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Let $K$ be a field, and $\phi: E_1\to E_2$ be an isogeny of elliptic curves over $K$. Given a prime $\ell$ different from the characteristic of $K$, $\phi$ induces an injection $T_\ell \phi: T_\ell E_1\to T_\ell E_2$. It is easy to check group theoretically that the cokernel of $T_\ell \phi$ is isomorphic to the $\ell$-primary part of $\ker\phi$. Can this isomorphism be made canonical?

Note: Over $\mathbb{C}$, this can be written down pretty straightforwardly. Let $E_i=\mathbb{C}/\Lambda_i$, where $\Lambda_i$ is a lattice. We make identify $\Lambda_1$ as a sublattice of $\Lambda_2$ via the isogeny. This gives an short exact sequence

$$0\to \Lambda_1 \to \Lambda_2 \to \Lambda_2/\Lambda_1\to 0,$$ where $\Lambda_2/\Lambda_1$ is naturally identified with $\ker{\phi}$. Tensoring with $\mathbb{Z}_\ell$ gives us $$ 0\to T_\ell E_1 \to T_\ell E_2 \to (\ker \phi)[\ell^\infty]\to 0.$$

But I am having trouble to do it over an arbitrary field.

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up vote 4 down vote accepted

Yes, this is canonical.

It is perhaps easiest to start with the $\ell$-divisible groups: we have the short exact sequence $$0 \to ker\phi[\ell^{\infty}] \to E_1[\ell^{\infty}] \to E_2[\ell^{\infty}] \to 0.$$ Now to get the Tate moduels, apply the functor $Hom_{\mathbb Z_{\ell}}(\mathbb Q_{\ell}/\mathbb Z_{\ell},\text{--})$, to obtain $$0 \to T_{\ell}E_1 \to T_{\ell} E_2 \to Ext^1_{\mathbb Z_{\ell}}(\mathbb Q_{\ell}/\mathbb Z_{\ell}, \ker(\phi)[\ell^{\infty}]) \to 0.$$ To compute the Ext^1, apply $Hom_{\mathbb Z_{\ell}}(\text{--},\ker(\phi)[\ell^{\infty}])$ to $$0 \to \mathbb Z_{\ell} \to \mathbb Q_{\ell} \to \mathbb Q_{\ell}/\mathbb Z_{\ell} \to 0$$ to find that $$Ext^1_{\mathbb Z_{\ell}}(\mathbb Q_{\ell}/\mathbb Z_{\ell}, \ker(\phi)[\ell^{\infty}]) = \ker(\phi)[\ell^{\infty}]$$ (where here "$=$" means "canonical isomorphism"). What I am using is the fact that $\ker(\phi)[\ell^{\infty}]$ is finite order, so that not only $Hom$s but also higher $Ext$s from $\mathbb Q_{\ell}$ vanish.

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