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## The Noetherian condition as compactness

Let’s think more about what an abstract theory of unique factorization of primes has to look like. One fundamental property it has to satisfy is that factorizations should be finite. Another way of saying this is that the process of writing elements as products of other elements (up to units) should end in a finite set of irreducible elements at some point. This condition is clearly not satisfied by sufficiently “large” commutative rings such as $\mathbb{C}[x, x^{ \frac{1}{2} }, x^{ \frac{1}{3} }, ... ]$, the ring of fractional polynomials.

Since we know we should think about ideals instead of numbers, let’s recast the problem in a different way: because we can write $x^{r} = x^{ \frac{r}{2} } x^{ \frac{r}{2} }$ for any $r$, the ascending chain of ideals $(x) \subset (x^{ \frac{1}{2} }) \subset (x^{ \frac{1}{4} }) \subset ...$ never terminates. In any reasonable theory of factorization writing $f = f_1 g_1$ and then comparing the ideals $(f) \subset (f_1)$, then repeating this process to obtain a chain of ideals $(f) \subset (f_1) \subset (f_2) \subset ...$, should eventually stabilize at a prime. This leads to the following definition.

Definition: A commutative ring $R$ is Noetherian if every ascending chain of ideals stabilizes.

Akhil’s posts at Delta Epsilons here and here describe the basic properties of Noetherian rings well, including the proof of the following.

Hilbert’s Basis Theorem: If $R$ is a Noetherian ring, so is $R[x]$.

Today we’ll discuss what the Noetherian condition means in terms of the topology of $\text{MaxSpec}$. The answer turns out to be quite nice.

Examples

The Noetherian condition is actually so strong that not every UFD satisfies it! For example, the ring $\mathbb{C}[x_1, x_2, ... ]$ of polynomials in countably many variables doesn’t satisfy the Noetherian condition because $(x_1) \subset (x_1, x_2) \subset (x_1, x_2, x_3) ...$ doesn’t stabilize, but this ring is clearly a UFD because the factors of any given polynomial lie in a finitely-generated subring. But from a geometric perspective we don’t really care about such rings, since $\text{MaxSpec } \mathbb{C}[x_1, x_2, ... ]$ contains $\mathbb{C}^{\infty}$. From a practical perspective any question concerning finitely many elements of this ring actually happens in a Noetherian subring.

As we saw above, the Noetherian condition implies that every non-zero non-unit is a product of irreducibles, although this representation is not necessarily unique. It follows that a Noetherian integral domain is a UFD if and only if every irreducible element is prime.

The Noetherian condition is equivalent to the condition that every ideal of $R$ is finitely generated, so it follows that PIDs are Noetherian, hence $\mathbb{Z}, \mathbb{C}[x]$ are Noetherian. Hilbert’s basis theorem together with a few other basic facts then tell us that any finitely-generated algebra over a Noetherian ring is Noetherian, so anything we want to do with polynomial rings or number rings will stay in the Noetherian regime.

Compactness

Remember that when we wanted to characterize the maximal ideals of the ring $C(X)$ of continuous functions on a compact Hausdorff space, we needed the compactness condition to take an ideal $I$ not contained in the ideals $m_x$ and extract finitely many elements $f_1, ... f_r$ of it. The Noetherian condition automatically lets us do this, so it’s reasonable to suppose that the Noetherian condition is equivalent to a kind of compactness. In fact, more is true.

Proposition: If a commutative ring $R$ is Noetherian, then every subset of $X = \text{MaxSpec } R$ is compact in the Zariski topology, i.e. $X$ is hereditarily compact, also known as Noetherian.

Proof. Let $S \subseteq X$. It is a general topological fact that to show that every open cover of a space has a finite subcover it suffices to do so for open covers by a basis of the topology, which here is given by the sets $U_f = \{ x | f \not \in m_x \}, f \in R$. So let $\{ U_{f_i} \}$ be such an open covering of $S$. The ideal $J$ generated by all the $f_i$ is finitely generated, say by $f_1, ... f_r$. For any point $x \in S$ there is some $f \in J$ such that $f \not \in m_x$, so the same must be true for the generators $f_1, ... f_r$, and this is equivalent to $U_{f_1}, ... U_{f_r}$ being a finite subcover.

Unfortunately, the converse is false. For example, let $R = \mathbb{C}[[x, xy, xy^2, ... ]]$. The “evaluation” homomorphism sending a power series to its constant term has kernel the maximal ideal $m = (x, xy, xy^2, ... )$, and since any element of $R - m$ is invertible in $R$, it follows that $m$ is the unique maximal ideal, so $\text{MaxSpec } R$ consists of a single point, i.e. $R$ is a local ring. But the ascending chain of ideals $(x) \subset (x, xy) \subset (x, xy, xy^2) \subset ...$ does not stabilize. (I believe replacing $\text{MaxSpec}$ with $\text{Spec}$, the prime spectrum, fixes this equivalence, but we won’t worry too much about the prime spectrum for now.)

Corollary: If $R$ is Noetherian and $X = \text{MaxSpec } R$ is Hausdorff, then $X$ is finite.

Zorn’s lemma

I sort of glossed over why we know that $\text{MaxSpec } R$ is non-empty. To prove this we need to know that every ideal of $R$ is contained in a maximal ideal. For general rings with unity this is a consequence of Zorn’s lemma, which I am reticent to apply since it is equivalent to the axiom of choice. However, for Noetherian rings one can get by with the axiom of countable choice, which is yet another reason to like Noetherian rings.

Next time we’ll see how the Noetherian condition allows us to decompose a variety into components.

### 7 Responses

1. on November 28, 2009 at 4:18 pm | Reply Theo

I’ve always been under the impression that even Spec doesn’t know enough when the ring is not Noetherian — for example, the power series ring k[[x]] has a unique non-zero prime ideal (any power-series with non-zero constant term is invertible, so the ideals are (1), (x), (x^2), …, 0), but geometrically the corresponding scheme _should_ be (the generic point and) a very large fuzzy point.

Similarly, for any n, I think that Spec(k[x]/x^n) is a single point, right? Again any polynomial that starts at 1 is invertible, so the ideals are (1), (x), …, (x^{n-1}). So the set-valued Spec (or topological-space-valued) really isn’t good enough; you need Spec to take values in locally-ringed spaces. And this is in the Noetherian regime — there are finitely many ideals.

But I’m not much of an algebraic geometer.

By the way, I’ve been very much enjoying the recent series of posts.

Are you planning on talking about commutative C* algebras? I really don’t know whether I believe that the correct algebraic data to attach to a (locally compact Hausdorff) topological space is the ring of all continuous functions or the ring of functions that vanish at infinity. The former seems better suited for an analogy with algebraic spaces; the later provides easier analytic tools.

• on November 28, 2009 at 4:31 pm | Reply Akhil Mathew

“I’ve always been under the impression that even Spec doesn’t know enough when the ring is not Noetherian”

Your first example is noetherian though–am I misunderstanding you?

• on November 28, 2009 at 6:44 pm | Reply Qiaochu Yuan

I agree that the structure sheaf clears up these issues, but I’m trying to stay away from machinery for now.

I don’t really know enough functional analysis to say anything meaningful about $C^{*}$-algebras, so probably not; my goal for the time being is just to understand Dedekind domains thoroughly.

2. on November 28, 2009 at 2:27 pm | Reply Akhil Mathew

“(I believe replacing with , the prime spectrum, fixes this equivalence, but we won’t worry too much about the prime spectrum for now.)”

One implication is true.

If $A$ is a valuation ring whose value group is $\mathbb{Q}$, then $A$ has only one prime ideal but is not noetherian.

I learned this by googling the question and finding these notes. But I disagree with the author’s assertion that this is because $Spec$ only gives information about prime ideals; knowing that the primes are finitely generated is enough to conclude that a ring is noetherian.

• on November 28, 2009 at 6:33 pm | Reply Qiaochu Yuan

Minor correction – valuation rings have two prime ideals, the unique maximal ideal and $(0)$.

What you can show is that $\text{Spec } R$ being Noetherian implies that every ascending chain of radical ideals terminates (and if $R$ is Jacobson it’s enough to show this for $\text{MaxSpec}$). The problem here is that one may have a chain of, say, primary ideals, all of which have the same radical, that does not.

• on November 28, 2009 at 7:10 pm | Reply Akhil Mathew

Hm, yes, you’re right about $(0)$.

But I think you can have more than two prime ideals if the value group is large enough (e.g. $\mathbb{Z} \oplus \mathbb{Z}$ with the lexicographic order—then take the ideal corresponding to elements with valuations $(m,n)$ for $m>n$; I think this is done systematically in Zariski-Samuels).

• on November 29, 2009 at 4:39 am | Reply Akhil Mathew

What I said in my previous comment is false; however, two non-maximal nonzero prime ideals in that valuation ring should arise from elements with valuations $(m,n)$ with, respectively, $m>0$, and $n>0$.