Determine if $\mathbb{Z}[\sqrt{2}i]$ is UFD.

Question

Determine if $\mathbb{Z}[\sqrt{2}i]$ is UFD.

I was trying to prove this statement but, I don't know how to do this, once I don't know any result that establishes a sufficient and necessary condition for an integral domain be UFD. Thanks a lot.

Answer 1

$\mathbb Z[i\sqrt2]$ is a UFD because it is a Euclidean domain. Consider the norm function $$N\colon \mathbb Z[i\sqrt2] \longrightarrow \mathbb Z,\quad a+i\sqrt2b\longmapsto \lvert a+i\sqrt2b\rvert^2 =a^2+2b^2.$$

Let now $x,y\in \mathbb Z[i\sqrt2]$ arbitrary, such that $y\neq 0$. Choose $q\in \mathbb Z[i\sqrt2]$ such that $$\left\lvert\frac xy -q\right\rvert \le \left\lvert \frac xy-q'\right\rvert,\qquad \forall q'\in \mathbb Z[i\sqrt2].$$ Set $r:= x-qy\in \mathbb Z[i\sqrt2]$. Then we have $\left\lvert \frac xy-q\right\rvert\le \frac{\sqrt3}2$ (look at the lattice $\mathbb Z[i\sqrt2]$ defines in the complex plane). It follows that $$\begin{align*} N(r) &= N(x-qy) = \lvert y\rvert^2\cdot \left\lvert \frac xy-q\right\rvert^2\\ &\le \lvert y\rvert^2\cdot \frac34 < N(y). \end{align*}$$ Hence $\mathbb Z[i\sqrt2]$ is a Euclidean domain and in particular a UFD.