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Theorem 1. If ring $B$ is an integral extension of ring $A$ and $P$ is prime ideal of $B$, then $P$ is maximal ideal of $B$ $\Leftrightarrow$ $A \cap P$ is maximal ideal of $A$.

Theorem 2. If an integral domain $B$ is integral extension of ring $A$, then for every prime ideal $p$ of $A$ there exists prime ideal $P$ of $B$, such that $p = A\cap P$.

These theorems I know to be correct. But I don't see how they implicate, that every prime ideal of $\mathcal {O_K}$ is maximal. I know that in $\mathbb Z$ every prime ideal is maximal. $\mathcal {O_K}$ is integral extension of $\mathbb Z$. So for every prime (maximal) ideal $p$ in $\mathbb Z$ there exists prime ideal $P$ of $\mathcal {O_K}$, such that $p$ = $\mathbb Z \cap P$. Now prime ideal $P$ is maximal when $\mathbb Z \cap P$ is maximal (prime) in $\mathbb Z$. As I understand that means that for every prime ideal $P$ of $\mathcal {O_K}$, $\mathbb Z \cap P$ is prime in $\mathbb Z$. Is my logic correct? And if, why last claim holds?

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If $P$ is a non-zero prime ideal of $\mathcal O_K$, then $P\cap\mathbb Z$ is a non-zero prime ideal of $\mathbb Z$ (why?). Then $P\cap\mathbb Z=p\mathbb Z$ for some prime $p\in\mathbb Z$. But $p\mathbb Z$ is maximal in $\mathbb Z$, so the conclusion.

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  • $\begingroup$ $P$ is non-zero prime ideal of $\mathcal {O_K}$ which means that if we multiply any element of this ideal by any element from $\mathcal {O_K}$ we will still have an element from $P$. $\mathbb Z \cap P$ is a subset of $P$ which contains only these elements of $P$, which are from $\mathbb Z$. If we multiply any element of this set by any integer then we will in result get an integer from $P$, so it will be in $\mathbb Z \cap P$. So $\mathbb Z \cap P$ is a non-zero prime ideal of $\mathbb Z$. $\endgroup$ – Jakub Majewski Dec 9 '15 at 21:10
  • $\begingroup$ @JakubMajewski I don't think your observations are enough for a proof of "$P\cap\mathbb Z$ is a non-zero prime ideal of $\mathbb Z$" provided $P\ne0$. $\endgroup$ – user26857 Dec 9 '15 at 21:19
  • $\begingroup$ So we know that $\mathbb Z \cap P$ is a non-zero ideal of $\,\mathbb Z$. To be prime it should satisfy: $(\mathbb Z \cap P) \neq \mathbb Z$ and $a,b\notin (\mathbb Z\cap P) \Rightarrow ab \notin (\mathbb Z\cap P)$, for $a,b \in \mathbb Z$. As $P \neq \mathcal O_K$, $P$ does not contain $1$, so $(\mathbb Z \cap P) \neq \mathbb Z$. $P$ is a non-zero prime ideal of $\mathcal O_K$, so $a,b\notin P \Rightarrow ab \notin P$, especially for $a,b \in \mathbb Z$, so the same is for $\mathbb Z\cap P$. $\endgroup$ – Jakub Majewski Dec 9 '15 at 22:29
  • $\begingroup$ @JakubMajewski The main point here is to show that $P\cap\mathbb Z$ is non-zero. The rest is more or less obvious. $\endgroup$ – user26857 Dec 9 '15 at 23:02
  • $\begingroup$ Because $P$ is non-zero prime ideal of $\mathcal O_K$, which is integral extension of $\mathbb Z$, there is an element $0\neq b\in P$, for which there exists minimal polynomial $f\in \mathbb Z[x]$, such that $f(b)=0$. So $f(b)=b^n+a_{1}b^{n-1}+\ldots +a_{n}=0$. Now by noticing that $a_{n}\neq 0$ (otherwise we could divide this polynomial by $0\neq b$ and we would have satysfing polynomial of lower degree than minimal, which is contradiction) we can subtract $b^n+\ldots +a_{n-1}b$ from both sides, and it becomes clear, that $a_{n}$ belongs also to $P$, so $\,0\neq a_{n}\in \mathbb Z\cap P$. $\endgroup$ – Jakub Majewski Dec 9 '15 at 23:53

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