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 Neronseveri group of $X$. By Severi's theorem of base, we know that the rank of the NeronSeveri 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 $$\#\{\mathcal{L} \in \mbox{Pic}(X)\chi(\mathcal{L}(m))=P(m) \mbox{ for } m \gg 0\}< \infty?$$

1$\begingroup$ What is $\mathscr{L}(m)$? Do you fix some ample class? Francesco's answer addresses a different question, taking as Hilbert polynomial $\chi (\mathscr{L}^{\otimes m})$. $\endgroup$ – abx Apr 9 '15 at 10:06

$\begingroup$ @abx: Yes, I do fix an ample line bundle. And as you guessed, $\mathcal{L}(m) \cong \mathcal{L} \otimes \mathcal{O}_X(m)$. $\endgroup$ – Ron Apr 9 '15 at 11:56

$\begingroup$ Oh, sorry. I misinterpreted your notation $\mathcal{L}(m)$ as $\mathcal{L}^m$ in additive notation, since you did not say that you were choosing an ample class. So my answer does not really answer your original question. $\endgroup$ – Francesco Polizzi Apr 9 '15 at 14:16

$\begingroup$ @Polizzi: May be I am wrong, it seems to me the Hilbert polynomial of all these $(1)$ curves are still the same (the formula is different from the one you mentioned). So, it seems to be still a counterexample. Am I wrong? $\endgroup$ – Ron Apr 9 '15 at 15:04

$\begingroup$ If you fix an ample divisor $H$, you cannot conclude in principle that $HE$ is independent from the $(1)$curve $E$ that you are choosing. $\endgroup$ – Francesco Polizzi Apr 9 '15 at 15:07
No.
For instance, there are rational surfaces containing infinitely many $(1)$curves. An example in given by the blowup $X$ 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 RiemannRoch theorem they have the same Hilbert polynomial, namely $$P(m)=\frac{mE(mEK_X)}{2}+ \chi(\mathcal{O}_X) = \frac{1}{2}(m^2m)+1.$$
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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.

$\begingroup$ @Polizzi: A stupid question, to contradict finiteness of the above mentioned set, is it sufficient to show that any two such $(1)$curves are not linearly equivalent? $\endgroup$ – Ron Apr 9 '15 at 8:36

1$\begingroup$ Well, if the curves are pairwise linearly inequivalent, the corresponding line bundles are pairwise nonisomorphic, hence you have infinitely many nonisomorphic invertible sheaves on $X$ with the same Hilbert polynomial, right? $\endgroup$ – Francesco Polizzi Apr 9 '15 at 8:39

$\begingroup$ @Polizzi: My confusion was, this is only showing pairwise nonisomorphic i.e., the cardinality of the set is greater than $2$. Now how do I conclude that the cardinality is not finite? $\endgroup$ – Ron Apr 9 '15 at 8:44

1$\begingroup$ Let $\mathcal{S}$ be the set of $(1)$curves and let $$f \colon \mathcal{S} \to \textrm{Pic}(X)$$ be the map sending each curve into its linear equivalence class. Since any two $E_1$ and $E_2$ in $\mathcal{S}$ are not linear equivalent, $f$ is an injective map. So its image is infinite, because $\mathcal{S}$ is infinite by construction. $\endgroup$ – Francesco Polizzi Apr 9 '15 at 8:48

$\begingroup$ @Polizzi: Thank you very much. Very helpful. Do you know of any criterion, under which the above mentioned set in my question will be finite? $\endgroup$ – Ron Apr 9 '15 at 8:51