The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 15.101.3. Let $R$ be a UFD. Then $\mathop{\mathrm{Pic}}\nolimits (R)$ is trivial.

Proof. Let $L$ be an invertible $R$-module. By Lemma 15.101.2 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. $\square$


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Comment #3239 by Dario WeiƟmann on

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