# Picard group vs class group

The question.

Let $$R$$ be a commutative ring. Let $$M$$ be an $$R$$-module with the property that there exists an $$R$$-module $$N$$ such that $$M\otimes_R N\cong R$$. Does there always exist an ideal $$I$$ of $$R$$ such that $$M$$ is isomorphic to $$I$$ as an $$R$$-module? (I suspect not in this generality)

The background.

Let $$R$$ be a commutative ring. Here are two groups that one could associate to $$R$$.

1. The "class group".

The first group is "inspired by number theory". One takes the ideals of $$R$$ and observes that they have a natural multiplication defined on them. One defines two ideals $$I$$ and $$J$$ to be equivalent if there exist nonzerodivisors $$s$$ and $$t$$ such that $$sI=tJ$$. This relation plays well with the multiplication, giving us a multiplication on the equivalence classes (unless I screwed up; my reference is "back of an envelope calculation"). This makes the equivalence classes into a commutative monoid, and one could define the class group of $$R$$ to be units of this monoid, i.e. elements with an inverse.

Note: one could instead use fractional ideals. The theory of fractional ideals is often set up only for integral domains, and if I did screw up above then maybe I should have restricted to integral domains. A fractional ideal is defined to be an integral ideal with a denominator so I don't think this changes the group defined here.

1. The Picard group.

The second group is "inspired by geometry" -- it's the Picard group of $$\operatorname{Spec}(R)$$. More concretely, take the collection (it's not a set) of isomorphism classes of $$R$$-modules $$M$$. This has a multiplication coming from tensor product, and satisfies the axioms of a monoid except that it's not a set. The units of this monoid however are a set, because another back of an envelope calculation seems to indicate that if $$M\otimes_R N\cong R$$ and we write $$1=\sum_i m_i\otimes n_i$$, a finite sum, then the $$m_i$$ generate $$M$$ as an $$R$$-module, giving us some control over the size of the units of the monoid -- they're all isomorphic to a quotient of $$R^n$$ so we have regained control in a set-theoretic sense. The units of the monoid are the second group.

The question comes from me trying to convince myself that these groups are not equal in general (for I don't really expect them to be equal in general). If $$R$$ is a Dedekind domain (so $$\operatorname{Spec}(R)$$ is a smooth affine curve) then we have here the classical definition and the fancy definition of the class group of $$R$$, and the answer to the question is "yes". This is because every rank 1 projective $$R$$-module is isomorphic to an ideal of $$R$$; if I recall correctly then more generally every rank $$n+1$$ projective $$R$$-module is isomorphic to $$I\oplus R^n$$ for some ideal $$I$$ (this is true at least for the integers of a number field) which enables you to compute the zero'th algbraic $$K$$-group (Grothendieck group) of $$R$$. But more generally than this I am not sure what is going on.

On the divisors Wikipedia page I read "Every line bundle $$L$$ on $$X$$ on an integral Noetherian scheme is the class of some Cartier divisor" which makes me think that the result might be true for Noetherian integral domains, but I don't see the proof even there (perhaps it's standard). The way it's phrased makes me wonder then whether there are non-Noetherian counterexamples.

• Unless I am missing some subtlety, the integral Noetherian scheme case is Prop 6.15 in Hartshorne's AG. Nov 5, 2020 at 16:53
• @PavelČoupek just integral in fact! And the preceding remark gives a reference to a counterexample in the non-integral case (but it's non-affine and in fact not even quasi-projective). Nov 5, 2020 at 17:06

Here is an attempt to generalize somewhat the above-mentioned proof from Hartshorne.

Claim: Let $$R$$ be a ring whose total ring of fractions $$R_{\mathrm{tot}}=S_{\mathrm{nzd}}^{-1}R$$ is Artinian. Then any invertible $$R$$-module is isomorphic to an invertible ideal.

(The hypothesis holds at least in the following two "natural" cases:

1. $$R$$ is a domain, which corresponds to the case of integral schemes,
2. $$R$$ is a Noetherian ring without embedded components, i.e. $$\mathrm{Ass}\,R$$ is precisely the set of minimal primes, in which cases the spectrum of $$R_{\mathrm{tot}}$$ consists precisely of these minimal primes, hence is $$0$$-dimensional.)

Proof: One proceeds as in the proof from Hartshorne. Given an invertible module $$M$$, this is a locally free module of constant rank $$1$$, and so is the $$R_{\mathrm{tot}}$$-module $$M \otimes_R R_{\mathrm{tot}}$$. As $$R_{\mathrm{tot}}$$ is Artinian, any locally free module of rank $$1$$ is actually a free module of rank $$1$$, and thus we have $$M=M\otimes_R R\hookrightarrow M\otimes_R R_{\mathrm{tot}}\simeq R_{\mathrm{tot}}.$$ This realizes $$M$$ as an $$R$$-submodule $$M'$$ of $$R_{\mathrm{tot}}$$. It is finitely generated (because $$M$$ is), let us call these generators $$a_1/s_1, \dots, a_n/s_n \in R_{\mathrm{tot}}.$$ But then $$s=s_1s_2 \dots s_n$$ is a non-zero divisor of $$R$$, and we have $$sM' \subseteq R$$. Thus, $$M$$ is isomorphic to the invertible ideal $$I:=sM'$$. $$\square$$

(I guess that the assumption can be relaxed a bit more, by assuming that $$R_{\mathrm{tot}}$$ is just a finite direct product of local rings (edit: Actually, even more by simply assuming that $$\mathrm{Pic}(R_{\mathrm{tot}})=1$$). But I don't know any new "natural" cases that this would provide.)

• So indeed it is almost always true. But we still have no counterexample to the general statement? Nov 6, 2020 at 9:01
• @KevinBuzzard For what it's worth, the setup of this question implies that such counterexamples (provably) exist (among Bézout rings); however, they seem quite inexplicit (which is actually the point of the cited question). Nov 6, 2020 at 17:57

I'll assume that $$R$$ is integrally closed in its fraction field. Let $$A$$ be the semilocalization of $$R$$ at all the maximal ideals where $$R$$ is not factorial. (That is, $$A=S^{-1}R$$ where $$S$$ is the complement of the union of all those maximal ideals.) Then $$Pic(R)$$ sits inside $$Cl(R)$$ and is in fact the kernel of the map $$Cl(R)\rightarrow Cl(A)$$.

This must be in a paper of Fossum somewhere, though I don't have the reference at hand.

• Might this lead to a counterexample to the general statement then? Nov 6, 2020 at 9:02