# Is $\mathbb{Z}[{ \sqrt 8 } ]$ a Euclidean domain?

Is $$\mathbb{Z}[{ \sqrt 8 } ]$$ a Euclidean domain ?

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

• How would you show $|a^2-8b^2|$ is a Euclidean function? For instance, how would you divide $\sqrt{8}$ by $2$? Aug 25 '19 at 19:35
• @Wojowu $(2\sqrt 2)^2= 8$ Aug 25 '19 at 19:36
• How does that answer my question? Aug 25 '19 at 19:37
• Consider the ideal $(2,\sqrt8)$. Aug 25 '19 at 19:38

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.

It is more straight forward to give a counter example. Since $$4=2 \cdot 2 = (\sqrt{8}+2)(\sqrt{8}-2)$$, $$\mathbb{Z}[\sqrt{8}]$$ is not a unique factorisation domain (UFD), hence not a Euclidean domain. Note that those factors are irreducible.

Suppose to the contrary that $$2$$ is not irreducible. There exists $$a,b \in \mathbb{Z}[\sqrt{8}]$$ with $$N(2)=4=2 \cdot 2=N(a)N(b)$$. It requires that $$a,b \notin U(\mathbb{Z}[\sqrt{8}])$$, the set of units. Therefore $$N(a)=N(b)=2$$.

Let $$a=u+v\sqrt{8}$$ with $$u,v \in \mathbb{Z}$$ and $$u,v$$ not both zero. It follows that $$N(a)=|u^2-8v^2|=2$$ $$\implies$$ $$8v^2-u^2= \pm 2$$. Therefore,

$$$$(2v)^2 = \frac{u^2}{2} \pm 1 .$$$$

LHS is even. If $$u$$ is even, RHS is odd; a contradiction. If $$u$$ is odd, $$\frac{u^2}{2}\notin \mathbb{Z}$$; another contradiction. Therefore $$2$$ is irreducible.

Note that $$2 \nmid \sqrt{8} \pm 2$$. Hence $$\mathbb{Z}[\sqrt{8}]$$ is not a UFD.

Link to a bigger image: Hasse diagram from rng to ED https://i.stack.imgur.com/jUcJX.png

• Good complete answer. Plus one. Jan 25 at 17:59

Learn what the property "integrally closed" means and why every UFD is integrally closed. In particular, every Euclidean domain is integrally closed (Euclidean domains are UFDs).

The ring $$\mathbf Z[\sqrt{8}] = \mathbf Z[2\sqrt{2}]$$ is not integrally closed since $$\sqrt{2}$$ is in its fraction field and is integral over $$\mathbf Z[\sqrt{8}]$$ (being a root of $$x^2-2$$) without being in $$\mathbf Z[\sqrt{8}]$$. Therefore $$\mathbf Z[\sqrt{8}]$$ is not Euclidean or even a UFD, without having to give a specific counterexample to the UFD property: we showed this ring fails to have a property that every UFD (in particular, every Euclidean domain) satisfies: being integrally closed.

By similar reasoning, if $$d$$ is a squarefree integer and $$m \geq 2$$, then $$\mathbf Z[m\sqrt{d}]$$ is not integrally closed and thus can't be Euclidean (or a UFD).

• My background: I audited (i.e. not a proper maths student) an introductory abstract algebra course for ten weeks. I fail to understand what does it mean by integrally closed. These two wikipedia entires, Integrally_closed_domain and Integral_element, are too technical for me. Will you please point me to a more elementary explanation? Are you saying the problem is that $\sqrt{2}+\sqrt{2}+\sqrt{2} \notin \mathbb{Z}[2\sqrt{2}]$? Does it mean that it is not even a ring? Since addition operation is not closed, $+: \mathbb{Z}[2\sqrt{2}] \times \mathbb{Z}[2\sqrt{2}] \rightarrow \mathbb{Z}[\sqrt{2}]$. Jan 26 at 14:36
• How $\mathbb{Z}[2\sqrt{2}]$ is different from $2\mathbb{Z}[\sqrt{2}]$? The textbook I use is R.B.J.T. Allenby (1991) Rings, Fields and Groups: An Introduction to Abstract Algebra. 2nd ed. Oxford: Butterworth-Heinemann. Jan 26 at 14:47
• The term "integrally closed" is a rather technical property. It is not about an operation like addition not being closed or something not being a ring. Focus on understanding the answers other people gave. Come back to this answer after you have a better understanding of abstract algebra (not just from auditing a course). The notation $\mathbf Z[\alpha]$ means polynomial expressions in $\alpha$ with integral coefficients. In particular, $\mathbf Z[2\sqrt{2}]$ contains $1$ while $2\mathbf Z[\sqrt{2}]$ does not contain $1$, just like $\mathbf Z$ vs. $2\mathbf Z$.
– KCd
Jan 26 at 15:10
• Thanks. One last question. I did attempt to understand lhf's answer but without success. What is $v_\pi(2)$? Jan 26 at 16:36
• Probably it means the highest power of a prime element $\pi$ that divides 2 (the $\pi$-adic valuation of 2). In $\mathbf Z$, $v_5(250) = 3$ since $250 = 5^3 \cdot 2$.
– KCd
Jan 26 at 16:42