This question already has an answer here:

I know that for $d=-1, -2$ the ring $\mathbb Z[\sqrt{d}]$ is an Euclidean Domain. I believe that it is not an Euclidean Domain for and $d \leq-3$.

I have been able to prove it for a handful of examples (like $d=-3$), showing that the resulting ring is not a UFD, but am not sure how to prove the claim in general. (And whether the resulting rings are not UFDs in the general case or merely not Euclidean.)


marked as duplicate by user26857, Norbert, Dennis Gulko, Sasha, tomasz Mar 2 '13 at 15:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


First of all, note that $\mathbb{Z}[\sqrt{d}] = \{ a + \sqrt{d}b \mid a,b \in \mathbb{Z} \}$ is not always equal to the usual ring of quadratic integers $\mathcal{O}(\sqrt{d})$, which consist of the largest subring of $\mathbb{Q}(\sqrt{d})$ for which its intersection with $\mathbb{Q}$ is the integers. If $d \equiv 1 \pmod{4}$, then $\mathcal{O}(\sqrt{d}) = \{ a + b \frac{1 + \sqrt{d}}{2} \mid a,b \in \mathbb{Z}\} = \{ a' + b' \sqrt{d}\}$, with either both $a', b'$ integers, or both $a',b'$ an integer plus $\frac{1}{2}$.

I will assume that the norm we use is $|a + b \sqrt{d} = a^2 - d b^2 = (a - \sqrt{d}b)(a + \sqrt{d}b)$, which is clearly multiplicative

For $\mathcal{O}(d)$, then it's a Euclidean domain for $d = -11,-7,-3,-2,-1$, but if you consider $\mathbb{Z}(\sqrt{d})$, it's indeed just $d = -1,-2$. This can actually be seen geometrically. These integers form a rectangular grid, and for any (not necessarily integer) point in this grid, there's always an integer point within a distance of at most $|\frac{1}{2} + \frac{1}{2}\sqrt{d}|$. For $d = -1,-2$, this gives $\frac{1}{2}$ and $\frac{3}{4}$. If we now divide a number $a = a_1 + a_2 \sqrt{d}$ by $b = b_1 + b_2 \sqrt{d}$, then we get a rational point $c = c_1 + c_2 \sqrt{d}$ in the grid, and there must exist an integer point at distance less than 1. We thus have $$ \frac{a}{b} = c_1 + c_2 \sqrt{d} = q_1 + q_2 \sqrt{d} + r_1 + r_2 \sqrt{d} $$ with $q_1,q_2$ integers, and $r_1,r_2 \in [-\frac{1}{2},\frac{1}{2}]$. Now $$ a = bq + br $$ and since $|br| = |b||r| < |b|$, the euclidean property is satisfied.

Now for $d = -3,-5,\dots$, then the point $1 + \sqrt{d}$ divided by 2 gives a point at the center of a lattice rect, and so the distance from this point to any integer point is at least 1. It's thus not possible to write $1 + \sqrt{d}$ as $2q + r$, with $|r| < 2$, which contradicts the definition of an Euclidean domain.


Not the answer you're looking for? Browse other questions tagged or ask your own question.