I’d like to go over associated primes in general rather than just the Noetherian ring form of primary decomposition of ideals. The natural generalization is to modules, since ideals are sub-modules over the ring treated as a module over itself. We’ll need to define a few things first.

Let M be an R-module. Let P be a prime ideal of R. Then we call this an associated prime of M if for some . The set of associated primes is denoted (since R will be understood, we’ll just drop that from here on).

Now suppose an ideal. Then the elements of are called the “prime divisors” of I.

Now we’ll get some basics out of the way. Suppose that R is Noetherian. First off, when . We show this by showing that any maximal element of the family is an associated prime. This is an important fact on its own.

Note that all we really are trying to show is that a maximal element of is prime as an ideal, since it will already be of the form . Let be a maximal element. Suppose and that and that . Then but . Thus . But by maximality, . Thus which means . Thus is prime and hence an associated prime of M.

The other fact we need is that the set of zero-divisors for M is precisely the set .

If , then this just says there is some such that and hence is a zero-divisor. The reverse inclusion just uses the previous fact. Let be such that for some . So we have . Then take a maximal element containing . By the last fact and hence .

For the theorem of the day, you may need a refresher on the spectrum of a ring, and on localization.

We no longer assume R is Noetherian. Let be multiplicative, and an -module. Viewing , then we have . In general, if is Noetherian, then for any R-module M, we have .

Let . Then . This is just because the elements of R that kill x, are just the fractions that kill x that are “actually” in R. This immediately gives us one inclusion, since if then .

Now suppose . Then there is some such that . Thus giving . Thus with . This proves the first statement.

We now show the second statement about M. Suppose . Thus we again get that and for some non-zero . Suppose that that in . Thus there is some such that in . But we’ve already noted that and , thus by primality . So . Thus giving one inclusion.

For the reverse, suppose . By clearing the denominator we can assume that for a non-zero we have . Let . Then . We have that is finitely generated since is Noetherian, so there is some such that . Thus which gives the reverse inclusion.

A nice little corollary is that for Noetherian rings a prime ideal if and only if .

December 21, 2009 at 5:41 am

I’m wondering about your setting. Is R supposed to be Noetherian from the very beginning? If not, I don’t know why Ass(M) must be nonempty, because in your proof, it seems unclear to me that has a maximal element.

December 21, 2009 at 5:42 am

oops, i meant .

December 21, 2009 at 6:38 am

Ack, yes that was supposed to be a hypothesis.