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Let $xy=z^n$ where $x$, $y$ and $z$ belong to a Dedekind domain $R$, with $n>1$, and $(x,y)=1$. We can also assume that the ideal class group of $R$ is torsion-free.

Then I’d like to show that $x=ur^n$, where $r\in R$ and $u$ is a unit in $R$. I thought of starting from the unique factorization of $(x)$ and $(y)$ into prime ideals, equating this to the prime ideal factorization of $(z)^n$. But I couldn’t get much further.

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Hint. For any prime divisor $P$ of $(x)$, we have $v_P(y)=0$ (why ?). One may deduce that $(x)=I^n$ for some ideal $I$ of $R$. Now use the assumption on the ideal class group to get that $I$ is a principal ideal. If $I=(r)$, we then get $(x)=(r^n)$, so...

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  • $\begingroup$ I think I can see why $v_P(y)=0$ for $P$ prime divisor of $(x)$ (because $(x,y)=1$, right?) but why exactly $(x)=I^n$? $\endgroup$
    – oc89
    Apr 19, 2022 at 21:48
  • $\begingroup$ What can you say about $v_P(x)$ , then ? $\endgroup$
    – GreginGre
    Apr 20, 2022 at 6:16
  • $\begingroup$ I’m not sure, to be honest. It must be $0 \mod n$? $\endgroup$
    – oc89
    Apr 20, 2022 at 12:59
  • $\begingroup$ Do you know how to proceed when $R=\mathbb{Z}$ ? The arguments are along the same lines. $\endgroup$
    – GreginGre
    Apr 20, 2022 at 14:47
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    $\begingroup$ Like in the accepted answer here, right? math.stackexchange.com/questions/1102067/… $\endgroup$
    – oc89
    Apr 20, 2022 at 16:10

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