Euclidean mapping question

I assume you follow the standard practice of requiring $delta$ to satisfy $delta(a)leqdelta(ab)$ for all nonzero $a,b$ instead of merely requiring existence of representative of each $b+(a)$, $bnotin(a)$, with a smaller $delta$ than $a$. (Otherwise on $mathbb{Q}$ you can define $delta(n)=n$, $delta(q)=1$ for all noninteger $q$).

So the only direction left to prove is "$A$ not a field $Rightarrowdelta$ has infinite range".

Suppose $A$ is not a field. There is some $xin A-{0}$ which is not invertible in $A$. Hence we have an infinite descending chain of ideals of $A$: $$tag{$dagger$} (x)supsetneq (x^2)supsetneq (x^3)supsetneqdots $$ $(x^{n+1})$ is an ideal, and it does not contain $x^n$. Hence there is some $qin A$ such that $x^n+x^{n+1}qneq 0$ and $delta(x^n+x^{n+1}q)<delta(x^n)$. Now $x^n+x^{n+1}qin(x^n)$ so $delta(x^n)leqdelta(x^n+x^{n+1}q)<delta(x^{n+1})$. Hence $delta(x)<delta(x^2)<delta(x^3)<dots$ and thus $delta$ has infinite range.

Note: In fact we can prove, in similar fashion, $delta(ab)=delta(a)$ iff $b$ is invertible.