# Noetherian Rings and Modules

Let  be a commutative ring with unit element. We will frequently work with -modules, which are like vector spaces but over a ring.

More precisely, an -module is an additive abelian group equipped with a map such that for all  and all we have , , , and . A submodule is a subgroup of that is preserved by the action of . An ideal in a ring is an -submodule , where we view as a module over itself.

Example 2.2.1   The set of abelian groups are in natural bijection with -modules.

A homomorphism of -modules is a abelian group homomorphism such that for any and we have . A short exact sequence of -modules

is a specific choice of injective homomorphism and a surjective homomorphism such that .

Example 2.2.2   The sequence

is an exact sequence, where the first map sends to , and the second is the natural quotient map.

Definition 2.2.3 (Noetherian)   An -module  is noetherian if every submodule of is finitely generated. A ring  is noetherian if  is noetherian as a module over itself, i.e., if every ideal of  is finitely generated.

Notice that any submodule  of a noetherian module  is also noetherian. Indeed, if every submodule of  is finitely generated then so is every submodule of , since submodules of are also submodules of .

Definition 2.2.4 (Ascending chain condition)   An -module satisfies the ascending chain condition if every sequences of submodules of  eventually stabilizes, i.e., there is some such that .

We will use the notion of maximal element below. If is a set of subsets of a set , ordered by inclusion, then a maximal element is a set so that no superset of is contained in . Note that it is not necessary that contain every other element of , and that could contain many maximal elements.

Proposition 2.2.5   If is an -module, then the following are equivalent:
1. is noetherian,
2. satisfies the ascending chain condition, and
3. Every nonempty set of submodules of contains at least one maximal element.

Proof. : Suppose is a sequence of submodules of . Then is a submodule of . Since is noetherian and is a submodule of , there is a finite set of generators for . Each must be contained in some , so there is an such that . But then for all , which proves that the chain of stabilizes, so the ascending chain condition holds for .

: Suppose 3 were false, so there exists a nonempty set  of submodules of  that does not contain a maximal element. We will use  to construct an infinite ascending chain of submodules of  that does not stabilize. Note that  is infinite, otherwise it would contain a maximal element. Let be any element of . Then there is an in that contains , otherwise would contain the maximal element . Continuing inductively in this way we find an in that properly contains , etc., and we produce an infinite ascending chain of submodules of , which contradicts the ascending chain condition.

: Suppose 1 is false, so there is a submodule of  that is not finitely generated. We will show that the set  of all finitely generated submodules of does not have a maximal element, which will be a contradiction. Suppose  does have a maximal element . Since  is finitely generated and , and is not finitely generated, there is an such that . Then is an element of  that strictly contains the presumed maximal element , a contradiction.

Lemma 2.2.6   If

is a short exact sequence of -modules, then  is noetherian if and only if both  and  are noetherian.

Proof. First suppose that  is noetherian. Then  is a submodule of , so  is noetherian. If is a submodule of , then the inverse image of in  is a submodule of , so it is finitely generated, hence its image is finitely generated. Thus  is noetherian as well.

Next assume nothing about , but suppose that both  and  are noetherian. If is a submodule of , then is isomorphic to a submodule of the noetherian module , so is generated by finitely many elements . The quotient is isomorphic (via ) to a submodule of the noetherian module , so is generated by finitely many elements . For each , let be a lift of to , modulo . Then the elements generate , for if , then there is some element such that is an -linear combination of the , and  is an -linear combination of the .

Proposition 2.2.7   Suppose  is a noetherian ring. Then an -module  is noetherian if and only if it is finitely generated.

Proof. If  is noetherian then every submodule of  is finitely generated so  is finitely generated. Conversely, suppose  is finitely generated, say by elements . Then there is a surjective homomorphism from to  that sends ( in th factor) to . Using Lemma 2.2.6 and exact sequences of -modules such as , we see inductively that is noetherian. Again by Lemma 2.2.6, homomorphic images of noetherian modules are noetherian, so  is noetherian.

Lemma 2.2.8   Suppose is a surjective homomorphism of rings and is noetherian. Then is noetherian.

Proof. The kernel of is an ideal in , and we have an exact sequence

with  noetherian. This is an exact sequence of -modules, where has the -module structure induced from (if and , then ). By Lemma 2.2.6, it follows that  is a noetherian -modules. Suppose  is an ideal of . Since  is an -submodule of , if we view  as an -module, then  is finitely generated. Since  acts on  through , the -generators of  are also -generators of , so  is finitely generated as an ideal. Thus  is noetherian.

Theorem 2.2.9 (Hilbert Basis Theorem)   If is a noetherian ring and is finitely generated as a ring over , then  is noetherian. In particular, for any  the polynomial ring and any of its quotients are noetherian.

Proof. Assume first that we have already shown that for any the polynomial ring is noetherian. Suppose is finitely generated as a ring over , so there are generators for . Then the map extends uniquely to a surjective homomorphism , and Lemma 2.2.8 implies that is noetherian.

The rings and are isomorphic, so it suffices to prove that if  is noetherian then is also noetherian. (Our proof follows [Art91, §12.5].) Thus suppose is an ideal of and that  is noetherian. We will show that is finitely generated.

Let be the set of leading coefficients of polynomials in . (The leading coefficient of a polynomial is the coefficient of highest degree, or 0 if the polynomial is 0; thus has leading coefficient .) We will first show that  is an ideal of . Suppose are nonzero with . Then there are polynomials  and  in  with leading coefficients  and . If , then is the leading coefficient of , so . Suppose and with . Then is the leading coefficient of , so . Thus is an ideal in .

Since is noetherian and is an ideal, there exist nonzero that generate  as an ideal. Since  is the set of leading coefficients of elements of , and the are in , we can choose for each an element with leading coefficient . By multipying the by some power of , we may assume that the all have the same degree .

Let be the set of elements of  that have degree strictly less than . This set is closed under addition and under multiplication by elements of , so is a module over . The module is the submodule of the -module of polynomials of degree less than , which is noetherian because it is generated by . Thus is finitely generated, and we may choose generators for .

We finish by proving using induction on the degree that every is an -linear combination of . If has degree 0, then , since , so is a linear combination of . Next suppose has degree , and that we have proven the statement for all elements of of degree . If , then , so  is in the -ideal generated by . Next suppose that . Then the leading coefficient  of  lies in the ideal  of leading coefficients of elements of , so there exist such that . Since has leading coefficient , the difference has degree less than the degree  of . By induction is an linear combination of , so is also an linear combination of . Since each and lies in , it follows that is generated by , so is finitely generated, as required.

Properties of noetherian rings and modules will be crucial in the rest of this course. We have proved above that noetherian rings have many desirable properties.

Subsections
William Stein 2012-09-24