# Is $\mathbb{Z}[{ \sqrt 8 } ]$ will form euclidean domain ? Yes/No

Is $$\mathbb{Z}[{ \sqrt 8 } ]$$ will form euclidean domain ? Yes/No

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

• How would you show $|a^2-8b^2|$ is a Euclidean function? For instance, how would you divide $\sqrt{8}$ by $2$? – Wojowu Aug 25 at 19:35
• @Wojowu $(2\sqrt 2)^2= 8$ – jasmine Aug 25 at 19:36
• How does that answer my question? – Wojowu Aug 25 at 19:37
• Consider the ideal $(2,\sqrt8)$. – Lord Shark the Unknown Aug 25 at 19:38
• @LordSharktheUnknown sir that mean im correct – jasmine Aug 25 at 20:12

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.