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is it true that any numerical polynomial , i.e. $f(t)\in \mathbb Q[t], f(n)\in\mathbb Z\ \forall n\in\mathbb Z\ $ can be presented as Hilbert polynomial of some variety?

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2 Answers 2

up vote 13 down vote accepted

This paper (Proposition 1.3) gives a necessary and sufficient condition for a polynomial to be the Hilbert polynomial of a projective scheme.

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Steven Landsburg's answer completely solves the problem for schemes. I notice that you asked for a variety, which is more restrictive. For example, $2t+2$ is the Hilbert polynomial of a pair of skew lines, but it is not the Hilbert polynomial of any irreducible variety. Such a variety would have to be a degree $2$ curve, by looking at the leading coefficient, and this curve would have to have genus $-1$ by Riemman-Roch, a contradiction. Similarly, $2 \binom{t+2}{2} -1 = t^2+3t+1$ is the Hilbert polynomial of two $\mathbb{P}^2$'s which meet transversely as a point and I believe that it is not the Hilbert polynomial of any irreducible variety.

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thank you very much! so, there is no answer for variety... –  Nikita Kalinin Mar 26 '12 at 19:58

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