This is from Silverman's Arithmetic of Elliptic Curves:
I have understood everything in this proof except the last line of part (b), which says that the $m$-torsion of $E(K)$ injects into $\tilde{E}(k)$, this is what I tried:
$mE(K)$ is contained in the kernel of the reduction map : $E_0(K)=E(K) \to \tilde{E}_{ns}(k) = \tilde{E}(k)$ here. And this kernel is $E_1(K)$. And since, we have the injection $$ E(K)/E_1(K) \to \tilde{E}(k)$$ It will induce an injection $$ E(K)/mE(K) \to \tilde{E}(k)$$
Am I right?
Any hints, corrections or suggestions are highly appreciated. Thank you.
$E_0(K) = \{P \in E(K) : \tilde{P} \in \tilde{E}_{ns}(k) \}$ $E_1(K) = \{ P \in E(K) : \tilde{P} = \tilde{O} \}$
A simple "yes, it holds as stated" could be the answer to the question in the OP, but to give more support i will insert some diagrams. First of all, by definitions, Proposition 2.1 and the part (a) of Proposition 3.1. in loc. cit. we have:
$\require{AMScd}$ $$ \begin{CD} 0 @>>> E_1(K) @>>> E_0(K) @>>> \tilde E_{ns}(k) @>>> 0 \\ @. @. @VVV \square @VVV @. \\ @. @. E(K) @>>> \tilde E(k) \end{CD} $$
(The first row is a short exact sequence, the vertical arrows are inclusions.) Now part (b) addresses the case when $\tilde E_{ns}(k) = \tilde E(k)$, so the vertical arrows are equalities. Now we consider the diagram with exact rows and columns:
$$ \begin{CD} @. 0 @. 0 @. 0 \\ @. @VVV @VVV @VVV\\ 0 @>>> \color{red}0@>>> E(K)[m] @>(!)>> \tilde E(k)[m] @>>> 0 \\ @. @VVV @VVV @VVV\\ 0 @>>> E_1(K) @>>> E(K) @>>> \tilde E(k) @>>> 0 \\ @. @VV m V @VV m V @VV m V\\ 0 @>>> E_1(K) @>>> E(K) @>>> \tilde E(k) @>>> 0 \\ @. @VVV @VVV @VVV\\ @. \vdots @. \vdots @. \vdots \end{CD} $$
(A diagram chasing or other arguments should show now that the map marked $(!)$ is a bijection.)
The red zero entry is insured in its column by Proposition 3.1. (a).