# why he take $d \equiv 2, 3 \mod 4$ is euclidean domain?

i have some confusion not getting in my head

this answer :Euclidean domain $\mathbb{Z}[\sqrt{d}]$

My attempt : if i take $$d= 10$$ then $$\mathbb{Z}[\sqrt 10]$$ is not euclidean domain but it satisfied $$10 \equiv 2 \mod 4$$

similarly if i take $$d= 13$$ then $$\mathbb{Z}[\sqrt 13]$$ is euclidean domain but $$13 \equiv 1 \mod 4$$ which is contradicts to above answer

Im confused why he take $$d \equiv 2, 3 \mod 4$$ is euclidean domain ?

• $R=\mathbb{Z}[\sqrt{13}]$ is not a Euclidean domain, because $\alpha=\frac{1+\sqrt{13}}{2} \notin R$, but $\alpha \in Frac(R)$, and $\alpha^2=3+\alpha$, so $R$ is not normal. On the other hand, the page gives you only a necessary condition, meaning that if $\mathbb{Z}[d^{1/2}]$ is an Euclidean domain, then $d-2$ or $d-3$ must be divisible by $4$, but the converse is not necessarily true. – Mindlack Aug 26 at 9:50

A necessary condition for a subring of a number field to be an Euclidean domain is to be integrally closed. If $$d\equiv 1 \pmod{4}$$, then the ring $$\mathbb{Z}[\sqrt{d}]$$ is not integrally closed, since $$\alpha=\frac{1+\sqrt{d}}{2}\in \mathbb{Q}(\sqrt{d})$$ verifies that $$\alpha^2=\alpha+\frac{d-1}4,$$ so $$\alpha$$ is a root of $$X^2-X+\frac{1-d}4\in \mathbb{Z}[X],$$ but $$\alpha$$ is not in $$\mathbb{Z}[\sqrt{d}]$$.
This is the reason that you need $$d\equiv 2,3\pmod{4}$$ for the ring $$\mathbb{Z}[X]$$ to be an Euclidean domain. Note, however, that this condition is not sufficient.