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I am looking for a short proof that if $L \supset K$ are finite extensions of the p-adic numbers $\mathbb{Q}_p$, then if $L/K$ is unramified, $L/K$ is Galois.

I think the proof is related to somehow injecting $Gal(L/K) \hookrightarrow Gal(k_L / k_K) $ where $k_L$ and $k_K$ are the respective residue fields (possibly using the Teichmuller map); then $f=[k_L:k_K] = [L:K]$ by the fact the extension is unramified, so we would get surjectivity by counting degrees. However, I can't quite put it all together.

I have seen a result somewhere about uniqueness of unramified extensions (adjoining a root of unity $\zeta_m$ or something along those lines), but I can't recall the result exactly. I would be very grateful for some help - thanks in advance.

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    I mentally equate "unramified extension of the p-adics" with "extension field of $\mathbf{F}_p$", the finite field of $p$ elements. So adjoining roots would give you the extensions -- every nonzero element of $\overline{\mathbf{F}}_p$ is a root of unity, and adjoining roots of unity to the $p-adics$ would amount to adjoining roots of unity to its residue field.2012-06-10
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    Let $n=[L:K]$ and pick $a \in {\mathcal O}_L$ so that $a \bmod \mathfrak m_L$ is prim. elt. over res. field of $K$. Let $F(x)$ be the min. poly. of $a$ over $K$, so $F(x)$ has coeff. in ${\mathcal O}_K$. Since $F(a) \equiv 0 \bmod {\mathfrak m}_L$ and $a \bmod {\mathfrak m}_L$ has degree $n$ over the res. field of $K$ (from $L/K$ being unram), $F(x) \bmod {\mathfrak m}_K$ is the min. poly. of $a \bmod {\mathfrak m}_K$ over the res. field of $K$. By Galois theory of finite fields, $F(x) \bmod {\mathfrak m}_L$ splits, so by Hensel $F(x)$ splits in $L[x]$. Thus $L = K(a)$ is Galois over $K$.2012-06-10
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    @KCd: So you're essentially saying that finite extensions of finite fields are separable, therefore the reduction of the minimal polynomial of $a$ mod $m_L$ is separable, therefore $L$ is the splitting field of $F$, i.e. the splitting field of a separable polynomial with coefficients in $K$? If so could you perhaps just clarify for me which part of that tells us that $L$ is the splitting field (and not just that the polynomial splits) over $K$? I'm sure it's obvious!2012-06-10
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    Since $L/K$ is unramified, $[L:K]$ equals the residue field degree. Notice where I wrote "from $L/K$ being unram" in my previous comment. Thus $[K(a):K] = n = [L:K]$, so $L = K(a)$. Therefore, since the min. poly. of $a$ over $K$ splits over $L$ and is generated by one root of that polynomial, $L$ is the splitting field of that polynomial over $K$. If you still aren't sure of the reasoning here, look in some *books* on local fields that discuss unramified extensions at an introductory level (Koblitz, Gouvea, Robert,...).2012-06-13
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    This is particular case of the question I answered here: http://math.stackexchange.com/questions/176532/unramified-extension-is-normal-if-it-has-normal-residue-class-extension/176551#1765512012-08-03

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