Determining if $\mathbb{Z}[a]$ is a discrete subring of $\mathbb{C}$. Let $a \in \mathbb{C}$ and consider the ring $\mathbb{Z}[a]$. Is there some nice criterion which will tell me whether $\mathbb{Z}[a]$ is discrete in the sense that there is some $\delta >0$ such that, whenever $x \in \mathbb{Z}[a] \setminus \{0\}$, one has $|x| \geq  \delta$?  It follows, then, that there is a smallest possible distance between any two points.
I have been able to show some pretty easy things, such as:

Fact 1: A subring $R \subset \mathbb{C}$ is discrete if and only if $\{ |x| : x \in R\} \subset [0,\infty)$ is order isomorphic to $\mathbb{N}$.
Fact 2: The only discrete (unital) subring of $\mathbb{R}$ is $\mathbb{Z}$.

Fact 2 makes the case $a \in \mathbb{R}$ pretty easy: we get $\mathbb{Z}[a]$ discrete if and only if $a \in \mathbb{Z}$.
The case where $a \in \mathbb{C}$ is pure imaginary is also pretty easy. You need $\mathbb{Z}[a^2] \subset \mathbb{Z}[a]$ to be discrete, so you need $a^2$ an integer, i.e. $a = \pm \sqrt{n}i$ for some positive integer $n$.  It's not hard to show that, for such $a$, the ring $\mathbb{Z}[a]$ is also discrete.
The next case I tried was the case where $a$ is a root of unity. If $a \neq 1$ is a third root of unity, things work out quite nicely: it turns out that $\mathbb{Z}[a]$ is a hexagonal lattice.
However, if $a \neq 1$ is a  fifth root of unity, it gets difficult to compute by bare hands, and I am not too sure what happens.
 A: If $\mathbb Z[a]$ is a discret, then $\mathbb Z$-module $\mathbb Z[a]$ finitely generated, hence $a$ must be an algebraic integer number. If $n$ is a degree of a minimal polynomial of $a$, then $\mathbb Z$-module $\mathbb Z[a]$ generated by $1,a,\ldots,a^{n-1}$ and they generate a discret subgroup of $\mathbb C^+$ ($\mathbb C$ by addition). Conversely, let $a$ be an algebraic integer of degree $n$ such, that $1,a,\ldots,a^{n-1}$ generate a discret subgroup of $\mathbb C^+$. Then $\mathbb Z[a]$ as a subfgroup of $\mathbb C^+$ is generated by $1,a,\ldots,a^{n-1}$, so, by assumption, $\mathbb Z[a]$ is a discret subgroup in $\mathbb C^+$.
It remains to understand in which cases $1,a,\ldots,a^{n-1}$ generate a discret subgroup of a $\mathbb C^+$. Besides that, should be considered, that $1,a,\ldots,a^{n-1}$ are linearly independent over $\mathbb Q$, and any discret subgroup in $\mathbb C^+$ has rank not greater 2, so $n\leq 2$. If $n=1$, then $a\in\mathbb Z$. Let $n=2$. If $a\not\in\mathbb R$, then $1,a$ generate over $\mathbb Z$ discret subgroup. If $a\in\mathbb R$, then $1,a$ generate discret subgroup of a $\mathbb R^+$ iff $\{1,a\}\subset t\mathbb Z$ for some $0\neq t\in\mathbb R$. Hence $1=ty$, $a=tx$ for some $x,y\in\mathbb Z$. Then $a=\frac{a}{1}=\frac{tx}{ty}=\frac{x}{y}\in\mathbb Q$ - contradiction, since $a$ has degree 2 over $\mathbb Q$.
In such a way, we get the answer: $\mathbb Z[a]$ is a discret subgroup in $\mathbb C^+$ iff $a\in\mathbb Z$ or $a$ is a nonreal quadratic algebraic integer (i.e. $a$ is a root of a polynomial $x^2+ux+v$, where $u,v\in\mathbb Z$ and discriminant is negative).
