Lemma 15.121.1. Let $R$ be a regular local ring. Let $f \in R$. Then $\mathop{\mathrm{Pic}}\nolimits (R_ f) = 0$.

Proof. Let $L$ be an invertible $R_ f$-module. In particular $L$ is a finite $R_ f$-module. There exists a finite $R$-module $M$ such that $M_ f \cong L$, see Algebra, Lemma 10.126.3. By Algebra, Proposition 10.110.1 we see that $M$ has a finite free resolution $F_\bullet$ over $R$. It follows that $L$ is quasi-isomorphic to a finite complex of free $R_ f$-modules. Hence by Lemma 15.119.1 we see that $[L] = n[R_ f]$ in $K_0(R)$ for some $n \in \mathbf{Z}$. Applying the map of Lemma 15.118.7 we see that $L$ is trivial. $\square$

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