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

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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## Comments (2)

Comment #3239 by Dario Weißmann on

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