# Can a ring of integers contain a $2$-dimensional noetherian normal integral domain?

Let $K$ be a number field with ring of integers $O_K$. Does there exist a $2$-dimensional subring $A\subset O_K$?

Clearly, if such a subring $A\subset O_K$ exists, we have that $A$ is an integral domain. Also, taking the integral closure (i.e. the normalization) we obtain a normal integral domain $A\subset O_K$ of dimension two.

Two problems remain:

1. How do you show that there exists such a subring?

2. How do you get a noetherian subring?

Also, I was wondering whether there is a "canonical" choice for such an $A$ (if it exists).

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If $A\subseteq\mathscr{O}_K$ is a subring, then necessarily $\mathbf{Z}\hookrightarrow A$. So $A$ is an integral extension of $\mathbf{Z}$. In general, if $R\rightarrow S$ is an integral extension of rings, then $\mathrm{dim}(R)=\mathrm{dim}(S)$, so, in our particular case, we get $1=\mathrm{dim}(\mathbf{Z})=\mathrm{dim}(A)$.
If $A \subset \mathcal O_K$, then in fact $\mathbb Z \subset A \subset \mathcal O_K$. Thus $A$ is finite over $\mathbb Z$, and hence is of dimension one.