Let $A= \{y \in \mathbb{C} :$ $y$ integral over $\mathbb{Z}$ }. Let $P\not=\{0 \}$ be a prime ideal of $A$. Prove that $P \cap \mathbb{Z} \not= \{0 \}$.
Iam totally stuck here, it is given that $P$ contains an element $a \not= 0$, I tried to assume that $P$ contains no integer, so $a$ must be a complex number. I have concluded that $A$ contains the whole $\mathbb{Z}$, and it contains all complex numbers of form $ki$ with $k \in \mathbb{Z}$ and if $A$ contains $a$, then it must contain its conjugate (its conjugate is the root of the same monic polynomial as $a$ is the root of)
Although I feel im close, I cant make it from here. Any hints please?
Hint:
Show that the constant term of the minimal polynomial of some nonzero element $z \in P$ also belongs to $P$.