2 added 814 characters in body; added 1 characters in body; added 1 characters in body; deleted 1 characters in body

Yes, it's just a linear algebra stuff. Probably, it's more convenient to look at the Weil pairing between the $\ell$-adic Tate modules of $A$ and its dual, taking into account that these modules are free $O\otimes Z_{\ell}$-modules of the same rank.

From the linear algebra point of view the situation is as follows.

Let $e: M \times N \to P$ be a perfect pairing of free $Z_{\ell}$-modules $M$ and $N$ of the same rank that takes values in a free $Z_{\ell}$-module $P$ of rank $1$. Assume also that $M$ and $N$ are free $O\otimes Z_{\ell}$-modules of the same rank. Then we get a natural $Z_{\ell}$-linear map

$M \times N \to Hom_{Z_{\ell}}(O\otimes Z_{\ell}, P),$ $(m,n) \mapsto [a \mapsto e(ax,y)]$ for all $a \in O$. Taking into account that $$Hom_{Z_{\ell}}(O\otimes Z_{\ell}, Z_{\ell})=D^{-1}\otimes Z_{\ell}$$ (via the trace map), we get the natural pairing

$$M \times N \to P \otimes_{Z_{\ell}}[D^{-1}\otimes Z_{\ell}].$$ Here $M$ and $N$ are the Tate modules of $A$ and its dual while $P$ is $Z_{\ell}(1)$ and $e$ is the Weil pairing.

1

Yes, it's just a linear algebra stuff. Probably, it's more convenient to look at the Weil pairing between the $\ell$-adic Tate modules of $A$ and its dual, taking into account that these modules are free $O\otimes Z_{\ell}$-modules of the same rank.