Let $p$ be a prime number, let $K$ denote a finite extension $\mathbb{Q}_{p}$ and let $\overline{K}$ be an algebraic closure of $K$. Let $A$ be an ellitpic curve over $K$ and denote by $T_{p}A$ its Tate module.

When $A$ has ordinary reduction then it is known that there is an exact sequence of $Gal(\overline{K}:K)$-modules: $$ 0 \rightarrow \mathbb{Z}_{p}(-1) \rightarrow T_{p}A \rightarrow \mathbb{Z}_{p} \rightarrow 0. $$ I would be interested in a characterization of the quotient $\mathbb{Z}_{p}$ and its applicability in the case of supersingular reduction. When $$ HT: H^{1}(A,{\cal O})\otimes_{K} \mathbb{C} _{p}\oplus H^{0}(A,\Omega ^{1})\otimes_{K} \mathbb{C} _{p}(-1) \simeq Hom_{\mathbb{Z} _{p}}(T_{p}A ,\mathbb{C} _{p}). $$ denotes the Hodge-Tate isomorphism, let $M \subset T_{p}A$ be the intersection of all kernels of homomorphisms in the image of $H^{1}(A,{\cal O})\otimes_{K} \mathbb{C} _{p}$ in $Hom_{\mathbb{Z} _{p}}(T_{p}A ,\mathbb{C} _{p})$ under $HT$. It is easy to see that the quotient $T^{Q}_{p}A:= T_{p}A/M$ is a free $\mathbb{Z} _{p}$-module of rank $\leq 2$, and that it gives rise to an exact sequence: $$0 \rightarrow M \rightarrow T_{p}A \rightarrow T^{Q}_{p}A \rightarrow 0. $$ If $A$ has ordinary reduction, then $T^{Q}_{p}A$ may be identified with $\mathbb{Z}_{p}$ from the first diagram.

If $A$ has supersingular reduction, is then $T^{Q}_{p}A$ also a free $\mathbb{Z}_{p}$-module of rank one? More generally, the quotient $T_{p}^{Q}A$ could be also defined when $A$ is an abelian variety of dimension $g$. Will $T_{p}^{Q}A$ be a free $\mathbb{Z}_{p}$-module of rank $g$?

`$\mathbb{Z}_{p}$`

-module`$T^{Q}_{p}A$`

of rank`$g$`

, together with a canonical surjection of`$\mathbb{Z}_{p}$`

-modules:`$f:T_{p}A \rightarrow T^{Q}_{p}A$`

. The canonical inclusion`$Hom_{\mathbb{Z}_{p}}(T^{Q}_{p}A, \mathbb{\C}_{p}) \subset $Hom_{\mathbb{Z}_{p}}(T_{p}A, \mathbb{\C}_{p})$`

should be equal to the image of`$ H^{1}(A,O) \otimes _{K} \mathbb{C}_{p}$`

. Can such a module be constructed in the non-ordinary case? (The morphism f does not need to compatible with Galois) $\endgroup$ – user23778 May 17 '12 at 23:24