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$\newcommand{\fp}{\mathfrak{p}} \renewcommand{\phi}{\varphi}$ I'm currently trying to understand a paper of Cassels [see page 189 for the relevant content, esp. (4.9)] and I've hit a little snag. The set up is as follows (although I've modernized notation and slightly changed the context):

Let $K$ be a number field, $\fp$ a (finite) prime of $K$, and $k_\fp$ the residue field. Fix an algebraic (resp. separable) closure $\bar{K}_\fp$ (resp. $\bar{k}_\fp$) of $K_\fp$ (resp. $k_\fp$). Let $E,E'$ be elliptic curves defined over $K_\fp$ with good reduction at $\fp$, and suppose we have an isogeny $\phi:E\to E'$ such that $\fp\nmid\deg\phi$. Let $\tilde{E},\tilde{E}'$ be the reduced curves over $k_\fp$. Then by some standard result (I think it's in Silverman somewhere) there is a separable isogeny $\tilde{\phi}:\tilde{E}\to \tilde{E}'$ which is the 'reduction' of $\phi$—that is, the obvious square commutes and the $\bar{k}_\fp$-points of the kernel of $\tilde{\phi}$ are the reduced $\bar{K}_\fp$-points of the kernel of $\phi$.

Now for the bit I don't get. According to Cassels, "because $\tilde{\phi}$ is separable, we have $$E'_1(K_\fp)\subseteq\phi(E(K_\fp))",$$ where $E'_1$ is the kernel of reduction $E'\to\tilde{E}'$. How does this follow?

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