Feeds:
Posts

## Irreducible components

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

Given a variety like $xy = 0$ we’d like to know if there’s a natural way to decompose it into its “components” $x = 0, y = 0$. 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 $\text{MaxSpec } R$ 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 $X$ is hereditarily compact if and only if every descending chain of closed subsets $X_1 \supset X_2 \supset ...$ stabilizes.

Proof. $\Rightarrow$: The proof is identical to the one in the previous post. The sets $X - X_1, X - X_2, ...$ form an open cover of $X - \bigcap X_i$, so we can choose a finite subcover. $\Leftarrow$: Let $S \subseteq X$. Any open cover of $S$ can be refined so that each open set contains the next, so their complements are a descending chain of closed subsets, which stabilizes. Hence $S$ has a finite subcover.

Irreducibility

What distinguishes a variety like $xy = 0$ from a variety like $x = 0$? Well, for one thing, the latter carries the cofinite topology, so all closed subsets are finite, whereas $x = 0$ and $y = 0$ 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 $X$ are closed.

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

c) Any Noetherian space $X$ has finitely many irreducible components.

Proof. 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 not proper.

b) The closure $X_1$ of any point $x$ is irreducible, and $X_1$ is either maximal or contained in another irreducible subset $X_2$, and so forth. A union of an ascending chain $X_1 \subset X_2 \subset ...$ of irreducible sets is itself irreducible, so Zorn’s lemma applies.

c) (Edited, 12/10/09:) This is “Noetherian induction.” If $X$ can’t be written as the union of finitely many irreducible closed subsets, then the set of closed subsets of $X$ that can’t be written as such a union has a minimal element $W$ since it satisfies the dcc. Since $W$ is not itself irreducible, it must be the union $W = W_1 \cup W_2$ of two of its proper closed subsets – but by minimality $W_1, W_2$ can be written as the union of finitely many irreducible closed subsets; contradiction. Now any irreducible closed subset of $X$ belongs to some irreducible component.

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

Irreducible components of $\text{MaxSpec}$

The ideal $(xy)$ in $\mathbb{C}[x, y]$ factors into the product of the prime ideals $(x), (y)$, so we might expect that there is a close relationship between prime ideals and irreducible subspaces of the maximal spectrum.

Proposition: a) Let $X$ be an irreducible subspace of $\text{MaxSpec } R$ and let $P = I(X) = \bigcap_{x \in X} m_x$. Then $P$ is prime.

b) If $P$ is a prime ideal, then $V(P) = \{ m | m \supseteq P \}$ is an irreducible subspace.

Proof. a) We’ll prove the contrapositive. If $P$ isn’t prime, there exist $f, g$ not in $P$, such that $fg \in P$, hence $V(f), V(g)$ are proper closed subsets of $X$ whose union is $X$.

b) We’ll prove the contrapositive. If $V(P) = X_1 \cup X_2$ where $X_1, X_2$ are closed, then $J = I(X_1) I(X_2)$ consists only of functions which vanish on $V(P)$, hence is contained in $P$, but $I(X_1), I(X_2)$ are not contained in $P$.

Unfortunately, not every prime ideal of a generic ring is an intersection of maximal ideals. For example, even for a Noetherian ring like $\mathbb{C}[[x, y]]$ the only maximal ideal is $(x, y)$, but both $(x)$ and $(y)$ 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 $\mathbb{Z}$ 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 $\text{Spec}$ can be recovered from $\text{MaxSpec}$, so we can do without the former (as we have been so far).

In a Jacobson ring $R$, the irreducible components of $\text{MaxSpec } R$ correspond to the minimal primes of $R$, 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 $\text{MaxSpec}$ it lies on, so there is a nice geometric interpretation here.

If $\text{MaxSpec } R$ has more than one component then $R$ 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 $\text{MaxSpec } R$ even looks “locally” like it has more than one component.

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

### 11 Responses

1. on April 17, 2013 at 12:06 pm | Reply paterz1

Another 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 | Reply paterz1

I 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 | Reply Qiaochu Yuan

I suppose there’s a little more that needs to be said here. I’ve written a Noetherian space $X$ as a union of finitely many irreducible components $\bigcup X_i$. Now, any other irreducible component must be one of the $X_i$ (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 $X$ into irreducible components must consist of precisely the same $X_i$.

• on April 17, 2013 at 12:26 pm | Reply paterz1

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 | Reply paterz1

• on April 17, 2013 at 12:38 pm | Reply Qiaochu Yuan

WordPress comments have limited nesting depth for some reason.

Yes, that’s what I had in mind.

2. on April 17, 2013 at 11:11 am | Reply paterz1

This 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 | Reply Qiaochu Yuan

You’re right. I don’t need to say anything about emptiness. I think I had in mind extra hypotheses on $X_1, X_2$ that are unnecessary.

3. on December 25, 2009 at 10:25 am | Reply soarerz

There’s a typo in the definition of $V(P)$ in the proposition.