Radical of the annihilator is the intersection of associated primes

[Barre]

1. Homework Statement
$R,M$ are Noetherian. Prove that the radical of the annihilator of an $R$-module$M$, $Rad(ann(M))$
is equal to the intersection of the prime ideals in the set of associated primes of $M$ (that is denoted so regretfully that I am not even allowed to spell it out by the system)

Homework Equations

This result is proved in exercises in Dummit and Foote by using tools like the support of $M$ and Zarski topology. In Hungerford, it appears early, just after definition of primary submodules, hence there should be a simple solution that I do not see.

Associated primes are such annihilators of elements of the module that the ideal is actually prime. Question comes up in the exercise section after introducing the notion of primary decomposition, without notion of ring spectrum, Zarski topology, module localizations and such.

The Attempt at a Solution

I am not looking for a solution, but just for a tip as where to start. It seems like the exercise should not be hard at all, but I've had some problems. One way inclusion is easy. Since annihilator of $M$ is the intersection of annihilators of all elements of $M$, annihilator of $M$ certainly is contained in every annihilator of an element, hence every associated prime. Radical of annihilator of $M$ is the set of prime ideals containing it, so it is at most as big as the intersection of associated primes.

And here I have my problem. I try a proof by contradiction, working with an element of the intersection of associated primes that is not in $Rad(ann(M))$. Because of the characterization of the radical of an ideal as the intersection of prime ideals containing it, if the chosen element is not in the radical, then there must exist a prime ideal that contains the radical, but not the element itself. I'm thinking of deriving a contradiction there, could this be a way to go? For example, the section introduced primary decomposition, and annihilator of M has such a decomposition. How would it look?

Could anyone point me, gently, in the right direction?

 

[micromass]

Is this exercise 10 on page 387??

The exercise there only wants you to prove it for rings satisfying the ascending chain condition. In particular, they want you to use 9(b).

 

[Barre]

Is this exercise 10 on page 387??

The exercise there only wants you to prove it for rings satisfying the ascending chain condition. In particular, they want you to use 9(b).

You are right about which exercise it is. R is supposed to satisfy ACC, yes. Else associated primes would not have to exist. I will look into what 9(b) (the filtration) implies for this problem, thanks. I sort of gave up on using 9b after I 'assumed' that the prime ideals associated with the filtration do not necessarily have to be associated primes of the module.

PS: Are you an algebraist? Not that I post very advanced questions, but you have answered/helped me with like every single problem I ever posted here :)

 

[micromass]

You are right about which exercise it is. R is supposed to satisfy ACC, yes. Else associated primes would not have to exist. I will look into what 9(b) (the filtration) implies for this problem, thanks. I sort of gave up on using 9b after I 'assumed' that the prime ideals associated with the filtration do not necessarily have to be associated primes of the module.

I think that the ideals in the filtration are exactly the associated primes of the module.

PS: Are you an algebraist? Not that I post very advanced questions, but you have answered/helped me with like every single problem I ever posted here :)

I'm not an algebraist at all actually. I like analysis more :tongue2:

 

[Barre]

I had to work with associated primes of the quotients of M to solve 9b if I remember correctly. I will see if I can prove they are in this case contained in the set of associated primes of M. Thanks.

 

[Barre]

Well let's see. We got that if $R,M$ are both Noetherian (satisfying ascending chain conditions on ideals/modules), then we can obtain a filtration $0 = M_0 \subset M_1 \subset M_2 \ldots \subset M_n = M$ where each factor $M_i/M_{i-1} \cong R/P_i$ for some prime ideal $P_i \subset R$. We construct this as follows. Let $M_0 = 0$. Pick $P_1 = ann(x) \in AP(M) = AP(M/M_0)$, the set of associated primes of $M$. Then there is a submodule of $M$ isomorphic to $R/P_1 \cong Rx$. Let this be $M_1$. Inductively, pick $P_i \in AP(M/M_{i-1})$ and let $M_i$ be the module such that $R/P_i \cong M_i/M_{i-1}$. Then $0 = M_0 \subset M_1 \subset \ldots$ is an ascending chain which must stabilize. By construction, this happens when for some $n, M_n = M$. Existence of associated prime for each quotient is a consequence of ACC property of the ring and module

Is it so that the prime ideals $P_i$ can be picked such that each $P_i \in AP(M)$ ? Equivalently, is $AP(M) \cap AP(M/N)$ always nonempty for arbitrary submodule $N$? If yes, then we could work by induction. The statement was that if $r \in R$ is contained in all associated primes of $M$, then for each $x \in M$, $r^nx = 0$ for some $n$.

For each element $x \in M$ there is a least index $i$ such that $x \in M_i$ but $x \not\in M_{i-1}$. Assume that for each $a \in M_{i-1}, r^na = 0$ for some $n$. If $a \in M_1 \cong R/P_1$, then $ra = 0$ and this is the base of induction. Let $x$ be any element on $M$, $x \in M_i, x \not\in M_{i-1}$. Then $x + M_{i-1} \neq M_{i-1}$. Because $M_i/M_{i-1} \cong R/P_i$, we see that $r \in P_i$ (by hypothesis) must annihilate $x + M_{i-1}$, hence $rx + M_{i-1} = M_{i-1}$, so $rx \in M_{i-1}$. By induction, $r^n(rx) = 0$ for some $n$, hence proof would be complete. Could this work? In any case, I still do not see if the prime ideals can be picked like I assumed. If not, can one construct a filtration in another way?