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

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

\begin{equation} (2v)^2 = \frac{u^2}{2} \pm 1 . \end{equation}

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.

Hasse diagram from rng to ED

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

  • $\begingroup$ Good complete answer. Plus one. $\endgroup$
    – Lubin
    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).

  • $\begingroup$ 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, $+: \mathbb{Z}[2\sqrt{2}] \times \mathbb{Z}[2\sqrt{2}] \rightarrow \mathbb{Z}[\sqrt{2}]$. $\endgroup$ Jan 26 at 14:36
  • $\begingroup$ 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. $\endgroup$ Jan 26 at 14:47
  • $\begingroup$ 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. $2\mathbf Z$. $\endgroup$
    – KCd
    Jan 26 at 15:10
  • $\begingroup$ Thanks. One last question. I did attempt to understand lhf's answer but without success. What is $v_\pi(2)$? $\endgroup$ Jan 26 at 16:36
  • $\begingroup$ 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 $250 = 5^3 \cdot 2$. $\endgroup$
    – KCd
    Jan 26 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.