On 11 May 2017 at 08:16, Vincent Delecroix <20100.delecr...@gmail.com> wrote:
> Hi,
>
> "primitive element" is meant as "generator for the multiplicative group
> GF(p)^*" and not the additive group GF(p). The OP question is about the
> former and Johan answer is about the latter.

Not really: generators of the additive group are coprime to p, not to p-1.

Perhaps Johan was thinking of the fact that if g is one multiplicative
generator (aka primitive root) then g^k is another if and only if
gcd(k,p-1)=1.

John Cremona

>
> For very large p such as what you asked for is likely to be delicate (but I
> am not a specialist).
>
> Vincent
>
>
> On 11/05/2017 08:45, Johan S. H. Rosenkilde wrote:
>>
>> Hi Panos
>>
>> In GF(p) then an element g is primitive if its embedding into ZZ is
>> coprime with p-1. Since Euclidean algorithm is so fast, you can test
>> this:
>>
>> sage: p = Primes().next(2^2048) #long
>> sage: g1 = 3
>> sage: gcd(g1, p-1)
>> 3
>> sage: g2 = 5
>> sage: gcd(g2, p-1)
>> 1
>>
>> So 3 is not a primitive element in GF(p) but 5 is. (Since 5 is also a
>> prime, you could also have done g2.divides(p-1) instead)
>>
>> Best,
>> Johan
>>
>>
>> Panos Phronimos writes:
>>
>>> Hello everyone,
>>>
>>> I am trying to calculate a primitive element (g) of a big Finite Field:
>>> GF(p) where p is prime number > 2^2048
>>>
>>> So then, i could share a secret integer (r) as: m=g^r, but it seems
>>> impossible to calculate it with function primitive_element()
>>> Is there another way i can use to calculate it?
>>>
>>> Thanks in advance,
>>> Panos
>>
>>
>
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