61.2 Some topology

Some preliminaries. We have defined spectral spaces and spectral maps of spectral spaces in Topology, Section 5.23. The spectrum of a ring is a spectral space, see Algebra, Lemma 10.26.2.

Lemma 61.2.1. Let $X$ be a spectral space. Let $X_0 \subset X$ be the set of closed points. The following are equivalent

1. Every open covering of $X$ can be refined by a finite disjoint union decomposition $X = \coprod U_ i$ with $U_ i$ open and closed in $X$.

2. The composition $X_0 \to X \to \pi _0(X)$ is bijective.

Moreover, if $X_0$ is closed in $X$ and every point of $X$ specializes to a unique point of $X_0$, then these conditions are satisfied.

Proof. We will use without further mention that $X_0$ is quasi-compact (Topology, Lemma 5.12.9) and $\pi _0(X)$ is profinite (Topology, Lemma 5.23.9). Picture

$\xymatrix{ X_0 \ar[rd]_ f \ar[r] & X \ar[d]^\pi \\ & \pi _0(X) }$

If (2) holds, the continuous bijective map $f : X_0 \to \pi _0(X)$ is a homeomorphism by Topology, Lemma 5.17.8. Given an open covering $X = \bigcup U_ i$, we get an open covering $\pi _0(X) = \bigcup f(X_0 \cap U_ i)$. By Topology, Lemma 5.22.4 we can find a finite open covering of the form $\pi _0(X) = \coprod V_ j$ which refines this covering. Since $X_0 \to \pi _0(X)$ is bijective each connected component of $X$ has a unique closed point, whence is equal to the set of points specializing to this closed point. Hence $\pi ^{-1}(V_ j)$ is the set of points specializing to the points of $f^{-1}(V_ j)$. Now, if $f^{-1}(V_ j) \subset X_0 \cap U_ i \subset U_ i$, then it follows that $\pi ^{-1}(V_ j) \subset U_ i$ (because the open set $U_ i$ is closed under generalizations). In this way we see that the open covering $X = \coprod \pi ^{-1}(V_ j)$ refines the covering we started out with. In this way we see that (2) implies (1).

Assume (1). Let $x, y \in X$ be closed points. Then we have the open covering $X = (X \setminus \{ x\} ) \cup (X \setminus \{ y\} )$. It follows from (1) that there exists a disjoint union decomposition $X = U \amalg V$ with $U$ and $V$ open (and closed) and $x \in U$ and $y \in V$. In particular we see that every connected component of $X$ has at most one closed point. By Topology, Lemma 5.12.8 every connected component (being closed) also does have a closed point. Thus $X_0 \to \pi _0(X)$ is bijective. In this way we see that (1) implies (2).

Assume $X_0$ is closed in $X$ and every point specializes to a unique point of $X_0$. Then $X_0$ is a spectral space (Topology, Lemma 5.23.5) consisting of closed points, hence profinite (Topology, Lemma 5.23.8). Let $x, y \in X_0$ be distinct. By Topology, Lemma 5.22.4 we can find a disjoint union decomposition $X_0 = U_0 \amalg V_0$ with $U_0$ and $V_0$ open and closed and $x \in U_0$ and $y \in V_0$. Let $U \subset X$, resp. $V \subset X$ be the set of points specializing to $U_0$, resp. $V_0$. Observe that $X = U \amalg V$. By Topology, Lemma 5.24.7 we see that $U$ is an intersection of quasi-compact open subsets. Hence $U$ is closed in the constructible topology. Since $U$ is closed under specialization, we see that $U$ is closed by Topology, Lemma 5.23.6. By symmetry $V$ is closed and hence $U$ and $V$ are both open and closed. This proves that $x, y$ are not in the same connected component of $X$. In other words, $X_0 \to \pi _0(X)$ is injective. The map is also surjective by Topology, Lemma 5.12.8 and the fact that connected components are closed. In this way we see that the final condition implies (2). $\square$

Example 61.2.2. Let $T$ be a profinite space. Let $t \in T$ be a point and assume that $T \setminus \{ t\}$ is not quasi-compact. Let $X = T \times \{ 0, 1\}$. Consider the topology on $X$ with a subbase given by the sets $U \times \{ 0, 1\}$ for $U \subset T$ open, $X \setminus \{ (t, 0)\}$, and $U \times \{ 1\}$ for $U \subset T$ open with $t \not\in U$. The set of closed points of $X$ is $X_0 = T \times \{ 0\}$ and $(t, 1)$ is in the closure of $X_0$. Moreover, $X_0 \to \pi _0(X)$ is a bijection. This example shows that conditions (1) and (2) of Lemma 61.2.1 do no imply the set of closed points is closed.

It turns out it is more convenient to work with spectral spaces which have the slightly stronger property mentioned in the final statement of Lemma 61.2.1. We give this property a name.

Definition 61.2.3. A spectral space $X$ is w-local if the set of closed points $X_0$ is closed and every point of $X$ specializes to a unique closed point. A continuous map $f : X \to Y$ of w-local spaces is w-local if it is spectral and maps any closed point of $X$ to a closed point of $Y$.

We have seen in the proof of Lemma 61.2.1 that in this case $X_0 \to \pi _0(X)$ is a homeomorphism and that $X_0 \cong \pi _0(X)$ is a profinite space. Moreover, a connected component of $X$ is exactly the set of points specializing to a given $x \in X_0$.

Lemma 61.2.4. Let $X$ be a w-local spectral space. If $Y \subset X$ is closed, then $Y$ is w-local.

Proof. The subset $Y_0 \subset Y$ of closed points is closed because $Y_0 = X_0 \cap Y$. Since $X$ is $w$-local, every $y \in Y$ specializes to a unique point of $X_0$. This specialization is in $Y$, and hence also in $Y_0$, because $\overline{\{ y\} }\subset Y$. In conclusion, $Y$ is $w$-local. $\square$

Lemma 61.2.5. Let $X$ be a spectral space. Let

$\xymatrix{ Y \ar[r] \ar[d] & T \ar[d] \\ X \ar[r] & \pi _0(X) }$

be a cartesian diagram in the category of topological spaces with $T$ profinite. Then $Y$ is spectral and $T = \pi _0(Y)$. If moreover $X$ is w-local, then $Y$ is w-local, $Y \to X$ is w-local, and the set of closed points of $Y$ is the inverse image of the set of closed points of $X$.

Proof. Note that $Y$ is a closed subspace of $X \times T$ as $\pi _0(X)$ is a profinite space hence Hausdorff (use Topology, Lemmas 5.23.9 and 5.3.4). Since $X \times T$ is spectral (Topology, Lemma 5.23.10) it follows that $Y$ is spectral (Topology, Lemma 5.23.5). Let $Y \to \pi _0(Y) \to T$ be the canonical factorization (Topology, Lemma 5.7.9). It is clear that $\pi _0(Y) \to T$ is surjective. The fibres of $Y \to T$ are homeomorphic to the fibres of $X \to \pi _0(X)$. Hence these fibres are connected. It follows that $\pi _0(Y) \to T$ is injective. We conclude that $\pi _0(Y) \to T$ is a homeomorphism by Topology, Lemma 5.17.8.

Next, assume that $X$ is w-local and let $X_0 \subset X$ be the set of closed points. The inverse image $Y_0 \subset Y$ of $X_0$ in $Y$ maps bijectively onto $T$ as $X_0 \to \pi _0(X)$ is a bijection by Lemma 61.2.1. Moreover, $Y_0$ is quasi-compact as a closed subset of the spectral space $Y$. Hence $Y_0 \to \pi _0(Y) = T$ is a homeomorphism by Topology, Lemma 5.17.8. It follows that all points of $Y_0$ are closed in $Y$. Conversely, if $y \in Y$ is a closed point, then it is closed in the fibre of $Y \to \pi _0(Y) = T$ and hence its image $x$ in $X$ is closed in the (homeomorphic) fibre of $X \to \pi _0(X)$. This implies $x \in X_0$ and hence $y \in Y_0$. Thus $Y_0$ is the collection of closed points of $Y$ and for each $y \in Y_0$ the set of generalizations of $y$ is the fibre of $Y \to \pi _0(Y)$. The lemma follows. $\square$

Comment #2531 by Brian Conrad on

You forgot to say $x \in U_0$ and $y \in V_0$ in the last paragraph of the proof of Lemma 52.2.1.

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