How to prove $\mathbb{Z}[\sqrt{2}i]=\{a+b\sqrt{2}i\mid a,b\in\mathbb{Z}\}$ is a principal ideal domain? I can prove it is unique factorization domain.
Moreover, how to prove $\mathbb{Z}[\sqrt{n}i]=\{a+b\sqrt{n}i\mid a,b\in\mathbb{Z}\}$ is not unique factorization domain for all $n\geq 3$ hence they are not principal ideal domain?
Let $N(x)=|x|^2$ be a Euclidean function.
you should check that for any nonzero $ a, b \in \mathbb{Z}[\sqrt{2}i]$, $N(a)\le N(ab)$. (Not hard. Just try!)
Next, for any $ a, b \in \mathbb{Z}[\sqrt{2}i], b\neq 0$, we write $\frac{a}{b}=c+d\sqrt{2}i$, where $c,d \in \mathbb{Q}$.
Then there exist 2 integers $m, n$ such that $|m-c|<\frac12$ and $|n-d|<\frac12$.
We have $a=b(\frac ab)=b(c+d\sqrt{2}i)=b[(c-m)+m+((d-n)+n)\sqrt{2}i]$
$=bq+r$ where $q=m+n\sqrt{2}i$ and $r=b[(c-m)+(d-n)\sqrt{2}i]$.
(Note that $q,r\in \mathbb{Z}[\sqrt{2}i]$)
Then you can check simply that $N(r)\le N(b)$ when $r\neq 0$ and that $N(r)=0$ when $r=0$.
Hereby, we show that the ring is actually a Euclidean domain. By a theorem in any abstract algebra text, the ring is PID.
For the first part, prove that it's a Euclidean domain with Euclidean function $N(x)=|x|^2$. Then it follows that it's a PID.
