On a PID that is not an Euclidean domain Let $\omega = \frac{1 + \sqrt{19}i}{2}$. The article here claims to prove that $\mathbb{Z}[\omega]$ is an example of a PID which is not a Euclidean domain. To prove that it is a PID, it takes an ideal $I$ and chooses a nonzero $b \in I$ as to minimize $|b|$ (the usual norm in $\mathbb{C}$). Then he supposes $a \in I \setminus (b)$ to obtain a contradiction.
According to him, it suffices to obtain $p, q \in \mathbb{Z}[\omega]$ such that $|ap - bq| < |b|$ to obtain a contradiction; but he neglects the case that $|ap - bq| = 0$, or $ap = bq$. Analyzing the proof, eventually he finds an integer $j \in \mathbb{Z}$ such that $|\frac{2a}{b} - \omega - j| < 1$, and multiplying by $|b|$ we get $|2a - b(\omega + j)| < |b|$. Hence by the minimality of $|b|$ we get $2a = b(\omega + j)$, but I cannot find a contradiction in this. We can see this implies $\frac{a}{b} = \frac{\omega + j}{2}$, and so $\operatorname{Im}(\frac{a}{b}) = \frac{\sqrt{19}}{4}$, and we chose $\frac{a}{b}$ for its imaginary part to lie in $[-\frac{\sqrt{19}}{4}, \frac{\sqrt{19}}{4}]$ but I cannot seem to deal away with these extremities.
Is there any simple way to fix this proof?

Note that $(2)\subset\mathbb{Z}[\omega]$ is prime, and that $\omega+j\notin(2)$ for any $j\in\mathbb{Z}$. 
From the fact that $2a=b(\omega+j)$ it follows that $b\in(2)$, i.e. $b=2c$ for some $c\in\mathbb{Z}[\omega]$. Then $a=c(\omega+j)$ and $2c,c(\omega+j)\in I$, hence either $c\omega\in I$ or $c(\omega-1)\in I$. Then also
$$c\omega(\omega-1)=-5c\in I,$$
and hence $c\in I$, contradicting the minimality of $|b|$.
