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...

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).

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.

Hi @Qiaochu; this is an interesting observation.

Yes you're right. So I guess that the key point is that a subgroup of an abelian variety is necessarily an abelian variety and a quotient of a linear group is linear. Thus the only linear group which is an abelian variety is the trivial group.

But doesn't it follow from the simpler observation that if $\theta\in End_k(G)$ and $\pi:G\rightarrow A$ then the map $\pi\circ\theta:K\rightarrow A$ is constant since $K$ is an affine variety and $A$ is projective?

Nice, so this answers completely my question.