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 $N(r) < N(b)$.


  • $\begingroup$ How do you define the norm? $\endgroup$
    – Bernard
    Jan 7, 2015 at 23:07

1 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.

  • $\begingroup$ I see. So if it is not given as part of the definition, it may not be true, but one can always define a new version of the norm that does increase on multiplication. And presumably still has the defining characteristics D&F give... I guess I need to think a little about why the division algorithm is still guaranteed to reduce the new norm. $\endgroup$
    – Leon Avery
    Jan 7, 2015 at 23:25
  • $\begingroup$ @Leon Right. The linked paper is very helpful in understanding the relationships between various definitions. $\endgroup$ Jan 7, 2015 at 23:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .