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?
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$P \cap \mathbb{Z}$ 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.
Martin Brandenburg
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The OP knows the first part of your answer because he asked it few hours ago; see here. – user26857 Oct 25 '14 at 14:38
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1Yes, your first part was also part of my original question. Thanks for the rest Martin, although I could have used a hint instead of solution :) – Oct 25 '14 at 14:48
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$A$ is integral over $\mathbb{Z}$, thus has the same dimension as $\mathbb{Z}$, which is $1$.
Amitai Yuval
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1Can you explain this, im not familiar with dimension theorems with maximal ideals. – Oct 25 '14 at 13:58
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