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

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.

• 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.
– Con
Aug 27 '19 at 10:50
• Yes, you are right Aug 27 '19 at 10:53

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

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

• Anyone care to explain the downvote? Aug 27 '19 at 12:59
• 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. Aug 27 '19 at 14:47
• @Youngsu I don't understand; could you explain what part requires that $Q$ is finitely generated over $Z$, and how? Aug 28 '19 at 7:44
• 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. Aug 28 '19 at 17:22