5
$\begingroup$

Let $K$ be a finite extension of $\mathbb{Q}$ (a number field) and $\mathcal{O}_K$ its ring of integers. One defines the norm of an element $\alpha\in K$ to be the determinant of the transformation $m_\alpha: K\to K$ of multiplication by $\alpha$ (where $K$ is considered as a vector space over $\mathbb{Q}$).

Now sometimes the integer ring is also a euclidean domain, i.e. it has a "euclidean norm" satisfying the defining property of division algorithm. My question is: in an integer ring which is also euclidean, will the norm defined above also serve as a euclidean norm?

Put otherwise: is there an example for a euclidean integer ring whose norm is not an euclidean norm?

  • 0
    http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.109.3835&rep=rep1&type=pdf2011-09-12

1 Answers 1

2

Wikipedia says that "$\mathbb Q(\sqrt {69})$ is Euclidean but not norm-Euclidean. Finding all such fields is a major open problem, particularly in the quadratic case."

  • 3
    It is always nice to find out you've stumbled unaware on a "major open problem".2019-03-13