Does there exist an Euclidean domain with no multiplicative Euclidean function?

An Euclidean domain, denoted $R$, is an integral domain

with an Euclidean function $d : R\setminus \{0\} \to \mathbb{N}$ such that

$1)\quad d(a) \leq d(ab)$, and

$2)\quad a = bq + r$ with either $r = 0$ or $d(r) < d(b)$.

I am interested in multiplicative Euclidean functions. That is, $d(ab) = d(a)d(b)$.

For example, one can choose

$\mathbb{Z}$ with $d(n) = |n|, \quad \mathbb{Z}[i]$ with $d(\alpha) = N(\alpha), \quad F[X]$ with $d(f) = 2^{\deg(f)}$ for a field $F$,

or, kind of a stupid example, any field $F$ with $d(a) = 1$.

I am new here. Thank you in advance.


2 Answers 2


This question has now been answered. There is a Euclidean domain but no Euclidean function for that ring into $\mathbb{N}$ is multiplicative. See my paper here.


There are various definitions in use for Euclidean domains. Generally one requires only condition (2) on the Euclidean function, and Euclidean functions satisfying the additional condition (1) are called submultiplicative. But, in fact, no generality is lost by assuming (1) since it can be show that every Euclidean domain admits a submultiplicative Euclidean function, viz. the minimal function

$$ d_{\min}(a) = \min\, \{d(a)\ :\ d\ \text{ is a Euclidean function on}\ R\}$$

However, it is currently unknown if one loses generality by replacing "submultiplicative" by "multiplicative", i.e. it is not known if every Euclidean domain admits a multiplicative Euclidean function (e.g. see the remark following Proposition 2.2 in Franz Lemmermeyer's excellent survey The Euclidean algorithm in algebraic number fields).


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