Does $a\mid b$ in a Euclidean domain imply $\operatorname{Norm}(b) \ge \operatorname{Norm}(a)$?

Question

Is the following proposition true?

Let $R$ be a Euclidean Domain, $a,b\in R$, $b\ne 0$, $a|b$. Then $N(b)\ge N(a)$.

I cannot (in my very limited knowledge) think of any counterexamples, but I haven't managed to come up with a proof, either.

EDIT: In response to comments, here's the definition I'm working with. (This is from Dummit and Foote Chapter 8.)

$N(0) = 0$, $N(r) \ge 0$, and if $a, b\in R$, $b \ne 0$, then there are $q, r \in R$ such that $a = bq + r$ and either $r = 0$ or $N(r) < N(b)$.

Thanks!

Answer 1

It depends on how you define Euclidean domains. The property $\rm N(a) \le N(ab)$ needn't be assumed in order to deduce all of the basic properties of Euclidean domains. It is true that any Euclidean function can be normalized to satisfy said property by defining $\rm\:\bar N(a) = min\: N(aR^*),\,\ R^* = R\backslash0.\:$

Compare also the analogous Dedekind-Hasse criterion for a PID. You may also find of interest the following paper, which gives an in-depth study and comparison of a dozen different definitions / axioms for Euclidean rings.

[1] Euclidean Rings. A. G. Agargun, C. R. Fletcher
Tr. J. of Mathematics, 19, 1995, 291 - 299.