In my research, I have to deal with the following cohomology:

Let $G$ be the semigroup of Hilbert polynomials, which as a subset of $\mathbb{Q}[x]$, consists of the Hilbert polynomials of all subschemes of $\mathbb{P}^N$ for various $N$, and the semigroup structure is given by product of polynomials. A theorem of Macaulay characterizes this semi-group completely. Consider $\mathbb{Z}/(2)$ as a trivial $G$-module. I have to calculate $H^2(G,\mathbb{Z}/(2))$.

I try to apply the universal coefficient theorem to do a reduction, but due to a lack of knowledge about $G$, I can't get further information.

Is it possible to calculate this group explicitly?

  • $\begingroup$ What does Macaulay's theorem say? $\endgroup$ Oct 8, 2017 at 11:47
  • $\begingroup$ Do you count the zero polynomial? $\endgroup$ Oct 8, 2017 at 11:49
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    $\begingroup$ If you don't include zero as a Hilbert polynomial you have a cancellative commutative semigroup and so its cohomology on a module with trivial actin is the same as the cohomology of its Grothendieck group (or group completion of group of fractions). $\endgroup$ Oct 8, 2017 at 13:00
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    $\begingroup$ @BenjaminSteinberg You can find Macaulay's theorem here comp.uark.edu/~ashleykw/SCATCooper.pdf Page 6. I do not assume 0 is there, so the problem is more tricky $\endgroup$ Oct 8, 2017 at 15:43
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    $\begingroup$ Your monoid consists of certain elements with consists ten 1 if I understood correctly. The constant term 1 polynomials form a free commutative monoid on the irreducible ones. And so this bigger monoid has a free abelian group of infinite rank as Grothendieck group. So your monoid also has a free abelian Grothendieck group. $\endgroup$ Oct 8, 2017 at 16:52


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