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Consider the maximal unramified extension $\mathbb{Q}_p^{nr}$ of the field $\mathbb{Q}_p$ of $p$-adic numbers. Its residue field is equal to $\overline{\mathbb{F}_p}$. Consider now some field $K$ satisfying $\mathbb{F}_p \subseteq K \subseteq \overline{\mathbb{F}_p}$ and consider the corresponding unramified extension $F$ of $\mathbb{Q}_p$, which is of course contained in $\mathbb{Q}_p^{nr}$.

Under what conditions on $K$ is the field $F$ dense in $\mathbb{Q}_p^{nr}$ for the topology given by the valuation?

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The maximal unramified extension is not the fraction field of the Witt vectors of $\bar{\mathbf{F}_p}$ because its valuation ring is not complete. The ring of Witt vectors of $\bar{\mathbf{F}_p}$ is the completion of the ring of integers of $\mathbf{Q}_p^{nr}$. So the question doesn't really make sense. – Keenan Kidwell May 25 '12 at 19:18
Sorry, that was sloppy. Better now? :-) – Evariste May 25 '12 at 21:46
up vote 2 down vote accepted

The question makes sense: if $K \subset \overline{\mathbb F}_p$, then $K^{\times}$ is a subgroup of $\overline{\mathbb F}_p$, a group of roots of unity. We can then take $F$ to be $\mathbb Q_p$ adjoint the corresponding group of roots of unity in $\mathbb Q_p^{ur}$.

But $F$ will then be closed in $\mathbb Q_p^{ur}$. (The $p$-adic completion $\hat{F}$ of $F$ is the fraction field of the ring of Witt vectors $W(K)$, and so is closed in the completion $\widehat{\mathbb Q_p^{ur}}$, which is the fraction field of the ring of Witt vectors of $\overline{\mathbb F}_p$. My claim then follows from the formula $F = \hat{F} \cap \mathbb Q_p^{ur}$.)

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