I am working on the proof of the main Theorem of complex multiplication states in "Advanced topics in the arithemtics of elliptic curves" of J.Silverman.


We have the following situation: $K$ is a quadratic imaginary field with ring of intgers $R\_K$, $E/C$ an elliptic curve with End$(E) \cong R\_K$. We fix an integer $m \geq 3$ and let $L/K$ a finite Galois extension satisfying $j(E) \in L$ und $E[m] \subset E(L)$. Then we define a Prime ideal $\mathcal{P}$ which does not divide $m$.
Now it says that we can use a proposition (VII, 3.1 in Arithmetic of elliptic curves from J.Silverman) which has as condition that the reduced curve is non-singular because of $\mathcal{P} \nmid m$. 


Q: How can I conclude that $E$ has good reduction on $\mathcal{P}$ because of $\mathcal{P} \nmid m$? 



How to conclude good reduction from $\mathcal{P} \nmid m$?
(Not sure this question is suitable, but I reply anyway).


Here is a simple counter-example. Take a cm curve like 27a1 over $\mathbb{Q}$ and adjoin the $2$-torsion points. This will be a sextic extension and the curve still has bad reduction at the place above $3$. So it does not suffice to add just $m$-torsion for any old $m$ to achieve good reduction everywhere.


To understand where a curve acquires good reduction everywhere, I think Serre-Tate is a good place to start.


Now, I think the problem with the question is that the intended question was different. My guess is that the question is about the proof of Theorem 8.2 in Silverman2. There $\mathfrak{P}$ is *chosen* such that the reduction is good by its property iv.


