Definition 15.102.1. Let $R$ be a ring. An $R$-module $M$ is *invertible* if the functor

is an equivalence of categories. An invertible $R$-module is said to be *trivial* if it is isomorphic to $A$ as an $A$-module.

We first define invertible modules as follows.

Definition 15.102.1. Let $R$ be a ring. An $R$-module $M$ is *invertible* if the functor

\[ \text{Mod}_ R \longrightarrow \text{Mod}_ R,\quad N \longmapsto M \otimes _ R N \]

is an equivalence of categories. An invertible $R$-module is said to be *trivial* if it is isomorphic to $A$ as an $A$-module.

Lemma 15.102.2. Let $R$ be a ring. Let $M$ be an $R$-module. Equivalent are

$M$ is finite locally free module of rank $1$,

$M$ is invertible, and

there exists an $R$-module $N$ such that $M \otimes _ R N \cong R$.

Moreover, in this case the module $N$ is (3) is isomorphic to $\mathop{\mathrm{Hom}}\nolimits _ R(M, R)$.

**Proof.**
Assume (1). Consider the module $N = \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ and the evaluation map $M \otimes _ R N = M \otimes _ R \mathop{\mathrm{Hom}}\nolimits _ R(M, R) \to R$. If $f \in R$ such that $M_ f \cong R_ f$, then the evaluation map becomes an isomorphism after localization at $f$ (details omitted). Thus we see the evaluation map is an isomorphism by Algebra, Lemma 10.22.2. Thus (1) $\Rightarrow $ (3).

Assume (3). Then the functor $K \mapsto K \otimes _ R N$ is a quasi-inverse to the functor $K \mapsto K \otimes _ R M$. Thus (3) $\Rightarrow $ (2). Conversely, if (2) holds, then $K \mapsto K \otimes _ R M$ is essentially surjective and we see that (3) holds.

Assume the equivalent conditions (2) and (3) hold. Denote $\psi : M \otimes _ R N \to R$ the isomorphism from (3). Choose an element $\xi = \sum _{i = 1, \ldots , n} x_ i \otimes y_ i$ such that $\psi (\xi ) = 1$. Consider the isomorphisms

\[ M \to M \otimes _ R M \otimes _ R N \to M \]

where the first arrow sends $x$ to $\sum x_ i \otimes x \otimes y_ i$ and the second arrow sends $x \otimes x' \otimes y$ to $\psi (x' \otimes y)x$. We conclude that $x \mapsto \sum \psi (x \otimes y_ i)x_ i$ is an automorphism of $M$. This automorphism factors as

\[ M \to R^{\oplus n} \to M \]

where the first arrow is given by $x \mapsto (\psi (x \otimes y_1), \ldots , \psi (x \otimes y_ n))$ and the second arrow by $(a_1, \ldots , a_ n) \mapsto \sum a_ i x_ i$. In this way we conclude that $M$ is a direct summand of a finite free $R$-module. This means that $M$ is finite locally free (Algebra, Lemma 10.77.2). Since the same is true for $N$ by symmetry and since $M \otimes _ R N \cong R$, we see that $M$ and $N$ both have to have rank $1$. $\square$

The set of isomorphism classes of these modules is often called the *class group* or *Picard group* of $R$. The group structure is determined by assigning to the isomorphism classes of the invertible modules $L$ and $L'$ the isomorphism class of $L \otimes _ R L'$. The inverse of an invertible module $L$ is the module

\[ L^{\otimes -1} = \mathop{\mathrm{Hom}}\nolimits _ R(L, R), \]

because as seen in the proof of Lemma 15.102.2 the evaluation map $L \otimes _ R L^{\otimes -1} \to R$ is an isomorphism. Let us denote the Picard group of $R$ by $\mathop{\mathrm{Pic}}\nolimits (R)$.

Lemma 15.102.3. Let $R$ be a UFD. Then $\mathop{\mathrm{Pic}}\nolimits (R)$ is trivial.

**Proof.**
Let $L$ be an invertible $R$-module. By Lemma 15.102.2 we see that $L$ is a finite locally free $R$-module. In particular $L$ is torsion free and finite over $R$. Pick a nonzero element $\varphi \in \mathop{\mathrm{Hom}}\nolimits _ R(L, R)$ of the dual invertible module. Then $I = \varphi (L) \subset R$ is an ideal which is an invertible module. Pick a nonzero $f \in I$ and let

\[ f = u p_1^{e_1} \ldots p_ r^{e_ r} \]

be the factorization into prime elements with $p_ i$ pairwise distinct. Since $L$ is finite locally free there exist $a_ i \in R$, $a_ i \not\in (p_ i)$ such that $I_{a_ i} = (g_ i)$ for some $g_ i \in R_{a_ i}$. Then $p_ i$ is still a prime element of the UFD $R_{a_ i}$ and we can write $g_ i = p_ i^{c_ i} g'_ i$ for some $g'_ i \in R_{a_ i}$ not divisible by $p_ i$. Since $f \in I_{a_ i}$ we see that $e_ i \geq c_ i$. We claim that $I$ is generated by $h = p_1^{c_1} \ldots p_ r^{c_ r}$ which finishes the proof.

To prove the claim it suffices to show that $I_ a$ is generated by $h$ for any $a \in R$ such that $I_ a$ is a principal ideal (Algebra, Lemma 10.22.2). Say $I_ a = (g)$. Let $J \subset \{ 1, \ldots , r\} $ be the set of $i$ such that $p_ i$ is a nonunit (and hence a prime element) in $R_ a$. Because $f \in I_ a = (g)$ we find the prime factorization $g = v \prod _{i \in J} p_ j^{b_ j}$ with $v$ a unit and $b_ j \leq e_ j$. For each $j \in J$ we have $I_{aa_ j} = g R_{aa_ j} = g_ j R_{aa_ j}$, in other words $g$ and $g_ j$ map to associates in $R_{aa_ j}$. By uniqueness of factorization this implies that $b_ j = c_ j$ and the proof is complete. $\square$

Recall that we have defined in Algebra, Section 10.54 a group $K_0(R)$ as the free group on isomorphism classes of finite projective $R$-modules modulo the relations $[M'] + [M''] = [M' \oplus M'']$.

Lemma 15.102.4. Let $R$ be a ring. There is a map

\[ \det : K_0(R) \longrightarrow \mathop{\mathrm{Pic}}\nolimits (R) \]

which maps $[M]$ to the class of the invertible module $\wedge ^ n(M)$ if $M$ is a finite locally free module of rank $n$.

**Proof.**
Let $M$ be a finite projective $R$-module. There exists a product decomposition $R = R_0 \times \ldots \times R_ t$ such that in the corresponding decomposition $M = M_0 \times \ldots \times M_ t$ of $M$ we have that $M_ i$ is finite locally free of rank $i$ over $R_ i$. This follows from Algebra, Lemma 10.77.2 (to see that the rank is locally constant) and Algebra, Lemmas 10.20.3 and 10.23.3 (to decompose $R$ into a product). In this situation we define

\[ \det (M) = \wedge ^0_{R_0}(M_0) \times \ldots \times \wedge ^ t_{R_ t}(M_ t) \]

as an $R$-module. This is a finite locally free module of rank $1$ as each term is finite locally free of rank $1$. To finish the proof we have to show that

\[ \det (M' \oplus M'') \cong \det (M') \otimes \det (M'') \]

whenever $M'$ and $M'"$ are finite projective $R$-modules. Decompose $R$ into a product of rings $R_{ij}$ such that $M' = \prod M'_{ij}$ and $M'' = \prod M''_{ij}$ where $M'_{ij}$ has rank $i$ and $M''_{ij}$ has rank $j$. This reduces us to the case where $M'$ and $M''$ have constant rank say $i$ and $j$. In this case we have to prove that

\[ \wedge ^{i + j}(M' \oplus M'') \cong \wedge ^ i(M') \otimes \wedge ^ j(M'') \]

the proof of which we omit. $\square$

Lemma 15.102.5. Let $R$ be a ring. There is a map

\[ c : \text{perfect complexes over }R \longrightarrow K_0(R) \]

with the following properties

$c(K[n]) = (-1)^ nc(K)$ for a perfect complex $K$,

if $K \to L \to M \to K[1]$ is a distinguished triangle of perfect complexes, then $c(L) = c(K) + c(M)$,

if $K$ is represented by a finite complex $M^\bullet $ consisting of finite projective modules, then $c(K) = \sum (-1)^ i[M_ i]$.

**Proof.**
Let $K$ be a perfect object of $D(R)$. By definition we can represent $K$ by a finite complex $M^\bullet $ of finite projective $R$-modules. We define $c$ by setting

\[ c(K) = \sum (-1)^ n[M^ n] \]

in $K_0(R)$. Of course we have to show that this is well defined, but once it is well defined, then (1) and (3) are immediate. For the moment we view the map $c$ as defined on complexes of finite projective $R$-modules.

Suppose that $L^\bullet \to M^\bullet $ is a surjective map of finite complexes of finite projective $R$-modules. Let $K^\bullet $ be the kernel. Then we obtain short exact sequences of $R$-modules

\[ 0 \to K^ n \to L^ n \to M^ n \to 0 \]

which are split because $M^ n$ is projective. Hence $K^\bullet $ is also a finite complex of finite projective $R$-modules and $c(L^\bullet ) = c(K^\bullet ) + c(M^\bullet )$ in $K_0(R)$.

Suppose given finite complex $M^\bullet $ of finite projective $R$-modules which is acyclic. Say $M^ n = 0$ for $n \not\in [a, b]$. Then we can break $M^\bullet $ into short exact sequences

\[ \begin{matrix} 0 \to M^ a \to M^{a + 1} \to N^{a + 1} \to 0,
\\ 0 \to N^{a + 1} \to M^{a + 2} \to N^{a + 3} \to 0,
\\ \ldots
\\ 0 \to N^{b - 3} \to M^{b - 2} \to N^{b - 2} \to 0,
\\ 0 \to N^{b - 2} \to M^{b - 1} \to M^ b \to 0
\end{matrix} \]

Arguing by descending induction we see that $N^{b - 2}, \ldots , N^{a + 1}$ are finite projective $R$-modules, the sequences are split exact, and

\[ c(M^\bullet ) = \sum (-1)[M^ n] = \sum (-1)^ n([N^{n - 1}] + [N^ n]) = 0 \]

Thus our construction gives zero on acyclic complexes.

It follows formally from the results of the preceding two paragraphs that $c$ is well defined and satisfies (2). Namely, suppose the finite complexes $M^\bullet $ and $L^\bullet $ of finite projective $R$-modules represent the same object of $D(R)$. Then we can represent the isomorphism by a map $f : M^\bullet \to L^\bullet $ of complexes, see Derived Categories, Lemma 13.19.8. We obtain a short exact sequence of complexes

\[ 0 \to L^\bullet \to C(f)^\bullet \to K^\bullet [1] \to 0 \]

see Derived Categories, Definition 13.9.1. Since $f$ is a quasi-isomorphism, the cone $C(f)^\bullet $ is acyclic (this follows for example from the discussion in Derived Categories, Section 13.12). Hence

\[ 0 = c(C(f)^\bullet ) = c(L^\bullet ) + c(K^\bullet [1]) = c(L^\bullet ) - c(K^\bullet ) \]

as desired. We omit the proof of (2) which is similar. $\square$

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.
$\square$

Lemma 15.102.7. A regular local ring is a UFD.

**Proof.**
Recall that a regular local ring is a domain, see Algebra, Lemma 10.105.2. 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 10.105.3, hence a domain by Algebra, Lemma 10.105.2, 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 10.119.6. 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 10.109.6. 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 10.77.2 we see that $\mathfrak p_ x$ is an invertible $R_ x$-module. By Lemma 15.102.6 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 10.119.3). 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 10.119.7. Since $(a_ i) \subset \mathfrak p$ and $\mathfrak p$ has height $1$ we conclude that $(a_ i) = \mathfrak p$ as desired.
$\square$

Lemma 15.102.8. 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 10.125.3. 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 15.76.4 we see that $M$ is perfect relative to $R$, in particular $M$ is pseudo-coherent as an $A$-module. By Lemma 15.71.9 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 15.102.5 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 15.102.4 we see that $L$ is trivial.
$\square$

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