Notice that both $A_\ell(N)$ and $B_\ell(N)$ are powers of $\ell$.

It is clear that the $\ell$-adic failure $A_\ell(N)$ can be nontrivial, that is, different from $1$. Suppose for example that $\alpha=\ell\beta$ for some $\beta\in E(K)$: then we have $$\begin{aligned}
K_{\ell^n,\ell^n}=K_{\ell^n}(\ell^{-n}\alpha)=K_{\ell^n}(\ell^{-n+1}\beta),\end{aligned}$$ and the degree of this field over $K_{\ell^n}$ is at most $\ell^{2(n-1)}$, so $\ell^2\mid A_\ell(N)$. In Example \[exa-17739g1\] we will show that the $\ell$-adic failure can be non-trivial also when $\alpha$ is strongly $\ell$-indivisible (see Definition \[def-divisib\]).

We now show that the adelic failure $B_\ell(N)$ can be non-trivial as well. Consider the elliptic curve $E$ over ${\mathbb{Q}}$ given by the equation $$\begin{aligned}
y^2 = x^{3} + x^{2} - 44 x - 84\end{aligned}$$ and with Cremona label 624f2 (see [@lmfdb [label 624f2](http://www.lmfdb.org/EllipticCurve/Q/624f2/)]). One can show that $E({\mathbb{Q}})\cong {\mathbb{Z}}\oplus ({\mathbb{Z}}/2{\mathbb{Z}})^2$, so that the curve has full rational $2$-torsion, and that a generator of the free part of $E({\mathbb{Q}})$ is given by $P=(-5,6)$. The $2$-division points of $P$ are given by $(1 + \sqrt{-3}, -3 + 7\sqrt{-3})$, $(-11 + 3\sqrt{-3},27 + 15\sqrt{-3})$, and their Galois conjugates, so they are defined over ${\mathbb{Q}}(\zeta_3)\subseteq {\mathbb{Q}}_3$, and we have $B_2(6):=[{\mathbb{Q}}_{2,2}\cap {\mathbb{Q}}_6:{\mathbb{Q}}_2]=[{\mathbb{Q}}(\zeta_3):{\mathbb{Q}}]=2$.

These computations have been checked with SageMath [@sagemath].

The torsion, Kummer and arboreal representations {#subsec:Representations}
