# Characteristic polynomial of endomorphism of the Tate module of an elliptic curve.

elliptic-curvesnumber theory

In Milne's book Elliptic Curves, he states (Corollary 3.23) that for any endomorphism $$\alpha$$ of $$E$$, we have the following facts about the induced endomorphism $$\alpha$$ of the Tate module $$T_\ell(E)$$:

1. The characteristic polynomial $$X^2+cX+d$$ of $$\alpha$$ has coefficients in $$\mathbb{Z}$$, and they are independent of $$\ell$$.

2. $$c^2\geq 4d$$

3. $$\alpha^2+c\alpha+d$$ is trivial, as an endomorphism of $$E$$.

These follow almost immediately after the definition of the Tate module, and the only nontrivial result between these is the fact that $$d$$ equals the degree of the morphism $$\alpha$$, which suffices to prove $$1$$, and $$3$$ follows easily from just the definition.

My issue is with part $$2$$, which Milne seems not to address in the proof of these statements. Ignoring the typo that the inequality should be reversed, I still don't see how this statement follows without extra technology.

Is there something silly I am missing from his proof? Does there exist an immediate proof of the "Weil Bound" for endomorphisms of elliptic curves, using only the definition of the Tate module and the determinant=degree result?

Okay it is possible to deduce the trace bound from just what we have here. For any endomorphism $$\alpha$$, we see that the subring generated by $$1$$ and $$\alpha$$ is an order in some quadratic extension of $$\mathbb{Q}$$, since it is a commutative subring of endomorphisms of the Tate module generated by one element. Then the degree equals determinant result gives us that every element of this subring has positive norm, so this is in fact an order in an imaginary quadratic extension of $$\mathbb{Q}$$. This then gives our result that any polynomial has complex conjugate roots, and we’re done.