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.
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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