Suppose $a,b$ are elements of an integral domain $D$ equipped with a Euclidean function $f:D\rightarrow \mathbb{Z}$. By definition, $f$ satisfies the following two properties: 1. For all $a\in D, f(ab)\geq f(a)$ for all $b\neq 0$; and 2. If $a,b\in D, b\neq 0$, then there exist $c,d\in D$ such that $a=bc+d$ and $f(d)<f(b)$.
Now I am trying to understand the following claim: If $a|b$ and $f(a)=f(b)$ then $a\sim b$.
Recall that $a\sim b$ if $a|b$ and $b|c$ (equivalently, if $a$ and $b$ differ by a factor of a unit). Using the above, we have $b=ac$ for some $c\in D$ and so $f(b)=f(ac)\geq f(a)$ and so $f(ac)=f(a)$, but I'm not sure how this helps.
Another approach is as follows: Suppose $a$ and $b$ are not associates. We then want to show that either $a\nmid b$ or $f(a)\neq f(b)$. Now, $a$ and $b$ not associates implies that either $a\nmid b$ or $b\nmid a$. If $a\nmid b$, we are done. So suppose $b\nmid a$. Then there exists $c,d\in D$ such that $a=bc+d$ and where $0<f(d)<f(b)$. Again I'm not sure how to proceed.
How can I continue?
Best Answer
If $f(a) =f(b) $ and $a\mid b$, then $$b=ac$$ for some $c$ and $$a=bd+e$$ for some $d, e$ with $f(e) <f(b)=f(a) $. If $e\neq 0$, then $$a=acd+e$$ and $$(1-cd)a=e$$ But if $1-cd\neq 0$ then $$f(a) >f(e)=f((1-cd)a)\geq f(a) $$ which is a contradiction. Thus $e=0$ and $cd=1$. In particular, $c$ is a unit, so $b$ is an associate of $a$.