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

## 15.121 A regular local ring is a UFD

We prove the result mentioned in the section title.

**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 $[L_ f] = n[R_ f]$ in $K_0(R_ f)$ for some $n \in \mathbf{Z}$. Applying the map of Lemma 15.118.7 we see that $L$ is trivial.
$\square$

Lemma 15.121.2. A regular local ring is a UFD.

**Proof.**
Recall that a regular local ring is a domain, see Algebra, Lemma 10.106.2. We will prove the unique factorization property by induction on the dimension of the regular local ring $R$. If $\dim (R) = 0$, then $R$ is a field and in particular a UFD. Assume $\dim (R) > 0$. Let $x \in \mathfrak m$, $x \not\in \mathfrak m^2$. Then $R/(x)$ is regular by Algebra, Lemma 10.106.3, hence a domain by Algebra, Lemma 10.106.2, hence $x$ is a prime element. Let $\mathfrak p \subset R$ be a height $1$ prime. We have to show that $\mathfrak p$ is principal, see Algebra, Lemma 10.120.6. We may assume $x \not\in \mathfrak p$, since if $x \in \mathfrak p$, then $\mathfrak p = (x)$ and we are done. For every nonmaximal prime $\mathfrak q \subset R$ the local ring $R_\mathfrak q$ is a regular local ring, see Algebra, Lemma 10.110.6. By induction we see that $\mathfrak pR_\mathfrak q$ is principal. In particular, the $R_ x$-module $\mathfrak p_ x = \mathfrak pR_ x \subset R_ x$ is a finitely presented $R_ x$-module whose localization at any prime is free of rank $1$. By Algebra, Lemma 10.78.2 we see that $\mathfrak p_ x$ is an invertible $R_ x$-module. By Lemma 15.121.1 we see that $\mathfrak p_ x = (y)$ for some $y \in R_ x$. We can write $y = x^ e f$ for some $f \in \mathfrak p$ and $e \in \mathbf{Z}$. Factor $f = a_1 \ldots a_ r$ into irreducible elements of $R$ (Algebra, Lemma 10.120.3). Since $\mathfrak p$ is prime, we see that $a_ i \in \mathfrak p$ for some $i$. Since $\mathfrak p_ x = (y)$ is prime and $a_ i | y$ in $R_ x$, it follows that $\mathfrak p_ x$ is generated by $a_ i$ in $R_ x$, i.e., the image of $a_ i$ in $R_ x$ is prime. As $x$ is a prime element, we find that $a_ i$ is prime in $R$ by Algebra, Lemma 10.120.7. Since $(a_ i) \subset \mathfrak p$ and $\mathfrak p$ has height $1$ we conclude that $(a_ i) = \mathfrak p$ as desired.
$\square$

Lemma 15.121.3. Let $R$ be a valuation ring with fraction field $K$ and residue field $\kappa $. Let $R \to A$ be a homomorphism of rings such that

$A$ is local and $R \to A$ is local,

$A$ is flat and essentially of finite type over $R$,

$A \otimes _ R \kappa $ regular.

Then $\mathop{\mathrm{Pic}}\nolimits (A \otimes _ R K) = 0$.

**Proof.**
Let $L$ be an invertible $A \otimes _ R K$-module. In particular $L$ is a finite module. There exists a finite $A$-module $M$ such that $M \otimes _ R K \cong L$, see Algebra, Lemma 10.126.3. We may assume $M$ is torsion free as an $R$-module. Thus $M$ is flat as an $R$-module (Lemma 15.22.10). From Lemma 15.25.6 we deduce that $M$ is of finite presentation as an $A$-module and $A$ is essentially of finite presentation as an $R$-algebra. By Lemma 15.83.4 we see that $M$ is perfect relative to $R$, in particular $M$ is pseudo-coherent as an $A$-module. By Lemma 15.77.6 we see that $M$ is perfect, hence $M$ has a finite free resolution $F_\bullet $ over $A$. It follows that $L$ is quasi-isomorphic to a finite complex of *free* $A \otimes _ R K$-modules. Hence by Lemma 15.119.1 we see that $[L] = n[A \otimes _ R K]$ in $K_0(A \otimes _ R K)$ for some $n \in \mathbf{Z}$. Applying the map of Lemma 15.118.7 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 (2)

Comment #6705 by 期中考试破大防 on

Comment #6908 by Johan on