It is easy to show that $Hom(\mathbb{Z}\_p,\mathbb{Z})=0$ (one just uses that each integer coprime to $p$ is a unit and therefore each image of such a homomorphism has infinitely many divisors). However, I wonder if this is true more generally: 


*Question:* Given a commutative ring $R$ and a maximal ideal $m \subseteq R$, is it true that $Hom\_R(\hat{R},R)=0$ where $\hat{R} = \varprojlim R/m^i$ ?


Of course, one has to require $R$ to be not complete with respect to $m$, since otherwise the Hom in question is simply the dual of $R$ and hence isomorphic to $R$. 



The (algebraic) dual of a completion 
Your example for $R=\mathbb{Z}$ (along with the proof) generalizes verbatim to the case $R$ a principal ideal domain and $m=(p)$ with a prime element $p$. 


As another example let $R$ be a commutative Noetherian local domain that is not complete. Then also $Hom\_R(\hat{R},R)=0$. A concrete example is $R=k[x\_1,...,x\_n]\_{(x\_1,...,x\_n)}$ with $k$ a field. 


This was shown by [Aldrich et. al: Derived Functors of Hom Relative to Flat Covers. Math. Nachr. 242(2002), 17-26], Lemma 3.3. 


Added: The following is used in the proof of the cited result: 


**Lemma:** If $R$ is a commutative Noetherian local ring then $\hat{R} \to Hom\_R(\hat{R},\hat{R}),\;r \mapsto r \cdot id$ is an isomorphism of $R$-modules. 


Proof: Let $E=E(R/m)$ be the injective hull of the residue field. By [Brodman, Sharp: Local Cohomology. Theorem 10.2.11] the map $\hat{R} \to Hom\_R(E,E), \; r \mapsto r\cdot id\_E$ is an isomorphism of $R$-modules (this is sometimes considered as part of Matlis duality). Hence we have $R$-module isomorphisms 
$$\begin{array}{lll}
Hom\_R(\hat{R},\hat{R}) & \cong & Hom\_R(\hat{R},Hom\_R(E,E))\cong Hom\_R(\hat{R} \otimes\_R E,E) \newline 
 & \cong & Hom\_R(E,E) \cong \hat{R}
\end{array}$$
where the 3rd isomorphism uses $\hat{R} \otimes\_R E \cong E$ (see [Matsumura: Comm. ring theory. Proof of 18.6 (iii)]. Composing the isomorphisms on element level now yields the map in the lemma. 


