Commutativity of $End(E)$ for $E$ elliptic curve

Question

If I consider an elliptic curve $E$ over a field of characteristic zero, is $End(E)$ always a commutative ring? If not, can someone give a counterexample?

Answer 1

It is classically known that the endomorphism ring of an elliptic curve $E/K$ is either $\mathbf Z$, an order in an imaginary quadratic field, or an order in a quaternion algebra; see e.g. Silverman, "The Arithmetic of Elliptic Curves", III, 9. A complete description of $End(E)$ can be found in

M. Deuring, "Die Typen der Multiplikatorenringe elliptischer Funktionen-körper", Abh. Math. Sem. Hamburg, 14, 197-272, 1941.

If char$(K)=0$, one says that $E$ has complex multiplication if $End(E) \neq \mathbf Z$; note that in characteristic $0$, then $End(E)\otimes \mathbf Q$ cannot be a quaternion algebra (the proof is analytic). In characteristic $p\neq 0$, there is a surprising connection between $End(E)$ and the subgroup $E[p]$ of $p$-torsion (over $\bar K$): if $K$ is perfect, $End(E)$ is an order in a quaternion algebra iff $E[p]=0$; $E$ is then called supersingular. If $K$ is a finite field, then $End(E) \neq \mathbf Z$, and there exist $E/\bar K$ with non commutative $End(E)$; more precisely, for $p$ odd, there are simple criteria which allow to determine the supersingular $E$ up to isomorphism.

Answer 2

Let $E$ be a supersingular elliptic curve (over a field of positive characteristic). Then its endomorphism algebra is the (unique) quaternion algebra over $\Bbb Q$ ramified at $p$ and at $\infty$. So the endomorphism algebra need not be commutative in general.