This is a rather technical question with no particular importance in any case of actual interest to me, but I've been writing up some notes on commutative algebra and flailing on this point for some time now, so I might as well ask here and get it cleared up.


I would like to define the Picard group of an arbitrary (i.e., not necessarily Noetherian) commutative ring $R$. Here are two possible definitions:


(1) It is the group of isomorphism classes of rank one projective $R$-modules under the 
tensor product.


(2) It is the group of isomorphism classes of invertible $R$-modules under the tensor product, where invertible means any of the following equivalent things [Eisenbud, Thm. 11.6]:


a) The canonical map $T: M \otimes\_R \operatorname{Hom}\_R(M,R) \rightarrow R$ is an isomorphism.  

b) $M$ is locally free of rank $1$ [**edit**: in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M\_{\mathfrak{p}} \cong R\_{\mathfrak{p}}$.]  

c) $M$ is isomorphic as a module to an invertible fractional ideal.


What's the difference between (1) and (2)? In general, (1) is stronger than (2), because projective modules are locally free, whereas a finitely generated locally free module is projective iff it is finitely presented. (When $R$ is Noetherian, finitely generated and finitely presented are equivalent, so there is no problem in this case. This makes the entire discussion somewhat academic.)


So, *a priori*, if over a non-Noetherian ring one used (1), one would get a Picard group that was "too small". Does anyone know an actual example where the groups formed in this way are not isomorphic? (That's stronger than one being a proper subgroup of the other, I know.) 


Why is definition (2) preferred over definition (1)? 



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


