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§4. Endomorphisms and Complex Multiplication 241

The special case of e N : N E × N E → µ N (k) is antisymmetric, and if M divides N,
then the following diagram is commutative:
e N
N E × N E −−−−→ µ N (k)
 
N:M
 
N:M
e M
M E × M E −−−−→ µ M (k).
Here ker(λ) is a scheme which is ´ etale for p deg(λ).

(3.9) Remark. Observe that the relation e N (λx, x ) = e N (x, ˆ λx ) shows that e N (λx,

λx ) = e N (x, ˆ λλx ) = deg(λ)e N (x, x ).
§4. Endomorphisms and Complex Multiplication
Now we study the ring End(E) = Hom(E, E) for an elliptic curve E over k. This
ring is equipped with the Rosati involution and the degree map which is possitive
and quadratic.

(4.1) Proposition. The ring End(E) has no zero divisors.
Proof. If µλ = 0, then 0 = deg(µλ) = deg(µ) deg(λ), and, therefore, either
deg(λ) = 0, so λ = 0, or deg(µ) = 0, so µ = 0.

For an alternative proof over C the ring End(E) is the set of complex numbers λ
with λL ⊂ L and, in particular, a subring of C, hence without zero divisors.

(4.2) Definition. For an element λ in End(E) the trace is T (λ = λ + ˆ λ and the
characteristic polynomial is

2
c λ (t) = t − T (λ)t + deg(λ).
Note that in terms of the trace the quadratic condition (c) of (3.5) becomes
deg(λ + µ) = deg(λ) + T ( ˆ λµ) + deg(µ).

(4.3) Proposition. The trace of λ in End(E) is in the subring Z of End(E) and c λ (t)
is in Z[t]. Further, c λ (λ) = 0.

Proof. We calculate that deg(1 + λ) = (1 + λ)(1 + ˆ λ) = 1 + (λ + ˆ λ) + λ ˆ λ is in Z,
and using the fact that deg(λ) is an integer, we deduce that T (λ) is an integer. Finally
2
we have c λ (λ) = λ − (λ + ˆ λ)λ + λ ˆ λ = 0.
By the previous proposition every element of End(E) satisfies a quadratic equa-
tion over the subring Z. This is a very strong restriction on End(E). Moreover, these
quadratic equations have the following additional positivity property.