Show that \(\operatorname{Frac}(R[t]) \cong \operatorname{Frac}(R)(t)\) for any integral domain \(R\).
Solution
Let \[ K=\operatorname{Frac}(R). \] Since \(R[t]\subseteq K[t]\), there is an inclusion \[ \operatorname{Frac}(R[t])\subseteq K(t). \]
Conversely, let \(P/Q\in K(t)\) with \(P,Q\in K[t]\) and \(Q\ne0\). Choose a nonzero \(d\in R\) clearing all denominators of the coefficients of both \(P\) and \(Q\). Then \[ dP,dQ\in R[t],\qquad dQ\ne0, \] and \[ \frac{P}{Q}=\frac{dP}{dQ}\in\operatorname{Frac}(R[t]). \] Hence \[ \boxed{\operatorname{Frac}(R[t])=K(t)=\operatorname{Frac}(R)(t).} \]