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I encountered the following interesting property of $\Bbb{Q}/\Bbb{Z}$ in one of my exams, many year ago:

Prove that there is no unital ring structure on $\Bbb{Q}/\Bbb{Z}$.

Well, if $\frac{a}{b}+\Bbb{Z}$ be as $1_{\Bbb{Q}/\Bbb{Z}}$, then $$b1_{\Bbb{Q}/\Bbb{Z}}= 1_{\Bbb{Q}/\Bbb{Z}}+\cdots+1_{\Bbb{Q}/\Bbb{Z}}=a+\Bbb{Z}=0$$ therefore, for any $\frac{p}{q}+\Bbb{Z}\in\Bbb{Q}/\Bbb{Z}$, $$b(\frac{p}{q}+\Bbb{Z})=b1_{\Bbb{Q}/\Bbb{Z}}(\frac{p}{q}+\Bbb{Z})=0.$$ But $$b(\frac{1}{b+1}+\Bbb{Z})=\frac{b}{b+1}+\Bbb{Z}\ne 0$$ so, if $\Bbb{Q}/\Bbb{Z}$ has a ring structure, it is non-unital.

Do you know, any interesting property about $\Bbb{Q}/\Bbb{Z}$ to share with us?

mathlander
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Qurultay
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    For each $n\in \Bbb N$ there is an element in $(\Bbb Q/\Bbb Z,+)$ of order $n$. – Sumanta Jun 30 '20 at 07:14
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    I don't think you even have a well-defined multiplication. For example, $(\frac12+\mathbb Z)\cdot(\frac12+\mathbb Z)=\frac14+\frac12\mathbb Z = (\frac14+\mathbb Z)\cup(\frac34+\mathbb Z)$. – celtschk Jun 30 '20 at 07:16
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    https://math.stackexchange.com/questions/122122/on-the-factor-group-bbb-q-bbb-z – Sumanta Jun 30 '20 at 07:18
  • @celtschk What about the groups $H^n(X;\Bbb{Q}/\Bbb{Z})$? do you have any information? – Qurultay Jun 30 '20 at 07:23
  • @Qurultay: I don't know what that notation means. – celtschk Jun 30 '20 at 07:24
  • @celtschk Sorry, I meant cohomology groups of topological space $X$ with coefficients in $\Bbb{Q}/\Bbb{Z}$. – Qurultay Jun 30 '20 at 07:26
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    $\Bbb Q/\Bbb Z$ is an injective $\Bbb Z$-module. – Sumanta Jun 30 '20 at 07:27
  • @Sumanta That was a good point, thanks – Qurultay Jun 30 '20 at 07:29
  • @Qurultay: Unfortunately I don't know anything about cohomology groups. – celtschk Jun 30 '20 at 07:30
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    $\text{Hom}\big(\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Q),\Bbb Z\big)\not\simeq \text{Hom}\big(\Bbb Q,(\Bbb Q/\Bbb Z\otimes\Bbb Z)\big)$. – Sumanta Jun 30 '20 at 07:33
  • On the multiplication... if the standard map $(x+\mathbb Z, y + \mathbb Z)\mapsto xy + \mathbb Z$ gave a ring structure, $\mathbb Z$ would be ideal. – Ennar Jun 30 '20 at 07:36
  • @Sumanta I know another one, without proof: $M$ is a flat $R$-module if and only if $\hom_{\Bbb{Z}}(M,\Bbb{Q}/\Bbb{Z})$ is injective $R$-module. – Qurultay Jun 30 '20 at 07:41
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    Yeah, you are right. The module $\text{Hom}_\Bbb Z(M,\Bbb Q/\Bbb Z)$ is called the character module of $M$. The proof of the statement you made is due to Lambek, and based on the fact $\Bbb Q/\Bbb Z$ is an injective $\Bbb Z$-module and adjoint isomorphism. – Sumanta Jun 30 '20 at 07:46
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    Quotient of a divisible group is divisible and $\text{Ext}(H,G)=0$ for two abelian groups $H,G$ with $G$ is divisible. Now, universal coefficients theorem for cohomology groups gives $H^n(C,G)\simeq \text{Hom}\big(H_n(C),G\big)$ for any divisible group $G$. So, for a pair of topological spaces $(X,A)$ we get $H^n(X,A; G)\simeq \text{Hom}\big(H_n(X,A);G\big)$. Now, consider $G=\Bbb Q/\Bbb Z$. – Sumanta Jun 30 '20 at 07:57
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    I think this question is too broad as is. That being said, the endomorphism ring of this abelian group is the profinite completion of the integers, $End(\mathbb Q/\mathbb Z) \simeq \widehat{\mathbb Z}$. See https://math.stackexchange.com/q/795543/96384 – Torsten Schoeneberg Oct 13 '22 at 00:42

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