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Let $\alpha$ be a nonreal algebraic number. I'm interested in conditions that imply that $\mathbb{Z}[\alpha]$ is dense in $\mathbb{C}$. I'm particularly interested algebraic integer $\alpha$.

This is what I know so far:

  • if there is a $n \in \mathbb{N}$ such that $\alpha^n \in \mathbb{R} \setminus \mathbb{Z}$, then $\mathbb{Z}[\alpha]$ is dense in $\mathbb{C}$;
  • algebraic integers of degree two don't satisfy the condition, although algebraic nonintegers of degree two may.

Many thanks in advance.

user153170
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1 Answers1

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If and only if either $\alpha$ has degree at least $3$ or has degree $2$ and is not an algebraic integer. The key lemma is that a subgroup $A$ of a finite-dimensional real vector space $V$ such that $A$ spans $V$ is either discrete or dense in a subspace of $V$, and in the first case it is isomorphic to $\mathbb{Z}^n$ where $n = \dim V$. Now observe that $\mathbb{Z}[\alpha]$ is isomorphic to $\mathbb{Z}^2$ if and only if $\alpha^2$ is an integer linear combination of $\alpha$ and $1$, and furthermore that $\mathbb{Z}[\alpha]$ cannot be dense only in a $1$-dimensional subspace because for any such subspace it must also be dense in $\alpha$ times that subspace.

Qiaochu Yuan
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