Chapter 0 — Preliminaries

§2. Irreducible Spaces. Noetherian Spaces

2.1. Irreducible spaces

(2.1.1) A topological space $X$ is called irreducible if it is nonempty and is not the union of two proper closed subspaces. Equivalently: $X\ne \varnothing$, and the intersection of any two (hence any finitely many) nonempty open subsets of $X$ is nonempty; or every nonempty open subset of $X$ is dense; or every closed subset $\ne X$ is nowhere dense [EGA: rare]; or every open subset of $X$ is connected.

(2.1.2) A subspace $Y$ of a topological space $X$ is irreducible if and only if its closure $\bar{Y}$ is irreducible. In particular, every subspace of the form $\bar{x}$ is irreducible; we shall express the relation $y\in \bar{x}$ (equivalently, $\bar{y}\subset \bar{x}$) by saying that $y$ is a specialization of $x$, or that $x$ is a generization of $y$. If an irreducible space $X$ contains a point $x$ with $X=\bar{x}$, we call $x$ a generic point of $X$. Every nonempty open subset of $X$ then contains $x$, and every subspace containing $x$ admits $x$ as a generic point.

(2.1.3) Recall that a topological space $X$ is called a Kolmogorov space if it satisfies the separation axiom:

(T₀) If $x\ne y$ are two distinct points of $X$, there is an open set containing exactly one of $x$, $y$.

If an irreducible Kolmogorov space admits a generic point, it admits exactly one, since every nonempty open set contains every generic point.

Recall that a topological space $X$ is called quasi-compact if every open cover of $X$ admits a finite subcover (equivalently, if every decreasing filtered family of nonempty closed sets has nonempty intersection). If $X$ is quasi-compact, every nonempty closed subset $A\subset X$ contains a minimal nonempty closed subset $M$ (since the set of nonempty closed subsets of $A$ is inductive for the relation $\supset$); if $X$ is moreover a Kolmogorov space, $M$ is necessarily reduced to a single point, called by abuse of language a closed point.

(2.1.4) In an irreducible space $X$, every nonempty open subspace $U$ is irreducible, and if $X$ admits a generic point $x$, then $x$ is also a generic point of $U$.

Let $(U_{\alpha })$ be a cover of a topological space $X$ by nonempty open sets, with nonempty index set. For $X$ to be irreducible, it is necessary and sufficient that each $U_{\alpha }$ be irreducible and that $U_{\alpha }\cap U_{\beta }\ne \varnothing$ for all $\alpha ,\beta$. The condition is plainly necessary; for sufficiency, it suffices to show that if $V$ is a nonempty open subset of $X$, then $V\cap U_{\alpha }\ne \varnothing$ for every $\alpha$, for then $V\cap U_{\alpha }$ is dense in $U_{\alpha }$ for every $\alpha$, and so $V$ is dense in $X$. Now there is at least one index $\gamma$ with $V\cap U_{\gamma }\ne \varnothing$; then $V\cap U_{\gamma }$ is dense in $U_{\gamma }$, and since $U_{\alpha }\cap U_{\gamma }\ne \varnothing$ for every $\alpha$, we have $V\cap U_{\alpha }\cap U_{\gamma }\ne \varnothing$.

(2.1.5) Let $X$ be an irreducible space and $f:X\to Y$ a continuous map into a topological space $Y$. Then $f(X)$ is irreducible, and if $x$ is a generic point of $X$, then $f(x)$ is a generic point of $f(X)$, hence also of $f\bar{X}$. In particular, if moreover $Y$ is irreducible with a unique generic point $y$, then $f(X)$ is dense in $Y$ if and only if $f(x)=y$.

(2.1.6) Every irreducible subspace of a topological space $X$ is contained in a maximal irreducible subspace, which is necessarily closed. The maximal irreducible subspaces of $X$ are called the irreducible components of $X$. If $Z_1$, $Z_2$ are distinct irreducible components of $X$, then $Z_1\cap Z_2$ is nowhere dense in each of $Z_1$ and $Z_2$; in particular, if an irreducible component admits a generic point (2.1.2), that point lies in no other irreducible component. If $X$ has only finitely many irreducible components $Z_i$ ($1\le i\le n$), and if $U_i=Z_i\cap \complement ({\bigcup }_{j\ne i}{Z}_j)$, then the $U_i$ are open, irreducible, pairwise disjoint, and their union is dense in $X$.

Let $U$ be an open subset of a topological space $X$. If $Z$ is an irreducible subset of $X$ meeting $U$, then $Z\cap U$ is open and dense in $Z$, hence irreducible; conversely, for every closed irreducible subset $Y$ of $U$, the closure $\bar{Y}$ of $Y$ in $X$ is irreducible and $\bar{Y}\cap U=Y$. There is therefore a bijective correspondence between the irreducible components of $U$ and the irreducible components of $X$ meeting $U$.

(2.1.7) If a topological space $X$ is the union of finitely many closed irreducible subspaces $Y_i$, then the irreducible components of $X$ are the maximal elements among the $Y_i$: for any closed irreducible $Z\subset X$, $Z$ is the union of the $Z\cap Y_i$, so $Z$ must be contained in some $Y_i$. If $Y$ is a subspace of a topological space $X$ and $Y$ has only finitely many irreducible components $Y_i$ ($1\le i\le n$), then the closures ${\bar{Y}}_i$ in $X$ are the irreducible components of $\bar{Y}$.

(2.1.8) Let $Y$ be an irreducible space with a unique generic point $y$, let $X$ be a topological space, and let $f:X\to Y$ be continuous. Then for every irreducible component $Z$ of $X$ meeting $f^{-1}(y)$, $f(Z)$ is dense in $Y$. The converse need not hold; however, if $Z$ admits a generic point $z$ and $f(Z)$ is dense in $Y$, then necessarily $f(z)=y$ (2.1.5). Moreover, $Z\cap f^{-1}(y)$ is then the closure of $z$ in $f^{-1}(y)$, hence irreducible; and since every irreducible subset of $f^{-1}(y)$ containing $z$ is contained in $Z$ (2.1.6), $z$ is a generic point of $Z\cap f^{-1}(y)$. Since every irreducible component of $f^{-1}(y)$ is contained in some irreducible component of $X$, one sees that if every irreducible component $Z$ of $X$ meeting $f^{-1}(y)$ admits a generic point, then there is a bijective correspondence between these components and the irreducible components of $f^{-1}(y)$, the generic points of $Z$ coinciding with those of $Z\cap f^{-1}(y)$.

2.2. Noetherian spaces

(2.2.1) A topological space $X$ is called Noetherian if the set of open subsets of $X$ satisfies the ascending chain condition (equivalently, if the set of closed subsets satisfies the descending chain condition). $X$ is called locally Noetherian if every $x\in X$ admits a neighborhood that is a Noetherian subspace.

(2.2.2) Let $E$ be an ordered set satisfying the descending chain condition, and let P be a property of elements of $E$ such that: if $a\in E$ and P(x) holds for every $x<a$, then P(a) holds. Under these conditions, P(x) holds for every $x\in E$ — "principle of Noetherian induction". Indeed, let $F=x\in E:\neg P(x)$; if $F$ were nonempty, it would have a minimal element $a$, and then P(x) would hold for every $x<a$, so P(a) would hold too, a contradiction. We shall apply this principle in particular when $E$ is a set of closed subsets of a Noetherian space.

(2.2.3) Every subspace of a Noetherian space is Noetherian. Conversely, every finite union of Noetherian subspaces of a topological space is Noetherian.

(2.2.4) Every Noetherian space is quasi-compact; conversely, a topological space in which every open subset is quasi-compact is Noetherian.

(2.2.5) A Noetherian space has only finitely many irreducible components, as one sees by Noetherian induction.