A Mind for Madness

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Associated Primes I

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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 P=ann(x) for some x\in M. The set of associated primes is denoted Ass_R(M) (since R will be understood, we’ll just drop that from here on).

Now suppose I\subset R an ideal. Then the elements of Ass(A/I) are called the “prime divisors” of I.

Now we’ll get some basics out of the way. Suppose that R is Noetherian. First off, Ass(M)\neq \emptyset when M\neq 0. We show this by showing that any maximal element of the family \mathcal{F}=\{ann(x) : 0\neq x\in M\} 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 \mathcal{F} is prime as an ideal, since it will already be of the form ann(x). Let A\in\mathcal{F} be a maximal element. Suppose A=ann(x) and that ab\in A and that b\notin A. Then abx=0 but bx\neq 0. Thus ann(x)\subset ann(bx). But by maximality, ann(x)\subset ann(bx). Thus ann(x)=ann(bx) which means a\in A. Thus A 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 \displaystyle \bigcup_{P\in Ass(M)}P.

If a\in \displaystyle \bigcup_{P\in Ass(M)}P, then this just says there is some x\in M such that a\in ann(x)\Rightarrow ax=0 and hence a is a zero-divisor. The reverse inclusion just uses the previous fact. Let a be such that ax=0 for some x. So we have a\in ann(x). Then take a maximal element P\in\mathcal{F} containing ann(x). By the last fact P\in Ass(M) and hence a\in \displaystyle \bigcup_{P\in Ass(M)}P.

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 S\subset R be multiplicative, and N an R_S-module. Viewing Spec(R_S)\subset Spec(R), then we have Ass_R(N)=Ass_{R_S}(N). In general, if R is Noetherian, then for any R-module M, we have Ass(M_S)=Ass(M)\cap Spec(R_S).

Let x\in N. Then ann_R(x)=ann_{R_S}(x)\cap R. 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 P\in Ass_{R_S}(N) then P\cap R\in Ass_R(N).

Now suppose Q\in Ass_R(N). Then there is some x\in N such that Q=ann_R(x). Thus x\neq 0 giving Q\cap S=\emptyset. Thus QR_S\in Spec(R_S) with QR_S=ann_{R_S}(x). This proves the first statement.

We now show the second statement about M. Suppose P\in Ass(M)\cap Spec(R_S). Thus we again get that P\cap S=\emptyset and P=ann_R(x) for some non-zero x\in M. Suppose that that (r/s)x=0 in M_S. Thus there is some t\in S such that trx=0 in M. But we’ve already noted that t\notin P and tr\in P, thus by primality r\in P. So PR_S=ann_{R_S}(x). Thus PR_S\in Ass(M_S) giving one inclusion.

For the reverse, suppose Q\in Ass(M_S). By clearing the denominator we can assume that for a non-zero x\in M we have Q=ann_{R_S}(x). Let Q^c=Q\cap R. Then Q=Q^cR_S. We have that Q^c is finitely generated since R is Noetherian, so there is some t\in S such that Q^c=ann_R(tx). Thus Q^c\in Ass_R(M) which gives the reverse inclusion.

A nice little corollary is that for Noetherian rings a prime ideal P\in Ass_R(M) if and only if PR_P\in Ass_{R_P}(M_P).

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Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

3 thoughts on “Associated Primes I

  1. 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 \frak{F} has a maximal element.

  2. oops, i meant \mathcal{F}.

  3. Ack, yes that was supposed to be a hypothesis.

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