Problem in solving a question related to Euclidean domain. The question is :
Let $R$ be the ring defined by $R = \{a + \frac {b(1 + i{\sqrt {19}})} {2} | a,b\in \mathbb {Z} \}$.Then $R$ is a principal ideal domain but not a Euclidean domain.
I have tried my best.But I fail.Please help me.
Thank you in advance.
 A: I denote $R$ your ring. For instance, I will show that $R$ is not euclidian. First, I want to prove that $U(R)$ = {$1,-1$}.
Let $x\in U(R)\Rightarrow \exists  y\in U(R) $ such that $x\cdot y =1\Rightarrow |x|^2\cdot|y|^2=1$. 
But $|x|^2\in \mathbb{Z}\Rightarrow x\in $ {$1,-1$}.
Let suppose that $R$ is euclidian. Then, there exists $\phi$ an euclidian norm.  Let $x\in R$ \ {$1,0,-1$} with $\phi(x)$ minim.  
Let $I=xR$ and $y\in R$ \ {$1,0,-1$}. There exists $q,r \in R$ such that $y=x\cdot q+r$ where $r=0$ or $\phi (r)< \phi(x)$. But $\phi(x)$ is minim. So $r=0\Rightarrow y\in I\Rightarrow R/I\subseteq$ {$\hat{1}, \hat{0}, \hat{-1}$}. 
I consider $\alpha=\frac{1+i\sqrt{19}}{2}$. I notice that $\alpha^2-\alpha+ 5= 0\Rightarrow  \hat{\alpha}^2-\hat{\alpha}+\hat{5}=\hat{0}$ in $R/I$.
But $R/I \subseteq ${$\hat{1}, \hat{0},\hat{-1}$}. 
For example, if I take $\hat{\alpha}=\hat{0}$, I obtain $\hat{5}=\hat{0}\Leftrightarrow 5\in I\Leftrightarrow 5=x\cdot z= (a+\frac{b\cdot(1+i\sqrt{19})}{2})\cdot (c+\frac{d\cdot(1+i\sqrt{19})}{2})$, where $a,b,c,d\in\mathbb{Z}$. 
I apply  the norm: 
$100= ((2a+b)^2+19b^2)\cdot((2c+d)^2+19d^2) $, where $a,b,c,d\in \mathbb{Z}$.
It's easy to show that the right side is divisible by $16$, but $100$ isn't. So, I have a contradiction. 
Now, I will prove that $R$ is principal. 
Lemma: Let $R$ a domain such that there exists  $f:R\rightarrow \mathbb{N}$ with properties: 
1) $f(a)=0\iff a=0$ and 
2) $\forall x,y\in R$, with $y$ doesn't divides $x$ there exists $a,b\in R$ such that $0<f(ax+by)< f(y)$.  
Then $R$ is a  principal ring.
In your problem $f$ is the norm function. You have to find $a, b$ which satisfy 2). Then you have to apply the lemma.
