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EDIT: somebody tried to close this question saying that this question similar to this. But I think my question is different,I want to show that $\mathbb{Z}[i\sqrt{3}]$ doesn't satisfy the basic definitions of Euclidean domain.

Let $D$ be the integral domain given by $D = \mathbb{Z}[i\sqrt{3}]=\{a+bi\sqrt{3} : a,b \in \mathbb{Z}\}$ and $v : D \to \mathbb{N}$ be the map defined by $v(a +bi\sqrt{3}) = a^2 + 3b^2$. We have to Show that $v$ is not a Euclidean valuation on $D$.

Now to show $v$ is not a Euclidean valuation I have to one of these property doesn't hold that

(i)$v(ab) \geq v(a)$ (and also $\geq v(b)$) for any $a,b \in \mathbb{Z}[i\sqrt{3}]$ with $a \neq 0$ and $b \neq 0$;

(ii) (Division algorithm) for any $a,b \in \mathbb{Z}[i\sqrt{3}]$ with $b \neq 0$, there exist $q,r \in \mathbb{Z}[i\sqrt{3}]$ (quotient and remainder) such that $a = qb + r$, where either $r = 0$ or $v(r) < v(b)$

I think both of these property which is definition of Euclidean domain is holding, how to show $v$ is not a Euclidean evaluation?

Alexander
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  • I think you can use the fact that $2\cdot 2 =(1+i\sqrt{3})(1-i\sqrt{3})$ to break the division algorithm. – B. Goddard Jun 08 '22 at 20:17
  • @B.Goddard What you are saying is its not Ufd but my question is does v saties the definition of Euclidean domain? The 2 property I have mentioned. – Alexander Jun 08 '22 at 20:27
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    I think if you divide $2$ by $1+i\sqrt{3}$ the valuation won't satisfy the inequality. – B. Goddard Jun 08 '22 at 20:31
  • @B.Goddard $r=0$ as $2.2=(1+i\sqrt{3})(1-i\sqrt{3})+0$, please clarify how it's not satify the inequality. – Alexander Jun 08 '22 at 20:39
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    You are not dividing $2$ by $1+i\sqrt{3}$ there, you are dividing $4$. You need to write $2=q(1+i\sqrt{3}) + r$ for some $q,r\in\mathbb{Z}[i\sqrt{3}]$ and with $r=0$ or $v(r)\lt v(1+i\sqrt{3}) = 4$. – Arturo Magidin Jun 08 '22 at 20:43
  • There is no $q$ such that $2=q(1+i\sqrt{3})$: that would require $q=2$ and $q=0$. So you need a nonzero remainder. If $r=a+bi\sqrt{3}$ is the remainder and $v(r)\lt 4$, then either $a^2=1$ and $b=0$; or $a=0$ and $b^2=1$ (otherwise the norm is at least $4$). But neither $2\pm 1$ nor $2\pm i\sqrt{3}$ is a multiple of $1+i\sqrt{3}$, so no such $q$ and $r$ exist. – Arturo Magidin Jun 08 '22 at 20:47
  • @ArturoMagidin Yeah. I calculated $q$ and for $2 \pm 1$ I got $q=\frac{1}{4}(1-i\sqrt{3}),\frac{3}{4}(1-i\sqrt{3})$ and for $2 \pm i\sqrt{3}$ got $q= \frac{1}{4}(5-i \sqrt{3}),\frac{-1}{4}(1+3i \sqrt{3}) $ but none of them belongs to $\mathbb{Z}[i\sqrt{3}]$. So no q exist. Thanks – Alexander Jun 08 '22 at 21:14

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