Proving $\{a+b\sqrt{-7}, a,b\in\frac{\mathbb{Z}}{2}\}$ is a euclidean domain

Question

I am having a bit of difficulty showing that the set $A=\{a+b\sqrt{-7},a,b\in\mathbb{Z}/2\}$ (i.e. using integers+half integers) is a euclidean domain. I see how a lot of the proofs work, like for $\mathbb{Z}[i]$ and $\mathbb{Z}[\sqrt{-2}]$. However, I am a bit confused, as it seems that the trick that happens doesn't work for me. In those proofs, you show that some remainder $r$ and elements $z,w$ such that $z=qw+r$. With this, you can show that $N(r)\leq\frac{3}{4}N(z)$. I am not getting this $\frac{3}{4}$, I am getting $\frac{8}{4}$, or just some constant that isn't less than 1. Furthermore, I've looked at some of the other posts when your $d=1$mod$4$, but it looks like they're using the same norm, so I don't know.

Here's my work: Let the norm of an element in $A$ be $N(a+b\sqrt{-7})=4(a^2+7b^2).$ Now I am not entirely sure about multiplying this by 4, but it seems that the norm function has to map into $\mathbb{N}\cup{0}$, so in order to guarantee that this happens I need the 4 to account for the half integers being squared (does it matter that $\mathbb{N}$ has a bijection with $\frac{\mathbb{N}}{4}$?

Let $z,w\in A$. Then there are real numbers $c,d$ such that $\frac{w}{z}=c+d\sqrt{-7}$. Let $q_1,q_2\in\frac{\mathbb{Z}}{2}$ be the closest elements to $c,d$ respectively. Then we have $|q_1-c|\leq\frac{1}{4}$ and $|q_2-d|\leq\frac{1}{4}$. Setting $q=q_1+q_2\sqrt{-7}$ and $r=\frac{w}{z}-q$, so $w=qz+rz$. Now it remains to show that $N(rz)=N(r)N(z)\leq N(z)$, and this is where my problem is. For $N(r)$, my first question is why in the world can I even apply the norm to $r$ since it must not even belong to my set $A$. Ignoring that technicality, we have that $r=(q_1-c)+(q_2-d)\sqrt{-7}$, so $N(r)=4[|q_1-c|^2+7|q_2-d|]\leq 4[(1/4)^2+7(1/4)^2=2.$ If my 4 from the norm wasn't part of my norm function, then I would be good to go, but I've explained why I'm not sure that works. Can someone help me figure out where I'm going wrong? Thanks.

Answer 1

Let $\omega = { 1 + \sqrt { -7 } \over 2}$, $A = {\mathbb Z}[ \omega]$, and $k = {\mathbb Q}[\omega]$. Let $$ N ( a + b \omega ) = (a+ b\omega) (a + b\bar \omega) = a^2 + ab + 2b^2= (a + b/2)^2 + 7/4 b^2, $$for $a, b\in \mathbb Q$. Then $N$ is multiplicative, maps $k$ to the positive rationals, and maps $A$ to $\mathbb N$.

Suppose $w$ and $z \in A$, and $z \ne 0$.Then $$ w /z = \mu + \nu \omega,$$ with $\mu,\nu \in {\mathbb Q}.$ Let $m, n \in \mathbb Z$, be such that $\epsilon = \mu -m$ and $\delta = \nu -n$, have absolute $\le 1/2$ (we'll need to be a bit more careful if both have absolute value exactly equal to a 1/2 - see below). Then, if $q = m + n\omega$, we have $$ w = q z + r,$$ where $$ r = z (\epsilon + \delta \omega).$$ $r \in A$, because $w$, $q$, and $z$, are, and $A$ is a ring, and $\rho = \epsilon + \delta \omega \in k,$ because $k$ is a field - in particular $\epsilon$ and $\delta \in \mathbb Q,$ so $N \rho$ makes sense. To show that $N r < N z$, we need only that $ N \rho < 1$, since $N r = N z N \rho$. But $$ N \rho = \epsilon^2 + \epsilon \delta + 2 \delta^2 = \left( \epsilon + {\delta \over 2}\right)^2 + {7\over 4} \delta^2.$$
Now, with the ('careless') choices above, $ 7/4 \delta^2 \le 7/16$, and $ (\epsilon + \delta/2)^2 \le 9/16$, so we only have that $N \rho \le 1 $, which isn't quite good enough (we need strict inequality). However, the problem (of no strict inequality) only occurs when the absolute values of $\epsilon$ and $\delta$ are equal to $1/2$. In that case, if we improve our choice of $\epsilon$ for it to have the opposite sign of $\delta$, and then all is well: $N \rho = 1/16 + 7/16 < 1$.