If it wasn’t clear, in this discussion all rings are assumed commutative.

Given a variety like we’d like to know if there’s a natural way to decompose it into its “components” . These aren’t its connected components in the topological sense, but in any reasonable sense the two parts are unrelated except possibly where they intersect. It turns out that the Noetherian condition is a natural way to answer this question. In fact, we will see that the Noetherian condition allows us to write uniquely as a union of a finite number of “components” which have a natural property that is stronger than connectedness.

**An alternate characterization**

First we need a second characterization of Noetherian spaces which is perhaps more natural than the one we gave before in light of its relationship to the Noetherian condition.

**Proposition:** A topological space is hereditarily compact if and only if every descending chain of closed subsets stabilizes.

*Proof.* : The proof is identical to the one in the previous post. The sets form an open cover of , so we can choose a finite subcover.

: Let . Any open cover of can be refined so that each open set contains the next, so their complements are a descending chain of closed subsets, which stabilizes. Hence has a finite subcover.

**Irreducibility**

What distinguishes a variety like from a variety like ? Well, for one thing, the latter carries the cofinite topology, so all closed subsets are finite, whereas and are proper closed subsets of the former. This leads to the following definition.

**Definition:** A topological space is called **irreducible** or hyperconnected if it is not the union of two of its proper closed subsets.

Be careful to distinguish this notion of irreducibility from the one for elements of rings. Now, irreducibility is a much stronger condition than connectedness; indeed, it’s equivalent to the condition that nonempty open subsets are dense, and implies both connectedness and local connectedness. Any cofinite topology is irreducible, so this approach seems promising since it implies that the maximal spectrum of a PID is irreducible.

**Definition:** An irreducible **component** of a topological space is an irreducible subspace that is not properly contained in any other irreducible subspace, i.e. it is **maximal.**

**Proposition:** a) The irreducible components of any topological space are closed.

b) Any element of is contained in at least one irreducible component.

c) Any Noetherian space has finitely many irreducible components.

*Proof.* a) It suffices to show that the closure of an irreducible subspace is irreducible. Given irreducible write where are closed in . Then the restrictions of to are closed in the subspace topology, so is reducible unless one of is disjoint from . Then the other must contain , and since it is closed it must be all of , so one of is not proper.

b) The closure of any point is irreducible, and is either maximal or contained in another irreducible subset , and so forth. A union of an ascending chain of irreducible sets is itself irreducible, so Zorn’s lemma applies.

c) (**Edited**, 12/10/09:) This is “Noetherian induction.” If can’t be written as the union of finitely many irreducible closed subsets, then the set of closed subsets of that can’t be written as such a union has a minimal element since it satisfies the dcc. Since is not itself irreducible, it must be the union of two of its proper closed subsets – but by minimality can be written as the union of finitely many irreducible closed subsets; contradiction. Now any irreducible closed subset of belongs to some irreducible component.

It follows that Noetherian spaces have a unique decomposition into finitely many irreducible components.

**Irreducible components of **

The ideal in factors into the product of the prime ideals , so we might expect that there is a close relationship between prime ideals and irreducible subspaces of the maximal spectrum.

**Proposition:** a) Let be an irreducible subspace of and let . Then is prime.

b) If is a prime ideal, then is an irreducible subspace.

*Proof.* a) We’ll prove the contrapositive. If isn’t prime, there exist not in , such that , hence are proper closed subsets of whose union is .

b) We’ll prove the contrapositive. If where are closed, then consists only of functions which vanish on , hence is contained in , but are not contained in .

Unfortunately, not every prime ideal of a generic ring is an intersection of maximal ideals. For example, even for a Noetherian ring like the only maximal ideal is , but both and are prime ideals. Rings such that every prime ideal is an intersection of maximal ideals are called Jacobson rings, and an extremely general form of the Nullstellensatz asserts that any finitely generated algebra over a Jacobson ring is Jacobson. Since has the property that every nonzero prime ideal is maximal and fields are Jacobson, it follows that any of the finitely-generated rings we care about are Jacobson. Geometrically, Jacobson rings are those for which can be recovered from , so we can do without the former (as we have been so far).

In a Jacobson ring , the irreducible components of correspond to the minimal primes of , so a Noetherian Jacobson ring has finitely many minimal primes. Just as every ideal is contained in a maximal ideal, every prime ideal (including every maximal ideal) contains a minimal prime ideal. The set of all minimal prime ideals contained in a given maximal ideal describes what components of it lies on, so there is a nice geometric interpretation here.

If has more than one component then cannot be an integral domain, which is one indication that the geometric perspective says something meaningful about prime factorization. Indeed, we’ll eventually show that bad things happen if even looks “locally” like it has more than one component.

Next time, we’ll describe more carefully the exact relationship between ideals and varieties.

on April 17, 2013 at 12:06 pm |paterz1Another issue I’m having… I can’t prove it on my own this time though.

“It follows that Noetherian spaces have a unique decomposition into finitely many irreducible components. ”

Why? You’ve only proven that the space can be written as a union of finitely many irreducible components, you don’t know if there are finitely many irreducible components at all, so we don’t know about uniqueness either… maybe I’m missing something but I don’t see it.

on April 17, 2013 at 12:18 pm |paterz1I think I figured it out, but it doesn’t seem so straightforward… I’ll wait for your answer though.

on April 17, 2013 at 12:21 pm |Qiaochu YuanI suppose there’s a little more that needs to be said here. I’ve written a Noetherian space as a union of finitely many irreducible components . Now, any other irreducible component must be one of the (since it must have non-empty intersection with one of the irreducible components above, and therefore must be identical to it), so any other decomposition of into irreducible components must consist of precisely the same .

on April 17, 2013 at 12:26 pm |paterz1Strangely I can’t reply below your answer.

Um is what you said equivalent to if $\mathcal M$ is an irreducible component of $X$, then $\mathcal M = \bigcup (\mathcal M \cap X_i)$, hence $\mathcal M \cap X_i = \mathcal M$ for some $i$, so that $\mathcal M \subseteq X_i$ hence $\mathcal M = X_i$ by maximality? Because you seem to “see it” but I wonder if you had the same details in mind or something even more straight forward.

on April 17, 2013 at 12:27 pm |paterz1(….) Guess that button did allow me to reply below your answer.

on April 17, 2013 at 12:38 pm |Qiaochu YuanWordPress comments have limited nesting depth for some reason.

Yes, that’s what I had in mind.

on April 17, 2013 at 11:11 am |paterz1This part :

“a) It suffices to show that the closure of an irreducible subspace is irreducible. Given S \subset X irreducible write \bar{S} = X_1 \cup X_2 where X_1, X_2 are closed in \bar{S}. Then the restrictions of X_1, X_2 to S are closed in the subspace topology, so S is reducible unless one of X_1, X_2 is disjoint from S. Then the other must contain S, and since it is closed it must be all of \bar{S}, so one of X_1, X_2 is empty. ”

feels a little wrong to me (I still agree with the statement though). I don’t understand how you get empty sets in there. My proof goes like this : if \bar{S} = X_1 \cup X_2, then S = (S \cap X_1) \cup (S \cup X_2) where (S \cap X_i) are closed in the subspace topology, hence S \cap X_1 = S or S \cap X_2 = S, i.e. S \subseteq X_1 or S \subseteq X_2. Since X_1 and X_2 are closed in \bar{S}, it means X_1 \cup X_2 = S \subseteq X_1 or X_1 \cup X_2 = S \subseteq X_2, i.e. S = X_1 or S = X_2. I don’t get any empty sets anywhere.

Can you either explain your proof more in detail or agree with my comment? I’m still a little confused with this irreducible component story…

on April 17, 2013 at 11:52 am |Qiaochu YuanYou’re right. I don’t need to say anything about emptiness. I think I had in mind extra hypotheses on that are unnecessary.

on April 17, 2013 at 11:55 am |paterz1Cool! Thanks.

on December 25, 2009 at 10:25 am |soarerzThere’s a typo in the definition of in the proposition.

on December 25, 2009 at 12:36 pm |Qiaochu YuanWhoops. Thanks for the correction!