# Primes in ring of Gauss integers - help

Primes in ring of Gauss integers - 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.

Hurkyl
Staff Emeritus
Gold Member
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)

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

Hurkyl
Staff Emeritus
Gold Member
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)

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)

Hurkyl
Staff Emeritus
Gold Member
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 Fp has a root of x^2 + x + 1, then there is a homomorphism from R onto Fp.

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.

Last edited:
mathwonk