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

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.

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.

• Adapted from math.stackexchange.com/a/1144437. – lhf Nov 20 '19 at 0:39
• @user23749 What would actually be helpful is if you would indicate which parts of this answer you don't understand rather than just writing it off wholesale. You know what a Euclidean domain is. Do you know what a unique factorization domain is? What an irreducible element is? The definition of the norm $\mathbb{Z}[\sqrt{-10}] \to \mathbb{Z}$? It's up to you to let us know what your background is in your question. You could also include where the question comes from, which might give an idea of the tools you have. Is it from a textbook? – Richard D. James Nov 20 '19 at 0:58
• @André3000 can you please check out my answer to this question? I'm having a hard time understanding this material. – user726063 Nov 23 '19 at 23:18
• @lhf can you check out my answer to this question? I'm not sure if it's correct; it appears this question was much easier than I initially thought. – user726063 Nov 23 '19 at 23:19

$$1.$$

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 $$N(y(z-w)). We have that $$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.