Let $R=\mathbb Z$ which is a ring and let $M=\mathbb Q/\mathbb Z$ which is an $R$-module.
Calculate $\operatorname{Ass}(M)$ and $\operatorname{Supp}(M)$.
Recall that
- $\operatorname{Supp}(M)=\{p\in\operatorname{Spec}(R):M_p\ne0\}$
- $\operatorname{Ass}(M)=\{p\in\operatorname{Spec}(R):\exists 0\ne m\in M:\operatorname{Ann}(m)=p\}$
- $\forall m\in M,\operatorname{Ann}(m)=\{r\in R:rm=0\}$
I have succeeded to show that $\operatorname{Ass}(M)=\{p\mathbb Z:p\text{ is prime}\}$.
EDIT:
I deleted some of the previous question and I would like to add an attempt. I would like if someone could verify my answer;
Let $p\in\mathbb N$ be a prime. $1\notin\langle p\rangle$. Hence $(\mathbb Q/\mathbb Z)_p\ni{{1\over p}+\mathbb Z\over 1}$. $$ {{1\over p}+\mathbb Z\over 1}={\mathbb Z\over 1}\iff\exists t\notin\ p\mathbb Z:\mathbb Z=t\cdot{1\over p}+\mathbb Z $$ So we conclude ${{1\over p}+\mathbb Z\over 1}\ne 0\Rightarrow p\mathbb Z\in\text{supp}(\mathbb Q/\mathbb Z)$.
Let us check weather $\langle0\rangle\in\text{supp}(M)$ where $M=\mathbb Q/\mathbb Z$.
Consider $p:=\langle0\rangle$;
$\mathbb Z_p=\mathbb Q $ because if ${a\over b}=_\mathbb Q{c\over d}\iff{a\over b }=_{\mathbb Z_p}{c\over d}$ and a similar calculation shows that $\mathbb Q\cong\mathbb Q_p$. Hence $M_p=(\mathbb Q/\mathbb Z)_p=\mathbb Q_p/\mathbb Z_p\cong \mathbb Q/\mathbb Q=0$. Hence $\langle0\rangle\notin\text{supp}(M)$.
We conclude: $\text{supp}(\mathbb Q/\mathbb Z)=\{p\mathbb Z:p\text{ is prime}\}$.
It's hard for me to understand what isn't compiling for you, but you're almost done. The question remaining is: for which primes $p\in\rm{Spec}(\mathbb{Z})$ do we have $\mathbb{Q}_p / \mathbb{Z}_p \neq 0$?
To sum up: $p\in\rm{Supp}(M)$ if and only if $\mathbb{Z}_p \neq \mathbb{Q}_p$. Does this happen for $p=(0)$? What about $p\neq (0)$?