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.