# Why do we only consider quadratic domains as Euclidean domains with squarefree integers?

I have been reading "Introductory algebraic number theory" by Alaca and Williams, and in the opening chapters they use the quadratic domains $\mathbb{Z}+\mathbb{z}(\sqrt{m})$ for non-square $m$ and $\mathbb{Z}+\mathbb{Z}\Big(\frac{1+\sqrt{m}}{2}\Big)$ where $m\equiv1 \mod{4}$ and non-square.

In the chapter on Euclidean domains, it defines and proves a number of results about the norms $\phi_m$ where for $x,y \in \mathbb{Q}, \phi_m(x+y\sqrt{m})=|x^2-my^2|$ where $m$ is now a squarefree integer, and go into much detail about conditions under which e.g. $\mathbb{Z}+\mathbb{Z}(\sqrt{m})$ is a Euclidean domain with respect to this function, but only for for squarefree $m$.

My question is: Why do we make this distinction? As far as I can tell, there is no reason to exclude, for example, $\mathbb{Z}+\mathbb{Z}(\sqrt{8})$. $\phi_8$ appears to behave in all the ways we want it to and seems to be a candidate for a euclidean function on the domain. I know that this is a subdomain of $\mathbb{Z}+\mathbb{Z}(\sqrt{2})$, but the book itself has exercises where it is shown that certain subdomains of norm-euclidean domains are not euclidean with respect to any function, so we can't always make observations about subdomains from our knowledge of larger ones.

## 2 Answers

The reason they do this is because $\mathbb{Z}+\mathbb{Z}(\sqrt{8})$ or more generally $R=\mathbb{Z}+\mathbb{Z}(k\sqrt{m})$, for $m$ squarefree and $k>1$ is not integrally closed. This means that there are elements $x$ of the fraction field Frac$R=K=\mathbb{Q}(\sqrt{m})$ that are not in $R$ which satisfy an equation of the form $$x^n +a_{n-1}x^{n-1}+\ldots+a_0=0.$$ for $a_i\in\mathbb{Z}$. In our case $x=\sqrt{m}$ is such an element satisfying $x^2-m=0$. However all euclidean domains are integrally closed as they are unique factorisation domains.

Added proof of the above fact: If $x\in K$ we have a prime decomposition $x=\prod_i p_i^{e_i}$ where the $e_i$ can also be negative. Suppose $e_1,\ldots,e_k$ are negative. If $x$ is integral over $R$ then $$x^n = -a_{n-1}x^{n-1}+\ldots+a_0.$$ After multiplying both sides by $(p_1^{e_1}\cdot\ldots\cdot p_k^{e_k})^n$ the left hand side is in $R$ and not divisible by $p_1$ while the right hand side is in $R$ and divisible by $p_1$, so we have a contradiction. So $x\in R$.

• Thanks a lot, I haven't yet done anything with Integral closures, but that would explain it! Thanks a lot. – Tom Oldfield Jun 25 '13 at 0:16

First, $\mathbb Q(\sqrt{8})$ is really just the same as $\mathbb Q(\sqrt{2})$, since $$\mathbb Q(\sqrt{8})=\{a+b\sqrt{8}\mid a,b\in\mathbb Q\}=\{a+b\cdot 2\sqrt{2}\mid a,b\in\mathbb Q\}=\{a+b'\sqrt{2}\mid a,b'\in\mathbb Q\}=\mathbb Q(\sqrt{2})$$ So, the fraction field of $\mathbb Z[\sqrt{8}]$ is just $\mathbb Q(\sqrt{8}) = \mathbb Q(\sqrt{2})$, whose ring of integers is $\mathbb Z[\sqrt{2}]\neq \mathbb Z[\sqrt{8}]$.

Hence, $\mathbb Z[\sqrt{8}]$ is not even integrally closed, let alone Euclidean, a PID or a UFD. Same goes for all non-squarefree integers.

• "whose integral closure is" should be "whose ring of integers is", no? – Mariano Suárez-Álvarez Jun 24 '13 at 22:24
• Of course! Thanks for the hint, I edited it. – Tomas Jun 24 '13 at 22:35