Let K be a field of characteristic 0, E/K an elliptic curve. The "Lefschetz principle" implies that $E[m] \simeq \mathbb{Z}/m \times \mathbb{Z}/m$, but for this to follow from the result for complex elliptic curves, one has to know that the $m$-torsion points of $E$ are in fact defined over $\overline{K}$.
Silverman gives an argument for this in his book on elliptic curves, which I don't understand. He says: "For $P \in E(\overline{K})$, the set $[m]^{-1}(P)$ is finite and invariant under $G_{\overline{K}/K}$, so every point in $[m]^{-1}(P)$ is defined over $\overline{K}$."
First, it seems to me that one has to take $P \in E(K)$ for the pre-image under $[m]$ to be invariant under $G_{\overline{K}/K}$: $P^{\sigma} = ([m](Q))^{\sigma} = [m](Q^{\sigma})$, so $Q^{\sigma}$ is still in the pre-image if and only if $P^{\sigma} = P$.
Second, I don't see how this implies the desired result, as one would have the same invariance under Galois if $\overline{K}$ were replaced by any field extension.
Can someone enlighten me?
I think it is not necessary to invoke Galois group. For $P\in E(K)$, $[m]^{-1}(P)$ is a finite algebraic variety over $K$, so all its points have coordinates in $\overline{K}$.