact if and only if $U'$ is quasi-compact, as pulling back by $c$ is a bijection between the opens of $X$ and $X'$ which commutes with unions. This in particular proves (1).

Proof of (2). It follows from the above that $X'$ has a basis of quasi-compact opens. Since $c^{-1}$ also commutes with intersections of pairs of opens, we see that the intersection of two quasi-compact opens $X'$ is quasi-compact. Finally, $X'$ is quasi-compact by (1) and sober by construction. Hence $X'$ is spectral.

Proof of (3). It is immediate that $X'$ is Noetherian as this is defined in terms of the acc for open subsets which holds for $X$. We have already seen in (2) that $X'$ is spectral. $\square$

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