Find all $\mathfrak{p}\in\operatorname{Spec}(\mathbb{Z}[\zeta])$ such that $\mathfrak{p}\cap\mathbb{Z}=7\mathbb{Z}$ where $\zeta=e^{(i\pi/3)}.$
I determined that the minimal polynomial is $f(x)=x^2-x+1$. Since $f(\zeta)=0$ and $f$ is an irreducible quadratic it must be the minimal polynomial.
Then $\mathbb{Z}[\zeta]\cong\mathbb{Z}[x]/(x^2-x+1)$. Since this is an integral extension of $\mathbb{Z}$ it has the same Krull dimension, namely $1$. Primes of $\mathbb{Z}[x]/(x^2-x+1)$ are primes of $\mathbb{Z}[x]$ that contain $(x^2-x+1)$ so all primes in $\mathbb{Z}[x]/(x^2-x+1)$ look like $(q,x^2-x+1)$ where $q\in\mathbb{Z}$ is prime. Thus there is only a single prime $\mathfrak{p}\in\operatorname{Spec}(\mathbb{Z}[\zeta])$ such that $\mathfrak{p}\cap\mathbb{Z}=7\mathbb{Z}$, namely $(7,x^2-x+1)$, which is the ideal $(7)\subset \mathbb{Z}[\zeta]$.
Somehow the wording of the problem "find all prime ideals and give generators" made me expect more... Is this correct? Many thanks for an answer.