why is $(2) \subseteq \mathbb{Z}[\zeta]$ a prime ideal? Previously I asked something about the ring $\mathbb{Z}[\zeta]$ where $\zeta$ is a cube root of $1$ in $\mathbb{C} \setminus \{1 \} $. Could you provide me hints for showing that $(2)$ is prime

Research effort
I tried to find some isomorphim with $\mathbb{Z}[X] / (X^3-1)$ with the substitution of $\zeta$, but this failed because $\ X-1 \mapsto 0 \ $, but $\ X-1 \notin (X^3+1) \ $.
I don't know if there is another useful isomorphism.
I also tried something straight forward.
Assume that $z \cdot z'=(a+b\zeta)(c+d\zeta) = 2m +2n\zeta$ for some $n,m \in \mathbb{Z}$. I calculated that 
$$z z' = ac+bd\zeta^2 + (ad+bc)\zeta = ac+bd (-1-\zeta) + (ad+bc)\zeta = (ac-bd)+(ad+bc-bd)\zeta$$. I took two elements $z, z' \notin (2)$ to show that $z \cdot z' \notin (2)$.
This leads to a lot of case-checking though, 9 cases if i'm not mistaking.
I hope that you can give me an approach that takes less time. If you think I should do the case-checking, I'm fine with it. Thank you for your answer.
 A: This is an answer/hint related to your initial attempt to find an isomorphism.

The rings $\mathbb Z[\zeta]$ and $\mathbb Z[X]/(X^3 - 1)$ are not isomorphic.  One way to see this is the left has rank two as a $\mathbb Z$-module, and the right one has rank three.  Another way is to see that $X^3 - 1$ is not irreducible, and so $\mathbb Z[X]/(X^3 -1)$ is not an integral domain. 
Do you know about minimal polynomials?  If so, what is the minimal polynomial of $\zeta$?  
Do you know how to factor $X^3 - 1$?  If so, and you compare with the minimal polynomial of $\zeta$, do you see a potentially helpful (and correct!) isomorphism? 
A: The said ring is isomorphic to $\mathbb Z[X]/(X^2+X+1)$. The ideal $(2)$ in it is prime iff the quotient by this ideal is an integral domain. That quotient would be $\mathbb F_2[X]/(X^2+X+1)$ where $\mathbb F_2$ is the field with two elements. Now it remains to prove that $X^2+X+1$ is an irreducible polynomial over $\mathbb F_2$. This should be doable for you, no?
A: The norm map $a+b\zeta\mapsto a^2-ab+b^2$ is multiplicative, integer valued,  and helpful.
