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 $\mathcal{F}\_s$?



Hilbert polynomials on a scheme
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".


