Give an example about associated primes where two containments are proper.

Question

I want to ask a question about associated primes. This question is just for my curiosity.

Let $R$ be a commutative ring with $1$. Let $M$ be a $R$-module. We use $\mathrm{Ass}_R(M)$ to denote the collection of associated primes of $M$. Let $L \subset M$ be a submodule. We set $N = M/L$. Then, the following containments are known. \[ \mathrm{Ass}_R(L) \subset \mathrm{Ass}_R(M) \subset \mathrm{Ass}_R(L) \cup \mathrm{Ass}_R(N). \tag{1} \] I want to ask you to tell me an example where the two containments on (1) are both proper in the same time.

My two attempts are as follows.

1. Let $R = \mathbb{Z}$ and let $p, q \in \mathbb{Z}$ be distinct positive primes. We set $M = \mathbb{Z}/(pq)$ and $L = (p) \subset M$ be an ideal. Then, we see that $\mathrm{Ass}_R(L) = \{ (q) \}$ and $\mathrm{Ass}_R(M) = \{ (p), (q) \}$ but $\mathrm{Ass}_R(L) \cup \mathrm{Ass}_R(N) = \{ (p), (q) \}$.
2. Let $R = \mathbb{Z}$ and let $p \in \mathbb{Z}$ be a prime. We set $M = \mathbb{Z}$ and $L = (p) \subset M$ be an ideal. Then, we see that $\mathrm{Ass}_R(L) = \mathrm{Ass}_R(M) = \{ (0) \}$ but $\mathrm{Ass}_R(N) = \{ (p) \}$.

Answer 1

We can go up one dimension. Take $R=\mathbb Z[x]$,

$M=R/(2x)$, $L=6M$ and $N=R/(6)=(\mathbb Z/6\mathbb Z)[x]$.

Then Ass$(M)=\{(2),(x)\}$, Ass$(L)=\{(x)\}$ and Ass$(N)=\{(2),(3)\}$.