Let $K$ be an imaginary quadratic field and $O_K$ be its ring of integers. We say $O_K$ is norm Euclidean if the norm is a Euclidean function. It is known from the classification of imaginary quadratic fields with class number 1 that $O_K$ is Euclidean if and only if it is norm Euclidean. Is there a more straightforward proof of this fact?


Assume that $O_K$ is Euclidean. The Motzkin Sets $_j$ are defined by $E_0 = \{0\}$, $E_1 = E_0 \cup O_K^\times$, $E_2$ is the set of elements of $O_K$ such that each residue class is represented by an element in $E_1$ etc. $O_K$ is Euclidean if every element of $O_K$ is in some $E_i$ (see Sect. 2.3 here). Since $E_1$ has $3$ elements, $O_K$ must have an element of norm $\le 3$, which implies that the discriminant of $K$ is $\ge -11$. All these rings are norm-Euclidean, and the converse is trivial.

This is classical and has nothing to do with the solution of the class number $1$ problem.

  • 1
    $\begingroup$ Motzkin has a short five page paper, titled the Euclidean algorithm, with a proof of the classical fact. His paper can be found in the references of the linked paper above. $\endgroup$ Jun 6 '16 at 23:29

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.