The Stacks project

Lemma 5.9.2. Let $X$ be a Noetherian topological space.

  1. Any subset of $X$ with the induced topology is Noetherian.

  2. The space $X$ has finitely many irreducible components.

  3. Each irreducible component of $X$ contains a nonempty open of $X$.

Proof. Let $T \subset X$ be a subset of $X$. Let $T_1 \supset T_2 \supset \ldots $ be a descending chain of closed subsets of $T$. Write $T_ i = T \cap Z_ i$ with $Z_ i \subset X$ closed. Consider the descending chain of closed subsets $Z_1 \supset Z_1\cap Z_2 \supset Z_1 \cap Z_2 \cap Z_3 \ldots $ This stabilizes by assumption and hence the original sequence of $T_ i$ stabilizes. Thus $T$ is Noetherian.

Let $A$ be the set of closed subsets of $X$ which do not have finitely many irreducible components. Assume that $A$ is not empty to arrive at a contradiction. The set $A$ is partially ordered by inclusion: $\alpha \leq \alpha ' \Leftrightarrow Z_{\alpha } \subset Z_{\alpha '}$. By the descending chain condition we may find a smallest element of $A$, say $Z$. As $Z$ is not a finite union of irreducible components, it is not irreducible. Hence we can write $Z = Z' \cup Z''$ and both are strictly smaller closed subsets. By construction $Z' = \bigcup Z'_ i$ and $Z'' = \bigcup Z''_ j$ are finite unions of their irreducible components. Hence $Z = \bigcup Z'_ i \cup \bigcup Z''_ j$ is a finite union of irreducible closed subsets. After removing redundant members of this expression, this will be the decomposition of $Z$ into its irreducible components (Lemma 5.8.4), a contradiction.

Let $Z \subset X$ be an irreducible component of $X$. Let $Z_1, \ldots , Z_ n$ be the other irreducible components of $X$. Consider $U = Z \setminus (Z_1\cup \ldots \cup Z_ n)$. This is not empty since otherwise the irreducible space $Z$ would be contained in one of the other $Z_ i$. Because $X = Z \cup Z_1 \cup \ldots Z_ n$ (see Lemma 5.8.3), also $U = X \setminus (Z_1\cup \ldots \cup Z_ n)$ and hence open in $X$. Thus $Z$ contains a nonempty open of $X$. $\square$


Comments (1)

Comment #9444 by on

In the second paragraph, first sentence, I think we should replace “which do not have finitely many irreducible components” by “which cannot be written as a finite union of irreducible closed subsets” (and in “by construction and are finite unions of their irreducible components,” change components by subsets). Otherwise, unless we appeal to Lemma 5.8.3, point 3 (which requires Zorn's lemma), we cannot write and as a union of irreducible closed subsets.

In the statement, point 2, one could add “and it is the union of these.”

At the end of second paragraph, the last sentence could be replaced by “since is empty, is a finite union of some irreducible subsets ; removing redundant 's, we win by Lemma 5.8.4.”

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