$\operatorname{Frac}(R[t]) = \operatorname{Frac}(R)(t)$

Let \(K=\operatorname{Frac}(R)\). Since \[ R[t]\subseteq K[t]\subseteq K(t) \] and \(K(t)\) is a field, the universal property of the fraction field gives \[ \operatorname{Frac}(R[t])\subseteq K(t). \]

Conversely, every element \(a/b\in K\) with \(a,b\in R\) and \(b\ne0\) is already a quotient of two elements of \(R[t]\), so \[ K\subseteq \operatorname{Frac}(R[t]). \] Also \(t\in R[t]\subseteq \operatorname{Frac}(R[t])\). Hence the field \(\operatorname{Frac}(R[t])\) contains the field generated by \(K\) and \(t\), namely \(K(t)\). Thus \[ K(t)\subseteq \operatorname{Frac}(R[t]). \] Therefore \[ \operatorname{Frac}(R[t])=K(t)=\operatorname{Frac}(R)(t). \]