Infinite direct sum of p-adic integers is not p-adic Studying Bousfield localization I stumbled upon this elementary example: we start with $\mathcal{D}$ the derived category of $p$-local abelian groups and we can consider the Bousfield class of $\mathbb{Z}/p$. The associated localization is the usual $p$-completion and the subcategory of $\mathbb{Z}/p$-local objects $\tilde{\mathcal{D}}$ should coincide with full subcategory generated by $\{ X : [\mathbb{Q},X]=0 \}$.
Now the fact is that $\mathcal{D}$ is a closed symmetric monoidal category with smash product the usual tensor product of chain complexes and $\mathbb{Z}_{(p)}$ the unit of such product is small. The localization functor sends $\mathbb{Z}_{(p)}$ to $\mathbb{Z}_p$ and this should not be small/finite. As proof of this claim we can observe that the coproduct of infinitely many copies of $\mathbb{Z}_p$ in $\tilde{\mathcal{D}}$ is not the usual direct sum $\bigoplus_\mathbb{N} \mathbb{Z}_p$ the the $p$-completion of this sum i.e. $\bigl(\bigoplus_\mathbb{N} \mathbb{Z}_p \bigr)^{}_p$. Then we observe that 
$\tilde{\mathcal{D}}[\mathbb{Z}_p, \bigl(\bigoplus_\mathbb{N} \mathbb{Z}_p \bigr)^{}_p]$ and $\bigoplus_\mathbb{N}\tilde{\mathcal{D}}[\mathbb{Z_p}, \mathbb{Z}_p]$ do not coincide.
My problem is that I cannot prove that $[\mathbb{Q},\bigoplus_\mathbb{N} \mathbb{Z}_p] \neq 0$ i.e. the direct sum is not $\mathbb{Z}/p$-local. To see that $[\mathbb{Q},\mathbb{Z}_p]=0$ is easy using the definition of the $p$-adic integers via limit and that $[\mathbb{Q},\mathbb{Z}/p^n]=0$. Therefore I have to produce a morphism of abelian groups $\phi \colon \mathbb{Q} \rightarrow \bigoplus_\mathbb{N} \mathbb{Z}_p$ which does not factorise through a finite number of copies of $\mathbb{Z}_p$, but I do not see how I can do this.
By the additivity condition we see easily $ m \cdot \phi(1/m)=\phi(1)$. Now in $\mathbb{Z}_p$ multiplication by $q \in \mathbb{Z}$ with $(q,p)=1$ is invertible and multiplication by $p$ is injective. Thus the same is true for the direct sum $\bigoplus_\mathbb{N} \mathbb{Z}_p$ since addition is done separately in each copy. Thus if $\phi(1)$ is located in the first $N$ terms of the sum it determines uniquely $\phi(1/q)$ and it should stay in the first $N$ copied of the $p$-adics. The condition $p \cdot \phi(1/p)=  \phi(1)$ just implies that $\phi(1)$ must be in the subgroup of multiples by $p$ but injectivity implies still $\phi(1)$ uniquely determines $\phi(1/p)$ and this is still in the first $N$ copies of $\mathbb{Z}_p$.
Is my reasoning about the factorisation wrong or is $\bigoplus_\mathbb{N} \mathbb{Z}_p$ actually $\mathbb{Z}/p$-local? Can you provide an explicit example of the map $\phi$ that I want?
 A: I think I have found the problem in my reasoning: to put it plainly I confused the sets of maps of $\mathbb{Z}_{(p)}$-modules with the hom set in the derived category.  I foolishly thought that if the former were zero then so would be the latter which is not absolutely the case. 
In the next section I will denote $[-,-]$ the set of morphisms in the derived category of $\mathbb{Z}_{(p)}$-modules and $F(-,-)$ the internal hom of this category, while $Hom(-,-)$ will be the set of morphisms of $\mathbb{Z}_{(p)}$-modules.
For an easy example: consider $\mathbb{Z}/p$ and $\mathbb{Z}_p$. It is easy to see $Hom(\mathbb{Z}/p, \mathbb{Z}_{p})=0$ since the p-adic integers have no $p$-torsion. But $F(\mathbb{Z}/p, \mathbb{Z}_{p})\cong F(\mathbb{Z}/p, \mathbb{Z}_{(p)})\otimes \mathbb{Z}_{p} \cong \Sigma^{-1} \mathbb{Z}/p \otimes \mathbb{Z}_{p}   \cong \Sigma^{-1} \mathbb{Z}/p$, where we used the fact that $\mathbb{Z}/p$ is a small object and it is self-dual up to desuspension. Therefore $[\mathbb{Z}/p, \mathbb{Z}_{p}]$ cannot be zero.
Now back to my question: why $\bigoplus \mathbb{Z}_{p}$ is not $\mathbb{Z}/p$-local? Let's try to write down its localization: we see that the localization is just the $p$-completion i.e. 
\begin{equation}
\biggl(\bigoplus \mathbb{Z}_{p} \biggr)_p = \text{lim} \biggl( \bigoplus \mathbb{Z}/p^n \biggr) = \text{lim} \biggl( \dots \rightarrow \bigoplus \mathbb{Z}/{p^3}\rightarrow \bigoplus \mathbb{Z}/{p^2} \rightarrow \bigoplus \mathbb{Z}/{p} \biggr)
\end{equation}
and this does not coincide with $\bigoplus \mathbb{Z}_{p}$ basically because limits and colimits do not commute in full generality. For example the sequence $(a_n)_n$ where $a_n=(1,p,p^2, \dots, p^n, 0, \dots)$ is defines an element in the limit to the right while on the left it would define a sequence $(b_m)_m$ where $b_m=p^{m} \in \mathbb{Z}_{p}$. But by definition of direct sum the sequence must have only finitely many non-zero terms.
Still I cannot understand why $[\mathbb{Q}, \bigoplus \mathbb{Z}_{p}]\neq0$. With computations similar to above we get $F(\mathbb{Q},  \mathbb{Z}_{p})=\text{lim}(\dots \mathbb{Z}_{p} \xrightarrow{p} \mathbb{Z}_{p} \xrightarrow{p}\mathbb{Z}_{p})$ should be zero, where we are using the fact that $\mathbb{Q} = \text{colim}(\dots \mathbb{Z}_{(p)} \xrightarrow{p} \mathbb{Z}_{(p)} \xrightarrow{p}\mathbb{Z}_{(p)})$ but I cannot see why $F(\mathbb{Q},  \bigoplus \mathbb{Z}_{p})=\text{lim}(\dots \bigoplus \mathbb{Z}_{p} \xrightarrow{p} \bigoplus \mathbb{Z}_{p} \xrightarrow{p}\bigoplus \mathbb{Z}_{p})=0$.
