most recent 30 from http://mathoverflow.net 2013-05-24T19:41:00Z http://mathoverflow.net/feeds/question/7490 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules Francisco Perdomo 2009-12-01T19:44:16Z 2010-01-07T19:14:15Z

<p>Let R be a normal noetherian domain.</p> <p>What is the difference between a finitely generated reflexive module and a finitely generated projective module?</p> <p>Can anybody recommend any references or make any suggestions about this? <hr></p> <p>Finitely generated projective modules can be identified with idempotents matrix...</p> <p>Finitely generated projective modules correspond with vector bundles over topological space...</p> <p>Are there similar results about reflexives modules?</p>

http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7496#7496 Gonçalo Marques 2009-12-01T20:09:44Z 2009-12-01T20:09:44Z

<p>According to Lam's "Lectures on Modules and Rings" every f.g. projective module is reflexive. See page 55, exercise 7. </p>

http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7503#7503 Mariano Suárez-Alvarez 2009-12-01T20:48:41Z 2009-12-01T21:18:14Z

<p>This is not an answer to your question, but I found it a few days ago looking for something else and did strike me as quite curious... :</p> <p>According to E. E. Enochs [A note on reflexive modules. Pacific J. Math. Volume 14, Number 3 (1964), 879-881.] a free module of infinite countable rank over a discrete valuation ring is reflexive iff the ring is <em>not</em> complete. </p> <p>A nice related example to keep in mind in this context is that a free abelian group of infinite countable rank <em>is</em> reflexive.</p>

http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7510#7510 Mariano Suárez-Alvarez 2009-12-01T21:50:01Z 2009-12-01T21:50:01Z

<p>More to the point: if $A$ is a left noetherian ring of global dimension at most $2$, then every finitely generated reflexive left $A$-module is projective. Indeed, if $M$ is a finitely generated module and $$P_1\to P_0\to M\to 0$$ is a projective presentation by finitely generated projectives, applying the functor $(\mathord-)^*=\hom_A(\mathord-,A)$ we get an exact sequence $$0\to M^*\to P_0^*\to P_1^*\to E\to 0,$$ when $E$ is just the cokernel of the map $P_0^*\to P_1^*$. Since $\mathrm{pdim}\,E\leq 2$ and since $P_0^*$ and $P_1^*$ are projective, the kernel of $P_0^*\to P_1^*$, namely $M^*$, must be projective. This shows that the dual of finitely generated module is projective, so a finitely generated reflexive, being isomorphic to the dual of its dual module, is projective.</p>

http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7519#7519 Graham Leuschke 2009-12-01T22:48:32Z 2009-12-01T22:48:32Z

<p>In general, there are many many more reflexive modules than projective modules. As pointed out in another answer, every second syzygy is reflexive, and not necessarily projective (unless your ring happens to be regular of dimension at most $2$). Thus the converse of that answer is, I think, more interesting.</p> <p>In particular, assume that $R$ is (semi)local. Then there are only finitely many indecomposable projectives. If there are finitely many indecomposable reflexives, then $R$ must have dimension at most $2$, and thus finite Cohen-Macaulay type (which is a very rare condition).</p> <p>The moral is that reflexivity is a (very) much weaker condition than projectivity. </p>

http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7588#7588 VA 2009-12-02T16:08:54Z 2009-12-02T16:08:54Z

<p>Well, the answer is well known of course. For a finitely generated module over a commutative normal Noetherian domain TFAE</p> <ol> <li>M is reflexive </li> <li>M is torsion-free and equals the intersection of its localizations at the codimension 1 primes </li> <li>M satisfies Serre's condition S2 and its support is Spec R. </li> <li>M is the dual of a f.g. module N</li> </ol> <p>As you say, a finite projective module is the same as a locally free sheaf on Spec R. Similarly, a finite reflexive module is the same as the push forward of a locally free sheaf from an open subset U of Spec R whose complement has codimension $\ge2$.</p> <p>So for an easy example take a Weil divisor D which is not Cartier, the associated divisorial sheaf (corresponding to an R-module of rank 1) is reflexive, not projective. Your example with a line on a quadratic cone is of this form.</p> <p>This stuff is standard and used all the time in higher-dimensional algebraic geometry around the Minimal Model Program. For an old reference covering some of this, see e.g. Bourbaki, Chap.7 Algebre commutative, VII.</p>

http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/11060#11060 David Speyer 2010-01-07T19:14:15Z 2010-01-07T19:14:15Z

<p>I <a href="http://mathoverflow.net/questions/4590/when-are-dual-modules-free" rel="nofollow">asked</a> a very similar question a few months ago and got some very good answers.</p>