A simplified proof of a BSD inequality in char $p$?

Question

One of the main results of the BSD conjecture in characteristic $p>0$ is the fact that the algebraic rank is less or equal than the analytic rank and that there is an equality iff the $\ell$-torsion ($\ell\neq p$) of the Tate-Shafarevich group is finite.

The proof that I know of this fact (which goes back to Tate) uses the elliptic surface associated to the elliptic curve, proves an analogue of this result for the surface and goes back to the elliptic curve.

Does there exist a proof of this result without using the elliptic surface?

Answer 1

Let me briefly sketch Tate's proof for context. If $X$ is a surface over $\mathbb{F}_q$ then Frobenius acts on the Neron-Severi group $NS(X)$ by multiplication by $q$. Using the cycle map from the theory of etale cohomology $NS(X)\otimes \mathbb{Q}_{\ell}$ injects into $H^2_{et}(\bar{X},\mathbb{Q}_{\ell})$ and, by the first statement, lands in the $q$-eigenspace of the action of Frobenius and the rank $r$ of $NS(X)$ is at most the multiplicity of this eigenvalue.

An obvious consequence of this is the inequality $r \le \dim H^2_{et}(\bar{X},\mathbb{Q}_{\ell}) = b_2(X)$. This latter inequality was proved by Igusa before Tate (Igusa, Jun-ichi, Betti and Picard numbers of abstract algebraic surfaces. Proc. Nat. Acad. Sci. U.S.A. 46 1960 724–726.). This proof uses a fibration of the surface as an essential tool and can perhaps be turned into a proof using elliptic curves. I don't know if you can do the same with the full Tate inequality. Whether one would want to do this is another question.