The notion of an Euclidean domain is defined using the auxiliary machinery of a Euclidean function. But, I wonder, is that auxiliary machinery actually needed? More precisely, in the language $\{+,-,*,0,1\}$ of rings, is the class $K$ of those rings which are Euclidean domains, first-order axiomatizable, and if so, is it finitely first-order axiomatizable? Also, if it is not first-order axiomatizable, is the first-order theory of the class $K$ finitely axiomatizable?

  • $\begingroup$ I think a nontrivial ultrapower of the integers is not a Euclidean domain. $\endgroup$ Commented Jan 23 at 1:44
  • $\begingroup$ I wonder if the first-order theory might be just the theory of gcd domains. $\endgroup$ Commented Jan 23 at 1:57
  • $\begingroup$ @DanielSchepler very nice question! (although I think it should probably be about Bézout domains rather than gcd domains? because the property that every finitely generated ideal is principal is first-order expressible) $\endgroup$ Commented Jan 23 at 3:18
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    $\begingroup$ @AtticusStonestrom I think Cohn discusses that in Bezout rings and their subrings $\endgroup$
    – rschwieb
    Commented Jan 23 at 12:07
  • $\begingroup$ @rschwieb thanks for the reference, I will have a look! $\endgroup$ Commented Jan 23 at 18:45

1 Answer 1


No, the class of Euclidean domains is not elementary.

Let $\mathcal{Z}$ be any nonstandard model of the theory of the ring of integers (e.g. a nontrivial ultrapower). By overspill, there are instances of the Euclidean algorithm implemented internally to $\mathcal{Z}$ which are nonstandardly long; these prevent any genuine Euclidean function for $\mathcal{Z}$ for existing, since they would force any such function to assign an infinite value to any of the numbers occurring in this instance.

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    $\begingroup$ +1; maybe it's worth additionally noting that a non-standard model of true arithmetic cannot even be a UFD, let alone a PID or euclidean domain $\endgroup$ Commented Jan 23 at 2:01

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