Homomorphisms between $ \mathbb{Z} $ modules. 
Calculate $\newcommand\Hom{\operatorname{Hom}}\Hom(\mathbb Z \oplus \mathbb Z_{p^\infty},\mathbb Z \oplus \mathbb Z_{p^\infty})$. Where $ \mathbb Z_{p^\infty}= \bigcup_{k=1}^{\infty}\bar{\langle\frac1{p^k}\rangle}$ for $p$ prime.

As $\mathbb Z$ and  $\mathbb Z_{p^\infty}$ are $\mathbb Z$-modules then applying the theorem $\Hom_A ( \bigoplus_{i \in I} M_i, \prod_{j \in J} N_j ) \cong\prod_{(i,j) \in I \times J} \Hom(M_i,N_j)$ because for finite indices have $\bigoplus_{i=1}^{n}M_i = \prod_{i=1}^n M_1$. Then the problem reduces to find those $\Hom$ where
$$ \Hom(\mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z} $$
$$ \Hom(\mathbb{Z},\mathbb{Z}_{p^\infty}) \cong \mathbb{Z}_{p^\infty}$$ 
but  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z})$ and  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z}_{p^\infty})$ not how to calculate them.
 A: To understand $\newcommand\Hom{\operatorname{Hom}}\Hom$, think about generators.  Since any generator of $\mathbb{Z}_{p^\infty}$ is torsion, it has to map to $0$ in $\mathbb{Z}$.  So,
$$
\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}) = 0.
$$
In order to understand $\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}_{p^\infty})$, consider the definition of $\mathbb{Z}_{p^\infty}$ as a direct limit.
$$
\mathbb{Z}_p \overset{p}{\to} \mathbb{Z}_{p^2} \overset{p}{\to} \mathbb{Z}_{p^3} \overset{p}{\to} \cdots \to \mathbb{Z}_{p^\infty}
$$
Any map from a direct limit is characterized by the commuting diagram of maps from each of the groups in the directed system.  So, if $x_n$ is a generator of $\mathbb{Z}_{p^n}$, then its image in $\mathbb{Z}_{p^{n+1}}$ is $px_{n+1}$ for some generator of $\mathbb{Z}_{p^{n+1}}$.  These are identified in the direct limit $\mathbb{Z}_{p^\infty}$, so for any well-defined $f \in \Hom(\mathbb{Z}_{p^\infty}, G)$ for any group $G$,
$$
f(x_n) = p f(x_{n+1}) = p^2 f(x_{n+2}) = \cdots
$$
Thus, $f(x_n)$ must be infinitely divisible: $f(x_n) = p^k f(x_{n+k})$ for any $k \ge 0$.  In fact, one can show that
$$
\Hom(\lim_{\to} A_n, G) \cong \lim_{\leftarrow}\Hom(A_n, G),
$$
for any abelian groups.  In your case,
$$
\begin{align}
\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}_{p^\infty}) &\cong \Hom(\lim_{\to} \mathbb{Z}_{p^n}, \mathbb{Z}_{p^\infty}) \\
&\cong \lim_{\leftarrow}\Hom(\mathbb{Z}_{p^n}, \mathbb{Z}_{p^\infty}) \\
&\cong \lim_{\leftarrow}\mathbb{Z}_{p^n}.
\end{align}
$$
The last group is an uncountable abelian group called the $p$-adic integers.
A: $\mathbb{Z}/p^{\infty}$ is divisible, this implies $\hom(\mathbb{Z}/p^{\infty},\mathbb{Z})=0$.
We have $\mathbb{Z}/p^{\infty}  \cong \mathrm{colim}_n ~\mathbb{Z}/p^n$, hence $\hom(\mathbb{Z}/p^{\infty},\mathbb{Z}/p^{\infty}) \cong \lim_n  \hom(\mathbb{Z}/p^n,\mathbb{Z}/p^{\infty})$
Now $\hom(\mathbb{Z}/p^n,\mathbb{Z}/p^{\infty})$ is isomorphic to the $p^n$-torsion of $\mathbb{Z}/p^{\infty}$, which is isomorphic to $\mathbb{Z}/p^n$. Hence, $\hom(\mathbb{Z}/p^{\infty},\mathbb{Z}/p^{\infty}) = \mathbb{Z}_p$, the ring of $p$-adic integers.
A: $\def\Z{\mathbb{Z}}\def\Hom{\mathrm{Hom}}$We have $\Hom(\Z_{p^\infty},\Z)=0$ because $\Z_{p^\infty}$ is divisible and $\Z$ is reduced.
On the other hand $\Hom(\Z_{p^\infty},\Z_{p^\infty})$ is the ring of $p$-adic integers. You find its description in any Algebra book (Cohn, for instance, or Atiyah-McDonald), or http://en.wikipedia.org/wiki/P-adic_number
