Let $E/k$ and $E'/k$ be isogenous over $k$.
We know:
- $E$ and $E'$ are isogenous if and only if $\# E(k) = \# E'(k)$ [Ex V.5.4, Sil86].
- Any non-constant morphism of curves is surjective [Thm. II.2.3, Sil86].
- The kernel of an isogeny can have more that one points in it, which means that the map may not be injective.
First, I didn't understand how $E$ and $E'$ have same number of points, but the isogeny between them is surjective and not injective.
In this thread, it's made clear with the answer
Surjective means here surjective from $E(\bar k)$ to $E′(\bar k)$ where $\bar k$ is the algebraic closure of $k$.
Does this mean that if $E/k$ and $E'/k$ are isogenous over $k$, the orders of $E(\bar k)$ and $E'(\bar k)$ might not be equal and the first one may be larger than the second?
PS: I'm not from math background and I humbly try to learn about isogenies.
[Sil86] Silverman, Joseph H., The arithmetic of elliptic curves, Graduate Texts in Mathematics, 106. New York etc.: Springer-Verlag. XII, 400 p. DM 148.00 (1986). ZBL0585.14026.