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Hi.

Can anyone answer the two following questions:

  1. For $n$-dimensional $X$ Cohen-Macaulay complex space, is it true that the sheaf of top degree homolorphic forms $\Omega^{n}_{X}$ has no torsion?

  2. For $f:X\rightarrow S$ Cohen-Macaulay morphism of reduced complex spaces, is it true that $\Omega^{n}_{X/S}$ has no torsion on $X$?

I think that these two questions have negative answers; but I don't know how to prove it.

In fact, if it is true then the "fundamental class morphism" would be injective!

Thank you.

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The answer to (1) is definitely "no" since there is always a nowhere-dense analytic set on whose complement the space $X$ is Cohen-Macaulay. One gets similarly many counterexamples to (2) beginning with any flat analytic map $f$. –  BCnrd Jul 26 '10 at 17:05
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Dear Mohamed: please try to not post duplicate questions. As it stands, one of yours got -1 and the other +1 (who knows what's up with that). –  BCnrd Jul 26 '10 at 17:07
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@BCnrd: I don't understand the relevance of the existence of an open dense Cohen-Macaulay subset. In any case it is easy to give an example: Consider the union of the coordinate axes in the plane, $xy=0$. Then we have the only relation $xdy+ydx=0$ so that that $xdy$ is a non-zero torsion element (killed by $x$ and $y$). –  Torsten Ekedahl Jul 26 '10 at 17:46
    
@Torsten: I was just pointing out that it should be ubiquitous that the requested condition fails, such as any situation where the sheaf is not torsion-free over the complement of a nowhere-dense analytic set (I was thinking of $X$ which is not generically reduced). –  BCnrd Jul 26 '10 at 18:53
    
@BCnrd: Got you. A non-reduced counterexample is of course not cheating yet it somehow feels like it.... –  Torsten Ekedahl Jul 26 '10 at 18:57

1 Answer 1

Thanks to Brian and Ekedahl.

Yes, in any case the answer is "NO"; After asking the question, I made some easy computation with the Whitney umbrella $\lbrace{(x,y,z)\in {\Bbb C}^{3}: x^{2}-zy^{2}=0}\rbrace$ which convinces me that 1) is not true.

For the second question, see Kunz-Waldi, Contem.Math 79 \$.5; the kernel of the relative fundamental class is almost never empty...

P.S: Excuse me, I don't know how to add some comment to the question.

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If one takes a reduced irreducible curve (hence Cohen-Macaulay), most of the time (and easy to check), the module of 1-forms will have torsion. In fact, a well known conjecture of Berger posits that if this module is torsion free then the curve is smooth. The conjecture is open to the best of my knowledge. –  Mohan Jul 26 '10 at 19:26
    
Hi Mohan, nice to see you here ! –  Chandan Singh Dalawat Jul 27 '10 at 3:03
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kaddar, you should accept this answer by clicking the checkmark to the left, if you're satisfied by this. That way the question will be marked as answered in the system. –  j.c. Sep 21 '10 at 3:19

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