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I'm studying basic Ring Theory. And in my textbook, the author states the definition of Euclidean domain:
An integral domain $R$ is called to be a Euclidean domain precisely when there is a function $f: R\setminus\{0\}\rightarrow\Bbb N_0$, called degree function of $R$, such that:
(i) If $a,b \in R\setminus\{0\}$ and there exists $c\in R$ such that $ac=b$ then $f(a)\le f(b)$.
(ii) $a,b\in R$ with $b\neq 0$, then there exist $q,r\in R$ such that $a=bq + r$ with $r=0$ or $r\neq 0$ and $f(r)\lt f(b)$.

I know the fact: all units in $R$ have smallest degree, and this question pops into my head:

I want to prove that all units have degree $0$.

Unfortunately I have no idea for it. Can anyone has an answer for my question or give me a readable explanation about it? I really appreciate !

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You can't prove that units have degree $0$: if $f\colon R\setminus\{0\}\to\mathbb{N}_0$ is a degree function, also $$ f_k\colon R\setminus\{0\}\to\mathbb{N}_0, \qquad f_k(a)=f(a)+k $$ is a degree function as well, for every integer $k\ge -f(1)$ (because $f(1)$ is the minimum value attained by $f$).

You can normalize your degree function, by defining $$ \hat{f}(a)=f(a)-f(1) $$ and now units will have degree $0$ with respect to $\hat{f}$.

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