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Thread: Noetherian rings and modules

  1. #1 Noetherian rings and modules 
    Forum Professor serpicojr's Avatar
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    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.

    Let be a chain of ideals, and let be the set of ideals in this chain. Then contains a maximal element, say . But then for any , we have . Since is maximal, we must have . Thus satisfies the ACC and is Noetherian.


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