## The division algorithm does not hold in the ring of integers of QQ(sqrt(-D)) with respect to the field norm when D is at most -15

Let $D \leq -15$ be a squarefree integer, let $K = \mathbb{Q}(\sqrt{D})$, and let $\mathcal{O}$ be the ring of integers in $K$. Show that the division algorithm does not hold in $\mathcal{O}$ with respect to the field norm on $K$.

We wish to find $\alpha, \beta \in \mathcal{O}$ such that, whenever $\alpha = \gamma\beta + \delta$, we have $N(\delta) \geq N(\beta)$.

If $D \not\equiv 1$ mod 4, then $\mathcal{O} = \{a+b\sqrt{D} \ |\ a,b \in \mathbb{Z}\}$. Let $\alpha = 2+\sqrt{D}$ and $\beta = 2$, and suppose there exist $\gamma = a+b\sqrt{D}$ and $\delta = h+k\sqrt{D}$ such that $\alpha = \beta\gamma + \delta$ and $N(\delta) < N(\beta)$. Then we have $h^2 - Dk^2 < 4$. Note that if $k \neq 0$, then $N(\delta)$ is too large; thus $k = 0$ and $h^2 < 4$. In fact, $h = \pm 1$, so that $2+\sqrt{D} = (2a \pm 1) + 2b\sqrt{D}$. Comparing coefficients, we have a contradiction mod 2. So no such $\gamma$ and $\delta$ exist.

If $D \equiv 1$ mod 4, then $\mathcal{O} = \{\frac{a}{2} + \frac{b}{2}\sqrt{D} \ |\ a,b \in \mathbb{Z}, a \equiv b \mod 2\}$. Let $\alpha = \frac{3}{2} + \frac{1}{2}\sqrt{D}$ and $\beta = 2$. Suppose there exist $\gamma = \frac{a}{2}+\frac{b}{2}\sqrt{D}$ and $\delta = \frac{h}{2}+\frac{k}{2}\sqrt{D}$ such that $\alpha = \beta\gamma + \delta$ and $N(\delta) < N(\beta)$. Now $h^2 - Dk^2 < 16$. Since $D \leq -15$, we have $(h,k) \in \{(0,\pm 1),$ $(\pm 1,0), (\pm 2,0),$ $(\pm 3,0)\}$. Since $h \equiv k$ mod 2, $(h,k) = (\pm 2,0)$. Now $\beta = 1/2$, a contradiction.