Let $K$ be a quadratic imaginary field, and $E$ an elliptic curve whose endomorphism ring is isomorphic to the full ring of integers of $K$. Let $j$ be its $j$-invariant, and $c$ an integral ideal of $K$. Consider the following tower:


$$K(j,E[c])\ \ /\ \ K(j,h(E[c]))\ \ /\ \ K(j)\ \ /\ \ K,$$


where $h$ here is any Weber function on $E$. (Note that $K(j)$ is the Hilbert class field of $K$). 


We know that all these extensions are Galois, and any field has ABELIAN galois group over any smaller field, EXCEPT POSSIBLY THE BIGGEST ONE (namely, $K(j,E[c]) / K$). 


Questions:


1. Does the biggest one have to be abelian? Give a proof or counterexample.


My suspicion: No, it doesn't. I've been trying an example with $K = \mathbf Q(\sqrt{-15})$, $E = C/O\_K$, and $c = 3$; it just requires me to factorise a quartic polynomial over $\bar{ \mathbf Q}$, which SAGE apparently can't do.


2. What about if I replace $E[c]$ in the above by $E\_{tors}$, the full torsion group?



Class Field Theory for Imaginary Quadratic Fields
Here is a case where it is non-Abelian. I use $K$ of class number 3. If I use the Gross curve, it is Abelian. If I twist in $Q(\sqrt{-15})$, it is Abelian for every one I tried, maybe because it is one class per genus. My comments are not from an expert.



```
> K<s>:=QuadraticField(-23);                                                    
> jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2);                             
> jrel:=PowerRelation(jinv,3 : Al:="LLL");                                      
> Kj<j>:=ext<K|jrel>;                                                           
> E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]);
> HasComplexMultiplication(E);
true -23                 
> c4, c6 := Explode(cInvariants(E)); // random twist with this j                
> f:=Polynomial([-c6/864,-c4/48,0,1]);                                          
> poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic                 
> R:=Roots(poly);                                                               
> Kj2:=ext<Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])>;                          
> KK:=ext<Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])>;                          
> assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here         
> f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1];  
> GaloisGroup(f); /* not immediate to compute */                                
Permutation group acting on a set of cardinality 12
Order = 48 = 2^4 * 3
> IsAbelian($1);                                                                
false

```

This group has $A\_4$ and $Z\_2^4$ as normal subgroups, but I don't know it's name if any.


PS. 5-torsion is too long to compute most often.


