A $\Bbb Z$-module that is not a direct sum of cyclic modules 
We would like to find a $\Bbb Z$-module that can not be written as a
  direct sum of cyclic modules.

My thought is: Let's consider the $\Bbb Z$-module $\Bbb Q$. In other words the abelian group $(\Bbb Q,+)$. This group is indecomposable. Indeed, let's assume that $\Bbb Q=\Bbb Za \oplus \Bbb Zb$, for some $a,b\in \Bbb Q$. But then, $ab \in \Bbb Za\cap \Bbb Zb$, so the intersection is not zero.  Generalising, if $\Bbb Q$ was a direct sum of more than one cyclic groups, then would not have that property and the argument is similar, ie the product if the generators would be in the intersection. 
And since this abelian group is not cyclic, we are done.
Questions: 
(1) Is this claim completely correct?
(2) Any other examples?
Thank you.
 A: Your argument is correct. 
Note that a $\mathbb{Z}$-module is an abelian group and by the fundamental theorem of finitely generated abelian groups any finitely generated abelian group is the direct sum of cyclic $\mathbb{Z}$-modules, so in particular we are looking for an abelian group that is not finitely generated. This naturally suggest to look at $\mathbb{Q}$.
A: Here's a hint that reflects what I think are the most important concepts here and doesn't depend on the structure theorem for finitely generated $\Bbb{Z}$-modules.
First a couple of standard definitions:


*

*A $\Bbb{Z}$-module $M$ is said to be torsion-free if whenever $x  \in M \setminus \{0\}$, $m \in \Bbb{Z}$ and $mx = 0$, then $m = 0$.

*For $m \in \Bbb{N}$ with $m > 1$, a $\Bbb{Z}$-module $M$ is said to be
$m$-divisible if for every $x \in M$, there is $y \in M$ such that $x = my$.


Now show that, if a sum $M = \bigoplus_i M_i$ of non-trivial cyclic $\Bbb{Z}$-modules is torsion-free, then each summand $M_i$ is torsion-free and deduce from this that $M$ is not $m$-divisible for any $m > 1$. But $\Bbb{Q}$ is torsion-free and $m$-divisible for every $m > 1$. (For a smaller example you can take the submodule of $\Bbb{Q}$ generated by the numbers $\frac{1}{m^i}$ for arbitrary $i \in \Bbb{N}$ and a fixed $m > 1$.)
