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Let $\mu_n$ denote the group of $n$-th roots of unity. In $\mathbb{C}$, this group has exactly $n$ elements. For a positive integer $n$ and a prime number $p$, using the canonical isomorphisms $\mu_{p^n} \cong \mathbb{Z}/p^n\mathbb{Z}$ for all $n$ as a direct system of factor groups, and the multiplication-by-$p$ homomorphisms $$\mathbb{Z}/p^n\mathbb{Z} \rightarrow \mathbb{Z}/p^{n+1}\mathbb{Z},$$ one sees that the direct limit of this system is the Prüfer group $$\mathbb{Z}(p^\infty) = \{x \in \mathbb{C}^\times: x^{p^n} = 1\,\, \mathrm{for\,\,some} \,\,n\}.$$

Similarly, as seen in this question Union of all finite cyclic groups, we can consider a direct system of cyclic group $\mathbb{Z}/n\mathbb{Z}$ and homomorphisms $\mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$ if $n|m$. The resulting direct limit is seen to be $\mathbb{Q}/\mathbb{Z}$, the set of rational numbers under addition modulo $1$. We have $$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}(p^\infty).$$

Question. Can we view $\mathbb{Q}/\mathbb{Z}$ as being isomorphic to $\mu_\infty = \{x \in \mathbb{C}^\times:x^n = 1\,\,\mathrm{for\,\,some}\,\, n\in \mathbb{Z}^+\}$, i.e., the subgroup of all roots of unity? If so, is there any way to define the isomorphism?

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1 Answer 1

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Yes, send $a/b \in \mathbb{Q}/\mathbb{Z}$ to $\zeta_{b}^{a}$, where $\zeta_{b}$ is a primitive $b$-th root of unity.

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  • $\begingroup$ Indeed... should've given some more thought before posting... Thanks! $\endgroup$
    – oleout
    Jun 27, 2021 at 5:31
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    $\begingroup$ This is $\frac ab\mapsto e^{2\pi i\frac ab}$. $\endgroup$
    – robjohn
    Jun 27, 2021 at 13:34
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    $\begingroup$ This is ill- or under-defined. You need to make an infinite number of arbitrary choices - a different primitive root $\zeta_b$ for each $b$ - and you need to synchronize the choices appropriately (so for example the primitive $9$th root cubed equals the primitive $3$rd root, etc.) Way simpler to just say $x\mapsto\exp(2\pi i x)$. $\endgroup$
    – anon
    May 22, 2022 at 9:28

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