# Strategy to prove the isogeny with arbitrary kernel is unique

I encountered a theorem saying that, Given an elliptic curve $E_1$ and arbitrary subgroup $H$, there only exist an isogeny $\phi:E_1\rightarrow E_2$ with $\ker\phi=H$ up to isomorphism of its image. Here's my question, I want to prove this theorem, or at least get some intuition.

Just kindly remind that, an isogeny is a morphism on elliptic curves that preserve both group structure and projective variety. So simply giving a group homomorphism that eliminate $H$ would not suffice.

But since uniqueness can be given simply by group structure of $E_1$, the theorem is actually stating the existence of such algebraic morphism. In other word, the theorem is saying that the canonical morphism given by $x\mapsto x+H$, is algebraic, which means that its image must be also another elliptic curve. Any idea?

• Velu gives an explicit formula for such an isogeny and proves that it is an elliptic curve but with lots of computations. The more intuitive proof actually relies on the isomorphism from the kernel to the galois group of the extension induced by the translation map. You can find the proof in Silvermann the arithmetic of elliptic curves. – Μάρκος Καραμέρης Jul 11 '18 at 13:03