Today we learned about Euclidean domains in class but I don't understand why we need one of the conditions stated in the definition. We called an integral domain $R$ a Euclidean domain if there exists a function $f$ from $R$ to strictly positive integers such that:

1) For $a,b$ non zero in $R$, $f(ab)\ge f(a)$.

2) If $a,b\in R$, $b\neq 0$, then we can write $a=bq+r$ with $q,r\in R$ such that either $r=0$ or $f(r)<f(b)$.

So from what I understand the whole point of a Euclidean domain is to be able to define a Euclidean algorithm, but I don't see why (1) is needed.
Furthermore later in the class we proved a Euclidean domain is a principal ideal domain and in the proof we didn't use the property (1), so my question is:

Why do we need (1) in the definition?

Thanks in advance.

  • 5
    $\begingroup$ Post in Wikipedia says that the condition 1) is superfluous. $\endgroup$ – Hanul Jeon Feb 19 '15 at 18:00
  • $\begingroup$ Thanks! good to clarify that $\endgroup$ – TheGeometer Feb 19 '15 at 18:16

The property $\rm\,f(ab)\ge f(a)\,$ need not be assumed in order to deduce all of the basic properties of Euclidean domains. In fact, any Euclidean function can be normalized to satisfy said property by defining $\rm\:\bar f(a)\, =\, min\: f(aR^*),\ R^* = R\backslash0.\:$

Compare also the analogous Dedekind-Hasse criterion for a PID. $ $And be sure to see this paper[1], 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.

  • $\begingroup$ Reference to Agargun's paper -- Euclidean Rings is really good $\endgroup$ – nature1729 Jan 24 '18 at 4:07

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