Lemma 15.102.6. 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 10.125.3. By Algebra, Proposition 10.109.1 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 15.102.5 we see that $[L] = n[R_ f]$ in $K_0(R)$ for some $n \in \mathbf{Z}$. Applying the map of Lemma 15.102.4 we see that $L$ is trivial.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)