A finitely generated, locally free module over a domain which is not projective? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:14:11Z http://mathoverflow.net/feeds/question/13817 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13817/a-finitely-generated-locally-free-module-over-a-domain-which-is-not-projective A finitely generated, locally free module over a domain which is not projective? Pete L. Clark 2010-02-02T11:52:03Z 2010-02-08T07:31:39Z <p>This is a followup to a previous question</p> <p><a href="http://mathoverflow.net/questions/13768/what-is-the-right-definition-of-the-picard-group-of-a-commutative-ring" rel="nofollow">http://mathoverflow.net/questions/13768/what-is-the-right-definition-of-the-picard-group-of-a-commutative-ring</a></p> <p>where I was worried about the distinction between invertible modules and rank one projective modules over an arbitrary commutative ring. I was worrying too much, because of the following theorem [Bourbaki, Commutative Algebra, Section II.5.2, Theorem 1]:</p> <p>Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. The following are equivalent:<br /> (i) $M$ is projective.<br /> (ii) $M$ is finitely presented and locally free in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}^{r(\mathfrak{p})}$.<br /> (iii) $M$ is locally free in the weaker sense and its rank function $\mathfrak{p} \mapsto r(\mathfrak{p})$ is locally constant on $\operatorname{Spec}(R)$.<br /> (iv) $M$ is locally free in the stronger sense: there exist $f_1,\ldots,f_n \in R$, generating the unit ideal, such that for each $i$, $M_{f_i}$ is a free $R_{f_i}$-module.<br /> (v) For every maximal ideal $\mathfrak{m}$ of $R$, there exists $f \in R \setminus \mathfrak{m}$ such that $M_f$ is a free $R_f$-module. </p> <p>This answers my previous question, because the rank function of an invertible module is identically one.</p> <p>In order to really feel like I understand what's going on here, I would like to see an example of a finitely generated locally free [in the weaker sense of (ii) above] module which is <em>not</em> projective. Thus $R$ must be non-Noetherian. The wikipedia article on <a href="http://en.wikipedia.org/wiki/Projective%5Fmodule" rel="nofollow">projective modules</a> contains some nice information, in particular sketching an example of such a module over a Boolean ring. For a Boolean ring though the localization at every prime ideal is simply $\mathbb{Z}/2\mathbb{Z}$, so it is not too surprising that there are more locally free modules than projectives.</p> <p>I would like to see an example with $R$ an integral domain, if possible. It would be especially nice if you can give a reference to one of the standard texts on commutative algebra which contains such an example or at least a citation of such an example.</p> http://mathoverflow.net/questions/13817/a-finitely-generated-locally-free-module-over-a-domain-which-is-not-projective/14111#14111 Answer by Clark Barwick for A finitely generated, locally free module over a domain which is not projective? Clark Barwick 2010-02-04T04:19:43Z 2010-02-04T14:28:42Z <p>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," <a href="http://www.numdam.org/numdam-bin/item?id=BSMF%5F1958%5F%5F86%5F%5F177%5F0" rel="nofollow">here</a>, 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.</p> <p>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.)</p> <p>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 <em>invariant factor</em> $I_n(M)$ is the annihilator of the $n$-th exterior power of $M$.</p> <p><strong>Lemma.</strong> (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 $\mathrm{Spec}(R)$.</p> <p>This openness translates to finite generation.</p> <p><strong>Proposition.</strong> If $M$ is a finitely generated flat $R$-module, then $M$ is projective iff its invariant factors are finitely generated.</p> <p><strong>Corollary.</strong> 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.</p> <p><strong>Corollary.</strong> Over an integral domain $R$, every finitely generated flat $R$-module is projective.</p> <p><strong>Corollary.</strong> A flat ideal $I$ of $R$ is projective iff its annihilator is finitely generated.</p> <p><strong>Example.</strong> 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.</p> <p>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 <code>$s=(s_1,s_2,\dots)$</code> of elements of <code>$\mathbf{F}_2$</code> that eventually stabilize at $0$.) Consider the ideal $I=(2+0)$.</p> <p>I first claim that for any prime ideal $\mathfrak{p}\in\mathrm{Spec}(R)$, the <code>$R_{\mathfrak{p}}$</code>-module <code>$I_{\mathfrak{p}}$</code> is free of rank $0$ or $1$. There are three cases: (1) If $x\notin\mathfrak{p}$, then <code>$I_{\mathfrak{p}}=R_{\mathfrak{p}}$</code>. (2) If $x\in\mathfrak{p}$ and $\mathfrak{p}$ does not contain $S$, then <code>$I_{\mathfrak{p}}=0$</code>. (3) Finally, if both $x\in\mathfrak{p}$ and $S\subset\mathfrak{p}$, then <code>$I_{\mathfrak{p}}$</code> is a principal ideal of <code>$R_{\mathfrak{p}}$</code> with trivial annihilator.</p> <p>It remains to show that $I$ is not projective as an $R$-module. But its annihilator is $S$, which is not finitely generated over $R$.</p> <p>[This answer was reorganized on the recommendation of Pete Clark.]</p> http://mathoverflow.net/questions/13817/a-finitely-generated-locally-free-module-over-a-domain-which-is-not-projective/14620#14620 Answer by Pete L. Clark for A finitely generated, locally free module over a domain which is not projective? Pete L. Clark 2010-02-08T07:31:39Z 2010-02-08T07:31:39Z <p>After mulling things over for a while, I realize that assuming the results from Bourbaki which I mentioned in the statement of the question, there is a very straightforward answer to my question. (I mean this as no slight to Clark Barwick's excellent answer, which came instantaneously and contains lots of other valuable information. Rather, I mean that had I thought more carefully I would not have needed to ask the question at all.)</p> <p>The key is the following simple result:</p> <p>Lemma: Let $R$ be a[n always commutative] ring, <code>$\mathfrak{p}_1 \subset \mathfrak{p}_2$</code> prime ideals of $R$, and $M$ a finitely generated locally free (in the weaker sense) $R$-module. Then <code>$r(\mathfrak{p}_1) = r(\mathfrak{p}_2)$</code>.</p> <p>The proof is obvious, once you realize that localizing at <code>$\mathfrak{p}_1$</code> is the same as localizing at <code>$\mathfrak{p}_2$</code> and then localizing at (the ideal in <code>$R_{\mathfrak{p}_2}$</code> naturally corresponding to) <code>$\mathfrak{p}_1$</code>. (This is the same argument that allows you to see that is enough to require $M_{\mathfrak{p}}$ to be free at every maximal ideal $\mathfrak{p}$.) Also I am using that the rank of a finitely generated free module over a [commutative!] ring is well-determined, as one sees by tensoring to the quotient field of some maximal ideal.</p> <p>[I had some kind of psychological block coming from a vague memory that the rank function was merely semicontinuous. As far as I can see now, semicontinuity does not come up anywhere in the study of the rank function. I had to see essentially this argument in print -- in Milnor's <em>Introduction to Algebraic K-Theory</em> just this evening -- in order to become unblocked.]</p> <p>Corollary: Let $R$ be a ring with a unique minimal prime ideal (e.g. an integral domain). Then any finitely generated locally free $R$-module has <em>constant</em> rank function and therefore (by the Bourbaki result above) is projective.</p> <p>This also illustrates why the most natural counterexamples come from zero-dimensional rings: the rank function is determined by its behavior on the minimal primes, so you might as well look at the zero-dimensional case. </p>