why Riemann Hypothesis over curves is easy but 'normal' hypothesis for the Riemann Zeta function $ \zeta (s) $ is hard ,

Question

despite both zetas $ \zeta (s,X) $ and $ \zeta (s)$ have the same functional equation, the same Euler prodcut and the same Riemann-Weil formula

why one of them is 'easy' and can be solved but the other is so hard?? the zeta one $ \zeta (s)$

Answer 1

One obvious difference, from a purely analytic point of view, is that $\zeta(s,X)$ is a function of $p^{-s}$, i.e. it is periodic by an imaginary number. This makes $\zeta(s,X)$ much simpler, e.g. the set of roots is also periodic, so there are only finitely many roots to consider. In fact the functional equation forces that $\zeta(s,X)$ is essentially a reciprocal polynomial in $p^{-s}$, which further simplifies the analysis. From a structural point of view, $\zeta(s,X)$ encodes a geometric object which comes with a natural topology, a Galois action, schemes and (co)homology groups derived from them; while a number field is not an algebraic variety, it has archimedean valuations, residue fields of every characteristic etc. Of course it might be the case that RH is not so hard, we just don't know.

Answer 2

An additional issue is that the adelic norm map in the function field has discrete image, so its Pontryagin dual is actually compact. In the case of a number field, it has non-discrete, non-compact image, so its Pontryagin dual is non-compact as well. Since Tate's thesis, we know that the Fourier analysis on these duals is intimately related to $L$-functions.

After-thought: I guess this boils down to the same issue, which GH already mentioned: we have not only one prime field for all residue fields, for number fields there are infinitely many. In fact, there is no residue field for archimedean fields (yet).