Can anyone gives me an example of $V$ such that there exists at least one pair of elements $ a,c(\neq 0 ,1) \in D$ such that $V(a) > V(ac)$

Question

$D$ is an Integral Domain.

$V$ only satisfies 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 either $r=0$ or $\, V(r) < V(b),\,$ where $V$ maps $D$ into (well-ordered) $\,\Bbb N$. ? Can anyone gives me an example of $V$ such that there exists at least one pair of elements $ a,c(\neq 0 ,1) \in D$ such that $V(a) > V(ac)$?

I have been really trying hard to construct one. But I can't.

Answer 1

Let $D=\Bbb Q[X]$ and $$ V(f)=\begin{cases} 2\deg f&\forall x\gg 0\colon f(x)>0 \\ 2\deg f+1&\forall x\gg 0\colon f(x)<0 \\ \text{whatever}&f=0\end{cases}$$

For $a,b\ne0$, we can find by polynomial division $q,r$ with $a=qb+r$ and either $r=0$ or $\deg r<\deg b$, in which case also have $V(r)<V(b)$. Thus the desired conditins hold. Now let $a=c=-1$ and note that $V(a)=1$, $V(ac)=0$.