Independent of endomorphism ring of elliptic curve

I can't seem to find the reference for the following fact

Suppose that $E$ is an elliptic curve defined over $k$ and $K, L$ are algebraically closed fields containing $k$. Then $\text{End}_K(E)$ is isomorphic to $\text{End}_L(E)$.

Where $\text{End}_K(E)$ is the ring of endomorphisms of $E(K)$. Of course, it suffices to show the result for $L$ being the algebraic closure of the ground field $k$.

• What is the problem exactly? By uniqueness of the algebraic closure, don't we have $K\cong L$?. From that the function fields should be isomorphic, and also the Galois groups as extensions of $k$. Then it doesn't seem much of a stretch that the endomorphism rings are isomorphic. Or am I missing something? – CurveEnthusiast Mar 20 '17 at 17:07
• @CurveEnthusiast The problem is that there are infinitely many algebraically closed field containing a particular field. For instance, both $\overline{\mathbb{Q}}$ and $\mathbb{C}$ are algebraically closed and contain $\mathbb{Q}$. What we are claiming is that if you have an elliptic curve $E$, say defined over $\mathbb{Q}$ then any endomorphism you can write over $\mathbb{C}$ are basically defined over $\overline{\mathbb{Q}}$. – An Hoa Mar 20 '17 at 21:38
• I think you want to show that there is a finite field extension of $k$ over which the morphism is defined. – user45878 Apr 6 '17 at 10:40