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For an $A$-module $M$, the support of $M$ is defined to be $$\mathrm{Supp}_A(M) := \{P \in \mathrm{Spec}(A) \mid M_p \neq \{0\}\}.$$ The associated primes are defined to be: $$\mathrm{Ass}_A(M):=\{P \in \mathrm{Spec}(A) \mid P=\textrm{ann}(m), m \in M\}.$$

My question: does there exist a module $M$ so that there are only finitely many minimal elements in $\mathrm{Supp}(M)$ with respect to inclusion, but infinitely many associated primes?

Thought 1: Every minimal element of $\mathrm{Supp}_A(M)$ is an associated prime.

Thought 2: If $A$ is Noetherian and $M$ a finite $A$-module, then $\mathrm{Ass}_A(M)$ is finite, so any example of a module that satisfies the above criteria cannot be of this form.

Thought 3: In essence, every element of the support will be a prime ideal $P \subset A$ so that $\mathrm{ann}(m) \subseteq P$ for some $m \in M$. If there are only finitely many of these that are minimal with respect to inclusion, can we still find infinitely many so that $P=\mathrm{ann}(m)$?

user26857
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Andres Mejia
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1 Answers1

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Choose your favorite ring $A$ with infinite spectrum, and your favorite closed subset $Z \subseteq \mathrm{Spec}\;A$ that has finitely many minimal elements (e.g. if $A$ is Noetherian, every closed subset should satisfy the second condition) but is infinite. Then take the module $$M=\bigoplus_{\mathfrak{p} \in Z}(A/\mathfrak{p}).$$

Then $\mathrm{Ass}\;M=\mathrm{Supp}\;M=Z$.

As you can see, this is very much non-finitely generated, but as you noted, any example (at least over $A$ Noetherian) would be, so it is OK.

Pavel Čoupek
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  • Should it be clear to me that there are only finitely many minimal elements in $\mathrm{Supp}M$? – Andres Mejia Nov 04 '16 at 19:07
  • Haha, indeed it is. Thank you, this was insightful. +1 – Andres Mejia Nov 04 '16 at 19:35
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    @AndresMejia I assumed it by choice of $Z$. You can always make it explicit: put $A=\mathbb{C}[x]$ and $Z=\mathrm{Spec},A,$ which has the unique minimal prime $(0)$. But it holds in general that a closed subset $Z$ of the spectrum of a Noetherian comm. ring has finite set of minimal primes: You can decompose it into a finite union of irreducible closed subsets. Each of those will have a unique minimal prime. So it's not terribly obvious, but it shouldn't be too hard to prove either. Unfortunately, I don't have a good elementary reference for this (it is mostly discussed in context of AG). – Pavel Čoupek Nov 04 '16 at 19:36
  • That is interesting, I just more or less "followed your directions" and used this construction on (interestingly) the same example. Thanks. – Andres Mejia Nov 04 '16 at 19:37