$\operatorname{Supp}(\mathbb{Q}/\mathbb{Z})$ as a $\mathbb{Z}$-module

Question

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.

Answer 1

Yes, it is correct that $\operatorname{Ann}(\Bbb{Q}/\Bbb{Z})=0$ and $\operatorname{Supp}(\mathbb{Q}/\mathbb{Z}) = \operatorname{Spec}(\mathbb{Z})$.

In general, for a $\Bbb{Z}$-module $M$ it is true that $\operatorname{Ann}(M)=\operatorname{exp}(M)$, the exponent of $M$, and hence that $$\operatorname{Supp}(M)=\{p\Bbb{Z}:\ p\mid\operatorname{exp}(M)\}=\operatorname{Spec}(\Bbb{Z}/\operatorname{exp}(M)\Bbb{Z}),$$ and in particular, if $M$ is a ring then $\operatorname{exp}(M)=\operatorname{char}(M)$.

Answer 2

Definition: For an $R$-module $M$, the set $\{P\ \mid $ $P$ prime ideal of $R$, $M_P\not=0\}$ is called the support of $M$ and is denoted by $\mathrm{Supp}(M)$.

Theorem: Let $R$ be a Noetherian ring and $M$ a finitely generated $R$-module. For any prime ideal $P$ of $R$, the following conditions are equivalent
$(i)$ $P \in \mathrm{Supp}(M)$.
$(ii)$ $P'\subseteq P$ for some $P' \in \mathrm{Ass}(M)$.
$(iii)$ $\mathrm{Ann}_R(M)\subseteq P$.

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, $\mathrm{Supp}(\mathbb{Q}/\mathbb{Z})=\mathrm{Max}(\mathbb{Z})=\mathrm{Spec}(\mathbb{Z})\setminus\{0\}$.