# Computing Hodge numbers by point counting

In the lecture note of Bhatt from Arizona winter school 2017, there is an exercise which claims if X is a proper smooth scheme defined over $$\mathbb{Z}[1/N]$$ and if there is a polynomial $$P$$ such that for every prime $$p$$ coprime to $$N$$ we have $$X(F_p)=P(p)$$ then the Hodge numbers $$h^{i,j}=0,i\not =j$$

I do not know how to attack this problem because if you want to use zeta functions and Weil conjectures you need the number of points of X over all finite fields. But I do not have any counter example. So is there a typo in this exercise or can someone hint how to prove the claim ?

• Let $H = \bigoplus_i H^i(X_{\overline{\mathbf{Q}}}, \mathbf{Q}_\ell)$, a graded Galois rep. The assumption on $X(\mathbf{F}_p)$ tells you the (graded) trace of ${\rm Frob}_p$ on $H$. By Chebotarev density, the conj classes ${\rm Frob}_p$ ($p>N$) are dense in the Galois group of $\mathbf{Q}$. Now using the polynomial $P$ you can build a graded Galois rep $H'$ which is a direct sum of $\mathbf{Q}_\ell(i)$'s which has the same Frobenius traces for $p>N$. Therefore $H$ and $H'$ have the same semisimplification. Finally, $p$-adic (for $p=\ell$ large) Hodge theory gives you the Hodge numbers. Sep 8, 2020 at 11:05
• @PitorAchinger I'm not familiar with graded representations. Is the equality of graded trace is enough to identify two such representations?if so why both of the notes and in the original argument of katz use weil conjectures
– ali
Sep 8, 2020 at 12:57
• yes, the Weil conjectures are needed to take care of the grading. Sorry for the sloppiness. Sep 8, 2020 at 14:18