On Nov 18, 2007, at 8:49 AM, Martin Albrecht wrote:

>
> On Sunday 18 November 2007, David Harvey wrote:
>> On Nov 18, 2007, at 4:16 AM, Robert Bradshaw wrote:
>>>> #1130
>>>
>>> This seems to rely on an earlier patch. (#1120?) See comments on  
>>> trac.
>>
>> I'm very concerned about this patch. It is not the case that the LCM
>> of the orders of all elements of E(GF(q)) will equal the order of E 
>> (GF
>> (q)). I haven't tried the code, but if I understand the code
>> correctly, it will go into an infinite loop on such cases, and it may
>> well give incorrect results in other cases.
>
> Yes, it should not go in, my bad, sorry. I quickly hacked to  
> together the
> algorithm in "Elliptic Curves" by Lawrence Washington and  
> apparently screwed
> up badly on the way. He writes:
>
> 7. If we are looking for the #E(F_q), then repeat steps (1)-(6)  
> [finding the
> order of a point, malb] with randomly chosen points in E(F_q) until  
> the
> greatest common multiple of the orders divides only one integer N  
> with q +
> 1 -2*sqrt(q) <= N <= q + 1 + 2*sqrt(q). Then N = #E(F_q).
>
> Apparently I overread the 'divides' part. Also, what is a   
> 'greatest common
> divisor'?

I still don't believe this algorithm.

Look at this example:

sage: K.<a> = GF(3^4)
sage: K.polynomial()
a^4 + 2*a^3 + 2
sage: E = EllipticCurve(K, [2*a^2 + 2*a + 2, 2*a^3 + 2*a + 1])
sage: points = E.points()
sage: len(points)
100
sage: LCM([P.order() for P in points])
10

The hasse bound says the the number of points must be in [64, 100].  
But if the best we can do is show divisibility by 10, that's not  
enough information: it could be 70, 80, 90, or 100.

Does Washington place any other restrictions on the finite field or  
on the curve?

david


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