Noetherian rings and modules

**serpicojr** · July 7th, 2008, 07:39 PM

For the purposes of this discussion, all rings are commutative with identity/unity. You can do all of this with noncommutative rings, but it'll be easier to just deal with commutative rings.

Let

be a ring. An (ascending) chain of ideals of

is a sequence of ideals

of

such that

. We say that

is Noetherian if

satisfies the ascending chain condition (ACC) on ideals. That is to say, given any ascending chain

of ideals of

, there exists an integer

(depending on the chain) such that for all

,

.

Note: I tend to say that the chain stabilizes in the above case, by which I mean it is eventually constant, where any sequence is eventually constant if it differs from a constant sequence by a finite number of terms.

Examples of Noetherian rings include the integers and polynomial rings over a field. Indeed, we have the

Proposition: Let

be a principal ideal ring. Then

is Noetherian.

Proof: Let

be a chain of ideals of

, and let

. Check that

is indeed an ideal. Since

is a principal ideal ring, there is an element

of

such that

. Since

, there must exist some

such that

. But then

. Clearly

, and so

. But then for any

, we have

, whence we clearly have

. Thus the ACC is satisfied, and

is Noetherian.

As the integers and any polynomial ring over a field are Euclidean domains, they are PID's and hence must be Noetherian.

There are some other useful characterizations of Noetherian rings. One of them is an important finiteness criterion which is useful in, say, algebraic geometry. Another is a property which allows us to pick out ideals maximal with respect to basically any given condition, which is what I will be using a lot in the algebraic number theory thread. To be specific, we have the

Theorem: The following are equivalent:
i.

is Noetherian;
ii. every ideal of

is finitely generated; and
iii. every nonempty set of ideals of

contains a maximal element (i.e., given a set

of ideals of

, there is an ideal

such that if

and

, then

).

Proof:

Let

be any ideal of

, and construct a chain of ideals as follows: let

be any element, and let

. If

, we are done. Otherwise, let

, and let

. Then

. If

, we are done. Otherwise, continue the process, throwing in another element of

to create a bigger ideal, stopping if the newly created ideal is equal to

and continuing otherwise. Since

satisfies the ACC on ideals, this process must stop at some point, as otherwise we would be able to construct a strictly increasing chain of ideals of

. Thus we must have

for some

, and clearly

is generated by

elements.

This proof is very similar to the proof that every principal ideal ring is Noetherian--just take a chain of ideals, look at the union (which must be an ideal and hence finitely generated), and then show that all of the elements generating the union must lie in some ideal in the chain.

Let

be a nonempty set of ideals of

. Construct a chain of ideals as follows: let

be any ideal. If

is maximal, we're done. Otherwise, there must exist

so that

. Continue the process: if you find a maximal element, stop; otherwise, you can find a bigger ideal and continue the chain. This process must stop, as otherwise we would be able to construct a strictly increasing chain of ideals of

. Thus our chain must stop at some maximal element.