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

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 ?

• you should say more about the restrictions on the ring and the exact properties of $\delta.$ If a Euclidean domain is meant,... – Will Jagy May 5 '20 at 15:03
• @WillJagy The only thing given in the exercise is $δ$ a euclidean degree function of a ring $R$ so I assume $R$ is a euclidean domain. Is that what you mean with restrictions on the ring ? – Moeee May 5 '20 at 15:05
• @Moee If it's a Euclidean domain, then your statement is not valid, because it is not a domain. – Matt Samuel May 5 '20 at 15:16
• @Moeee Which disqualifies $\mathbb Z_4$ from being a Euclidean domain. – Matt Samuel May 5 '20 at 15:21
• My original answer assumed $\delta(-1)=0$, which need not be the case. This is where we need to use the assumption that $x$ is not invertible. – Matt Samuel May 5 '20 at 20:05

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.
• Perhaps make it clear that we may assume that $\delta(a) \le \delta (ab)$ for all nonzero $a,b \in R$. – lhf May 5 '20 at 16:09