MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
As a non-commutative person, let me add that one can also consider the invertible $R$-$R$-bimodules, and/or the group of self-equivalences of the category of, say, left $R$-modules. – Mariano Suárez-Alvarez Feb 2 '10 at 1:45
I can't help but wonder what makes a person non-commutative. Were you born like that? – Cam McLeman Feb 2 '10 at 1:58
Now (2) is a group. But is (1) a group? – VA. Feb 2 '10 at 2:13
@VA: I think so. The dual of a finite rank projective module is again finite rank projective. Is there something else to check? – Pete L. Clark Feb 2 '10 at 2:20
(...and you have to check that the tensor product of finite rank projectives is finite rank projective. Unless I have made some silly mistake, this seems to come out immediately from the characterization of such a guy as a direct summand of a finitely generated free module.) – Pete L. Clark Feb 2 '10 at 2:24
up vote 12 down vote accepted

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.)

share|cite|improve this answer
Bourbaki (and Clark) to the rescue. What a surprise. An acquaintance of mine was visiting Paris, and apparently they cite Bourbaki there up to the theorem number in lectures there. And of course, "Soit C un corps commutatif." – Harry Gindi Feb 2 '10 at 11:54
The Bourbaki books (some more than others; CA is still widely read nowadays) are certainly excellent references for basic material, the more so if you have internet access to a savant who can quote them chapter and verse. The problem (for me) comes when I try to read them in the usual linear manner: they cover the trivial and the important in equal detail, and the end product is about five times as long as it should be. – Pete L. Clark Feb 2 '10 at 12:08
I doubled my there there. Oops. – Harry Gindi Feb 2 '10 at 13:03

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.)

share|cite|improve this answer

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.

share|cite|improve this answer
I believe this is not quite correct, depending upon what you mean by "finite rank". It is true though if the rank is finite and constant, which is the situation I asked about. – Pete L. Clark Feb 2 '10 at 3:31
hm? I don't need any constancy. – Martin Brandenburg Feb 2 '10 at 8:25
Bourbaki, Section II.5.2: For an A-module P, TFAE: (a) P is finitely generated projective. (c) P is finitely generated, for each p in Spec(A), P_{p} is free, and the rank function p |-> rank(P_{p}) is locally constant on Spec(A). – Pete L. Clark Feb 2 '10 at 8:40
I suppose it also depends on what you mean by locally free: I meant that the localization at each prime ideal is free. If instead you mean "locally in the Zariski topology" -- your condition about the f_i's above -- then that implies the local constancy of the rank (same theorem in Bourbaki). – Pete L. Clark Feb 2 '10 at 11:19
The "of course" doesn't make sense to me, because that's not what was meant in the standard text that I referenced (Eisenbud). The point is that it's subtle whether the weaker sense of locally implies the stronger sense (yes for Noetherian rings or with local constancy of the rank; no in general). – Pete L. Clark Feb 2 '10 at 12:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.