Prove if $R$ is not a field then the value set $δ(R)$ is infinite, $δ$ a Euclidean degree function.

Question

I was trying to solve this exercise in my course notes, but the statement didn't seem right to me. When looking at the ring $\mathbb{Z}/4\mathbb{Z}$, it is clearly not a field since $2 + \mathbb{Z}$ does not have an inverse. But the set $\mathbb{Z}/4\mathbb{Z}$ , which is $\{0+ \mathbb{Z};1+ \mathbb{Z};2+ \mathbb{Z};3+ \mathbb{Z}\}$ is finite, and we could easily define a Euclidean degree function on it with a finite value set ( by letting $δ(1) = δ(3) = 1$ , $δ(2) = 2$ and $δ(0) = -\infty$).

So this is a ring that is not a field with a Euclidean degree function on it with a finite value set. I'm I wrong or does the ring have to be infinite or none of the previous ?

Also would the contraposition of the statement be : $δ(R)$ finite $\Rightarrow$ $R$ a field ?

Thanks in advance.

Answer 1

We may assume that $\delta(a)\leq \delta(ab)$ for all $a,b\neq 0$. Using this, we show that if $\delta(x)>0$ and $x$ is not a unit then $\delta(x)<\delta(x^2)$.

Let $\delta(x)>0$ for some $x$ that is not a unit. Then there exist $q$ and $r$ such that $$x^3 + x = qx^2+r$$ with $\delta(r)<\delta(x^2)$. Thus $$x(x^2+1) = qx^2+r$$ $$x(x^2+1-qx) = r$$ If $r=0$, then $qx=x^2+1$, so $$x(x-q)=-1$$ which is a contradiction since in such a situation $x$ has an inverse (namely $q-x$). Thus $r\neq 0$, $\delta(r)<\delta(x^2)$, and $\delta(r)\geq \delta(x)$ since $\delta(x(x^2+1-qx))\geq \delta(x)$. It follows that $\delta(x)<\delta(x^2)$.

We can thus recursively construct an increasing sequence of values $x_0,x_1,\ldots$ such that $x_0=x$ and $$x_i=x_{i-1}^2$$ By the same argument, $\delta(x_{i-1})<\delta(x_i)$ for all $i$, hence the number of values is infinite.