Let $K$ be a finite extension field of $\mathbb{Q}\_p$. Let us consider a semiabelian variety $G$ defined over $K$, i.e there exists an extension of an abelian variety $B$ and a torus $T$ defined over $K$


$$0 \rightarrow T \rightarrow G \rightarrow A \rightarrow 0$$


It is know [that](https://mathoverflow.net/questions/112710/is-the-n-torsion-of-an-extension-of-an-abelian-variety-by-a-torus-finite-and-fl) $G[n]$ is finite flat. My question is whether we can have an exact sequence
$$0 \rightarrow T[p^\infty](L) \rightarrow G[p^\infty](L) \rightarrow A[p^\infty](L) \to 0 \;?$$
where $X[p^\infty](L)$ is $p$-power torsion points of $X$ over any extension $L$ of $K$.


In particular, do we have that if $G[p^\infty](L)$ is finite then is $A[p^\infty](L)$?



$p$-power torsion of semiabelian variety
$\newcommand{\Spec}{\mathrm{Spec}}\newcommand{\oL}{\overline{L}}\newcommand{\bG}{\mathbb{G}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\cL}{\mathcal{L}}$Not in general. The sequence of $p$-divisible groups $0\to T[p^{\infty}]\to G[p^{\infty}]\to A[p^{\infty}]\to 0$ is exact as a sequence of sheaves of abelian groups on the
étale site of $\Spec\, K$ hence there is an exact sequence


$$0\to T[p^{\infty}](L)\to G[p^{\infty}](L)\to A[p^{\infty}](L)\xrightarrow{\delta} H^1(G\_L,T[p^{\infty}](\oL))$$


I will describe the connecting homomorphism $\delta$ in terms of the Kummer map in the case $T=\bG\_m$ and then will say how to witness that it does not vanish in general.


Recall the Kummer map: the short exact sequence $0\to A[p^n]\to A\to A\to 0$ of fppf sheaves on $\Spec K$ gives rise to the exact sequence $0\to A[p^n](L)\to A(L)\xrightarrow{p^n}A(L)\to H^1(G\_L,A[p^n](\oL))$ and hence an embedding $\kappa\_{A,p^n}: A(L)/p^n\to H^1(G\_L,A[p^n](\oL))$.


The data of an extension $0\to \bG\_m\to G\to A\to 0$ is equivalent to the choice of a degree zero line bundle $\cL$ on $A$ (to a line bundle $\cL$ corresponds the semi-abelian variety $G\_{\cL}$ which is the total space of $\cL$ equipped with the unique group structure compatible with the projection down to $A$). Consider the corresponding point $[\cL]\in (\mathrm{Pic}^0A)(L)=A^{\vee}(L)$ of the dual abelian variety, and send it along the Kummer map $\kappa\_{A^{\vee},p^n}$ to get a Galois cohomology class $\kappa\_{A^{\vee},p^n}([\cL])\in H^1(G\_L,A^{\vee}[p^n](\oL))$


**Lemma.** The homomorphism $\delta|\_{A[p^n](L)}$ is the cuprpoduct with the class $\kappa\_{A^{\vee},p^n}([\cL])$ followed by the Weil pairing:


$$A[p^n](L)\xrightarrow{\kappa\_{A^{\vee},p^n}([\cL])\,\cup\,-}H^1(G\_L,A^{\vee}[p^n](\oL)\otimes A[p^n](\oL))\to H^1(G\_L,\mu\_{p^n}(\oL))\to H^1(G\_L,\mu\_{p^{\infty}}(\oL))$$


Before sketching a proof, let me remark that the important information that we will get from this lemma is that $\delta$ varies non-trivially when variying the line bundle $L$, thanks to the Weil pairing being perfect; this is the point of rewriting $\delta$ like this, even if the formula appears more complicated than the definition of $\delta$.


Idea of the proof of the lemma: The Weil pairing identifies $A^{\vee}[p^n]$ with the internal group scheme of homomorphisms $Hom(A[p^n],\mu\_{p^n})$ and it follows from the construction of the pairing in Theorem 8.1.3 [here](http://virtualmath1.stanford.edu/%7Econrad/249CS15Page/handouts/abvarnotes.pdf) that for a degree zero line bundle $\cL$ the class of the extension $0\to\mu\_{p^n}\to G\_{\cL}[p^n]\to A[p^n]\to 0$ of etale sheaves of $\bZ/p^n$-modules in $\mathrm{Ext}^1\_{L\_{et}}(A[p^n],\mu\_{p^n})=H^1(G\_L,Hom(A[p^n],\mu\_{p^n})(\oL))=H^1(G\_L,A^{\vee}[p^n](\oL))$ coincides with the image of $[\cL]$ under the Kummer map. This implies the statement because, by definition, $\delta$ is the cup-product with this class in $H^1(G\_L,Hom(A[p^n],\mu\_{p^n})(\oL))$


We can now give an example when $\delta$ is non-zero and therefore the map $G[p^{\infty}](L)\to A[p^{\infty}](L)$ is not surjective. Fix an abelian variety $A$ and an integer $n$ and take $L$ to be a large enough finite extension of $K$ such that $A[p^n](L)=A[p^n](\overline{K})$ and $A^{\vee}[p^n](L)=A^{\vee}[p^n](\overline{K})$. Note that this condition says that the Galois actions of $G\_L$ on $A[p^n](\oL)$ and $A^{\vee}[p^n](\oL)$ are trivial (and therefore so is the action on $\mu\_{p^n}(\oL)$).


It remains to choose the line bundle $\cL$ appropriately: if our goal was to prove only that the composition $A[p^n](L)\xrightarrow{\kappa\_{A^{\vee},p^n}([\cL])\,\cup\,-}H^1(G\_L,A^{\vee}[p^n](\oL)\otimes A[p^n](\oL))\to H^1(G\_L,\mu\_{p^n}(\oL))$ is non-zero, we could have taken any $\cL\in A^{\vee}(L)$ whose image under the Kummer map is non-zero. Indeed, by the triviality of the Galois action on $A[p^n](\oL)$ and $A^{\vee}[p^n](\oL)$ we factor them out of Galois cohomology: $H^1(G\_L,A^{\vee}[p^n](\oL)\otimes A[p^n](\oL))=H^1(G\_L,\bZ/p^n)\otimes A^{\vee}[p^n](L)\otimes A[p^n](L)$ and the claim follows from the Weil pairing being non-degenerate.


To make sure that the image of $A[p^n](L)$ in $H^1(G\_L,\mu\_{p^{\infty}}(\oL))$ is still non-zero we need to choose $\cL$ with a bit more care as the map $H^1(G\_L,\mu\_{p^n}(\oL))\to H^1(G\_L,\mu\_{p^{\infty}}(\oL))$ has a non-zero kernel given by $\mu\_{p^{\infty}}(L)/(p^n\cdot \mu\_{p^{\infty}}(L))$. But $\mathrm{rk}(A(L)\otimes\bZ\_p)=\dim A\cdot [L:\mathbb{Q}\_p]$ while that kernel is a cyclic group, so we can always choose $\cL$ that will do the job.


