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Chapter 15: More on Algebra > Section 15.96: Picard groups of rings

Lemma 15.96.3. Let $R$ be a UFD. Then $\mathop{\mathrm{Pic}}\nolimits(R)$ is trivial.

Proof. Let $L$ be an invertible $R$-module. By Lemma 15.96.2 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 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 28443–28446 (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 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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