How about this:
sage: E=EllipticCurve('75a1')
sage: P=E.heegner_point(-56)
sage: P1 = P.point_exact()
sage: K = P1[0].parent()
sage: E = P1.curve()
sage: G = K.automorphisms()
sage: def apply(sigma, pt):
....: E = pt.curve()
....: return E([sigma(c) for c in pt])
sage: sum([apply(sigma,P1) for sigma in G], 0)
(0 : 1 : 0)
sage: r56 = K(-56).sqrt()
sage: sum([apply(sigma,P1) for sigma in G if sigma(r56)==r56], 0)
(47/14 : 363763/587699796992*a^7 + 363763/41978556928*a^6 +
2404005/2998468352*a^5 + 80502545/10494639232*a^4 +
49062308611/146924949248*a^3 + 138252123705/73462474624*a^2 +
31053597881/18365618656*a - 9514261313/4591404664 : 1)
That last point is defined over Q(sqRT-56)):
sage: tP = sum([apply(sigma,P1) for sigma in G if sigma(r56)==r56], 0)
sage: tP[1].minpoly()
x^2 + x + 20007/2744
On Sun, 16 Feb 2020 at 09:07, kevin lui <[email protected]> wrote:
> I think this has to do with the fact that when you call `point_exact`,
> Sage is mapping sending all the conjugates to the same point in the Hilbert
> class field.
>
> # As expected, only one conjugate is equal to the original.
> [Q == P for Q in P_cong]
> # Output: [True, False, False, False]
>
> # Annoyingly, these are all equal.
> [Q.point_exact() == P.point_exact() for Q in P_cong]
> # Output: [True, True, True, True]
>
> We can get around this by working numerically.
>
> P_cong_lambda = [Q.map_to_complex_numbers() for Q in P_cong]
> P_trace_lambda = sum(P_cong_lambda)
> P_trace_E_CC = E.period_lattice().elliptic_exponential(P_trace_lambda)
> X, Y, Z = P_trace_E_CC; P_trace_E_CC
> # Output: (3.35714285714284 + 2.94598962112462e-14*I : -0.500000000000106
> - 2.65352233041626*I : 1.00000000000000)
>
> Now we can use `algdep` to get a point on in an imaginary quadratic field.
>
> fX = algdep(X ,2); fX
> # Output: 14*x - 47
> fY = algdep(Y, 2); fY.disc().squarefree_part()
> # Output: (2744*x^2 + 2744*x + 20007, -14)
> -14 is also the squarefree part of -56 so that's a good sign! Now put this
> point on an elliptic curve.
>
> L.<b> = NumberField(fY)
> E_L = E.change_ring(L)
> E_L([47/14, b, 1])
>
> No errors so I think that's an actual point on E. I think there's a smart
> way to identify Y as a point on K=QQ(sqrt(-56)) but it's late and I'm
> blanking on it so I just left it in L.
>
>
>
>
>
>
> On Friday, February 14, 2020 at 10:37:16 PM UTC-8, Ahmed Matar wrote:
>>
>> Sage has the ability to compute the Heegner points over ring class
>> fields. I would like to compute the trace of the Heegner point defined over
>> the Hilbert class field down to K. Apparently this is not implemented in
>> sage, but coding it should be easy. However I'm having an issue with the
>> code below
>>
>> E=EllipticCurve('75a1')
>> P=E.heegner_point(-56)
>> # Note that K=Q(sqrt(-56)) has class number 4
>> P_conj=P.conjugates_over_K() #These are the 4 conjugates of the Heegner
>> point P
>>
>> P_trace=P_conj[0].point_exact()+P_conj[1].point_exact()+P_conj[2].point_exact()+P_conj[3].point_exact()
>>
>> P_trace should be the desired trace of P down to K. When one prints
>> P_trace we see that it is a point in projective space whose 3rd coordinate
>> is 1 so the 1st and second coordinates should be in K. However if we try
>> P_trace[0].minpoly('x'), we get a polynomial of degree 4. What am I doing
>> wrong?
>>
>> Ahmed
>>
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