Is $\mathbb{Z}[\sqrt{2}]$ a lattice?

Question

Is $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}\mid a,b\in\mathbb{Z}\}$ a discrete subgroup of $\mathbb{R}$?

How to prove that?

please recall the definition – janmarqz Jun 10 '15 at 21:25 — janmarqz, Jun 10 '15 at 21:25

score 5 · Answer 1 · answered Jun 10 '15 at 21:31

5

No. It fails to be discrete. In fact $\mathbb Z[\sqrt 2]$ is dense in $\mathbb R$.

It suffices to observe that the sequence $(\sqrt 2-1)^n\in \mathbb Z[\sqrt 2] $ converges to $0$.

answered Jun 10 '15 at 21:31

Hagen von Eitzen

How to show that it is not discrete. – Nemo Jun 10 '15 at 21:35
@Nemo, what's your definition of discrete? One definition is that there exists an constant $\delta > 0$ such that no two members of the set are closer than $\delta$ together. That definition is violated here, as the example given by Hagen demonstrates. – Simon S Jun 10 '15 at 21:42
@simon: Your definition is correct, and $\mathbb{Z}[\sqrt{2}]$ is not a discrete subgroup of $\mathbb{R}$ (therefore, not a lattice), as Hagen showed. – Nemo Jun 10 '15 at 21:51

1 Answers1