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.

Hello,

In general $K(j,E[c])$ will not be abelian over $K$ (the reason being that $K(j,h(E[c]))$ is the ray class field of $K$ of conductor $c$, therefore maximal for abelian extensions of conductor $c$). However, $K(j,E[c])$ is always abelian over $K(j)$. In particular, if the class number of $K$ is $1$, the answer is yes to both your questions, because $K(j)=K$.

For more on this, see Silverman's "Advanced topics in the AEC". In particular, see pages 135-138, and Example 5.8 discusses briefly this question.

Alvaro

The extension K(j,E_{tors}) is abelian over the Hilbert class field of K, hence over K if K has class number 1. Silverman (Advanced topics, p. 138) says that, in general, the extension is not abelian. For getting a counterexample, looking at ${\mathbb Q}(\sqrt{-15})$ is the right idea. Instead of factoring the quartic you might simply want to compute its Galois group, which you probably can read off the discriminant and the cubic resolvent.

The field generated by $E_{tors}$ is the union of the fields generated by the $E[c]$, so for $c$ large enough you should see the nonabelian group already there.