# The set of the form $m+n\sqrt{-3}$, $m, n\in\mathbb Z$ or $m, n$ are both halves of odd integers is a Euclidean domain

Let $$D$$ be the set of complex numbers of the form $$m+n\sqrt{-3}$$ where $$m$$ and $$n$$ are either both in $$\mathbb Z$$ or are both halves of odd integers. Show that $$D$$ is a Euclidean domain relative to $$\delta(m+n\sqrt{-3})=m^2+3n^2$$.

This is an exercise on Page 149, Basic Algebra I, Jacobson. It is easy to show that $$D$$ is a domain. And we have already known that $$\mathbb Z[\sqrt{-3}]$$ is not a Euclidean domain (See: Is $$\mathbb Z[\sqrt{-3}]$$ Euclidean under some other norm? ). However, $$\mathbb Z [\sqrt 3]$$ is a Euclidean domain (See: https://math.stackexchange.com/a/1154089/549397 ).

I have checked that $$\delta$$ is a homomorphism from $$D$$ to $$\mathbb Z$$. My question is:

Why the method of proving $$\mathbb Z[\sqrt 3]$$ is a Euclidean domain is not applicable to $$D$$, even, not applicable to $$\mathbb Z[\sqrt -3]$$ ?

• Well, why do you assume it isn't? – Eric Wofsey Jan 16 at 6:29
• @EricWofsey You mean the method proving $\mathbb Z [\sqrt 3]$ ? Since $\mathbb Z [\sqrt -3]$ is not a Euclidean domain – Bach Jan 16 at 6:33
• Right, that makes sense for $\mathbb{Z}[\sqrt{-3}]$, but what about $D$? – Eric Wofsey Jan 16 at 6:34

Well, the way to answer this question is just to go through the proof for $$\mathbb{Z}[\sqrt{3}]$$ step by step and see where something special about its norm is used. The only place that happens is in the red inequality below from the end of the proof you linked: \begin{align*} N(\gamma) &= N(\beta \cdot \theta) \\ &= N(\beta) \cdot N(\theta) \\ &= N(\beta) \cdot \vert (p - r)^2 - 3 (q - s)^2 \vert \\ &\color{red}\leq N(\beta) \cdot \max\{ (p - r)^2, 3(p - s)^2\} \\ &\leq\frac34 N(\beta)\\ &< N(\beta). \end{align*}
Going through these steps with your function $$\delta$$ on $$D$$ or $$\mathbb{Z}[\sqrt{-3}]$$, we would instead have \begin{align*} \delta(\gamma) &= \delta(\beta \cdot \theta) \\ &= \delta(\beta) \cdot \delta(\theta) \\ &= \delta(\beta) \cdot \vert (p - r)^2 \color{red}+ 3 (q - s)^2 \vert. \\ \end{align*} That plus sign instead of a minus sign makes all the difference, since we can no longer say that $$\vert (p - r)^2+ 3 (q - s)^2 \vert$$ is at most $$\max\{ (p - r)^2, 3(p - s)^2\}$$. Instead the best bound we can get on $$\vert (p - r)^2+ 3 (q - s)^2 \vert$$ using $$|p-r|\leq 1/2$$, $$|q-s|\leq 1/2$$ is $$1$$ which lets us conclude $$\delta(\gamma)\leq \delta(\beta)$$ instead of $$\delta(\gamma)<\delta(\beta)$$ as we need.
Note, though, that this doesn't mean there isn't any way to adapt the proof for $$D$$; you'll just have to make more changes. In particular, you'll need to use the fact that the coefficients of elements of $$D$$ can sometimes be half-integers to make a better choice of $$p$$ and $$q$$ in some cases.