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

Notice that any submodule of is Noetherian, because if every submodule of is finitely generated then so is every submodule of , since submodules of are also submodules of .

- is Noetherian,
- satisfies the ascending chain condition, and
- Every nonempty set of submodules of contains at least one maximal element.

**
:** 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.

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 .

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 along with 0. If are nonzero with , then there are polynomials and in with leading coefficients and . If , then is the leading coefficient of , so . If and with , then is the leading coefficient of , so . Thus is an ideal in , so since is Noetherian there exists 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 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 .

Suppose is an arbitrary element. We will show by induction on the degree of that is an -linear combination of . Thus suppose this statement is true for all elements of of degree less than the degree of . If the degree of is less than , then , so is in the -ideal generated by . Next suppose that has degree . Then the leading coefficient of lies in the ideal of leading coefficients 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.