Line bundles on a principal localization of a regular local ring are trivial

If $R_f=0$, the assertion is vacuous. Otherwise $L$ is finitely presented: its dual gives finite elements $l_i\in L$, ${\lambda }_i\in L^{\vee }$ with ${\sum }_i{\lambda }_i(l)l_i=l$ for every $l$, exhibiting $L$ as a direct summand of a finite free module. Choose a finite presentation matrix for $L$ over $R_f$. Multiply its finitely many columns by powers of $f$ to lift that matrix to $R$; its cokernel $M$ is finite over $R$, and $M_f\cong L$.

Let $d=\dim R$. By [F1], a minimal finite-rank free resolution of $M$ has projective $d$-th syzygy. A finite projective module $P$ over a local ring is free: lift a basis of $P\mathrm{/}\mathfrak{m}P$, yielding a surjection $R^r\to P$ by [F2]; it splits by projectivity, and its kernel is finite with zero reduction modulo $\mathfrak{m}$, so it vanishes by [F2]. Truncate the resolution using a finite free module for its last syzygy. Thus $M$ has a bounded resolution by finite free modules. Localization gives such a resolution of $L$ over $R_f$.

Since $L$ is projective, the surjection from the degree-zero free module splits, making its kernel finite projective. Inductively every subsequent short exact sequence in the localized resolution splits. For a split sequence $0\to P\to Q\to T\to 0$ of finite projective modules of constant ranks, exterior multiplication gives $\det Q\cong \det P\otimes \det T$: after any localization choose bases and concatenate them, and the resulting transition determinants multiply, so the local identifications glue independently of the chosen splitting. All these modules have constant ranks, since they are direct summands in the finite free resolution and $R_f$ is a domain. Multiplying the determinant identities with alternating signs gives $L=\det L\cong {⨂}_i(\det (F_i{)}_f{)}^{(-1{)}^i}$. Every $F_i$ is free, so the last invertible module is trivial. Consequently $L\cong R_f$. AC is inherited from the resolution and regularity suppliers.