Let $R$ be a commutative noetherian ring. I know that an $R$-module is invertible iff it is finitely generated and locally free of rank one. I presume then that there are examples of non-finitely generated, rank one projective modules which are not invertible. Can someone help me out by providing a specific example? Additionally, is there any extension of the notion of the Picard group that includes non-finitely generated rank one projectives?
[Math] An example of a rank one projective R-Module that is not invertible
ac.commutative-algebraag.algebraic-geometryexamples
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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.)
It is impossible to produce an example of a finitely generated flat $R$-module that is not projective when $R$ is an integral domain. See: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," here, Appendice, Lemme 5, p. 249. Also see Bourbaki Algèbre Chapitre X (Algèbre Homologique, "AH") X.169 Exercise Sect. 1, No. 13. I also sketch an alternate proof that there are no such examples for $R$ an integral domain below.
Observe that, for finitely generated $R$-modules $M$, being locally free in the weaker sense is equivalent to being flat [Bourbaki, AC II.3.4 Pr. 15, combined with AH X.169 Exercise Sect. 1, No. 14(c).]. ($R$ doesn't have to be noetherian for this, though many books seem to assume it.)
There's a concrete way to interpret projectivity for finitely generated flat modules. We begin by translating Bourbaki's criterion into the language of invariant factors. For any finitely generated flat $R$-module $M$ and any nonnegative integer $n$, the $n$-th invariant factor $I_n(M)$ is the annihilator of the $n$-th exterior power of $M$.
Lemma. (Bourbaki's criterion) A finitely generated flat $R$-module $M$ is projective if and only if, for any nonnegative integer $n$, the set $V(I_n(M))$ is open in $\mathrm{Spec}(R)$.
This openness translates to finite generation.
Proposition. If $M$ is a finitely generated flat $R$-module, then $M$ is projective iff its invariant factors are finitely generated.
Corollary. The following conditions are equivalent for a ring $R$: (1) Every flat cyclic $R$-module is projective. (2) Every finitely generated flat $R$-module is projective.
Corollary. Over an integral domain $R$, every finitely generated flat $R$-module is projective.
Corollary. A flat ideal $I$ of $R$ is projective iff its annihilator is finitely generated.
Example. Let me try to give an example of a principal ideal of a ring $R$ that is locally free in the weak sense but not projective. Of course my point is not the nature of this counterexample itself, but rather the way in which one uses the criteria above to produce it.
Let $S:=\bigoplus_{n=1}^{\infty}\mathbf{F}_2$, and let $R=\mathbf{Z}[S]$. (The elements of $R$ are thus expressions $\ell+s$, where $\ell\in\mathbf{Z}$ and $s=(s_1,s_2,\dots)$ of elements of $\mathbf{F}_2$ that eventually stabilize at $0$.) Consider the ideal $I=(2+0)$.
I first claim that for any prime ideal $\mathfrak{p}\in\mathrm{Spec}(R)$, the $R_{\mathfrak{p}}$-module $I_{\mathfrak{p}}$ is free of rank $0$ or $1$. There are three cases: (1) If $x\notin\mathfrak{p}$, then $I_{\mathfrak{p}}=R_{\mathfrak{p}}$. (2) If $x\in\mathfrak{p}$ and $\mathfrak{p}$ does not contain $S$, then $I_{\mathfrak{p}}=0$. (3) Finally, if both $x\in\mathfrak{p}$ and $S\subset\mathfrak{p}$, then $I_{\mathfrak{p}}$ is a principal ideal of $R_{\mathfrak{p}}$ with trivial annihilator.
It remains to show that $I$ is not projective as an $R$-module. But its annihilator is $S$, which is not finitely generated over $R$.
[This answer was reorganized on the recommendation of Pete Clark.]
Best Answer
A rank one projective module $M$ over a commutative noetherian ring is necessarily finitely generated. Indeed assume otherwise. Let $a_i$ be the minimal idempotents of $R$ (there are finitely many since $R$ is commutative noetherian), so that $R=\bigoplus_i a_iR$. Then $a_iM$ is projective over the connected (=indecomposable) noetherian ring $a_iR$. Bass (Illinois Math J, 1963) showed that a infinitely generated projective module over a connected noetherian commutative ring is necessarily free. So here $a_iM$ is in addition free of rank one over $a_iR$. If follows that $M=\bigoplus_i a_iM$ is finitely generated (actually: free of rank one), contradiction.
Remarks: 1) I assume any reasonable definition of "[locally free of] rank one", for instance the (weak) assumption that $M\otimes R_P$ is free of rank one for every prime $P$ in $R$.
2) Tom Goodville gave a nice example of a module, locally free of rank one, that is not projective. [Recall that a f.g. locally free module has to be projective.] This is the $\mathbf{Z}$-submodule of $\mathbf{Q}$ consisting of fractions $a/b$ with squarefree $b$. It is locally free in the above weak sense (localization at primes), but not in the stronger sense, where the strong sense of locally free would be: free over every open subset in some open covering of the spectrum (is this equivalent to being projective? I'm not sure in the infinitely generated case).