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Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ?

Why $v(a) \leq v(ab)$ is not needed? How we can deduce from the first one?

cmi
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  • [Euclidean domain](https://en.wikipedia.org/wiki/Euclidean_domain#Definition). – MSDG Dec 07 '18 at 14:48
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    It would be good if you could elaborate exactly which part of the construction you find to be "not comprehensible". – MSDG Dec 07 '18 at 15:12
  • How the new function satisfies the division algorithm?@MisterRiemann – cmi Dec 07 '18 at 15:13
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    I'm not really an algebraist, so I'm learning this as we speak. Here is [another reference](http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf), which seems to a bit more in-depth about this. (In particular, see Theorem 2.1). – MSDG Dec 07 '18 at 15:20

2 Answers2

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Suppose $V$ satisfies only the first Euclidean property, i.e. for all $\,a,b\in D\,$ if $\,b\neq 0\,$ then there are $\,q,r\in D\,$ such that $\, a = qb +r\,$ with $\, V(r) < V(b),\,$ where $V$ maps $D$ into (well-ordered) $\,\Bbb N.\,$ We show how to construct from $V$ another Euclidean function $v$ that satisfies $\, v(a) \le v(ab)\,$ if $\,ab\neq 0$.

Derive $\,v\,$ from $\,V\,$ as follows

$$\begin{align} v(0) &= V(0)\\ v(a) &= {\rm min}\{ V(b)\ :\ b\in aD\backslash 0\} \end{align}$$

Note $\,v(a)\le V(a)\,$ since it is clear if $\,a = 0,\,$ else it follows by $\, a\in aD\backslash 0$

$v$ is also a Euclidean function: if $\,a,b\in D\,$ and $\,b\neq 0\,$ then $\,v(b) = V(bc)\,$ for $\,0\neq c\in D.\,$ Since $\,V\,$ is a Euclidean function there are $\,q,r\in D\,$ such that $\, a = qbc + r\,$ and $\,V(r) < V(bc) = v(b).\,$ But by above we know $\,v(r)\le V(r)\,$ thus $\,v(r) < v(b),\,$ so $\,v\,$ is a Euclidean function.

Note $\, v(a) \le v(ab)\,$ if $\,ab\neq 0\,$ since $\,aD\backslash 0\supseteq abD\backslash 0$ $\,\Rightarrow\,{\rm min}\,V(aD\backslash 0) \le {\rm min}\, V(abD\backslash 0)$

Remark $ $ See the paper cited here by Agargun & Fletcher for a comprehensive study of the logical relationships between various common definitions of Euclidean domains and rings.

Bill Dubuque
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The second property is superfluous because only the first one is needed to prove that every ideal of a Euclidean domain is principal.

Geoffrey Trang
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