Difference between $\mathrm{ass}(M)$ and $\mathrm{supp}(M)$

Question

I'm reading through Lang's chapter about Noetherian modules and rings in Algebra. More specifically, subchapter about associated primes.

Let $A$ be a commutative ring, and $M$ its module. If $\mathfrak p$ is an associated prime of $M$, by definition, that means that there exists $x \in M$ such that $\mathrm{ann}(x) = \mathfrak p$. Then, surely, $M_{\mathfrak p} \neq 0$ because $x/1 \neq 0$ in it.

This shows that $\mathrm{ass}(M) \subseteq \mathrm{supp}(M)$. (You can read the definition of $\mathrm{supp}(M)$ here).

I was searching for an example of $\mathrm{ass}(M) \neq \mathrm{supp}(M)$, but all my simple examples failed me. Can someone provide an example of this situation?

Answer 1

Very simple: take $A$ a noetherian domain. $\operatorname{Ass}A=\{0\}$ and $\operatorname{Supp}A=\operatorname{Spec}A$. They're different unless $A$ is a field.

Note: If $A$ is reduced, $\operatorname{Ass}A=\operatorname{Min}A$, which is different from $\operatorname{Spec}A$ if the Krull dimension of $A$ is positive.