discrete subrings of $\mathbb{R}^r\times\mathbb{C}^s$ Given a number field $K$ of degree $r+2s$, its ring of integers $\mathcal{O}_K$ is a discrete subring of $\mathbb{R}^r\times\mathbb{C}^s$ under the diagonal embedding
$$
\alpha\mapsto(\sigma_i(\alpha))_i,
$$
where $\sigma_1,\ldots,\sigma_r$ are the $r$ real embeddings and $\sigma_{r+1},\ldots,\sigma_{r+s}$ are choices from the $s$ conjugate pairs of complex embeddings.
My question is about the converse, does every discrete subring of $\mathbb{R}^n\times\mathbb{C}^m$ (product topology and product ring structure) come from number rings in some fashion (maybe allowing $\oplus$, $\otimes$, or restricting to domains, or restricting to rings where the projections are all isomorphisms)?
 A: Basically, yes.
Let $R$ be such a ring. The assumption that $R$ is discrete means that $R$ is a
free $\mathbf{Z}$-module of rank at most $n + 2m$. The fact that $R$ is a subring implies that both $R$ and $S = R \otimes \mathbf{Q}$ is reduced.
So $S$ is a finite dimensional algebra over $\mathbf{Q}$ which is reduced. It's certainly also artinian because the ideals will also be vector spaces. I claim that any artininan and reduced algebra over a field $k$ is a product of finite extensions of $k$. 
Proof: by http://stacks.math.columbia.edu/tag/00KJ (1)=>(8) we deduce that $S$ is a finite product of local Notherian $k$-algebras whose maximal ideals are nilpotent. Because $S$ is reduced, these maximal ideals are actually trivial, and hence $S$ is a product of fields (which are finite dimensional $k$-vector spaces).
If $S = \prod F_i$, I claim that $R$ is a finite index subring of $\prod \mathcal{O}_{F_i}$. Otherwise, if $\alpha = (\alpha_i)$ with some $\alpha_i \in F_i$ not integral, then the image of $R$ inside $F_i$ would contain $\mathbf{Z}[\alpha_i]$, which has infinite rank as an abelian group exactly when $\alpha_i$ is not integral. On the other hand, once one has a containment of $R$ inside this product, the fact that both rings are free abelian groups of the same rank implies that the index is finite.
Note, however, that $R$ itself need not decompose as a product even when $S$ does. For example, take $R$ to be the subring of $(a,b) \in \mathbf{Z} \oplus \mathbf{Z}$ such that $a-b$ is divisible by some fixed integer $N$. More generally, there are many complicated subrings of $\mathbf{Z}^n$. 
If one restricts to domains, of course, then $S = F$ and $R$ is exactly a finite index subring (order) of $\mathcal{O}_F$.
