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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?

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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

1 Answer 1

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.

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