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I have a question on the definition of an Euclidean domain. By definition, a ring $R$ is Euclidean if there is norm function $N:R\to \mathbb{N}_{\geq 0}\cup \{ -\infty \}$ such that for all $a$ and $b\neq 0$, there exists $q$ and $r$ such that $a=bq+r$ and $(N(r)<N(b)$ or $r=0)$. My question is on the utility of the "$r=0$". For example, in the regular euclidean division in $\mathbb{Z}$, where $N$ is the absolute value, we can write $a=bq+r$ and ensure that $|r|<|b|$, without worying about the case $r=0$. Also, in $F[x]$ where F is field, with $N$ being the degree of a polynomial, this is neither a problem as if $r=0$, $N(r)=-\infty$ and $N(r)<N(b)$ as $b\neq 0$ anyway.

Thanks in advance !

DagunDagun
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  • This is just added to ensure that $0$ is always a valid remainder. Some books don't even require the norm to be defined at the point $0$. (or you can simply define it on zero to be $-\infty$, like in the polynomial ring) – Mark Apr 10 '23 at 21:30
  • I'm not sure I understand your question. Are you asking why aren't we absorbing $r=0$ into some more general expression like in the division in $\mathbb Z$ where it is absorbed into $|r|<|b|$? – Randy Marsh Apr 10 '23 at 21:31
  • Yes this is precisely my question @RandyMarsh. I feel like the "$r=0$ condition is useless. – DagunDagun Apr 10 '23 at 21:42
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    The paper I cite here by Agargun and Fletcher examines the relationship between all $12$ possible common definitions of Euclidean rings - including the variations you mention. – Bill Dubuque Apr 10 '23 at 22:10
  • Ok thanks a lot @BillDubuque, this helps ! – DagunDagun Apr 10 '23 at 23:18

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