This question provides the def of $\operatorname{Supp}(M)$ and a simple example
I want to find $\operatorname{Supp}(\mathbb{Q}/\mathbb{Z})$ as a $\mathbb{Z}$-module.
My attempt. $\mathbb{Z}$ is a PID. $\operatorname{Ann}(\mathbb{Q}/\mathbb{Z})=(0)$, because there is no number $x \neq 0$ such that $\mathbb{Q}(x) \subset \mathbb{Z}$. All prime ideals contain (0). This means that $\operatorname{Supp}(\mathbb{Q}/\mathbb{Z}) = \operatorname{Spec}(\mathbb{Z})$
Is it right?
UPD: the answer is $\operatorname{Spec}(\mathbb{Z}) \setminus (0)$ as the zero ideal does not belong to the support.
Best Answer
But $\mathbb{Q}/\mathbb{Z}$ is not a finitely generated as a $\mathbb{Z}$-module.
Let $P$ be a prime ideal of $\mathbb{Z}$. We have two cases:
1) If $P=0$, then $(\mathbb{Q}/\mathbb{Z})_P\cong\mathbb{Q}_P/\mathbb{Z}_P\cong\mathbb{Q}_0/\mathbb{Z}_0\cong\mathbb{Q}/\mathbb{Q}=0$. Thus $0\not\in \mathrm{Supp}(\mathbb{Q}/\mathbb{Z})$.
2) If $P\not=0$, then $(\mathbb{Q}/\mathbb{Z})_P\cong\mathbb{Q}_P/\mathbb{Z}_P$. Now since $\mathbb{Q}_P\not=\mathbb{Z}_P$, we have $P\in \mathrm{Supp}(\mathbb{Q}/\mathbb{Z})$.
Therefore, $\mathrm{Supp}(\mathbb{Q}/\mathbb{Z})=\mathrm{Max}(\mathbb{Z})=\mathrm{Spec}(\mathbb{Z})\setminus\{0\}$.