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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.

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  • $\begingroup$ Annihilators cannot be empty as they are ideals. What about $x = 0$ for $\mathbb{Q} \cdot x = 0$? But maybe you just wanted to say that the annihilator is the zero ideal and not empty? That would also match with your latter statement. $\endgroup$
    – Con
    Aug 27 '19 at 10:50
  • $\begingroup$ Yes, you are right $\endgroup$ Aug 27 '19 at 10:53
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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, $\mathrm{Supp}(\mathbb{Q}/\mathbb{Z})=\mathrm{Max}(\mathbb{Z})=\mathrm{Spec}(\mathbb{Z})\setminus\{0\}$.

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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 $\operatorname{exp}(M)=\operatorname{char}(M)$.

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  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ Aug 27 '19 at 12:59
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    $\begingroup$ I believe your reason requires that $Q$ is finitely generated as a module over $Z$. But I find the rest of your argument interesting and helpful. $\endgroup$
    – Youngsu
    Aug 27 '19 at 14:47
  • $\begingroup$ @Youngsu I don't understand; could you explain what part requires that $Q$ is finitely generated over $Z$, and how? $\endgroup$ Aug 28 '19 at 7:44
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    $\begingroup$ Hi Servaes. The place where you have that for a module $M$ over a ring $R$, $Supp (M) = V(ann(M))$ needs $M$ to be finitely generated as a module over $R$. In this question, $Q/Z$ is not a finitely generated $Z$-module. $\endgroup$
    – Youngsu
    Aug 28 '19 at 17:22

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