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?
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.