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
 A: It is more straight forward to give a counter example. Since $4=2 \cdot 2 = (\sqrt{8}+2)(\sqrt{8}-2)$, $\mathbb{Z}[\sqrt{8}]$ is not a unique factorisation domain (UFD), hence not a Euclidean domain. Note that those factors are irreducible.
Suppose to the contrary that $2$ is not irreducible. There exists $a,b \in \mathbb{Z}[\sqrt{8}]$ with $N(2)=4=2 \cdot 2=N(a)N(b)$. It requires that $a,b \notin U(\mathbb{Z}[\sqrt{8}])$, the set of units. Therefore $N(a)=N(b)=2$.
Let $a=u+v\sqrt{8}$ with $u,v \in \mathbb{Z}$ and $u,v$ not both zero. It follows that $N(a)=|u^2-8v^2|=2$ $\implies$ $8v^2-u^2= \pm 2$. Therefore,
\begin{equation}
(2v)^2 = \frac{u^2}{2} \pm 1
.
\end{equation}
LHS is even. If $u$ is even, RHS is odd; a contradiction. If $u$ is odd, $ \frac{u^2}{2}\notin \mathbb{Z}$; another contradiction. Therefore $2$ is irreducible.
Note that $2 \nmid \sqrt{8} \pm 2$. Hence $\mathbb{Z}[\sqrt{8}]$ is not a UFD.

Link to a bigger image:
Hasse diagram from rng to ED https://i.stack.imgur.com/jUcJX.png
A: 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).
A: Every euclidean domain is a UFD.
Now, let $\alpha=\sqrt 8$. Then $2^3=\alpha^2$. If $ \mathbb{Z}[{ \sqrt 8 } ]$ is a UFD, we have${}^*$ $$2=\beta^2=(a+b\sqrt8)^2=(a^2+8b^2)+2ab\sqrt8$$
However, this equation has no solutions with $a,b \in \mathbb Z$.
Therefore, $ \mathbb{Z}[{ \sqrt 8 } ]$ is not a UFD and so cannot be an euclidean domain.
${}^*$ $2^3=\alpha^2$ implies $3v_\pi(2) = 2v_\pi(\alpha)$ for every prime $\pi$. Then $v_\pi(2)$ must be even and so $2$ is a square.
