# Proof of Theorem 6.1(b) in Silverman's AEC

I'm learning about the construction of the dual isogeny in Silverman's Arithmetic of Elliptic Curves. In particular, I'm reading the proof for Theorem 6.1 from Chapter III. There is a (probably easy) fact used in Line 4 of the proof for Theorem 6.1(b) that I'm not seeing immediately. I was wondering if someone can help me see why it's true.

Let $$\phi:E_1\to E_2$$ be a non-constant isogeny. Let $$Q\in E_2$$. Then (why is it true that) $$\sum_{P\in\phi^{-1}(Q)}P = [\#\phi^{-1}(Q)] P \tag{for any P\in \phi^{-1}(Q)}$$ Here, addition is for points on an elliptic curve, and $$[k]: E_1\to E_1$$ is the multiplication by $$k$$ map for any $$k\in\mathbb{Z}$$.

The summation they are subtracting in the proof matters too; fix $$P$$ and prove then that $$\phi^{-1}(Q)=\{P+T\mid T\in\phi^{-1}(O)\}$$
• Thanks, that was very helpful. I will write the solution here for completion. Since $\phi^{-1}(Q)$ is a coset of the kernel $\phi^{-1}(O)$, group theory tells us that for any representative $P\in \phi^{-1}(Q)$ we have $\phi^{-1}(Q) = P+\phi^{-1}(O)$. Then, simply rewrite the sum $\sum_{P\in \phi^{-1}(Q)} P = \sum_{T\in \phi^{-1}(O)} P+ T$ (for a fixed $P\in\phi^{-1}(Q)$) to complete the deduction. May 20, 2021 at 4:42