Let $\zeta$ := $e^{2\pi i/3}$ =$-\frac{1}{2} + \frac{\sqrt{-3}}{2} \in \mathbb{C}$, and let $\mathcal{\mathbb{Z}}[\zeta] \subset \mathbb{C}$ be the subring generated by $\zeta$ and $\mathbb{Z}$ and let $f(X)$ be the polynomial $X^2 + X +1 \in \mathcal{\mathbb{Z}}[\zeta]$
- Show that $f(\zeta)=0$ and that the homomorphism of rings $\psi_\zeta :\mathcal{\mathbb{Z}}[x]\rightarrow \mathbb{C}$, defined by $g(X)\mapsto g(\zeta)$ induces an isomorphism of rings $\mathcal{\mathbb{Z}}[x]/(f(X)) \cong \mathcal{\mathbb{Z}}[\zeta]$.
- Prove that $X^2+X+1$ is irreducible in $\mathbb{Z}_2[X]$.
- Deduce that 2 is a prime element of $\mathcal{\mathbb{Z}}[\zeta]$.
- Prove that 3 is not a prime element of $\mathcal{\mathbb{Z}}[\zeta]$ but is contained in a unique prime ideal $P$ of $\mathcal{\mathbb{Z}}[\zeta]$.
I solved the first three points but the last one partially solved. I need to show that 3 is contained in a single prime ideal $P$ of $\mathcal{\mathbb{Z}}[\zeta]$ but I don't know how to do it
For $\alpha = a+b\zeta \in \mathbb{Z}[\zeta]$, with $a,b\in\mathbb{Z}$, define $$ N(\alpha) = \alpha \overline{\alpha} = a^2-ab+b^2. $$ It is clear that if $\alpha,\beta\in\mathbb{Z}[\zeta]$ then $N(\alpha\beta) = N(\alpha) N(\beta)$. It is easy to see that with this norm, $\mathbb{Z}[\zeta]$ is an euclidean domain, and hence a principal ideal domain. Also, if $\alpha$ is an unit, then $N(\alpha) = 1$. You can show that the only units in $\mathbb{Z}[\zeta]$ are $\pm 1$, $\pm(1 + \zeta)$ and $\pm \zeta$.
Using this find all divisors of $3$ up to units in $\mathbb{Z}[\zeta]$. If $\alpha$ is a divisor of $3$ then $3=\alpha\beta$ for some $\beta$ and $N(\alpha)N(\beta) = 9$. If $\alpha$ nor $\beta$ are units, then $N(\alpha) = N(\beta) = 3$. Write $\alpha=a+b\zeta$ as above and then you have to solve the diophantine equation $a^2-ab+b^2 = 3$.
For solving this equation, consider separatedly the cases when $a$ and $b$ have the same sings and opposite sings. You will find that the only possibilities are $\alpha=\pm (1-\zeta)$, $\alpha=\pm (2+\zeta)$ and $\alpha = \pm(1+2\zeta)$. But all of them are associated of $1-\zeta$ (for example $2+\zeta = (1-\zeta)(1+\zeta)$ and $1+\zeta$ is an unit). Thus the only divisor of $3$ up to units is $1-\zeta$.
Finally: Let $P$ be a prime ideal in $\mathbb{Z}[\zeta]$ such that $3\in P$. As $3$ is not prime, then $P\neq (2)$, and because $P$ is a principal ideal domain, we have $P=(a)$ for some $a$. But theb $a$ is a divisor of $3$ thus $a=u(1-\omega)$ for some unit $u$. Thus $P=(1-\zeta)$ is the only prime ideal containing $3$.