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Lang in his book Elliptic functions writes in Chapter 7, §1. that there is a natural isomorphism $$\mathbb{Q}^2/\mathbb{Z}^2\cong \amalg_p \mathbb{Q}_p^2/\mathbb{Z}_p^2$$ which should correspond to the primary decomposition.

How does this isomorphism work?

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    $\begingroup$ My first guess would be that $\mathbf{Z}$ is the kernel of the natural mapping $\iota:\mathbf{Q}\to\amalg_p\mathbf{Q}_p/\mathbf{Z}_p$, $q\mapsto (\overline{q},\overline{q},\overline{q},\ldots)$. Furthermore, the mapping is surjective (most likely by the Chinese remainder theorem). And $\mathbf{Q}_p/\mathbf{Z}_p$ is isomorphic to the $p$-power torsion of $\mathbf{Q}/\mathbf{Z}$, no? Assuming that $\amalg_p$ stands for a direct sum, where only finitely many components are non-zero. $\endgroup$ – Jyrki Lahtonen Jul 13 at 10:14
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Concretely any rational number is of the form $$x = \sum_j \frac{a_j}{p_j^{e_j}}+n, \qquad p_j \nmid a_j, e_j \ge 1, a_j \in 0 \ldots p^{e_j}-1$$ in a unique way and $$\amalg_p (\frac{a_p}{p^{e_p}}, \frac{b_p}{p^{d_p}}) \mapsto (\sum_p\frac{a_p}{p^{e_p}},\sum_p \frac{b_p}{p^{d_p}})$$ is an isomorphism $$\amalg_p \mathbb{Q}_p^2/\mathbb{Z}_p^2 \to \mathbb{Q}^2/\mathbb{Z}^2$$

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