Non-zero prime ideals are maximal in the ring of algebraic integers

Question

Let $A= \{y \in \mathbb{C} :$ $y$ integral over $\mathbb{Z}$ }. Let $P\not=\{0 \}$ be a prime ideal of $A$. I am supposed to prove that $P$ is also a maximal ideal. But I cant make it, is this really even true?

Answer 1

$P \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. It cannot be zero: Choose $p \in P \setminus \{0\}$. There is some polynomial equation $p^n+z_{n-1} p^{n-1} + \dotsc + z_0=0$ with $z_i \in \mathbb{Z}$. Choose $n$ minimal. Then we cannot have $z_0 = 0$. Thus, $z_0 \in P \cap \mathbb{Z}$ and $z_0 \neq 0$.

Hence, $P \cap \mathbb{Z} = (p)$ for some prime number $p$. Then $\mathbb{Z}/(p) \to A/P$ is an injective homomorphism. Every element of $A/P$ is integral over $\mathbb{Z}/(p)$, which is a field. This implies (see below) that $A/P$ is a field and hence $P$ is maximal.

Fact. If $R \to S$ is an injective integral homomorphism of integral domains and $R$ is a field, then $S$ is a field. Proof is an exercise.

Answer 2

$A$ is integral over $\mathbb{Z}$, thus has the same dimension as $\mathbb{Z}$, which is $1$.