Prove that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain and $\mathbb{Z}[\sqrt{-10}]$ is not

Question

Prove that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain and $\mathbb{Z}[\sqrt{-10}]$ is not.

I know that in general, to prove something is a Euclidean domain, I must prove the existence of a division algorithm involving a norm. In the case of $\mathbb{Z}[\sqrt{-2}],$ the norm is $a^2 + 2b^2.$ I know how to prove that the Gaussian integers $\mathbb{Z}[\sqrt{-1}]$ is a Euclidean domain, but I'm not sure if the proof for that relates to this proof.

Also, proving that $\mathbb{Z}[\sqrt{-10}]$ is not a Euclidean domain involves determining which ideals are not principal, but I'm not sure how to find a non-principal ideal. I think it should be generated by at least two elements of $\mathbb{Z}[\sqrt{-10}]$ though.

I'm new to abstract algebra, so I'd like more than just a simple hint, if possible.

Answer 1

If $\mathbb{Z}[\sqrt{-10}]$ were an Euclidean domain, then it would be a UFD.

By considering norms, we see that $2$, $5$, and $\sqrt{-10}$ are irreducible in $\mathbb{Z}[\sqrt{-10}]$.

Since $10 = (-1)(\sqrt{-10})^2 = 2 \times 5$ are two distinct factorizations into irreducibles, $\mathbb{Z}[\sqrt{-10}]$ is not a UFD and so cannot be an Euclidean domain.

Answer 2

$1.$

Since $\mathbb{Z}[\sqrt{-2}]$ is an integral domain, it suffices to prove that it has a division algorithm via $N(x) = a^2+2b^2,$ where $x=a+b\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}].$ Let $R=\mathbb{Z}[\sqrt{-2}]$. We want to check that $\forall a,b\in R,b\neq 0,\exists q,r$ such that $a=bq +r$, where $r=0$ or $N(r) < N(b)$. Let $x,y\in R, y\neq 0$. Since $\mathbb{R}(i)$ is a subfield of $\mathbb{C}$, we know that every nonzero element $x\in\mathbb{R}(i)$ has a multiplicative inverse. Consider $z=xy^{-1}\in \mathbb{R}(i)$, where $x,y\in\mathbb{Z}[\sqrt{-2}]$. Take $w=c+d\sqrt{-2}\in\mathbb{Z}[\sqrt{-2}]$. Say $z=a+b\sqrt{-2},a,b\in\mathbb{R}$, where $|a-c|\leq \dfrac{1}{2}$ and $|b-d|\leq \dfrac{1}{2}$. Notice that $z = w + (z-w)$. Since $z=xy^{-1}$, we have that $x=yw +y(z-w)$. Since $y,w\in\mathbb{Z}[\sqrt{-2}], yw\in\mathbb{Z}[\sqrt{-2}]$. As well, $y(z-w)=yz-yw = x-yw\in\mathbb{Z}[\sqrt{-2}]$ We want to show that $N(y(z-w))<N(y)$. We have that $$N(y(z-w)) = N(y)N(z-w) = N(y)N((a-c)+(b-d)\sqrt{-2})\\ =N(y)[(a-c)^2+2(b-d)^2]\\ \leq (\dfrac{1}{4}+\dfrac{1}{2})N(y)=\dfrac{3}{4}N(y) < N(y)$$ as $y\neq 0\Rightarrow N(y) > 0$. Thus, $\mathbb{Z}[\sqrt{-2}]$ is an integral domain.

$2.$

Since every Euclidean domain is a principal ideal domain, if $\mathbb{Z}[\sqrt{-10}]$ has an ideal that is not principal, then it is not a Euclidean domain. We will show that the ideal $\langle 2,\sqrt{-10}\rangle$ is not principal. First, notice that every element of this ideal is of the form $2x+\sqrt{-10}y,x,y\in\mathbb{Z}[\sqrt{-10}].$ Hence $x=k_1 + k_2\sqrt{-10}$ and $y= k_3+k_4\sqrt{-10}$ for $k_1,k_2,k_3,k_4\in\mathbb{Z}$. Subbing this into the form $2x+\sqrt{-10}y$ we have $2(k_1+k_2\sqrt{-10})+\sqrt{-10}(k_3+k_4\sqrt{-10})=2(k_1-5k_4) + (2k_2+k_3)\sqrt{-10})$. Since $k_1-5k_4, 2k_2+k_3\in\mathbb{Z}$ and $\forall x,y\in\mathbb{Z}, 2x+\sqrt{-10}y \in\mathbb{Z}$, we have that every element of the ideal $\langle 2,\sqrt{-10}\rangle$ is of the form $2x+y\sqrt{-10},x,y\in\mathbb{Z}$. Now, suppose that $\langle 2,\sqrt{-10}\rangle =\langle d\rangle$ for some $d\in\mathbb{Z}[\sqrt{-10}]$. Then $d \mid 2$ and $d \mid \sqrt{-10}\Rightarrow 2 = x_2d$ and $\sqrt{-10} = x_3d\Rightarrow N(2) = N(x_2)N(d)\;(1)$ and $N(\sqrt{-10} = N(x_3)N(d)\;(2).$ Since $\langle d\rangle = \langle 2,\sqrt{-10}\rangle,$ it is of the form $2x_2+\sqrt{-10}x_3$. Thus, $N(d) = N(2x_2+x_3\sqrt{-10}) = (4x_2^2 + 10x_3^2)$. So $(1)$ becomes $4 = N(x_2)(4x_2^2 + 10x_3^2).$ But, since $4x_2^2 + 10x_3^2$ is nonnegative and cannot equal $1$ or $2$, it must be $4,$ in which case $x_2 = 1$ and $x_3=0\Rightarrow N(d)=4$. Subbing this into $(2)$ gives $10=N(x_3)N(d)$, which is impossible as $4N(x_3)=10$ has no integer solutions. Hence, $\langle 2,\sqrt{-10}\rangle$ is not principal so $\mathbb{Z}[\sqrt{-10}]$ is not a Euclidean domain.

Answer 3

In fact $\mathbb{Z}[\sqrt{-10}]$ is not a UFD: indeed one may write $10= (-1) \times (\sqrt{-10})^2=2 \times 5$, so $\mathbb{Z}[\sqrt{-10}]$ is not Euclidean.