Well, if $P$ is not the only prime above $p$, then $O_P$ cannot be a finitely-generated $\mathbb{Z}_{(p)}$-module for the following reason. Suppose $Q$ is another prime ideal above $p$ and select $\beta\in Q\setminus P$. Then $\beta^{-1}\in O_P$. If $O_P$ were finitely-generated as a module over $\mathbb{Z}{(p)}$, \mathbb{Z}_{(p)}$, then it would be integral over $\mathbb{Z}{(p)}$, \mathbb{Z}_{(p)}$, and hence would be contained in the integral closure of $\mathbb{Z}_{(p)}$ in $K$, which is $O_p$. But then $\beta^{-1}\in O_p$, so $1/\beta=\alpha/m$ for some integer $m$ not divisible by $p$. This means that $m=\alpha\beta\in Q$, whence $m\in Q\cap\mathbb{Z}=(p)$, a contradiction.
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Well, if $P$ is not the only prime above $p$, then $O_P$ cannot be a finitely-generated $\mathbb{Z}_{(p)}$-module for the following reason. Suppose $Q$ is another prime ideal above $p$ and select $\beta\in Q\setminus P$. Then $\beta^{-1}\in O_P$. If $O_P$ were finitely-generated as a module over $\mathbb{Z}{(p)}$, then it would be integral over $\mathbb{Z}{(p)}$, and hence would be contained in the integral closure of $\mathbb{Z}_{(p)}$ in $K$, which is $O_p$. But then $\beta^{-1}\in O_p$, so $1/\beta=\alpha/m$ for some integer $m$ not divisible by $p$. This means that $m=\alpha\beta\in Q$, whence $m\in Q\cap\mathbb{Z}=(p)$, a contradiction. |
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