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Let $X$ be a projective variety over $\mathbb{C}$. If $L$ is an ample line bundle, then $h_L$ denotes the Hilbert polynomial.

Is it true that, if $L$ and $L'$ are ample line bundles which are equal in the Neron-Severi group, then $h_L = h_{L'}$?

Does this imply, together with finite generation of Neron-Severi groups, that the set of polynomials $\{h_L \ | \ L $ ample line bundle on $X \}$ is finite?

Can anyone recommend a text (book or article) in which these things are explained (to some extent)?

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    $\begingroup$ As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 13, 2018 at 15:29
  • $\begingroup$ @R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)? $\endgroup$
    – Paolo Avi
    Commented Oct 13, 2018 at 15:54
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    $\begingroup$ Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 13, 2018 at 15:55

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Yes for the first question, by Riemann--Roch.

No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).

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    $\begingroup$ You don't really need Riemann-Roch, only that $\chi$ is constant in an algebraic family. $\endgroup$
    – abx
    Commented Oct 13, 2018 at 15:28
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    $\begingroup$ Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety. $\endgroup$
    – Paolo Avi
    Commented Oct 13, 2018 at 15:53
  • $\begingroup$ @abx: Of course. $\endgroup$
    – Sasha
    Commented Oct 13, 2018 at 15:59
  • $\begingroup$ @Paolo Avi: Right, I corrected this. $\endgroup$
    – Sasha
    Commented Oct 13, 2018 at 15:59

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