How to compute a cyclic subgroup of a class group?
 
  
 
  
 
 
 
 
 
 
    
        
        asked 7 years ago  
 anonymous user 
Anonymous
 
 
 
 
    
        
        updated 7 years ago  
 
 FrédéricC gravatar image  
 
 
 
 
 
Hello, I am quite new to computing in Sage and I am stuck at this point. Please help me out. I am going to use algebraic number theory notation without explaining them in detail. All computations are in quadratic fields.
 
What I have :
 
     
  • Δp=p2ΔK where p is a prime, ΔK=−pq and ΔK≡1 mod 4
    •  
  • I computed the class group, C(Δp) in Sage
    •  
 
What I want :
 
     
  • Set f=[(p2,p)] in C(Δp) where (p2,p) is the standard representation of an ideal of norm p2
    •  
  • Set F=⟨f⟩
    •  
 
My question is :
 
     
  • How to get that cyclic subgroup F?
    •  
 
I have been looking at AbelianGroup and ClassGroup in Sage but I am not quite understanding how to use these to actually get what I want.
 
Any help in this matter would be greatly appreciated.
 
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Comments
Could you please provide the Sage code you already has ?
tmonteil gravatar imagetmonteil (
    
        7 years ago
    )
Let us see if your notation matches the guess of my sage and number theory impressions.
The following code initializes the imaginary quadratic number field of discriminant −pq, p prime. (And q too in my case.)
sage: p = 17
sage: q = 23
sage: DK = -p*q
sage: DK
-391
We initialize the quadratic field with the above discriminant and some order...
sage: K.<a> = QuadraticField( DK )
sage: K.discriminant()
-391
sage: Order = K.order( p*a )
sage: Order.discriminant().factor()
-1 * 2^2 * 17^3 * 23
sage: p
17
sage: Order.discriminant() == 4 * p^2 * DK
True
Well, we've got some factor four. If you do not expect it, please give us some explicit values for p,q.
Note: The class group is so far implemented only for the max order.
dan_fulea gravatar imagedan_fulea (
    
        7 years ago
    )
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