Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to relate two (maybe not) different decompositions of a noetherian topological space into irreducible subsets, given in Ravi Vakil's notes on algebraic geometry.

Exercise 4.6.N : Let $X$ be a topological space, then any point is contained in a an irreducible component.

It follows that any space $X$ is the union of its (closed) irreducible components, but there is not statement of uniqueness.

Proposition 4.16.14 : Let $X$ be a noetherian topological space, and $Z \subseteq X$ a closed (non-empty) subset. Then there is a unique decomposition $Z = Z_1 \cup \cdots \cup Z_n$ where the $Z_i$'s are irreducible closed subsets, none containing another.

So in particular $X$ is a finite union of irreducible closed sets, but here it seems like they may not be its components.

The exercise is just a small exercise, while the proposition seems to be more important. So my question is : Why is the decomposition in the proposition more important than the one in the exercise ?

Is it because of the uniqueness statement ? But I think the decomposition of $X$ into its irreducible closed components also satisfies this uniqueness condition.

Is it because it applies to any closed subset $Z$ ? But $Z$ also has the decomposition of the exercise. However, the subsets are irreducible in $Z$ and not in $X$, so this is the point ?

Thank you for your help.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

I think the most important thing about the proposition is the finiteness part: any Noetherian topological space can be written as a finite union of irreducible closed subsets.

This is not true of, say, $\mathbb{R}$ with the usual topology. In fact, the only irreducible closed subsets of $\mathbb{R}$ are the singletons, so while it is true that $\mathbb{R}$ is a union of its irreducible closed sets (points), it is an uncountable union of these sets. No information about the topology of $\mathbb{R}$ is gained by knowing this decomposition.

If you want to think about things algebraically, you probably know that in algebraic geometry most Noetherian spaces you deal with are spaces that are (at least locally) given by the spectrum of a Noetherian ring. The Noetherian condition for rings is important in commutative algebra because it is a finiteness condition, saying that the ring can't be too crazy. It then shouldn't be surprising that the usefulness of Noetherianity on the geometry side of things is that it imposes a finiteness condition on the spaces you deal with.

The uniqueness part of the proposition is also important, of course, but once you know that a Noetherian space can be written as a finite union of irreducible closed subsets, it doesn't take any deep reasoning to derive the uniqueness.

share|improve this answer
    
Thanks for the further explanation, everything you said makes perfect sense to me. The thing is, we can always write $X$ as a union of its irreducible components. If $X$ is noetherian, we also have the decomposition from the proposition, where apparently the subsets are not necessarily components. But then, the decomposition in components should have less subsets (since they are maximal irreducible), and thus also finite. So why is the decomposition from the proposition not the decomposition in components ? Why is it better ? –  Bogdan Nov 29 '12 at 2:13
    
@redfiloux: If $X$ is a non-Noetherian topological space, then it is true that $X$ can be written as a union of irreducible closed subsets, but I don't think the terminology "the irreducible components of $X$" makes sense in general. If $X$ is Noetherian, then it makes sense to talk about irreducible components of $X$, and in this case the $Z_i$ from the proposition are the irreducible components of $X$. –  froggie Nov 29 '12 at 2:25
    
It always makes sense to talk about the irreducible component, it is just an irreducible subset maximal for this property (wrt inclusion), see for example here en.wikipedia.org/wiki/Irreducible_component. Well, if the $Z_i$ are the irreducible components of $X$, that is what I was looking for. I think it follows from uniqueness that these are actually the components ! So maybe in general, for any closed $Z \subseteq X$, the $Z_i$'s are the irreducible components of $Z$. Btw, are those components disjoint ? Not necessarily right ? I'll try some examples. –  Bogdan Nov 29 '12 at 2:36
1  
@redfiloux: The irreducible components of a Noetherian space are not necessarily disjoint. A good example of this is the space that is the union of two intersecting lines in the Zariski topology. If $X$ is a Noetherian space and $Z\subset X$ is closed, then $Z$ can also be written uniquely as a union of finitely many irreducible components (this follows from the proposition and the fact that $Z$ is itself a Noetherian topological space). When $X$ is not Noetherian, I wouldn't be surprised if $X$ didn't have any irreducible components, though I don't have an example right now.... –  froggie Nov 29 '12 at 2:42
    
Thanks for the example. I begin to see more clear in this. The last sentence is not correct : exercise 4.6.N is exactly to show that in any topological space $X$, every point $x$ is contained in an irreducible component. It uses Zorn's Lemma though on, I think on a subset of the poset of irreducible closed subsets containing $x$. –  Bogdan Nov 29 '12 at 2:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.