## Tag `0BCH`

Chapter 15: More on Algebra > Section 15.97: Picard groups of rings

Lemma 15.97.3. Let $R$ be a UFD. Then $\mathop{\mathrm{Pic}}\nolimits(R)$ is trivial.

Proof.Let $L$ be an invertible $R$-module. By Lemma 15.97.2 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 exists a $a_i \in R$, $a_i \not \in (p_i)$ 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. $\square$

The code snippet corresponding to this tag is a part of the file `more-algebra.tex` and is located in lines 29027–29030 (see updates for more information).

```
\begin{lemma}
\label{lemma-UFD-Pic-trivial}
Let $R$ be a UFD. Then $\Pic(R)$ is trivial.
\end{lemma}
\begin{proof}
Let $L$ be an invertible $R$-module. By Lemma \ref{lemma-invertible}
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
$\varphi \in \Hom_R(L, R)$ 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 exists a $a_i \in R$,
$a_i \not \in (p_i)$ 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.
\medskip\noindent
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 \ref{algebra-lemma-cover}). 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.
\end{proof}
```

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