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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?

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Wikipedia indeed says the first constraint is unnecessary in a certain sense: en.wikipedia.org/wiki/Euclidean_domain#Definition However, dropping it doesn't give you any extra generality. –  Qiaochu Yuan May 9 '12 at 8:23
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@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

1 Answer 1

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 http://en.wikipedia.org/wiki/Euclidean_domain 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.

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Great! Thank you! –  hxhxhx88 May 10 '12 at 5:43

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