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Let $K$ be a quadratic imaginary field. Let $L$ be a number field which contains $K$ and let $E/L$ be an elliptic curve defined over $L$ with complex multiplication by $K$, i.e. such that $End_{\overline{L}}(E)\otimes_{\mathbf{Z}}\mathbf{Q}\simeq K$. One may associate to $E$ a Groessencharacter of $L$ taking values in $K$ $$ \psi:\mathbb{A}_L^{\times}/L^{\times}\rightarrow K^{\times}. $$,

Q1: Is there an obvious upper bound for the conductor of $\psi$ depending only on $L$ (which seems, a priori, to be almost equivalent to ask for an upper bound for the conductor of $E/L$)?

Q2: If $K$ has class number one, then we may choose $L=K$. Can we give an explicit formula for the conductor of $\psi$ in this very special situation?

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    $\begingroup$ The conductor of $E$ is the square of the conductor of $\psi$. Even if you fix the $j$-invariant, there are infinitely many iso classes of CM elliptic curves over $L$, and only finitely many of bounded conductor (by Shafarevich's theorem, for example). $\endgroup$ – Ari Shnidman Jul 14 '14 at 16:19
  • $\begingroup$ So if we look at the set of CM elliptic curves by $K$defined over $\overline{\mathbf{Q}}$, then since they are all isogeneous, they share the same set of primes of good reduction. So is it possible to determine the set of primes of bad reduction (necessarily additive since CM) stricly in terms of $K$. For example if $K$ has class number one a naive guess would be to say that the conductor is supported on primes dividing $12\cdot disc(K)$.... But this is probably to naive. $\endgroup$ – Hugo Chapdelaine Jul 14 '14 at 19:07
  • $\begingroup$ An isogeny over $\overline{\mathbb{Q}}$ need not be defined over $K$. Another way to see that the answer to both of your questions is `no' is to take a single CM elliptic curve over $L$ and look at the family of its quadratic twists (in which the (norms of the) conductors of its members are unbounded). $\endgroup$ – Kestutis Cesnavicius Jul 14 '14 at 23:19
  • $\begingroup$ I see, so then may be I should add an hypothesis like semi-stability and then this means the $E/L$ should have good reduction everywhere... $\endgroup$ – Hugo Chapdelaine Jul 15 '14 at 3:18
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The explicit formula should be: For each prime $p$ of $L$, take the local ring $\mathcal O_p$ of $p$ and look at the induced map $\mathcal O_p^{\times} \to \mathbb A_L^\times \to K^\times$. Let $\pi$ be a uniformizer and take the smallest $n$ such that the image of $1+ \pi^n \mathcal O_p$ along this map is trivial, then take $\# \mathcal O_p /\pi^n$.

This is the local factor, then you take the product over all $p$.

This follows from class field theory.

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  • $\begingroup$ Yes sure, but it seems to me that it should be possible to bound the support of the conductor of $\psi$ just in terms of the number field $L$ (the discriminant of $L$, its degree etc). $\endgroup$ – Hugo Chapdelaine Jul 15 '14 at 3:12
  • $\begingroup$ So the point is as @Cesnavicius wrote, the support of the conductor won't be bounded. $\endgroup$ – Hugo Chapdelaine Jul 15 '14 at 3:27
  • $\begingroup$ Yes, you can see this very easily by multiplying the Grossencharacter with a quadratic character $\mathbb A_L^\times \to \mu_2$. $\endgroup$ – Will Sawin Jul 15 '14 at 3:29

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