I suspect that since Euclid uses the Euclidean Algorithm to perform division on constructible numbers in Elements, the ring of integers of the constructible numbers are a Euclidean Domain, but I have not been able to find a proof for or against this.

It is clear that the ring of integers of some quadratic extensions that are a subset of the constructible numbers, such as $\mathbb Q[\sqrt 2]$ or $\mathbb Q[\sqrt 3]$ are Euclidean domains (see the Euclidean Domain Wikipedia article).

  • $\begingroup$ $\sqrt 2$ and $1$ are constructible, and yet the Euclidean algorithm fails with these inputs. $\endgroup$ Aug 18, 2015 at 14:19
  • 2
    $\begingroup$ The constructible numbers $K$ form a field, and a field is trivially a Euclidean domain (everything divides everything else). I think a more interesting question to ask would be whether the ring of integers of $K$ is Euclidean. $\endgroup$
    – D_S
    Aug 18, 2015 at 14:21
  • $\begingroup$ Yes that is what I meant to ask. I will edit the question. $\endgroup$
    – hatch22
    Aug 18, 2015 at 14:31

1 Answer 1


The answer to your question is no.

  1. If a ring $R$ is Euclidean with respect to some function $f$, it is also Euclidean with respect to some submultiplicative function $g$ satisfying $g(a) \le g(ab)$.
  2. Any ring $R$ Euclidean with respect to some submultiplicative function is either a field or possesses irreducible elements (see https://en.wikipedia.org/wiki/Euclidean_domain for both claims).
  3. The ring of constructible numbers does not possess irreducible elements: if $\alpha$ belongs to this ring, then so does $\sqrt{\alpha}$, and $\alpha = \sqrt{\alpha} \cdot \sqrt{\alpha}$ is a nontrivial factorization.

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