Severi's theorem of base and Hilbert polynomial Let $X$ be a smooth projective variety over $\mathbb{C}$ satisfying $H^1(\mathcal{O}_X)=0$. Fix $i:X \to \mathbb{P}^n$ a closed immersion and let $\mathcal{O}_X(1)$ be the corresponding very ample line bundle. This means that the Picard group is isomorphic to the Neron-severi group of $X$. By Severi's theorem of base, we know that the rank of the Neron-Severi group is finite. Does this mean that for a fixed Hilbert polynomial $P$, there are only finitely many invertible sheaves (upto isomorphism) on $X$ with Hilbert polynomial $P$ i.e., is $$\#\{\mathcal{L} \in \mbox{Pic}(X)|\chi(\mathcal{L}(m))=P(m) \mbox{ for } m \gg 0\}< \infty?$$
 A: No. 
For instance, there are rational surfaces containing infinitely many $(-1)$-curves. An example in given by the blow-up $X$ of $\mathbb{P}^2$ at nine points that are the base locus of a general pencil of cubics: indeed, $\textrm{Aut}(X)$ contains a copy of $\mathbb{Z}^8$ generated by translations by differences of the nine sections of the elliptic fibration coming from the pencil, and the orbit of a $(-1)$-curve by this subgroup gives infinitely many of them.  
Two such $(-1)$-curves are not linearly equivalent, since any $(-1)$-curve $E$ is isolated into its linear equivalence class, however by the Riemann-Roch theorem  they have the same Hilbert polynomial, namely $$P(m)=\frac{mE(mE-K_X)}{2}+ \chi(\mathcal{O}_X) = -\frac{1}{2}(m^2-m)+1.$$
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ADDED. I was confused by the OP's notation in the first version of the question and I  took as "Hilbert polynomial" the quantity $\chi(\mathcal{L}^m)$. It appears from the comments (and the edit) above that the OP was actually thinking to a more standard Hilbert polynomial of type $\chi(\mathcal{L} \otimes H^m)$, were $H$ is a (fixed) ample class. Then my answer does not provide a counterexample in this case.   
