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.

  • $\begingroup$ Unless I am missing some subtlety, the integral Noetherian scheme case is Prop 6.15 in Hartshorne's AG. $\endgroup$ Nov 5 '20 at 16:53
  • $\begingroup$ @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). $\endgroup$ Nov 5 '20 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.)

  • $\begingroup$ So indeed it is almost always true. But we still have no counterexample to the general statement? $\endgroup$ Nov 6 '20 at 9:01
  • $\begingroup$ @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). $\endgroup$ Nov 6 '20 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.

  • $\begingroup$ Might this lead to a counterexample to the general statement then? $\endgroup$ Nov 6 '20 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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