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 $P=\mathrm{ann}(m)$?
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.