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serpicojr · Posted: Mon Jul 07, 2008 5:39 pm Post subject: Noetherian rings and modules

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 I_n

of

such that $I_1 \subset I_2 \subset I_3 \subset \cdots$ . We say that

is Noetherian if

satisfies the ascending chain condition (ACC) on ideals. That is to say, given any ascending chain $I_1 \subset I_2 \subset I_3 \subset \cdots$ of ideals of

, there exists an integer

(depending on the chain) such that for all $n, m \geq N$ , I_n = I_m

.

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 $I_1 \subset I_2 \subset I_3 \subset \cdots$ be a chain of ideals of

, and let $I = \cup_{n=1}^\infty I_n$ . Check that

is indeed an ideal. Since

is a principal ideal ring, there is an element

of

such that

. Since $a \in I$ , there must exist some

such that $a \in I_N$ . But then $I = aR \subset I_N$ . Clearly $I_N \subset I$ , and so I_N = I

. But then for any $n \geq m \geq N$ , we have $I_N \subset I_m \subset I_n \subset I = I_N$ , whence we clearly have I_m = I_n

. Thus the ACC is satisfied, and

is Noetherian. $\Box$

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 $\mathcal{S}$ of ideals of

, there is an ideal $I \in \mathcal{S}$ such that if $J \in \mathcal{S}$ and $I \subset J$ , then I = J

).

Proof: $(i.\Rightarrow ii.)$ Let

be any ideal of

, and construct a chain of ideals as follows: let $x_1 \in I$ be any element, and let I_1 = (x_1)

. If

, we are done. Otherwise, let $x_2 \in I\setminus I_1$ , and let I_2 = (x_1,x_2)

. Then $I_1 \subsetneq I_2$ . If I_2 = I

, 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 I = I_n

for some

, and clearly I_n

is generated by

elements.

$(ii. \Rightarrow i.)$ 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.

$(i. \Rightarrow iii.)$ Let $\mathcal{S}$ be a nonempty set of ideals of

. Construct a chain of ideals as follows: let $I_1 \in \mathcal{S}$ be any ideal. If I_1

is maximal, we're done. Otherwise, there must exist $I_2 \in \mathcal{S}$ so that $I_1 \subsetneq I_2$ . 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.

$(iii. \Rightarrow i.)$ Let $I_1 \subset I_2 \subset I_3 \subset \cdots$ be a chain of ideals, and let $\mathcal{S}$ be the set of ideals in this chain. Then $\mathcal{S}$ contains a maximal element, say $I_N \in \mathcal{S}$ . But then for any $n \geq N$ , we have $I_N \subset I_n$ . Since I_N

is maximal, we must have I_N = I_n

. Thus

satisfies the ACC and is Noetherian. $\Box$