Bogomolov's paper "Sur l'algébricité des représentations l-adiques" proves that the image of the $\ell$-adic Galois representation associated to an abelian variety over a number field $K$ is open in the $\mathbb{Q}\_
{\ell}$ points of its $\mathbb{Q}\_{\ell}$-Zariski closure. 


The Comptes-Rendus announcement has only sketched the proof. My French is limited, but I have tried to capture the key step below. 


In the abelian case, it proceeds as follows. Let $H\_{\ell}$ be the $\mathbb{Q}\_
{\ell}$ Zariski-clsure of the image of the absolute Galois group. Replacing $K$ by a finite extension, we assume that the is connected. If $H\_{\ell}$ is commutative, then $\rho$ is abelian. We apply properties of Hodge-Tate representations to deduce that $H\_{\ell}$ is a torus. For otherwise, 
 $H\_{\ell}$ contains a factor of $\mathbb{G}\_a$. Composing with $\rho$ would give an infinite unramified abelian extension of $K$. [\*\* Why? \*\*]


I do not understand the last step used to obtain an infinite unramified abelian extension. Perhaps somebody else can see it? 


The Russian version is at <http://www.mathnet.ru/links/c33c60182f2bcfa61bc2cb3a0b684c62/im1843.pdf>, but I cannot read a word of Russian. 



An abelian Hodge-Tate representation lands in a torus
What Bogomolov proved is the algebraicity of the corresponding $\ell$-adic representation. What you are trying to do is to prove its semisimplicity, which was done by Faltings 4 years later. The presence of additive factors $G\_a$ in the Zariski closure of the Galois image is not an issue as far as you are interested only in the algebraicity. In fact, since every linear nilpotent Lie algebra (in char 0) is algebraic (coincides with the Lie algebra of a certain linear algebraic group), one may replace the original representation by its semisimplification and deal with a certain semisimple $\ell$-adic representation $\rho$ of the absolute Galois group $Gal(K)$ of a number field $K$. Replacing $K$ by its suitable finite algebraic extension, one may assume that the inertia groups of places not lying over $\ell$ act as unipotent operators and $H\_{\ell}$ is connected. So, if $\rho$ is semisimple abelian then it is unramified outside divisors of $\ell$. Since it is of Hodge-Tate type, the images (under $\rho$) of the inertia groups of divisors of $\ell$ are open in their Zariski closures $U\_{\lambda}$ (see Serre's Abelian $\ell$-adic representations and elliptic curves), i.e., the corresponding local $\ell$-adic representations are algebraic. Now, if $U$ is is the algebraic subgroup of $H\_{\ell}$ that is generated by all $U\_{\lambda}$'s, then the images of all the inertia groups generate (after taking the closure) an open subgroup in $U(\mathbf Q\_{\ell})$. Now it suffices to check that $H\_{\ell}=U$. 


If $U$ is a proper subgroup of $H\_{\ell}$ then the composition 
$$Gal(K)\to H\_{\ell}(\mathbf Q\_{\ell}) \twoheadrightarrow H\_{\ell}(\mathbf Q\_{\ell})/U(\mathbf Q\_{\ell})$$ 
gives rise to a continuous surjective homomorphism from $Gal(K)$ to $\mathbf Z\_{\ell}$ that corresponds to an infinite everywhere unramified abelian extension of $K$. Since such extensions do not exist, we get the desired contradiction.


By the way, Bogomolov's Izvestiya paper is available in English
<http://mr.crossref.org/iPage/?doi=10.1070%2FIM1981v017n01ABEH001329> .


