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$K/F$ is an abelian galois extension of a number field. $\mathcal{O}_F$ (resp. $\mathcal{O}_K$) is the integer ring of $F$ (resp. $K$). Let $\mathfrak{P}$ be a prime of $\mathcal{O}_K$ and $\mathfrak{P}\cap \mathcal{O}_F=\mathfrak{p}$ is a prime of $\mathcal{O}_F$. $I$ denote the inertia group of $K/F$ at $\mathfrak{p}$. $K^I$ is the field fixed by $I$. Prove: $K^I$ is the maximal field which is unramified at $\mathfrak{p}$.

I solved it in finite case by comparing the ramification index of $K^I$ and the maximal $\mathfrak{p}$-unramified field. But the index may become $\infty$ in infinite case.

Is there any hint?

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  • $\begingroup$ You can reduce to the finite case: Take an arbitary $\alpha$ in $K^I$ and consider the finite extension $F(\alpha)/F$. $\endgroup$
    – Jeff
    Jun 27 '15 at 12:56
  • $\begingroup$ @Jeff Do you mean prove that every $F(\alpha)/F$ is unramified? But how it help to prove the maximality? $\endgroup$
    – show5
    Jun 28 '15 at 3:32
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The fact that the extension is abelian means that every subextension is Galois. Let $\alpha \in K$. We want to show that $F(\alpha)/F$ is unramified at $\mathfrak p$ if and only if $\alpha\in K^I$. Let $G$ be the Galois group of $F(\alpha)/F$.

The result follows from the finite case by checking that the inertia group of $F(\alpha)/F$ at $\mathfrak p$ is $I\cap G$.

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