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In his answer to this question What is a concrete example to demonstrate that $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is NOT a norm-Euclidean domain? Robert Soupe essentially looks up in a map to try to find remainders for the Euclidean algorithm. But how do you actually do the equivalent of the floor function in this domain?

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  • $\begingroup$ How do you want to order $\mathbb Q(\sqrt{-19})$? I'm sure you know that there is no total order consistent with the field operations? $\endgroup$ – AlexR Aug 17 '15 at 21:23
  • $\begingroup$ @AlexR Just for the record, speaking only for myself: when I gave that answer, I did not think about total order at all, though I had in the back of my mind that the proof that I've seen for $\mathbb{Z}[\sqrt{-2}]$ being Euclidean invokes some kind of division and flooring or ceilinging. I'll give the citation for that proof later. $\endgroup$ – Robert Soupe Aug 18 '15 at 12:26
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It is quite clear that the concern here is using the Euclidean algorithm, or determining if the algorithm can be used. But since this is not explicitly stated in the question, I will not make any promises as to the suitability of my answer to the algorithm.

The integers of an imaginary quadratic field form either a rectangular or lozenge lattice. Therefore if a number in the field is not an integer, it falls within a rectangular or lozenge-shaped region that has integers for corners. For example: $$\frac{10}{\frac{3}{2} + \frac{\sqrt{-19}}{2}} = \frac{15}{7} - \frac{5 \sqrt{-19}}{7}.$$ This is not an algebraic integers, but it falls within a lozenge-shaped region that has the integers $$\frac{3}{2} - \frac{5 \sqrt{-19}}{2}, 2 - 3 \sqrt{-19}, \frac{5}{2} - \frac{5 \sqrt{-19}}{2}, 2 - 2 \sqrt{-19}$$ for corners. (Someone doublecheck my math or geometry as the case may be, please).

It's hard to say that any of these is closer to some negative infinity than the others, but you can say that exactly one of them is closer to $0$ than the others. I suggest trying that one for your "floor."

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