Why the proof of III 7.4 of J. Silverman's AEC is not way easier?

Question

In Chapter 3 of "Arithmetic of Elliptic Curves" Theorem 7.4 says: given $E_{1}, E_{2}$ elliptic curves, the map: $Hom(E_{1},E_{2})\otimes \mathbb{Z}_{l}\to Hom(T_{l}(E_{1}),T_{l}(E_{2}))$ obviously defined, so that $\Phi\mapsto \Phi_{l}$ is injective. By $l$ we denote a prime other than the characteristic of the field of definition and $T_{l}(E_{i})$ are the corresponding Tate modules. The proof is quite technical but I have no problems with it. Yet I think it might be considerably shortened and made almost trivially easy. As the book points out on the contrary that the introduction of the "preperiodic" module $M^{div}$ is crucial I would like to ask what is wrong with my "stupid" proof. Here it is: let us take $\Phi\in Hom(E_{1},E_{2})\otimes \mathbb{Z}_{l}$ such that $\Phi_{l}=0$. By the definition itself of tensor product one can write $\Phi=\alpha_{1}\Phi_{1}+...+\alpha_{t}\Phi_{t}$ where $\alpha_{i}\in \mathbb{Z}_{l}$ and $\Phi_{i}\in Hom(E_{1},E_{2})$. By the definition of inverse limit $\alpha_{i}=\{a_{i}(n)\}_{n\in \mathbb{N}}$, $a_{i}(n)\in \mathbb{Z}/l^{n}\mathbb{Z}$ and $res_{n+1\to n}(a_{i}(n+1))=a_{i}(n)$ ("compatible" string). As $\Phi_{l}=0$ then $(a_{1}\Phi_{1}+...+a_{t}\Phi_{t})(E_{1}[l^{n}])=0$. Therefore $Ker([l^{n}])\subset Ker(a_{1}\Phi_{1}+...+a_{t}\Phi_{t})$ which by III 4.11 yelds the existence of the unique $\lambda\in Hom(E_{1},E_{2})$ such that $a_{1}\Phi_{1}+...+a_{t}\Phi_{t}=[l^{n}]\circ \lambda$. Therefore $\lambda=b_{1}\Phi_{1}+...+b_{t}\Phi_{t}$ for $a_{i}=l^{n}b_{i}$ for $i=1, ..., t$. As $\Phi_{1}, ..., \Phi_{t}$ do not depend on $n$ and $a_{i}(n)\equiv 0$ mod$(l^{n})$ for all $i$ and all $n$, $\alpha_{i}=0$ for all $i$, which would prove the statement.

Answer 1

$\Phi_i$s are not necessary a basis for $Hom(E_1,E_2)$(because we don't know yet $Hom(E_1,E_2)$ has finite rank) and so we don't know that $\lambda$ is a linear combination of $\Phi_i$s.