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First, let $K$ be a field. Then the field of rational functions $K(x)$ is a projective $K[x]$ module. I can prove this by showing that $K(x)$ is not a free $K[x]$-module because $K[x]$ is a principal ideal domain, which means that free implies projective.

My question is, if we were to replace $K$ by an integral domain $R$, could $S^{-1}(R[x])$, the field of fractions of $R[x]$, be a projective $R[x]$-module? Since $R[x]$ is not necessarily a PID, the above strategy would no longer apply.

Poko
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  • What does $R(x)$ mean? The field of fractions of $R[x]$? – Eric Wofsey Sep 18 '17 at 06:21
  • Yes, I'm sorry I wasn't clear. I'll edit my original post. – Poko Sep 18 '17 at 06:23
  • What do you mean by "$R(x),$ the field of fractions of $R[x]$"? $R(x)$ is not a field of fraction of $R[x]$. The field of fraction of $R[x]$ is $Q(R)(x)$ where $Q(R)$ is the field of fractions of $R$. – Krish Sep 18 '17 at 06:29
  • OK, what I meant was the set $S^{-1}(R[x])$ where $S=R[x]\setminus {0}$. Does this make sense? – Poko Sep 18 '17 at 06:34
  • If $S=R[x] \setminus {0}$, then $S^{-1}R[x]$ is same as $Q(R)(x)$. You are also inverting non-zero the elements of $R$. – Krish Sep 18 '17 at 06:38
  • My apologies. I am not completely familiar with the notation. – Poko Sep 18 '17 at 06:42

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