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 $k\in\mathbb{Z}$.

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1 Answer 1


The summation they are subtracting in the proof matters too; fix $P$ and prove then that $\phi^{-1}(Q)=\{P+T\mid T\in\phi^{-1}(O)\}$

  • $\begingroup$ 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. $\endgroup$
    – klein4
    May 20, 2021 at 4:42

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