Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the definition of domain, we first define a degree function $\vartheta: R^\times \rightarrow \mathbb{N}$ with such two constraints:

(1) $\vartheta(f)\leq \vartheta(fg)$ for all $f,g\in R^\times$.

(2) for all $f,g\in R$ with $f\in R^\times$, there exist $q,r\in R$ with $g=qf+r$ and either $r=0$ or $\vartheta(r)<\vartheta(f)$.

I wonder why we need the first constraints? I think with only the second constraint, it is enough to prove the theorem: every Euclidean ring is a PID.

Can anyone give me a example where the first constraint is used?

share|cite|improve this question
Wikipedia indeed says the first constraint is unnecessary in a certain sense: However, dropping it doesn't give you any extra generality. – Qiaochu Yuan May 9 '12 at 8:23
@QiaochuYuan That looks like an answer to me. – Alex Becker May 9 '12 at 8:31
@QiaochuYuan I see, thank you! – hxhxhx88 May 9 '12 at 12:07
up vote 1 down vote accepted

From my sci.math post on 2009/7/2: The property $\rm V(a) \le V(ab)$ 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 $\rm\:v(a) = min\: V(aD^*),\ D* = D\backslash0.\:$ This is so well-known it is even in the Wikipedia Compare also the analogous Dedekind-Hasse criterion for a PID. And be sure to see this paper[1]. It 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.

share|cite|improve this answer
Great! Thank you! – hxhxhx88 May 10 '12 at 5:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.