A subring of $\mathbb{C}$ having $\mathbb{Z}$ as a direct summand as additive $\mathbb{Z}$-modules Let $R$ be a ring such that $\mathbb{Z}\subseteq R\subseteq \mathbb{C}$ (or $\mathbb{R}$). I want to see how the following statements are related:
1) $R\cap \mathbb{Q}=\mathbb{Z}$ and
2) $R=\mathbb{Z} + S$, where $+$ stands for internal direct sum of (additive) $\mathbb{Z}$-submodules $\mathbb{Z}$ and $S$ of $R$, and $=$ does NOT mean merely isomorphic but means that they are exactly same as sets. 
I saw the condition 2) on a book on the character theory of finite groups, and I am trying to see when this happens. 
It is easy to see that 2) implies 1): if $R=\mathbb{Z} + S$ for some subgroup $S$, then any element $q\in R\cap \mathbb{Q}$ can be written uniquely as $n+s$ where $n\in \mathbb{Z}$ and $s\in S$. Then it follows that $s=q-n\in\mathbb{Q}\cap S$, so $ms\in \mathbb{Z}\cap S$ for some nonzero integer $m$. But $\mathbb{Z}\cap S=\{0\}$, so it follows that $s=0$.
But for the converse, I don't even know if it is true or not. Does 1) imply 2)? If not, can you show me some examples of $R$ which satisfies 1) but not 2), or examples of conditions which implies (or equivalent to) 2)? What if we drop the assumption that $R$ is a ring and do this for general additive $\mathbb{Z}$-submodule of $\mathbb{C}$ (or $\mathbb{R}$)? I also welcome any reference suggestions for this type of questions. 
 A: Here is a counterexample.  Fix a transcendental number $c\in\mathbb{C}$ and an element $\alpha\in\hat{\mathbb{Z}}\setminus\mathbb{Z}$, where $\hat{\mathbb{Z}}$ is the profinite completion of $\mathbb{Z}$.  Let $R$ be the subring of $\mathbb{Q}[c]$ consisting of elements of the form $f(c)/n$ where $f\in\mathbb{Z}[x]$, $n\in\mathbb{Z}$, and $f(\alpha)$ is divisible by $n$ in $\hat{\mathbb{Z}}$.  Note that $R\cap\mathbb{Q}=\mathbb{Z}$ since if $f(c)/n\in R\cap\mathbb{Q}$ then $f$ must be an integer constant which is divisible by $n$.
Now suppose $\mathbb{Z}$ were a direct summand of $R$ and let $p:R\to\mathbb{Z}$ be the associated projection map.  For each nonzero $n\in\mathbb{Z}$, let $\alpha_n$ be an integer with the same mod $n$ residue as $\alpha$.  Then $\frac{c-\alpha_n}{n}\in R$, so applying $p$ we find that $p(c)-\alpha_n$ must be divisible by $n$ in $\mathbb{Z}$.  That is, $p(c)$ has the same mod $n$ residue as $\alpha$.  This means that $p(c)$ is an integer which maps to $\alpha$ under the natural inclusion $\mathbb{Z}\to\hat{\mathbb{Z}}$, which contradicts our choice that $\alpha\in\hat{\mathbb{Z}}\setminus\mathbb{Z}$.
(You may also be interested in my answer to Is $\mathbb{Z}$ the only totally-ordered PID that is "special"? which discusses some other properties of these rings.)
