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In Example of a $ \mathbb{Q} $-factorial, CM normal, projective, Mori dream space $ Z $ such that $ \operatorname{Cox}(Z) $ is integral and not CM I asked for an example of a Cohen Macaulay, normal, projective, Mori dream space $ Z $ such that $ \operatorname{Cox}(Z) $ is integral and not Cohen Macaulay.

Jason Starr pointed out that if $ Z $ is a very general Abelian variety of dimension $ g>1 $ should have a cyclic class group with ample generator $ \mathcal{L} $ so that $ \operatorname{Cox}(Z) $ should be graded and isomorphic to the section ring $ \oplus_{n \in \mathbb{N}} H^{0}(Z, \mathcal{L}^{\otimes n}) $. Then the result that such Abelian varieties are not Arithmetically Cohen Macaulay should be an example. Does someone have a reference that a) very general Abelian varieties of dimension $ g>1 $ have a cyclic class group with ample generator and b) that such varieties exist.

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    $\begingroup$ I think you may be conflating the expressions "very general" and "of general type", which have little to do with each other. Only the first was meant. $\endgroup$
    – Will Sawin
    Commented Dec 16, 2023 at 13:19
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    $\begingroup$ Also, what you call the "class group" is the Néron-Severi group. $\endgroup$
    – abx
    Commented Dec 16, 2023 at 15:27
  • $\begingroup$ @WillSawin you are right. Looking back at that post I realized that I was conflating the notions. I will change it. $\endgroup$
    – Schemer1
    Commented Dec 16, 2023 at 15:50
  • $\begingroup$ @WillSawin the change has been made. $\endgroup$
    – Schemer1
    Commented Dec 16, 2023 at 15:52
  • $\begingroup$ @abx No, I mean the class group. The Neron-Severi group is the group $ \operatorname{Div}(Z) $ modulo numerical equivalence. The class group is the group $ \operatorname{Div}(Z)/\operatorname{PDiv}(Z) $. The Cox ring of $ Z $ is graded by the class group and not the Neron-Severi group unless there are some circumstances present under which the two are isomorphic. I may be misunderstanding Jason Starr's argument if his argument necessitates that it be the Neron-Severi group. $\endgroup$
    – Schemer1
    Commented Dec 16, 2023 at 15:56

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