7. Noetherian Rings.

7.1 Proposition. Let R be a commutative ring. Then the following are equivalent:

(1) Every ideal of R is finitely generated.

(2) Every ascending chain of ideals of R is of finite length.

(3) Every non-empty set of ideals of R has a maximal element.

Proof. (1 ⇒ 2) and (2 ⇒ 3) are easy. (3 ⇒ 1) Let I be an ideal of R and S the set of finitely generated ideals contained in I. S (being non-empty because it contains (0)) has a maximal element J. Suppose J ≠ I and pick i ∈ I \ J. Then iR + J ⊆ I is finitely generated and strictly contains J, a contradiction. □

7.2 Definition. A commutative ring satisfying one and hence all of the equivalent conditions of the preceeding theorem is called Noetherian.

Condition (2) above is called “the ascending chain condition” (ACC) (for ideals). Similar conditions for objects other than ideals, or for special kinds of ideals, often play a rôle. For instance, a generalization of Noetherian domains is given by requiring the ascending chain condition only for divisorial ideals; the resulting domains are called Mori domains.

7.3 Definition. Let R be a commutative ring, M an R-module. M is called Noetherian if every R submodule of M is finitely generated.

Since the ideals of a ring R are precisely the R-submodules of R, a ring R is Noetherian (as a ring) if and only if it is a Noetherian R-module.

7.4 Remark. As in Proposition 7.1, the following are equivalent:

(1) M is a Noetherian R-module

(2) Every ascending chain of R-submodules of M is of finite length.

(3) Every non-empty set of R-submodules of M contains a maximal element.

7.5 Lemma. Let M be am R-module and L a submodule of M. Then M is Noetherian if and only if both L and M/L are Noetherian.

Proof. If M is Noetherian, then L is Noetherian, since every submodule of L is s submodule of M. Also, M/L is Noetherian, since every submodule of M/L is of the form π(N), where N is a submodule of M and π: M → M/L the canonical projection, and π(N) is generated by the images under π of a set of generators of N.

Conversely, suppose L and M/L are Noetherian. Given a submodule N of M, let a n_{1}, … n_{s} be generators of N ∩ L and m_{1} + L, …, m_{t} + L generators of (N + L)/L with m_{1}, …, m_{t} ∈ N. Then N is generated by n_{1}, … n_{s}, m_{1}, …, m_{t}. □

7.6 Proposition. For every Noetherian ring R, every finitely generated R-module is Noetherian.

Proof. We only need to show that every free R-module of finite rank is Noetherian, since every R-module generated by n elements is a homomorphic image of the free R-module of rank n.

Induction on the rank of M: For n = 1, M = R, and R is Noetherian. For rank(M) = n > 1, M = R ⊕ L, L a free R-module of rank n − 1. L is Noetherian by induction hypothesis, M/L = R is Noetherian, and therefore M is Noetherian by the previous Lemma. □

7.7 Theorem (Hilbert’s basis theorem). Let R be a Noetherian ring. Then R[x] is Noetherian.

Proof. Let J ≠ (0) be an ideal of R[x]. For n ∈ ℕ, let I_{n} be the set of leading coefficients of polynomials of degree n or less in J and set I = ⋃_{n∈ℕ} I_{n}. The I_{n} are ideals of R with I_{0} ⊆ I_{1} ⊆ I_{2} ⊆ … and therfore I is an ideal of R too. The fact that both I and all I_{n} are finitely generated now implies that J is finitely generated: let g_{1}, …, g_{m} ∈ J be polynomials whose leading coefficients generate I and let N = max_{1≤k≤m} deg g_{k}. To g_{1}, …, g_{m} we add for each n < N a finite set of elements of J of degree n or less whose leading coefficients generate I_{n}. That the resulting finite set of polynomials generates J is now easily seen by induction on the degree of an element of J. □

7.8 Corollary. Let R be a Noetherian ring. Then R[x_{1}, …, x_{n}] and therefore every finitely generated R-algebra (as a homomorphic image of a polynomial ring R[x_{1}, …, x_{n}]) is Noetherian.

7.9 Lemma. Let R be a commutative ring and P an ideal of R that is not finitely generated and which is maximal with respect to this property. Then P is prime.

Proof. Suppose ab ∈ P and a ∉ P, b ∉ P. By maximality of P, P + (a) is finitely generated, by elements p_{1} + r_{1}a, …, p_{k} + r_{k}a (with p_{i} ∈ P, r_{i} ∈ R), and similarly, P + (b) is finitely generated, by elements q_{1} + s_{1}a, …, q_{m} + s_{m}a (with q_{i} ∈ P, s_{i} ∈ R).

Also, the ideal (P : a) = {r ∈ R | ra ∈ P } properly containes P (because P ⊂ P + (b) ⊆ (P : a)), and is therefore finitely generated, by j_{1}, …, j_{h}, say.

If c∈P, then c∈P +(a) and there exist t_{1}, …, t_{k} ∈R with c=t_{1}p_{1}+…+t_{k}p_{k}+ (t_{1}r_{1} + … + t_{k}r_{k})a. Let d = t_{1}r_{1} + … + t_{k}r_{k}, then da = c − (t_{1}p_{1} + … + t_{k}p_{k}) ∈ P and therefore d ∈ (P : a). Let t_{1}′, …, t_{h}′ ∈ R such that d = t_{1}′j_{1} + … + t_{h}′j_{h}.

Then c = t_{1}p_{1} + … + t_{k}p_{k} + t_{1}′j_{1}a + … + t_{h}′j_{h}a, and p_{1}, …, p_{k} as well as j_{1}a, …, j_{h}a are in P. We have shown that p_{1}, …, p_{k}, j_{1}a, …, j_{h}a generate P, a contradiction. □

7.10 Theorem (Cohen). Let R be a commutative ring. If every prime ideal of R is finitley generated, then R is Noetherian.

Proof. The set of non-finitely-generated ideals of a commutative ring clearly satisfies the hypothesis of Zorn’s lemma. Therefore, if this set is non-empty, it has a maximal element by Zorn’s lemma. Such a maximal element is a prime ideal by the previous lemma. Therefore, every non-Noetherian ring has a prime ideal that is not finitely generated. □