rational circle group is coproduct

Proposition

Let \(\mathbb{Q}/\mathbb{Z}\) be the rational circle group.
Then there exists a canonical isomorphism

\begin{align*} \mathbb{Q}/\mathbb{Z} \cong \bigoplus_{p \in \mathbb{P}} \mathbb{Z}[\frac{1}{p}]/\mathbb{Z} \end{align*}

Proof

We define a map as follows:
Let \(\frac{a}{b} \in \mathbb{Q}/\mathbb{Z}\) be arbitrary, w.l.o.g. \(a < b\).
Then \(b = \prod_{i=1}^n p_i^{n_i}\) has some prime decomposition and \([a] \in \mathbb{Z}/b \mathbb{Z}\) corresponds - under the Chinese Remainder Theorem - to some element \((a_1,...,a_n) \in \bigoplus_{i=1}^n \mathbb{Z}/p^i \mathbb{Z}\).

Now map

\begin{align*} f: \mathbb{Q}/\mathbb{Z} \rightarrow& \prod_{p \in \mathbb{P}} \\ (\frac{a}{b}) \mapsto& (\frac{a_i}{p^{n_i}}) \\ \end{align*}

and observe that this map - if well-defined - restricts to a map \(f: \mathbb{Q}/\mathbb{Z} \rightarrow \prod_{p \in \mathbb{P}}\).

well-defined

Let \(\frac{a}{b} = \frac{a'}{b'}\), then w.l.o.g. we may assume that \(\frac{a'}{b'} = \frac{p a}{p b}\).
Suppose \(p \mid b\) and \(p = p_1\) in the prime decomposition of \(b\), then the chinese remainder theorem decomposes

\begin{align*} pa = (p \cdot a_1,a_2,...) \end{align*}

Suppose \(p \not\mid b\), then \(p = p_{n+1}\) and the chinese remainder theorem decomposes this as

\begin{align*} pa = (a_1,...,a_n,p) \end{align*}

and then \(\frac{p}{p} \in \mathbb{Z}[\frac{1}{p}]/\mathbb{Z}\) is zero.
Hence in both cases \(f( \frac{a}{b}) = f(\frac{a'}{b'})\).

group homomorphism

follows as the chinese remainder theorem is additive

surjective

follows from the chinese remainder theorem

injective

follows from the chinese remainder theorem