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I am trying to prove the following: if $R$ is a PID and $Q$ is the field of fractions of $R$, then any element of $Q$ has an expression of the form $f/g$ where $(f,g)=(1)$. I proved the following: In a PID, if $x$ and $y$ have gcd $d$, then $(x,y)=(d)$ and there exists $a,b\in R$ such that $ax+by=d$ and $(a,b)=(1)$. Here is my work so far

Take any $x/y\in Q$. Since $R$ is a PID, we can find a gcd of $x$ and $y$ and write $(x,y)=(d)$, so that $fx+gy=d$ for some $f,g\in R$ and $(f,g)=(1)$. I claim that $x/y\sim f/g\iff gx=fy$ . This is where I am stuck. Any hints would be greatly appreciated.

yastown
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  • Do you have prime decomposition in a PID yet? If so, that would help a lot. – Doge Chan Feb 23 '22 at 00:42
  • No, we don't have prime decomposition yet. – yastown Feb 23 '22 at 00:46
  • $d$ divides $x$ and $y$, so there are elements $u,v$ of $R$ such that $x=ud$ and $y=vd$. Can you find a way two use these elements $u,v$ with the properties of the PID such that you can get your conclusion? – Doge Chan Feb 23 '22 at 00:53
  • As in $\Bbb Q$ simply cancel the gcd $,d=(x,y),$ from $x = d\bar x,\ y = d\bar y,$ to get $, x/y = (d\bar x)/(d\bar y)\sim \bar x/\bar y$ and $(\bar x, \bar y) = 1$ by the linked post (e.g. by the gcd distributive law). – Bill Dubuque Feb 23 '22 at 02:49

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