Let $K$ be a number field, let $A$ be an abelian variety over $K$, and let $H$ be a torus over $K$. For a prime $l$, we have the natural map
$$\mathrm{Ext}^1(A, H) \otimes\_{\mathbb{Z}} \mathbb{Z}\_l \rightarrow \mathrm{Ext}^1\_{\mathrm{Gal}(\overline{K}/K)}(T\_l A, T\_l H),$$
where the first $\mathrm{Ext}$ is in the category of commutative algebraic groups over $K$ and the second one is in the category of continuous representations of $\mathrm{Gal}(\overline{K}/K)$ on finite free $\mathbb{Z}\_l$-modules. Is the displayed map injective and, if so, why?



Extensions of an abelian variety by a torus vs. extensions of their $\ell$-adic Tate modules
It is affirmative (up to a compatibility check at the end of the argument given here). If $K'/K$ is a finite Galois splitting field of $H$ and $T' = H\_{K'}$ is the associated split $K'$-torus then we have an exact sequence of $K$-tori $$0 \rightarrow H \rightarrow {\rm{R}}\_{K'/K}(T') \rightarrow T \rightarrow 0.$$
But ${\rm{Hom}}(A,T)=0$ since $T$ is affine and ${\rm{Hom}}\_K(T\_{\ell}(A),T\_{\ell}(T))=0$ by consideration of Frobenius eigenvalues. Thus, the natural maps
$${\rm{Ext}}^1\_K(A,H) \rightarrow {\rm{Ext}}^1\_K(A,{\rm{R}}\_{K'/K}(T')),\,\,\,
{\rm{Ext}}^1\_K(T\_{\ell}(A),T\_{\ell}(H)) \rightarrow {\rm{Ext}}^1\_K(T\_{\ell}(A),T\_{\ell}({\rm{R}}\_{K'/K}(T'))$$
are injective. It therefore suffices to treat ${\rm{R}}\_{K'/K}(T')$ in place of $H$.


Using pushout along ${\rm{R}}\_{K'/K}(T')\_{K'} \rightarrow T'$ and pullback along $A \rightarrow {\rm{R}}\_{K'/K}(A\_{K'})$, the Ext's are **compatibly** identified with ${\rm{Ext}}^1\_{K'}(A\_{K'},T')$ and ${\rm{Ext}}^1\_{K'}(T\_{\ell}(A\_{K'}),T\_{\ell}(T'))$, so this reduces us (upon renaming $K'$ as $K$) to treating the case when $H$ is a split torus, and then even $H = {\rm{GL}}\_1$ by direct sum compatibility in $H$. So far, so good.


But in the special case when $H={\rm{GL}}\_1$ we naturally have
$${\rm{Ext}}^1\_K(A,{\rm{GL}}\_1) \simeq A^{\vee}(K)$$
for the dual abelian variety $A^{\vee}$ (Barsotti's formula, valid over any field), and likewise the Galois Ext is ${\rm{H}}^1(K,T\_{\ell}(A^{\vee}))$. The $\ell^n$-power Kummer sequence of $A^{\vee}$ defines an injective map
$$\delta:A^{\vee}(K)/(\ell^n) \rightarrow {\rm{H}}^1(K,A^{\vee}[\ell^n]).$$
Letting $S$ be a non-empty finite set of places containing the archimedean places and the bad places of $A$ over $K$ and the $\ell$-adic places, $\delta$ is seen to factor through ${\rm{H}}^1(K\_S/K,A^{\vee}[\ell^n])$ where $K\_S/K$ is the maximal extension unramified outside $S$. 


Since ${\rm{Gal}}(K\_S/K)$ has good cohomological finiteness properties, it is legitimate to identified the injective map between 
 inverse limits over $n$ with a map
$$A^{\vee}(K) \otimes\_{\mathbf{Z}} \mathbf{Z}\_{\ell} \rightarrow {\rm{H}}^1(K\_S/K,T\_{\ell}(A^{\vee})).$$
The correctness of the left side uses that $A^{\vee}(K)$ is finitely generated (so its $\ell$-adic completion is given by tensoring against $\mathbf{Z}\_{\ell}$). Composing with the injective inflation map then defines an *injection*
$${\rm{Ext}}^1\_K(A,{\rm{GL}}\_1) \otimes\_{\mathbf{Z}} \mathbf{Z}\_{\ell} \rightarrow
{\rm{Ext}}^1\_K(T\_{\ell}(A), T\_{\ell}({\rm{GL}}\_1)).$$
The final step is to check that this injection is actually the original map of interest (in the special case $H = {\rm{GL}}\_1$); this is left to the reader to sort out (maybe it is correct up to a harmless sign or something like that). 


