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Let $\Bbb{Q}_p⊆L_0⊆L$, where $L$ is extension of $\Bbb{Q}_p$ and $L_0$ be maximal unramified sub extension of $L/\Bbb{Q}_p$. Let $O_0$ be ring of integers of $L_p$, $O$ be ring of integers of $L$. Then, why $O$ is finitely generated free module over $O_0$ ?

If this claim follows from some famous fact, I want to know the fact. Thank you in advance.

Poitou-Tate
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    You mean that $O_0$ is the rong of integers of $L_0$ and $O$ the ring of integers of $L$, right? You’re asking why $O$ is free finitely generated over $O_0$, correct? Freeness follows from “finitely generated+flat over a DVR”, $O_0$ being a DVR is well-known (if not its literal definition) and thus flatness is equivalent to torsion-freeness (which is immediate). So we just need to show “finitely generated”. – Aphelli Apr 21 '22 at 06:41
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    The key is that the absolute value topology on $L$ is the $L_0$-vector space topology. Now $O$ is compact, so if we have some open $U$, then $O \subset t U $ for some $t \in L_0^{\times}$. Take now a $L_0$ basis $(u_1,\ldots,u_n)$ of $L$ and take $U=O_0u_1 + \ldots + O_0u_n$. This injects $O$ in a finite $O_0$-module. As $O_0$ is Noetherian, $O$ is finite over $O_0$. – Aphelli Apr 21 '22 at 06:43
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    The ring $O_0$ is a subring of $O = O_L$, so it makes no sense to think of $O_o$ as an $O$-module. You mean the reverse: why is $O$ a finitely generated $O_0$-module? These rings are PIDs of a special type (they're DVRs), and we're in fields of characteristic $0$, so from an algebraic number theory look up the proof that the ring of integers of a number field is a finitely generated $\mathbf Z$-module. The same proof should work in the $p$-adic setting too, since $L/\mathbf Q_p$ is separable. Usually it's discriminant argument: put $O$ between two finite free $O_0$-modules of equal rank. – KCd Apr 21 '22 at 23:53

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