# Intersection of closed sets in $\mathbb{Q}^2$

My question has connection with this question. Let $k>0$ be an integer without square factor. We consider the ring $\mathbb{Z}[\sqrt{k}]$. Let $N(a+b\sqrt{k}):= |a^2-b^2k|$ for $(a,b) \in \mathbb{Q}^2$. We suppose $(\mathbb{Z}[\sqrt{k}],N)$ is not euclidian.

Let $F:=\{(x,y) \in \mathbb{Q}^2 | N(x+y\sqrt{k})\geq 1\}$.

We consider the set $$E=\bigcap_{(a,b)\in \mathbb{Z}^2} (F+(a,b)).$$

The complement of $E$ in $\mathbb{Q}^2$ is dense.

Is $E$ locally finite ?

Remark: If $(\mathbb{Z}[\sqrt{k}],N)$ is euclidian, $E$ is empty. Because if $\frac{a}{d}+\frac{b}{d}\sqrt{k} \in \mathbb{Q}[\sqrt{k}]$, we have the euclidian division $a+b\sqrt{k}=(q_1+q_2\sqrt{k})d+r_1+r_2\sqrt{k}$ with $N(r_1+r_2\sqrt{k})<N(d)$. So $(\frac{a}{d},\frac{b}{d}) \notin (F+ (q_1,q_2))$

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Just a slight nitpick: $k$ must be non-square. Also, by definition of $N$, isn't $F=\Bbb Q^2$? – Cameron Buie Jul 3 '12 at 14:56
At a guess, I'd say the modulus in the definition of $N$ shouldn't be there... – Ben Millwood Jul 3 '12 at 15:15
@BenMillwood The modulus is probably meant to be there. It's an "adjusted" norm that sometimes an integer quadratic domain becomes Euclidean with respect to. A Euclidean function can not be negative, so it can't be Euclidean with respect to the usual norm. But certainly, some definition here is off, as Cameron mentioned what we have implies $F=\mathbb{Q}^2$ and this problem seems odd. – Ragib Zaman Jul 3 '12 at 15:22
Sorry, I made a mistake. I have edited. It's $N(x+y \sqrt{k})\geq 1$ in the definition of $F$. – francis-jamet Jul 3 '12 at 20:05