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Let R be a normal noetherian domain.

What is the difference between a finitely generated reflexive module and a finitely generated projective module?

Can anybody recommend any references or make any suggestions about this?


Finitely generated projective modules can be identified with idempotents matrix...

Finitely generated projective modules correspond with vector bundles over topological space...

Are there similar results about reflexives modules?

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6 Answers

up vote 6 down vote accepted

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... :

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 not complete.

A nice related example to keep in mind in this context is that a free abelian group of infinite countable rank is reflexive.

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It´s a very interesting example! All projective modules are flats, and then they are divisorial. This is a near concept to reflexive modules: for fin. gen. modules (more generally for lattices), divisorial modules are reflexive modules. I never saw before an example that showed that the finite condition is essential. Thanks! –  Francisco Perdomo Dec 1 '09 at 22:06
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Well, the answer is well known of course. For a finitely generated module over a commutative normal Noetherian domain TFAE

  1. M is reflexive
  2. M is torsion-free and equals the intersection of its localizations at the codimension 1 primes
  3. M satisfies Serre's condition S2 and its support is Spec R.
  4. M is the dual of a f.g. module N

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 $\ge2$.

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.

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.

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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.

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That is the case of regular local rings with Krull dimension less or equal 2; for regular domains (no necessary local) with Krull dimension less or equal 2 reflexive modules are locally free, and then projective. Thank you for your nice answers –  Francisco Perdomo Dec 2 '09 at 15:07
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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.

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).

The moral is that reflexivity is a (very) much weaker condition than projectivity.

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The reflexivity is a weaker condition, but in integrally closed noetherian domains that are enough to work. You spoke about every second syzygy is reflexive. I know that result for Krull domains (ker of homomorphism of divisorial are divisorial), but I think you speak about every type of ring. Isn´t it? Please, Can you give me references? –  Francisco Perdomo Dec 1 '09 at 23:14
    
I don't know what "in integrally closed noetherian domains that are enough to work" means. Even for Cohen-Macaulay (or Gorenstein) isolated singularities, reflexivity is a much much weaker condition. For the other part, it's certainly true if $R$ is locally Gorenstein at the minimal primes, and I dimly remember it's always true. See Auslander-Bridger for the generically-Gorenstein case. –  Graham Leuschke Dec 1 '09 at 23:31
    
When I say "in integrally closed noetherian domains that are enough to work" It means that the work you can do in general rings with fin. gen. projective go on in Krull domains with divisorial lattices. Thanks for references –  Francisco Perdomo Dec 2 '09 at 15:06
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According to Lam's "Lectures on Modules and Rings" every f.g. projective module is reflexive. See page 55, exercise 7.

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Yes! All f. g. projective modules are reflexives. But it isn´t a equivalence: for example in R=k[x^2,xy,y^2] the module M=(x^2,xy) is reflexive but it isn´t projective. I´m looking for extras –  Francisco Perdomo Dec 1 '09 at 21:54
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I asked a very similar question a few months ago and got some very good answers.

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