# 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)?

• Apparently you are the type of person who asks others for help without the intent to follow even the most minimal form of social conventions, namely, to acknowledge any effort that has gone into answering your question. Charming. – Gordon Jul 18 '17 at 0:54
• @Gordon Thank you for taking the time to respond to my question. I'm sorry that my sporadic use of this site offends you personally. In my experience, answering a question is usually its own reward (otherwise no one would answer questions here, or pursue mathematics period). – yoyo Jul 19 '17 at 22:56
• "answering a question is usually its own reward." What a faux high minded response. Many people answer questions here are generous with their time and hope that their efforts allow the OP and others to learn some mathematics. Did you ask the question because you were interested in the answer, or because you thought it was such a brilliant question that even if you never returned to bother looking at the answer you had made the world a better place? Don't bother answering, since the question is rhetorical, and I will certainly not bother interacting with you in the future. – Gordon Jul 20 '17 at 1:12

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 $\mathcal{O}_F$.