Euclidean mapping question Give $A$ is Euclidean domain and $\delta: A\setminus \left\{0\right\} \to \mathbb{N}$ is Euclidean mapping.
I have prove that "$A$ is a field if and only if $\delta$ is constant".
So, I wonder that "$A$ is not a field if and only if $\delta(A\setminus \left\{0\right\})$ is infinite subset of $\mathbb{N}$"?
My attempt:
($\Leftarrow$) Suppose that $\delta(A\setminus \left\{0\right\})$ is infinite subset of $\mathbb{N}$.
By contradiction, we have $A$ is a field. This yields, $\delta(A\setminus \left\{0\right\})$ have one value (conflict with what we suppose).
($\Rightarrow$) I try to use contradiction to show something nonsense but I still can't find the answer.
 A: I assume you follow the standard practice of requiring $\delta$ to satisfy $\delta(a)\leq\delta(ab)$ for all nonzero $a,b$ instead of merely requiring existence of representative of each $b+(a)$, $b\notin(a)$, 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 $\Rightarrow\delta$ has infinite range".
Suppose $A$ is not a field.  There is some $x\in A-\{0\}$ which is not invertible in $A$.  Hence we have an infinite descending chain of ideals of $A$:
$$\tag{$\dagger$}
(x)\supsetneq (x^2)\supsetneq (x^3)\supsetneq\dots
$$
$(x^{n+1})$ is an ideal, and it does not contain $x^n$.  Hence there is some $q\in A$ such that $x^n+x^{n+1}q\neq 0$ and $\delta(x^n+x^{n+1}q)<\delta(x^n)$.  Now $x^n+x^{n+1}q\in(x^n)$ so $\delta(x^n)\leq\delta(x^n+x^{n+1}q)<\delta(x^{n+1})$.
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.
