We first define invertible modules as follows.
Definition 15.117.1. 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.
\[ \text{Mod}_ R \longrightarrow \text{Mod}_ R,\quad N \longmapsto M \otimes _ R N \]
Lemma 15.117.2. 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
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
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
because as seen in the proof of Lemma 15.117.2 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 15.117.3. Let $R$ be a UFD. Then $\mathop{\mathrm{Pic}}\nolimits (R)$ is trivial.
Proof. Let $L$ be an invertible $R$-module. By Lemma 15.117.2 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
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. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)