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For a scheme $X$ of finite type over $k$, and a coherent sheaf $\mathcal{F}$ on $X$, the Hilbert polynomial of $\mathcal{F}$ is defined by $\Phi(n)=\chi(\mathcal{F}(n))$.

And for a scheme $X$ over $S$ (with some suitable conditions), I was told that we can define Hilbert polynomials by defining on each fibers. That is, for each point $s\in S$, we consdier the Hibert polynomial $\Phi_s$ of the restriction $\mathcal{F}_s$ of the sheaf on the fiber $X_s$.

But I don't understand why can we define the Hilbert polynomials of $\mathcal{F}_s$?

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    $\begingroup$ I'm not sure what your question is. When discussing Hilbert polynomials it is usually understood that $X$ is projective with very ample line bundle $\mathcal O(1)$. In the relative setting, you would need a projective $X/S$ with a very ample line bundle $\mathcal O(1)$ on $X$ relative to $S$. This restricts to a very ample line bundle on each fiber, and you define the Hilbert polynomial in terms of the restricted line bundle. $\endgroup$
    – Jack Huizenga
    Commented Aug 7, 2013 at 9:30
  • $\begingroup$ Regarding your flag, I am somewhat hesitant to delete this question, because the answer is rather informative. You are welcome to edit your question to make it clear exactly how you were confused. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 8, 2013 at 8:18

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Seconding Jack, it is not clear what question you are asking. However, my guess is that you want to know why, or rather when, the Hilbert polynomial is well-defined independently of the geometric point $s$ of $S$. With the hypotheses that Jack listed, if you also assume that $\mathcal{F}$ is an $S$-flat, locally finitely presented, quasi-coherent $\mathcal{O}_X$-module, then the Hilbert polynomial functions, $s\mapsto \chi(X_s,\mathcal{F}_s(n))$, is a locally constant function from $S$ to the Abelian group of numerical $\mathbb{Q}$-modules. One reference for this is the corollary on p. 50 of Mumford's "Lectures on curves on an algebraic surface".

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