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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.

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    $\begingroup$ you should say more about the restrictions on the ring and the exact properties of $\delta.$ If a Euclidean domain is meant,... $\endgroup$ – Will Jagy May 5 '20 at 15:03
  • $\begingroup$ @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 ? $\endgroup$ – Moeee May 5 '20 at 15:05
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    $\begingroup$ @Moee If it's a Euclidean domain, then your statement is not valid, because it is not a domain. $\endgroup$ – Matt Samuel May 5 '20 at 15:16
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    $\begingroup$ @Moeee Which disqualifies $\mathbb Z_4$ from being a Euclidean domain. $\endgroup$ – Matt Samuel May 5 '20 at 15:21
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    $\begingroup$ 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. $\endgroup$ – Matt Samuel May 5 '20 at 20:05
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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.

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

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