6 clarified reference, fixed typo

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 Homologique Chapitre X (AH) 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 aver over $R$.

[This answer was reorganized on the recommendation of Pete Clark.]

5 reorganized on the recommendation of Pete Clark

Let me try

It is impossible to give produce an example of a principal ideal of a ring finitely generated flat $R$ that is locally free in the weak sense but not projective.

(I tried briefly, but I couldn't think of an example where R$-module when$R$is an integral domain. I suspect there are no such examples: every flat cyclic module over an integral domainis projective, and I think an "invariant factor" argument like the one below shows that it's enough to check on cyclic modules. EDIT: It is indeed impossible to produce such an example; seeSee: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," here, Appendice, Lemme 5, p. 249.) Incidentally249. Also see Bourbaki Algèbre Homologique (AH) X.169 Exercise Sect. 1, No. 13. I also sketch an alternate proof that there are no such examples for$R$an integral domain below. Observe that, for finitely generated$R$-modules$M$, being locally free in the weaker sense is the same as equivalent to being flat [Bourbaki Bourbaki, AC II.3.4 Pr. 15]15, combined with AH X.169 Exercise Sect. 1, No. 14(c).]. ($R$doesn't have to be noetherian for this; , though some many books seem to assume this.it.) There's a concrete way to interpret projectivity for finitely generated flat modules. We begin by translating Bourbaki's criterion into the language of invariant factors. For any finitely generated flat$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$is a finitely generated , and projective flat$R$-module, then$M$is projective iff its invariant factors are finitely generated. (It looks to me as though the converse also true i.e., that Corollary. The following conditions are equivalent for a ring$R$: (1) Every flat cyclic$R$-module is projective. (2) Every finitely generated flat$R$-module is projectiveiff its invariant factors are . Corollary. Over an integral domain$R$, every finitely generated flat$R$-module is projective.Is that right? Is it well-known?) Corollary. A flat ideal$I$of$R$is projective only if (and, I think, if) iff its annihilator is finitely generated. Example. Let me try to give an example of a principal ideal of a ring$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$I=(2+0)$. [This answer was reorganized on the recommendation of Pete Clark.] 4 added 241 characters in body 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. (I tried briefly, but I couldn't think of an example where$R$is an integral domain. I suspect there are no such examples: every flat cyclic module over an integral domain is projective, and I think an "invariant factor" argument like the one below shows that it's enough to check on cyclic modules.modules. EDIT: It is indeed impossible to produce such an example; see: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," here, Appendice, Lemme 5, p. 249.) Incidentally, for finitely generated$R$-modules$M$, being locally free in the weaker sense is the same as being flat [Bourbaki AC II.3.4 Pr. 15]. ($R$doesn't have to be noetherian for this; though some books seem to assume this.) Lemma. If$M$is a finitely generated, and projective$R$-module, then its invariant factors are finitely generated. (It looks to me as though the converse also true — i.e., that a finitely generated flat$R$-module is projective iff its invariant factors are finitely generated. Is that right? Is it well-known?) Corollary. A flat ideal$I$of$R$is projective only if (and, I think, if) its annihilator is finitely generated. Example. 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 aver$R\$.

3 clarified the structure of R
2 fixed notation, spelling
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