Definition 15.115.1. 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 $R$ as an $R$-module.

We first define invertible modules as follows.

Definition 15.115.1. 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 $R$ as an $R$-module.

Lemma 15.115.2. 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 $N$ in (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 10.23.2. 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 10.78.2). 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 15.115.2 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 15.115.3. Let $R$ be a UFD. Then $\mathop{\mathrm{Pic}}\nolimits (R)$ is trivial.

**Proof.**
Let $L$ be an invertible $R$-module. By Lemma 15.115.2 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 10.23.2). 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. $\square$

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