Why is the Noetherian hypothesis necessary for this fact about associated primes?

Question

In a Noetherian ring, the set of associated prime ideals of an ideal is the set of primes which can be written as $(I:z)$.

I'm new to associated primes, and I was wondering why the Noetherian hypothesis here is necessary. In Matsumura's book Commutative Ring Theory, the associated primes of a module $M$ are defined as primes that occur as $\operatorname{ann}(m)$ for some $m\in M$. In this base, a prime $P$ would have to annihilate some element in $A/I$, which means $P$ is the set of elements $p$ such that $px\subset I$ for some $x\in A/I$, which is exactly the definition of $(I:x)$.

What am I missing? Why must we assume $A$ is Noetherian for this to hold?

(The quote comes from @wxu's answer to this question. The question itself has a Noetherian hypothesis. I am trying to figure out why it is necessary.)

score 5 · Accepted Answer · answered Jul 11 '13 at 18:39

5

Swanson's notes on primary decomposition gives an explanation for this. In particular, go to Remark 3.11 at the bottom of page 9.

For people who don't want to click through the link, here is the gist: Sure, you can define associated primes like that without the Noetherian condition. But there are many other reasonable definitions as well. It happens that for Noetherian rings all these definitions agree so it is more convenient to assume that extra condition.

answered Jul 11 '13 at 18:39

Ragib Zaman

35,127

Thanks. Does this mean the Noetherian hypothesis in the question I linked is not necessary? – Potato Jul 11 '13 at 18:43
@Potato Yes, if you use that definition that is commonly used for Noetherian modules as the general definition, the result still holds. – Ragib Zaman Jul 11 '13 at 18:51
Thank you. That clears everything up. – Potato Jul 11 '13 at 18:51

Why is the Noetherian hypothesis necessary for this fact about associated primes?

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