Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x$ and $y$ be nonzero elements of an integral domain $D$. Then $x$ and $y$ are associates if and only if $x = yd$ for some unit $d \in D$.

  • I am done proving the $\Leftarrow$ part.

  • For $\Rightarrow$ what I did was: If $x$ and $y$ are associates then $x \mid y$ and $y \mid x$ so $x \mid y$ implies that $xs = y$ for some $s \in D$ and $y \mid x$ implies that $yt = x$ for some $t \in D$. That is, $yts=y$ which implies $ts=1$. Therefore $s$ is a unit and $t$ is a unit.

Am I on the right track? Can I say that the proof is complete?

share|cite|improve this question
Your argument is fine, and clearly written. I am assuming that $x\mid y$ and $y\mid x$ is the given definition of associates. – André Nicolas Nov 28 '12 at 16:37
Really? Thanks so much Sir. :) – ghet Nov 28 '12 at 16:38

This community wiki solution is intended to clear the question from the unanswered queue.

Your reasoning for $(\Longleftarrow)$ appears to be correct. Note that the reason you can cancel: $y = yts \implies 1 = ts$ is because we're in an integral domain.

For $(\Longrightarrow)$, the reasoning would be something like this: Since $x = yd$ where $d$ is a unit, then $y \mid x$, we can rewrite $x = yd$ as $y = xd^{-1}$ since $d$ is a unit and hence $x \mid y$. Since $x \mid y$ and $y \mid x$, then $x$ and $y$ are associates.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.