Question

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)?

Answer 1

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.)

Answer 2

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.)

Answer 3

There is no difference. If $M$ is locally free of finite rank, then $M$ is of finite presentation (and projective).

Take a partition of unity $f_1,...,f_n$, such that $M_{f_i}$ is free over $R_{f_i}$. Since $R \to R_{f_1} \oplus ... \oplus R_{f_n}$ is faithfully flat, it suffices to show the properties for $M_{f_1} \oplus ... \oplus M_{f_n}$, which is very easy.

Definition (2) is prefered because it reveals the geometric content: classification of line bundles.