## How to tell if a ring is Noetherian August 9, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
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I briefly outlined the definition and first properties of Noetherian rings and modules a while back.  There are several useful and well-known criteria to tell whether a ring is Noetherian, as I will discuss in this post.  Actually, I’ll only get to the first few basic ones here, though these alone give us a lot of tools for, say, algebraic geometry, when we want to show our schemes are relatively well-behaved.  But there are plenty more to go.

Hilbert’s basis theorem

It is the following:

Theorem 1 (Hilbert) Let ${A}$ be a Noetherian ring. Then the polynomial ring ${A[X]}$ is also Noetherian.

Pick an ideal ${I \subset A[X]}$; we must show ${I}$ is finitely generated.

For each ${n}$, consider the set of all polynomials in ${I}$ of degree ${n}$, and let ${J_n}$ denote the set of their leading coefficients. Now ${J_n}$ is actually an ideal in ${A}$ because ${I=AI}$; note also that ${J_n \subset J_m}$ if ${n, because one can multiply an element of ${I}$ by ${X^{m-n}}$ and stay in ${I}$. Since ${A}$ is Noetherian, the ${J_n}$ eventually stabilize at some ${J}$, which must be finitely generated. Let ${a_1, \dots, a_k}$ be generators; by definition, these are the leading coefficients of some polynomials ${P_1, \dots, P_k \in I}$. Set ${d := \mathrm{max}(deg \ P_i)}$.

The idea now is that given any polynomial ${P \in I}$, we can subtract multiples of ${P_1, \dots, P_k}$ to bring the degree down to less than ${d}$. Suppose ${D := deg \ P > d}$. Then the leading coefficient ${p_0}$ of ${P}$ lies in ${J}$, say

$\displaystyle p_0 = c_1 a_1 + \dots c_k a_k;$

then

$\displaystyle P - \sum c_i X^{D - deg \ P_i} P_i \in I$

has degree strictly less than that of ${P}$, and we proceed inductively.

So the ${P_i}$‘s aren’t yet generators of ${I}$, but almost there: we just have to find ${A}$-generators of the ${A}$-module ${M := \{ Q \in I: deg \ Q < d \}}$, and we can pool these with the ${P_i}$‘s to get ${A[X]}$-generators of ${I}$. But ${M}$ is a submodule of the finitely generated ${A}$-module ${A^d}$, hence finitely generated since ${A}$ is Noetherian. This completes the proof.

The basis theorem is actually often used in the following form:

Corollary 2 Let ${A}$ be a Noetherian ring and ${B \supset A}$ a finitely generated ${A}$-algebra. Then ${B}$ is Noetherian.

Since the quotient ring of a Noetherian ring is Noetherian, we reduce to ${B}$ a polynomial ring; then it follows from Hilbert’s theorem and induction.

In algebraic geometry, this is important because it implies for instance that a scheme of finite type over a field is Noetherian, and is thus fairly well-behaved. As another example, it shows that a subvariety of ${A^n_k}$—which by definition is defined by the zero set of some ideal in the polynomial ring ${k[X_1, \dots, X_n]}$—can be cut out by a finite number of polynomial equations: just take generators of that ideal.

Localization Criteria

Localization preserves the property of a ring being Noetherian:

Proposition 3 If ${A}$ is a Noetherian ring and ${S}$ a multiplicative subset, then ${S^{-1}A}$ is a Noetherian ring.

This follows from general facts about localization, namely that any ideal of ${S^{-1}A}$ is of the form ${S^{-1}I}$ for ${I \subset A}$ an ideal; hence since ${I}$ is finitely generated, so is ${S^{-1}I}$.

There is a more interesting converse: If sufficiently many localizations are Noetherian, so is the original ring.

Theorem 4 Let ${A}$ be a ring with elements ${f_1, \dots, f_n}$ generating the unit ideal. If each localization ${A_{f_i}}$ is Noetherian, so is ${A}$.

For each ${i}$, there are localization maps ${\phi_i: A \rightarrow A_{f_i}}$. Given an ideal ${I \subset A}$, we associate the ideals ${I_{f_i} \subset A_{f_i}}$. By definition, these are defined as ${\phi_i(I) A_{f_i}}$.

I claim that

$\displaystyle I = \bigcap_i \phi_i^{-1}( I_{f_i} ).\ \ \ \ \ (1)$

Indeed, the inclusion ${\subset}$ is clear from the definitions, but the other inclusion requires more checking. Suppose ${x \in A}$ is such that ${\phi_i(x) \in I_{f_i}}$ for all ${i}$; this means that there are powers ${f_i^{m_i}}$ with

$\displaystyle f_i^{m_i} x \in I.$

So, consider the ideal ${J := \{ y\in A: yx \in I \}}$; then ${f_i^{m_i} \in J}$. If we show that ${J=(1)}$, then ${1x = x \in I}$, and my claim will be proved.

We use:

Lemma 5 If ${f_1, \dots, f_n}$ generate the unit ideal in ${A}$, so do ${f_1^m, \dots, f_n^m}$ for any ${m}$.

Indeed, given a relation

$\displaystyle \sum d_i f_i = 1 ,$

raise it to a high power ${M}$ to get

$\displaystyle \left ( \sum d_i f_i \right)^M = 1,$

and every term on the left, when expanded, will lie in the ideal generated by ${f_1^m \dots, f_n^m}$ if ${M >> m}$ is very large.

Return to the proof of the theorem. If we have a sequence ${I^{(1)}, I^{(2)}, \dots}$ of ideals, then for each ${i}$, the ${I^{(j)}_{f_i}}$ each stabilize since ${A_{f_i}}$ is Noetherian; thus by (1), so do the ${I^{(j)}}$.

This result implies the following (see e.g. Hartshorne, II.3):

Corollary 6 If ${\mathrm{Spec} \ A}$ is a locally Noetherian scheme, then ${A}$ is a Noetherian ring.

This is interesting because being locally Noetherian says something about an open affine cover via Noetherian rings.  But this result says that any open affine subscheme comes from a Noetherian ring.

So, we got to some of the basic criteria.  But there are other questions that arise.  For instance, is a subring of a Noetherian ring Noetherian?  In general, no, but there are important cases when we can say yes.   I’ll discuss this in the next post.

1. Harrison - August 10, 2009

So, I’m not an expert in computational ring theory, but I feel like in general, isn’t it hard to tell whether a given ring is isomorphic to a polynomial ring? These are nice ways to construct new Noetherian rings, but given, say, an oracle for a ring I don’t see how either of these is useful.

Still a fine post; I sort of love the proof of Hilbert’s basis theorem, so that’s good.

Harrison - August 10, 2009

By the way, I should make it clear that I’m not implying that these results aren’t incredibly useful — I know enough algebraic geometry to know they are — just that the title is a little misleading.

Akhil Mathew - August 10, 2009

I’m pretty sure that the main reason people care about Hilbert’s basis theorem is actually Corollary 2. Of course, one doesn’t usually find polynomial rings themselves, but instead quotients: the affine varieties in classical algebraic geometry come from finitely generated algebras over a field. One could more generally consider schemes of finite type over a field, which are thus Noetherian. Since algebraic geometry is often easier within Noetherian schemes (e.g. Hartshorne Chapter III focuses mainly on this case), knowing that classical affine (or projective) varieties are of this form is important.

Besides, there are criteria for when an algebra is finitely generated. For instance, if $G$ is a finite group acting via algebra-automorphisms on a finitely generated algebra $A$, then the fixed points $A^G$ form a f.g. algebra, as I’ll probably discuss in a post either today or tomorrow. This can be generalized.

2. Anonymous - August 10, 2009

1. Minor typo: for the inductive step in HBT, when you subtracted off terms to get a lower degree polynomial you forgot to write the factor of P_i.

2. With regards to equation (1): do we need to assume that none of the f_i are zero-divisors?

Anonymous - August 10, 2009

2. Nevermind! I was worried that something in would be killed by localizing; and, not reading carefully, I thought (1) took place in something like the ring of total quotients, not in A.

Akhil Mathew - August 10, 2009

Thanks for the correction!

3. The Noetherian condition as compactness « Annoying Precision - December 17, 2009

[...] posts at Delta Epsilons here and here describe the basic properties of Noetherian rings well, including the proof of the [...]