Is an unramified cover of the p-adics determined by its degree?

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...

• Dear Nicole, As Zarrax has indicated, the answer is "yes" (let's assume the degree is finite, just to avoid irrelevant annoyances), and this is a fundamental fact about unramified extensions. As you indicate in your question, the key point is that all the extension is coming just from extending the residue field (and this extension is in turn determined by its degree). One way to "constructively" see the uniqueness is in terms of Witt vectors: if the residue field of the unramified extension $K$ is equal to $k$, then the ring of integers of $K$ can be described explicitly in terms of $k$ ... – Matt E Dec 22 '10 at 4:26
• ... 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 $W(k)$. – Matt E Dec 22 '10 at 4:27

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 $K_2$.
• Right, but any two infinite unramified extensions of $\mathbb{Q}_p$ have the same degree: $\aleph_0$. So the result is not true for infinite extensions. – Pete L. Clark Dec 21 '10 at 23:39