9
$\begingroup$

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?

Improve this question
$\endgroup$

6 Answers 6

Reset to default
17
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Another reference is SP, Tag 0AVB. $\endgroup$
    – Minseon Shin
    Commented Oct 31, 2018 at 2:39
10
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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! $\endgroup$
    – Hideyuki Kabayakawa
    Commented Dec 1, 2009 at 22:06
  • $\begingroup$ @Francisco Pedromo I believe Enochs has done a fair amount of work with reflexive modules. If you have access to MathSciNet you could check out his publications. You might find something of interest. $\endgroup$
    – Chris Leary
    Commented Jun 15, 2014 at 3:19
  • 1
    $\begingroup$ @Francisco Pedromo You can look up Enochs' page at the University of Kentucky website, math.as.uky.edu. I believe he has a CV and list of publications there. $\endgroup$
    – Chris Leary
    Commented Jun 15, 2014 at 3:23
8
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ 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 $\endgroup$
    – Hideyuki Kabayakawa
    Commented Dec 2, 2009 at 15:07
  • $\begingroup$ Hi, I'm no algebraist and I look for a reference that I can cite in a paper about what you prove above. Or is it so trivial that I can simply assert it? $\endgroup$
    – Loïc Teyssier
    Commented Mar 28, 2023 at 13:17
  • $\begingroup$ @LoïcTeyssier, I don't know of a direct reference (this or something similar is probably done in Auslander and Bridger's paper called Stable module theory, but that is an atomic bomb to kill this particular mosquito!) Commutative people will probably know more direct references $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 28, 2023 at 14:32
6
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Hideyuki Kabayakawa
    Commented Dec 1, 2009 at 23:14
  • $\begingroup$ 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. $\endgroup$
    – Graham Leuschke
    Commented Dec 1, 2009 at 23:31
  • $\begingroup$ 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 $\endgroup$
    – Hideyuki Kabayakawa
    Commented Dec 2, 2009 at 15:06
3
$\begingroup$

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

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – Hideyuki Kabayakawa
    Commented Dec 1, 2009 at 21:54
3
$\begingroup$

I asked a very similar question a few months ago and got some very good answers.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .