Skip to main content
1 of 6
Georges Elencwajg
  • 47.5k
  • 14
  • 159
  • 241
  1. It is not true in the second definition that for an arbitrary ring a) is equivalent to c).
    Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring $B$ and a projective module of rank $1$ over $B$ which is not isomorphic to an invertible fractional ideal of $B$. This does not contradict Eisenbud's Theorem 11.6 because $R$ is explicitly supposed noetherian there.

  2. I think the only reasonable definition of $\operatorname {Pic(R)}$ valid for any commutative ring is to define it as the Picard group of the affine scheme $X=\operatorname {Spec}(R)$.
    As is the case for any locally ringed space $(X,\mathcal O_X)$ the Picard group consists of isomorphism classes of locally free $\mathcal O_X$-Modules of rank one.
    This is exactly the definition used with much success for general, non-affine, schemes but also for topological spaces, differential manifolds, etc.

  3. In our special case $X=\operatorname {Spec}(R)$ the definition in 2) translates in purely algebraic terms to:
    $\operatorname {Pic(R)}$ consists of isomorphism classes of $R$-modules $M$ such that there exist finitely many elements $r_1,\cdots,r_n\in R$ with:
    $\alpha$) $\sum Rr_i=R$
    $\beta$) $M_{f_i}$ is a free $R_{f_i}$-modules of rank $1$ for all $i$.
    (Remark: $\beta$) implies that $M$ is finitely generated over $R$)

  4. The modules $M$ in 3) can also be characterized as the finitely generated modules over $R$ such that equivalently:
    i) $M$ is projective of rank $1$
    ii) The modules $M_\frak m$ are free of rank $1$ over $R_\frak m$ for all maximal ideals $\mathfrak m\subset R$
    iii) The canonical $R$-linear map $M\otimes_RM^*\to R:m\otimes \phi\mapsto\phi(m)$ is bijective
    Note that these are pleasant algebraic characterizations of the modules generating $\operatorname {Pic(R)}$, but the better, conceptual definition is that given in 2) and 3).
    A final caveat:
    The implication ii)$\implies$ i) above is false if one does not assume that $M$ is finitely generated over $R$.
    A counterexample is to be found in Bourbaki Algèbre commutative, Chapitre II, Exercices §5, 7) b)

Georges Elencwajg
  • 47.5k
  • 14
  • 159
  • 241