Processing math: 100%

The Stacks project

Lemma 5.8.4. Let X be a topological space and suppose X = \bigcup _{i = 1, \ldots , n} X_ i where each X_ i is an irreducible closed subset of X and no X_ i is contained in the union of the other members. Then each X_ i is an irreducible component of X and each irreducible component of X is one of the X_ i.

Proof. Let Y be an irreducible component of X. Write Y = \bigcup _{i = 1, \ldots , n} (Y \cap X_ i) and note that each Y \cap X_ i is closed in Y since X_ i is closed in X. By irreducibility of Y we see that Y is equal to one of the Y \cap X_ i, i.e., Y \subset X_ i. By maximality of irreducible components we get Y = X_ i.

Conversely, take one of the X_ i and expand it to an irreducible component Y, which we have already shown is one of the X_ j. So X_ i \subset X_ j and since the original union does not have redundant members, X_ i = X_ j, which is an irreducible component. \square


Comments (1)

Comment #9443 by on

In the second paragraph, it seems that “take one of the and expand it to an irreducible component ” needs Lemma 5.8.3, point (3), which requires Zorn's lemma. It can be done without the axiom of choice: the second paragraph may be replaced by

Suppose , where is some irreducible subset of . By the argument in the first paragraph, for some . Since the original union does not have redundant members, . Therefore is maximal in the set of irreducible subsets of , that is, is an irreducible component of .

There are also:

  • 9 comment(s) on Section 5.8: Irreducible components

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.