Proof of $\mathbb Z[a]$ is not a lattice, where $a\not\in \mathbb Z$

Question

I wanted to proof that $\mathbb Z[\sqrt{2}]$ is not a lattice and I came up with the following proof:

Suppose that the set is a lattice. Since $\mathbb Z[\sqrt{2}]$ is a ring, we know that $\sqrt{2}-1\in \mathbb Z[\sqrt{2}]$. Using the same fact again we know that $a_n=(\sqrt{2}-1)^n\in \mathbb Z[\sqrt{2}]$ for all $n\in \mathbb N.$

Now notice that $\lim_{n\rightarrow \infty}a_n=0$, hence $0\in \mathbb Z[\sqrt{2}]$ is an accumulation point of the lattice. So we conclude that the lattice is not discrete, which is a contradiction.

Then I noticed that I can apply this proof also to rings of the form $\mathbb Z[\frac{p}{q}]$, although I thought that they are lattices. Now I am pretty sure that this is not true due to the above proof. Can someone confirm that I am right/wrong?

Thanks.

Answer 1

The ring $R=\Bbb Z[p/q]$ contains $1/q$. You can use the same trick; $a_n=(1/q)^n\in R$ and $a_n\to0$.

Answer 2

$\mathbb{Z}+ \beta \mathbb{Z},\beta \in \mathbb{C}$ is a lattice in $ \mathbb{C}$ (a free abelian group of rank $2$ spanning $ \mathbb{C}$ seen as a $ \mathbb{R}$ vector space) iff $\beta \not \in \mathbb{R}$.

And hence $\mathbb{Z}[\alpha]$ is a lattice iff $\mathbb{Z}[\alpha] = \mathbb{Z}+ \beta \mathbb{Z}$ for some $\beta \in \mathbb{Z}[\alpha],\beta \not \in \mathbb{R}$, ie iff $\alpha$ is a non-real quadratic integer, which means $\alpha = n+m \sqrt{-D}, m \ne 0$ or $\alpha = \frac{(2n+1)+(2m+1) \sqrt{-D}}{2},D \equiv 3 \bmod 4$.