Why is $v(a) \leq v(ab)$ superfluous in Euclidean domain definition?

Question

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?

Answer 1

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 $\,v\,$ satisfying $\, v(a) \le v(ab)\,$ 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.

Answer 2

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