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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

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

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