Is $\mathbb{Q}/\mathbb{Z}$ decomposable? I would like to see if this $\mathbb{Q}/\mathbb{Z}$ as a $\mathbb{Z}$ module is decomposable. 
I'm trying to come up with a decomposition that looks like: 
$$ \mathbb{Q}/\mathbb{Z} = \bigoplus_{p} \left(\prod_{k} \frac{1}{p^k} \mathbb{Z} \right). $$
Is this correct? Is there a theoretical way to see this? I know that $\mathbb{Q}$ is not decomposable and both $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ are injective modules but one is torsion free and the latter is a torsion abelian group. But I'm unclear how injectivity, torsion-ness and decomposition could be connected. 
 A: Any torsion $\mathbb{Z}$-module $A$ splits as a direct sum $\bigoplus_p A_p$, where $A_p$ is the submodule of elements whose order is a power of $p$.  This is essentially a version of the Chinese remainder theorem: the case $A=\mathbb{Z}/n$ is just the classical Chinese remainder theorem for modular arithmetic, and the proof for general $A$ uses similar ideas.  (Here's a sketch: for any $a\in A$ of order $n$, the classical Chinese remainder theorem says that $a$ can be written as a sum of elements whose orders are prime powers dividing $n$, since the subgroup generated by $a$ is isomorphic to $\mathbb{Z}/n$.  This shows that $A=\sum_p A_p$, and the sum is direct because any element of $A_{p_1}+\dots+A_{p_n}$ has order which is a product of powers of $p_1,\dots,p_n$ and in particular cannot be in $A_q$ for any other prime $q$.)
In the case $A=\mathbb{Q}/\mathbb{Z}$, the submodule $A_p$ consists of (the $\mathbb{Z}$-cosets of) fractions whose denominator is a power of $p^k$.  You can think of $A_p$ as the union (or more precisely, direct limit) of the groups $\mathbb{Z}/p^k$ where you think of $\mathbb{Z}/p^k$ as the subgroup of $\mathbb{Z}/p^{k+1}$ consisting of multiples of $p$. (Here $\mathbb{Z}/p^k$ corresponds to the subgroup of $\mathbb{Q}/\mathbb{Z}$ consisting of fractions with denominator $p^k$.)
