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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 $\mathcal{P} \nmid m$?

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    $\begingroup$ Do you mean the "discriminant" with "gradient"? Not sure if i understood what you say. where do i need then the condition? $\endgroup$
    – anama
    Commented Jun 12, 2017 at 15:50
  • $\begingroup$ I meant the elliptic curve is up to a rational map $zy^2 = 4x^3 - gxz^2-gz^3,g=\frac{27 j}{j-1728}$ that you can reduce $\mod \mathcal{P}$ and check it is non-singular. Maybe it meant the points in $E[m]$ have coordinates $\not \in\mathcal{P}$ ? $\endgroup$
    – reuns
    Commented Jun 12, 2017 at 16:25
  • $\begingroup$ @reuns "up to rational map" is a very bad idea if you want to talk about reduction. If you change a curve by a quadratic twist, the $j$-invariant stays the same but the reduction property changes. $\endgroup$
    – Chris Wuthrich
    Commented Jun 13, 2017 at 9:17

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(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.

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  • $\begingroup$ Thanks for your answer. Your argument that this prime is chosen so by its property iv convince me, because I thougt so too. But what for we use then the fifth property which is that $\mathcal{P}$ does not divide $m$? $\endgroup$
    – anama
    Commented Jun 13, 2017 at 9:48
  • $\begingroup$ To be certain that the $m$-torsion injects as in VII.3.1 in Silverman1, otherwise there could be $m$-torsion in the formal group. $\endgroup$
    – Chris Wuthrich
    Commented Jun 13, 2017 at 12:29

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