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If two varieties over $\mathbb{Q}$ have the same number of $\mathbb{F}_p$-points for all but finitely many primes do they have the same number of $\mathbb{F}_{p^n}$-points for all $n>1$ and for all but finitely many primes?

If they are both smooth proper do we in fact have $H^i(X_{\overline{\mathbb{Q}}}, \mathbb{Q}_l)\cong H^i(Y_{\overline{\mathbb{Q}}}, \mathbb{Q}_l)$ as Galois modules?

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    $\begingroup$ Chebotarev applied to the virtual representation $\sum_i (-1)^i \left(H_c^i(X_{/\overline{\mathbb{Q}}}, \mathbb{Q}_\ell) - H_c^i(Y_{/\overline{\mathbb{Q}}}, \mathbb{Q}_\ell)\right)$ (but lemme know if I’ve said anything stupid! I only think about curves admittedly…). $\endgroup$
    – alpoge
    Commented Sep 19, 2021 at 21:18
  • $\begingroup$ The trace does not determine the entire characteristic polynomial does it? $\endgroup$
    – lkx
    Commented Sep 20, 2021 at 3:36
  • $\begingroup$ The way it goes is: the tr. of all but fin. many $\mathrm{Frob}_p$'s on said virtual rep. is $0$. The trace function is a cont. function on the abs. Gal. gp. valued in $\mathbb{Q}_\ell$. By Chebotarev that set of Frobenii is dense. So the fun. is $0$ on the whole gp., hence $0$ on $\mathrm{Frob}_p^n$ for all $n$. I prefer to think of it via mod-$\ell^N$ rep.s: given $p$ and $n$, for all $N$ the action of $\mathrm{Frob}_p^n$ on the mod-$\ell^N$ rep. matches the action of some $\mathrm{Frob}_q$ by usual Chebotarev for finite Gal. ext.s, so the trace is $0$ mod $\ell^N$, now take $N\to\infty$. $\endgroup$
    – alpoge
    Commented Sep 20, 2021 at 3:51
  • $\begingroup$ btw I just saw your edit aka second question and I suspect it's unknown (again, not my expertise) because it's not known if the rep.s are semisimple in general. That said we can at least read off the char. poly. of $\mathrm{Frob}_p$ on $H^i$ by purity ($p$ large and via your hypotheses of smoothness and properness) and so I think it at least follows in your situation that the semisimplifications of the $H^i$'s agree as Galois modules. $\endgroup$
    – alpoge
    Commented Sep 20, 2021 at 4:12
  • $\begingroup$ But we only know the trace so when there is many eigenvalues it should fail right? There probably should be a counterexample with surfaces. Are there non-isogenous abelian surfaces over a finite field with the same number of points? $\endgroup$
    – lkx
    Commented Sep 20, 2021 at 5:07

1 Answer 1

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The question was essentially answered by alpoge.

For the first question note that the function from the absolute Galois group of $\mathbb{Q}$ to $\mathbb{Q}_l$ sending an automorphism to the trace of the matrix it acts by on $\sum\limits_{i\geq 0}(-1)^i[H^i_c]$ is the same for the Frobeniuses of all but finitely many primes. The Frobeniuses of all but finitely many primes are in fact a dense subset of the absolute Galois group so two continuous functions agreeing on them must be equal. Then the traces of the powers of Frobeniuses are also equal.

For the second question the Brauer-Nesbitt theorem (formulated for profinite groups) shows that $H^i_c$ have isomorphic semisimplifications and then we would be done if we knew that smooth proper varieties have semisimple cohomology. But the latter is a hard conjecture so this is where we are.

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  • $\begingroup$ i agree with everything except: in the answer to the first question because you don't assume the varieties are smooth and proper i dunno how to extract the trace on just the $H^i_c$'s from only the information of all point counts (when they're smooth proper you can use purity) --- but im not an expert. anyway otherwise i agree with the summary! $\endgroup$
    – alpoge
    Commented Sep 20, 2021 at 5:55
  • $\begingroup$ @alpoge if we look at the trace on the alternating sum of $H^i_c$ then we don't have to pick them apart right? $\endgroup$
    – lkx
    Commented Sep 20, 2021 at 6:09
  • $\begingroup$ yeah exactly! that’s why i did that in the first comment above —- i was just correcting your summary $\endgroup$
    – alpoge
    Commented Sep 20, 2021 at 6:13
  • $\begingroup$ Thank you, corrected $\endgroup$
    – lkx
    Commented Sep 20, 2021 at 6:17
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    $\begingroup$ whatcha mean? i was just thinking that the zeta function of $X$ mod $p$ is rational and an alternating product of the $L$-functions of the $H^i$’s, and you can pick out the $i$-th guy by taking the polynomial in the factorization (there’s no cancellation by purity) whose roots are all the roots/poles of the zeta function of $X$ mod $p$ of absolute value $p^{i/2}$. $\endgroup$
    – alpoge
    Commented Sep 20, 2021 at 6:40

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