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.

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