let $w=-1/2+\sqrt{3}/2i$. Show that $\Bbb{Z}[w]$ is a PID.
I am trying to get a proof to above. My idea is to use the proof idea of $\Bbb{Z}[i]$ which is:
We define the norm function as $$ N:\Bbb{Z}[i]\rightarrow \Bbb{Z}^{\ge 0}, N(z)=|z|^2.$$ Note that $$N(z)=0\iff z=0.$$ Now, to show the existence of $q,r$ such that $z=qw+r,N(r)<N(w)$ Let $\frac{z}{w}=t+is$ where $t,s\in \Bbb{Q}$ after rationalising the denominator. Choose integers such that $$|t-a|\le \frac{1}{2}, |s-b|\le \frac{1}{2}.$$ So $$z=w(t+is)=w(a+ib)+w((t-a)+i(s-b)= wq+r.$$ To show the norm condition is satisfied, note $$N(r)=N(w)N((t-a)+i(s-b))=N(w)(1/4+1/4)=N(w)/2<N(w).$$
Since it is an euclidean domain, it is also a PID.
So for $\Bbb{Z}[w]$, I am using the same norm function defined above but I am not sure how to show that euclidean algorithm is working here.
Any ideas/solutions?
