What is the right definition of the Picard group of a commutative ring?

Although this question has already been answered, I would like to point out that the assertion also follows from a little bit of category theory (which does not seem to be discussed in the Bourbaki reference).

Claim: Let $R$ be any commutative ring, and let $M$ be an $R$-module which is invertible for the tensor product. Then $M$ is finitely generated and projective.

Proof: The functor from $R$-modules to $R$-modules given by tensoring with $M$ is an auto-equivalence. Since being projective is a property completely internal to the categorical structure, it is preserved by auto-equivalences. In particular, since $R$ is projective, so is $R \otimes_R M \simeq M$.

Similarly, one sees that $M$ is finitely presented, because the finitely presented $R$-modules are exactly the compact objects of the category.

(More generally: Given any symmetric monoidal category, if the unit object satisfies some categorical property, then so does any invertible object. This is useful in other contexts. Example: any invertible object in the stable homotopy category has to be a finite spectrum, because finite spectra are the compact objects; from here it's not too hard to conclude that the invertible objects in spectra are the spheres.)

For what it's worth, I think in Bourbaki's Algèbre Commutative, this is chapter II, section 5.4 (or so), but I don't have a copy in front of me. (Pete confirms that it's II.5.4, Theorem 3.)

1) About the second definition:
$\alpha$) It is not true 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.
$\beta$) It is also not true that b) is equivalent to c):
Take $R=\mathbb Z$ and $\mathbb Z\subsetneq M=\bigcup \frac {1}{p_1\cdots p_i}\mathbb Z\subsetneq \mathbb Q$ where $p_i$ is the $i$-th prime.
Then for all prime $\mathfrak p\subset \mathbb Z$ the $\mathbb Z_\mathfrak p$-module $M_\frak p$ is free of rank $1$, since $$M_{(0)}=\mathbb Z_{(0)}(=\mathbb Q) \quad \operatorname {and}\quad M_{(p_i)}=\frac {1}{p_i}\mathbb Z_{(p_i)}$$ However the $\mathbb Z$-module $M$ is neither finitely generated nor projective (over $\mathbb Z$, projective=free)

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 $f_1,\cdots,f_n\in R$ with:
$\alpha$) $\sum Rf_i=R$
$\beta$) $M_{f_i}$ is a free $R_{f_i}$-modules of rank $1$ for all $i$.

These modules are called locally free of rank one.
Remark: $\beta$) implies that locally free modules of rank one are finitely generated over $R$, since "finitely generated" is a local condition

4) The locally free modules of rank one defined in 3) can also be characterized as the modules $M$ over $R$ such that equivalently:
i) The module $M$ is finitely generated, projective and for all primes $\mathfrak p\subset R$ the (necessarily!) free $R_\mathfrak p$- module $M_\mathfrak p$ has rank $1$
ii) The module $M$ is finitely generated and 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, but the conceptual definition is that given in 2) and 3).

Edit: WARNING
The confusion is made worse by Bourbaki's unfortunate decision to define a projective module of rank $1$ as a finitely generated module $P$ for which $M_\mathfrak p$ is free of rank $1$ over $R_\mathfrak p$ for all primes $\mathfrak p\subset R$.
As my example 1) $\beta$) shows, omitting to require that $P$ be finitely generated [as is done in Pete's condition (2) b)] means accepting modules which aren't even projective, and which don't satisfy (2) a) nor (2) c) of the question.