On Thursday, 17 October 2013 11:12:42 UTC+1, Georgi Guninski wrote:
>
> E=EllipticCurve(QQ,[-100,0]) 
> sha=E.sha() 
> sha.an_padic(13) 
>
> > padic_prec = max(bounds[1:]) + 5 
> ValueError: max() arg is an empty sequence 


Oooops. That is clearly a bug in something I have written. I will fix this, 
but I won't be able to do so before the end of the month I fear.

You can still get the result in two ways. Either :

sage: E = EllipticCurve([-100,0])
sage: sha = E.sha()
sage: sha.an_padic(13,use_twists=False)
1 + O(13^2)

or redoing what the function actually does :

sage: E = EllipticCurve([-100,0])
sage: l = E.padic_lseries(13)
sage: f = l.series(3)
sage: f
9 + 8*13 + 7*13^2 + 5*13^3 + 8*13^4 + O(13^5) + (12 + O(13^2))*T + (5 + 
O(13^2))*T^2 + (2 + 8*13 + O(13^2))*T^3 + (10 + 13 + O(13^2))*T^4 + O(T^5)
sage: s = f[0]
sage: s = s * E.torsion_order()^2 / E.tamagawa_product() 
sage: s/ ( 1-1/l.alpha() )^2
1 + O(13^5)

Tough if you change 100 for a larger square it will quickly become very 
slow without using the twist (which is exactly was is broken here).

Final note. The valuation part of the p-adic BSD actually holds although it 
is not implemented here (since E has complex multiplication). These 
computations prove that the 13-torsion part of Sha is trivial.

Chris.

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