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Quoted from this question.

We called an integral domain $R$ a Euclidean domain if there exists a function $f$ from $R$ to strictly positive integers such that:
1) For $a,b$ non zero in $R$, $f(ab)\ge f(a)$.
2) If $a,b\in R$, $b\neq 0$, then we can write $a=bq+r$ with $q,r\in R$ such that either $r=0$ or $f(r)<f(b)$.

One of the answers below say that the definition is equivalent if we remove 1) above. To prove this, we take $$ g(a)=\min_{x\in R^* a} f(x), R^*=R\backslash \{0\}. $$

Obviously $g$ satisfies 1). Now I am struggling to show that the new function $g$ satisfy 2), given that $f$ satisfies 2). Any hints?

Ma Joad
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1 Answers1

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Let $a,b\in R$ with $b\neq 0$. Let $g(b)=f(cb)$ for some $c\neq 0$(note that the minimum is always attained). Since $bc\neq 0$, we can write $a=bcq+r$ with $q,r\in R$ such that either $r=0$ or $f(r)<f(cb)$. Let $q_1:=cq$ and $r_1:=r$. So $a=bq_1+r_1$. If $r=r_1=0$, there is nothing to prove. Suppose $r_1\neq 0$. Then $$g(r_1)\leq f(r_1)=f(r)<f(cb)=g(b).$$

cqfd
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