This is a problem from Aluffi's book, chapter V 2.17.

"Let $R$ be a Euclidean Domain that is not a field. Prove that there exists a nonzero, nonunit element $c$ in $R$ such that $\forall a \in R$, $\exists q$, $r \in R$ with $a = qc + r$, and either $r = 0$ or $r$ a unit."

Ok, I know that if $c\mid a$ then $r=0$, but if $c\nmid a$, not sure about what to do. I took the classic Euclidean Domain $\mathbb{Z}$ as example, and in $\mathbb{Z}$ I know that $c = 2$ ( also $-2$). Then I tried to generalize this.

I did $c = unit + unit$, but this didn't help and exercise 2.18 showed me that $c$ is not always $unit+unit$. I'm out of ideas, need some help.


  • $\begingroup$ So Aluffi doesn't use the terminology of universal side divisor. Interesting. $\endgroup$ – Xam Feb 11 '17 at 14:56

If $R$ is not a field, then it has nonunits. Consider the set $S=\{\varphi(a)\mid a\text{ is not a unit, and }a\neq 0\}$, where $\varphi$ is the Euclidean function.

It is a nonempty set of positive integers. By the Least Element Principle, it has a smallest element. Let $c\in R$ be a nonunit, nonzero, such that $\varphi(c)$ is the smallest element of $S$.

Edited. I claim that $c$ satisfies the conditions of the problem. Let $a\in R$. Then we can write $a = qc + r$, with either $r=0$ or $\varphi(r)\lt \varphi(c)$. If $r=0$, we are done. If $r\neq 0$, then $\varphi(r)\lt \varphi(c)$, then $\varphi(r)\notin S$, hence $r$ does not satisfy the condition

$r$ is not a unit, and $r\neq 0$.

Since $r\neq 0$, it follows that $r$ must be a unit.

Thus, for every $a\in R$, there exist $q,r\in R$ such that $a=qc+r$, and either $r=0$ or $r$ is a unit, as desired.

  • $\begingroup$ In that case we have $a = qc + r$, with $r = 0$ or $\varphi(r)\leq\varphi(c)$ for all $a \in R$. Using the Least Element Principle i know that $r = 0$ or $\varphi(r) = \varphi(c)$, cant be smaller. $\endgroup$ – Integral Jun 15 '12 at 17:10
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    $\begingroup$ @Integral: The definition of "Euclidean domain" requires that the $q$ and $r$ satisfy $r=0$ or $\varphi(r)\lt\varphi(c)$; must be strictly smaller, cannot be equal. So either $r=0$ or $\varphi(r)\lt \varphi(c)$. Also, notice that $\varphi(c)$ is not the smallest value that $\varphi$ can take overall, it's just the smallest value that it can take when the input is not a unit. $\endgroup$ – Arturo Magidin Jun 15 '12 at 17:11
  • $\begingroup$ ops... In that case we have a=qc+r, with r=0 or φ(r)<φ(c) for all a∈R. Using the Least Element Principle i know that r=0 , cant be smaller. But how this let me know that r could be 1 ? $\endgroup$ – Integral Jun 15 '12 at 17:13
  • $\begingroup$ @Integral: no. You have the definition of Euclidean domain wrong. And you are not looking carefully at the definition of $S$. There is no a priori reason why something cannot have $\varphi$ smaller than $\varphi(c)$. Look carefully at the definition of $S$, don't just repeat the same false statement you did three minutes ago. $\endgroup$ – Arturo Magidin Jun 15 '12 at 17:14
  • $\begingroup$ Yes, my fault. I need some time to process. $\endgroup$ – Integral Jun 15 '12 at 17:18

Try an element $c$ of smallest value greater than $1$. Then division by $c$ has a remainder of zero or value $1$, which is therefore an unit.


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