Primes in ring of Gauss integers - help

[T-O7]

Primes in ring of Gauss integers - help!

I'm having a very difficult time solving this question, please help!
So I'm dealing with the ring R=\field{Z}[\zeta] where
\zeta=\frac{1}{2}(-1+\sqrt{-3})
is a cube root of 1.
Then the question is:
Show the polynomial x^2+x+1 has a root in F_p if and only if p\equiv1 (mod 3).

I thought i could show this in two steps, by showing that:
a) a solves x^2+x=-1(mod p) if and only if a is an element of order 3 in F^x_p.
b)F^x_p contains an element of order 3 if and only if p\equiv1 (mod 3).

I've proved part b, but i can't seem to get a hold of a.
Please help

 

[Hurkyl]

Well, first I think you need to review the problem and definitions... the things you've said don't seem to connect to one another.


Anyways, I think this might help: note that

x^3 - 1 = (x - 1) (x^2 + x + 1)

 

[T-O7]

Hmm...sorry, i don't see what you mean.

 

[Hurkyl]

It means that if a is a root of x^2 + x + 1, then it is also a root of x^3 - 1. (i.e. it is a cube root of 1)

 

[T-O7]

ah, so a^3=1, and a has order 3. We can apply the argument backwards, and that will prove a). I see, thanks Hurkyl!
There's also a second part to this problem, which says (p) is prime ideal in R if and only if p=-1 (mod 3)
Apparently the first part of this problem applies, but i'll have to think about this more.

 

[Hurkyl]

Ah, so that's why you mentioned R.

(Incidentally, I think you meant algebraic integers, not Gaussian integers)

I don't know if it will help, but note that if F_p has a root of x^2 + x + 1, then there is a homomorphism from R onto F_p.

 

[robert Ihnot]

I was looking at a somewhat similar problem, https://www.physicsforums.com/showthread.php?t=60863

You use quadratic reciprocity on X^2\equiv-3 Mod p to discover that p is of the form 3k+1. Thus, this prime splits over a field with\sqrt{-3},so it would not generate a prime ideal.

Example: 7=2^2+2+1=(2-\zeta)(2-\zeta^2)

By the way, as Hurkyl points out, this is an algebratic number theory problem.

 

[mathwonk]

this is elementary. look at the group homomorphism from the multiplicative group Fp - {0} to itself defined by cubing. then if there is a primitive cube root of 1, the map is 3 to 1, and has image of order 1/3 the order of the group, i.e. then 3 divides p-1. on the other hand if 3 divides the order of the group, it is elementary group theory that there exists an element of order 3.