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)?

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    $\begingroup$ 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. $\endgroup$ Commented Feb 2, 2010 at 1:45
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    $\begingroup$ I can't help but wonder what makes a person non-commutative. Were you born like that? $\endgroup$ Commented Feb 2, 2010 at 1:58
  • $\begingroup$ Now (2) is a group. But is (1) a group? $\endgroup$
    – VA.
    Commented Feb 2, 2010 at 2:13
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    $\begingroup$ @CamMcLeman, re, maybe he's born with it … maybe it's non-Abelian? $\endgroup$
    – LSpice
    Commented Sep 30, 2021 at 19:44
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    $\begingroup$ @LSpice It took 30 years for all the ingredients of that joke to mature from inception to delivery, but I think we can all agree it was worth the wait. $\endgroup$ Commented Oct 3, 2021 at 21:20

4 Answers 4


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.)

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    $\begingroup$ 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." $\endgroup$ Commented Feb 2, 2010 at 11:54
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    $\begingroup$ 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. $\endgroup$ Commented Feb 2, 2010 at 12:08
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    $\begingroup$ I doubled my there there. Oops. $\endgroup$ Commented Feb 2, 2010 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.)


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).

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.

  • $\begingroup$ So if we call (1), (2a), (2b), (2c) the four conditions in Peter L. Clark's original question, we have (1⇔2a) by the reference to Bourbaki cited in Clark Barwick's answer, but (2c) is not equivalent to (1/2a) nor to (2b) by the counterexamples you cite. Did I get this right? But are (1/2a) and (2b) equivalent or not? Furthermore, since you propose yet another definition of the Picard group, we now have potentially FOUR different Picard groups (or at least, Picard sets — maybe they aren't all groups) and I am tempted to ask about all possible maps between them. This is getting really messy. $\endgroup$
    – Gro-Tsen
    Commented Mar 27, 2018 at 12:30
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    $\begingroup$ @Gro-Tsen. I have added an edit which points to one source of the messiness. I think that nobody doubts that 2) and 3) in my answer are the correct definitions. The trouble begins with alternative definitions which are thought to be equivalent but aren't. In Pete's question (1) and (2)a) are equivalent (and equivalent to what I have described as the correct definition) iff , as in Bourbaki, "rank one" includes the condition that $M$ be finitely generated. As to (2) b), it is not equivalent to (1) nor 2)a) nor 2)c) because it doesn't even imply projective. (to be continued) $\endgroup$ Commented Mar 27, 2018 at 19:27
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    $\begingroup$ Finally (2)c) only describes some of the modules in (2)a). So (2)c) is not equivalent to (2)a) in general, but it is equivalent under some mild conditions, for example if $R$ is a domain. $\endgroup$ Commented Mar 27, 2018 at 19:27

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.

  • $\begingroup$ 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. $\endgroup$ Commented Feb 2, 2010 at 3:31
  • $\begingroup$ hm? I don't need any constancy. $\endgroup$ Commented Feb 2, 2010 at 8:25
  • $\begingroup$ 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). $\endgroup$ Commented Feb 2, 2010 at 8:40
  • $\begingroup$ 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). $\endgroup$ Commented Feb 2, 2010 at 11:19
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    $\begingroup$ 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). $\endgroup$ Commented Feb 2, 2010 at 12:23

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