Is $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain? 
Is  $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain ?

I have some  confusion  that is   what is difference between  euclidean  domain and euclidean Norms ?
My attempt : I thinks  yes
i know  that  $d( a+b \sqrt 8) = |a^2 - 8b^2 | $ as i can show it is  euclidean domain  by same pattern $\mathbb{Z}[{ \sqrt 2 } ]$ is  euclidean  domain

Learn what the property "integrally closed" means and why every UFD is integrally closed. In particular, every Euclidean domain is integrally closed (Euclidean domains are UFDs).
The ring $\mathbf Z[\sqrt{8}] = \mathbf Z[2\sqrt{2}]$ is not integrally closed since $\sqrt{2}$ is in its fraction field and is integral over $\mathbf Z[\sqrt{8}]$ (being a root of $x^2-2$) without being in $\mathbf Z[\sqrt{8}]$.
Therefore $\mathbf Z[\sqrt{8}]$ is not Euclidean or even a UFD, without having to give a specific counterexample to the UFD property: we showed this ring fails to have a property that every UFD (in particular, every Euclidean domain) satisfies: being integrally closed.
By similar reasoning, if $d$ is a squarefree integer and $m \geq 2$, then $\mathbf Z[m\sqrt{d}]$ is not integrally closed and thus can't be Euclidean (or a UFD).
