I believe I've read this before, but I can't find it despite hours of searching on Google.

I've know the common definition of isogeny in elliptic curves, as $\phi:E_1 \rightarrow E_2$ a nonconstant morphism of curves satisfying $\phi(0)= 0$, however somewhere else I read a really lovely intuitive explanation that basically stated that if a curve $E_1$ is isogenic with curve $E_2$, then, not only will points, such as $p_1$, $q_1$ map to $p_2$, $q_2$ but moreover, the mapping of the point $E_1(p_1+q_1)$ to $E_2$, will result in the same point on $E_2$ as when the points $E_1(p_1)$ & $E_1(q_1)$ are first mapped to $E_2$ and then added on the $E_2$ curve.

I hope this makes sense, am I confusing this or does anyone have a resource that shows this detail?

  • 1
    $\begingroup$ Actually, preserving the group operation is what "morphism" means; that $\phi(p_1+q_1) = \phi(p_1) + \phi(q_1)$. Just a random mapping that isn't constrained to preserve the group properties wouldn't be very interesting... $\endgroup$
    – poncho
    Jun 17, 2020 at 20:39
  • $\begingroup$ Ok so my take is correct? Do you have any link to docs that show this characteristic? I can’t find it anywhere. Thanks @poncho $\endgroup$
    – Woodstock
    Jun 17, 2020 at 20:47
  • $\begingroup$ Other than the Wikipedia article on "morphism", well, no, I don't have one available. It's one of those ideas that's so "in the water" that it's usually just assumed everyone knows (and so there's little point in spelling it out) $\endgroup$
    – poncho
    Jun 17, 2020 at 21:12
  • 2
    $\begingroup$ See theorem 4.8 on page 71 of The Arithmetic of Elliptic Curves by Silverman. It proves that an isogeny between elliptic curves is a group homorphism. $\endgroup$
    – user69015
    Jun 17, 2020 at 21:42
  • $\begingroup$ Thanks all, appreciate it. $\endgroup$
    – Woodstock
    Jun 17, 2020 at 21:43

1 Answer 1


Given two elliptic curves $E_1$ and $E_2$, an isogeny from $E_1$ to $E_2$ is an algebraic map $\phi$ (it is basically defined from rational functions) that sends the neutral element from $E_1$ to the neutral element of $E_2$.

As stated in Theorem 4.8 in page 71 of The Arithmetic of Elliptic Curves by J.H. Silverman:

Let $$ \phi : E_1 \longrightarrow E_2 $$ be an isogeny. Then $$ \phi(P+Q) = \phi(P) + \phi(Q)\quad \text{for all } P, Q \in E_1.$$

In other words, the isogeny is a group homomorphism (the group law of the curve $E_1$ is preserved after the transfer to the curve $E_2$).

The proof involves the divisor class groups of the curves, using the fact that the isogeny induces a group homomorphism between them.

  • 1
    $\begingroup$ thank you! marked as answered $\endgroup$
    – Woodstock
    Jun 22, 2020 at 11:39

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