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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.

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  • $\begingroup$ 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
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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.)

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  • $\begingroup$ 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
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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.

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

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