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I'm sort of stuck on this problem and I could use a hint:

Let $R$ be a Euclidean domain with a Euclidean function $d$ such that, for all non-zero $a$ and $b$,

  1. $d(a) \leq d(ab),$
  2. $d(a+b) \leq \max\{d(a), d(b)\}.$

For $a, b$ non-zero, prove that $d(a) = d(ab)$ if and only if $b$ is invertible.

I've proven that if $b$ is invertible then $d(a) = d(ab)$, but I'm not sure how to prove the converse. Do I start by assuming that $d(ab) = d(a)$ and then prove that $b$ must have an inverse?

  • 2
    Sir$z$h, I suggest that i$f$ you now understand how to do the problem, that you post your solution as an answer. Then, after a while, you can accept it.2012-11-22

1 Answers 1

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$\exists q, r \in R$ such that $a = abq + r$ with $r = 0$ or $d(r) < d(ab)$
If we can prove that $r = 0$ then we have our proof. So lets disprove that $d(r) < d(ab)$:
$d(a) = d(ab) > d(r) = d(a-abq) = d(a(1-bq) \geq d(a) $, Contradiction. Hence $r=0$ and $a = abq$, so $a(1-bq)= 0$.