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An integral domain $R$ is a Euclidean domain if there is a degree function $$\deg : R-\{0\} \to \mathbb{Z}_{\ge 0} $$ such that

  • For every $a,b\in R$ with $b\ne 0$ there are $q,r\in R$ such that $$ a=bq+r$$ in which either $r=0$ or $\deg(r)<\deg(b)$.
  • For every nonzero $a,b\in R$ we have $\deg(a)\le \deg(ab)$.

Now the question is:

Let $R$ be a Euclidean domain with degree function $\deg$. Is it true that in the division algorithm $$ a=bq+r$$ we can always choose some remainder $r$ so that either $a-r=0$ or $\deg (a-r) \le \deg a$ ?

This is certainly not true of all possible remainders in a division, but for example in $\mathbb{Z}$ there is always at least one remainder with the above property. The same is true about the ring of polynomials over a field $F[x]$ and the ring of Gaussian integers $\mathbb{Z}[i]$. So is it true in any Euclidean domain? Or are there any counterexamples? And if so, is there any extra property of the domain or its degree function that guarantees the existence of such remainders (for example $\deg$ being multiplicative)?

Edit: I included the definition of Euclidean domains for more clarity. Also, the degree function is not required to be defined at $0$, so the trivial case of $r=a$ and $q=0$ is mentioned separately.

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  • $\begingroup$ What do you require of $\deg$? (Different authors impose rather different axioms.) $\endgroup$ Commented Aug 27, 2023 at 14:35
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    $\begingroup$ I don't think it's true for $\mathbb{Z}\left[\sqrt{-2}\right]$ then, with the usual Euclidean norm (= square of the absolute value). If you look at the four radius-$1$ circles covering the fundamental domain (a rectangle with sidelengths $1$ and $\sqrt 2$), you will see that all four circles are needed (no three cover the whole domain), so that $a-r$ might be further away from the origin than $a$. $\endgroup$ Commented Aug 27, 2023 at 15:15
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    $\begingroup$ This paper may be useful. $\endgroup$ Commented Aug 27, 2023 at 23:52
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    $\begingroup$ @GerryMyerson The degree of an integer is its absolute value. $\endgroup$ Commented Aug 28, 2023 at 15:40
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    $\begingroup$ I'd prefer to keep this question open, as I'm curious which quadratic rings $\mathbb Z[\alpha]$ satisfy this stronger version of Euclideanness. $\endgroup$ Commented Aug 28, 2023 at 18:05

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