I scoured Silverman's two books on arithmetic of elliptic curves to find an answer to the following question, and did not find an answer:
Given an elliptic curve E defined over H, a number field, with complex multiplication by R, and P is a prime ideal in the maximal order of H and E has good reduction at P. Is it legitimate to reduce an endomorphism of E mod P?
In the chapter "Complex Multiplication" of the advanced arithmetic topics book by Silverman, a few propositions and theorems mention reducing an endomorphism mod P.
A priori, this doesn't seem trivial to me. Sure, the endomorphism is comprised of two polynomials with coefficients in H. But I still don't see why if a point Q is in the kernel of reduction mod P, why is phi(Q) also there. When I put Q inside the two polynomials, how can I be sure that P is still in the "denominator" of phi(Q)?
(*) I looked at the curves with CM by sqrt(-1), sqrt(-2) and sqrt(-3), and it seems convincing that one can reduce the CM action mod every prime, except maybe in the case of sqrt(-2) at the ramified prime.
Best Answer
I'm not sure if there's a trivial way to see this. One answer is to use the fact that every rational map from a variety X / $\mathbb{Z}_p$ to an abelian scheme is actually defined on all of X (see for instance Milne's abelian varieties notes). Here, since the generic fiber is open in X you can apply this by viewing the map you started with as a rational map.