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If $K_1$ and $K_2$ are subfields of a pre-chosen $\overline{\mathbb{Q}_p}$, and if they're both unramified at $p$, and $[K_1:\mathbb{Q}_p]=[K_2:\mathbb{Q}_p]$, does that imply that $K_1=K_2$?

My intuition says that this is true because all that's happening is that we're extending the residue field, and there's only one way to do that if we fix the degree of the extension. But that's not a proof...

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    ... it is the ring of Witt vectors of $k$. (Typically denoted $W(k)$.) Serre's *Local fields* will give you the construction. The field $K$ itself is then determined as the fraction field of $W(k)$.2010-12-22

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This is slightly outside my domain.. but Proposition 5.4.11 of Gouvea's (excellent) book on p-adics says "For each $f$ there is exactly one unramified extension of degree $f$. It can be obtained by adjoining to ${\mathbb Q}_p$ a primitive $p^f - 1$-st root of unity". So in your case $K_1$ and $K_2$ would both have to be extensions of ${\mathbb Q}_p$ by a primitive $p^f - 1$-st root of unity. Since any two such extensions contained within the same $\bar{{\mathbb Q}_p}$ must be the same, $K_1$ must be $K_2$.

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    Right you are. I guess the right definition of degree here should be profinite size.2010-12-22