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One way the Weil pairing for an Abelian Variety $A/k$ is phrased is the following (for simplicity, let me only deal with the multiplication by $m$ map ($[m]: A\to A$) instead of arbitrary isogenies):

Define $P$ to be the group of line bundles $\mathscr L$ on $A$ such that $[m]^*\mathscr L$ is trivial and $A[m]$ to be the $m$-torsion points. Then, we define a map: $$\langle -,-\rangle : A[m] \times P \to k^\times$$ where given $\mathscr L \in P, a\in A[m]$, we let $D$ be a divisor corresponding to $\mathscr L$ and define $g(x)$ so that $\operatorname{div}(g) = [m]^*D$ and define $\langle a, \mathscr L\rangle = g(x+a)/g(x)$.

This reminds me a lot of the way that torsors are identified with cocyles in descent theory and indeed we can reinterpret the above in the following way:

Consider the Galois étale extension $[m]: A \to A$ and we want to classify $\mathscr O_A$ torsors (=$P$) for this extension. Well, they are classifed by $H^1(A[m],k^\times)$ since $A[m]$ is the Galois group of the extension and $\operatorname{Aut}(\mathscr O_A) = k^\times$. Further, since $A[m]$ acts trivially on $k^\times$, this cohomology group is simply $\operatorname{Hom}(A[m],k^\times)$, so we have established the bijection: $$P \to \operatorname{Hom}(A[m],k^\times)$$ and following the construction through, we see that this is the same thing as the Weil pairing above.

My Question: Does this observation lead to further insights? I don't know the theory of Weil pairings on Abelian varieties very well but they are certainly quite a useful thing on Elliptic Curves.

Conversely, can this connection be used in the opposite way, to gain more insight into Galois descent in general? Basically, this seems like a very natural formulation, so I am sure this has been discussed before in the literature but I can't find anything. Where should I look?

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  • $\begingroup$ I believe this is essentially the approach Mumford takes either in the appendix of his book on Abelian Varieties or in his series of papers "On the Equations Defining Abelian Varietes" $\endgroup$ Commented Dec 19, 2017 at 18:22

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