I am trying to solve a problem about Euclidean Domain. The problem is:
Let $R$ be an Euclidean Domain, $a,b \in R-\{0\}, a|b$ and $d(a)=d(b)$. Show that $a$ and $b$ are associates.
I am trying to show it by showing that $b|a$. The idea is to get a contradiction if we suppose $b$ does not divide $a$. However, the definition of Euclidean Domain that I am working does not involve the $d-$inequality, that is, if $a,b \in R-\{0\}$ then $d(a) \leqslant d(ab)$. I know that there exist a way to show that we can always consider this inequality, but my question is: can someone give me a hint about how to solve it without using this inequality?
There are another question about similar thinks here, but I did not find any of them talking about my specific question and this is why I am asking about it. If there is another question regarding this, please, let me know.
Thank you so much everyone!
This isn't true without assuming the $d$-inequality. For instance, let $R=\mathbb{Z}$ and define $d(a)=a$ if $a>0$ and $d(a)=-2a$ if $a<0$. Then you still have division with remainder for this $d$, since you can always choose your remainder to be nonnegative. But $a=-2$ and $b=4$ satisfy $a\mid b$ and $d(a)=d(b)$ and are not associates.