Timeline for When are dual modules free?
Current License: CC BY-SA 2.5
12 events
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| Jul 17, 2016 at 19:01 | comment | added | Jason Starr | As others have noted, the answer contains a mistake. A correct statement is in the following MO post: mathoverflow.net/questions/244514/… | |
| May 23, 2014 at 20:22 | comment | added | Emerton | @GrahamLeuschke: Dear Graham, I think that the first half is okay provided that $M$ is assumed a priori to have finite projective dimension. (One can proceed by induction on the projective dimension.) Regards, | |
| Sep 15, 2012 at 20:08 | comment | added | Keenan Kidwell | Dear @Greg Muller, Do you have a reference for your Ext vanishing criteria for reflexivity? I only need it for the ring $R=\mathbf{Z}_p[[T]]$ if it helps (regular local of dimension $2$). In the book Cohomology of Number Fields by Neukirch, Schmidt, and Wingberg, they characterize the kernel and cokernel of the map from $M$ to its double dual (for $M$ f.g. over $R$) as $\Ext^1_R(DM,R)$ and $\Ext_R^2(DM,R)$, where $D$ is a certain functor on the ``homotopy category" of $R$-modules, but I can't find anything connecting this with $\Ext^i_R(M,R)$, $i=1,2$. | |
| Jan 8, 2010 at 20:58 | comment | added | Mariano Suárez-Álvarez | Notice that your claim about characterizing projective modules with the vanishing of Ext to the ring is actually independendent of ZFC, as its special case over the integers is the Whitehead problem. | |
| Jan 8, 2010 at 20:52 | history | edited | Greg Muller | CC BY-SA 2.5 |
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| Nov 10, 2009 at 14:57 | comment | added | Graham Leuschke | I'm not sure about this part: a module is projective if Ext^i(M,D)=0 for all i>0, and it's reflexive if Ext^i(M,D)=0 for i=1, 2. (I'm assuming D=R, the domain, yes?) For the second half, I think you mean something like Ext^i(D(M),D)=0, where now D(-) is Auslander's transpose. The first half is problematic too: there are lots of modules (e.g., so-called "totally reflexive" modules, which have no extensions against the ring but are not necessarily projective. | |
| Nov 10, 2009 at 14:54 | comment | added | Kevin Buzzard | They're not editable but they are deletable. | |
| Nov 8, 2009 at 21:46 | vote | accept | David E Speyer | ||
| Nov 8, 2009 at 15:19 | comment | added | Greg Muller | Ha, yes, that is a good point. Unfortunately, comments aren't editable, so that lapse can be documented for all eternity. | |
| Nov 8, 2009 at 13:27 | comment | added | David E Speyer | That's not a UFD! It's locally factorial, but not globally. A variety with nontrivial Pic can't be a UFD. But the rest of your answer looks very useful, thanks! | |
| Nov 8, 2009 at 5:46 | comment | added | Greg Muller | I should mention, this should provide examples of UFDs where there are non-free duals. Take the ring C[x,y]/y^2-x(x-1)(x-2), so the ring of functions on a smooth elliptic curve minus a point at infinity. Since it is a smooth curve, its global dimension is 1, and so reflexive=projective. However, this ring has non-free projectives. To see this, consider the sheaf of ideals which vanish at some point on the curve. Its a line bundle and so its projective. However, no matter how you complete it to the missing point, you can't get a principle divisor, and so it can't be free. | |
| Nov 8, 2009 at 5:23 | history | answered | Greg Muller | CC BY-SA 2.5 |