Unramified p-adic extension implies Galois 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.
 A: Let $A$ and $B$ be the rings of integers of $K$ and $L$ respectively.
Let $\mathfrak{p}$ and $\mathfrak{P}$ the unique maximal ideals of $A$ and $B$ respectively.
Let $F = A/\mathfrak{p}$, $F' = B/\mathfrak{P}$.
For any $\alpha \in B$, we denote by $\bar \alpha$ the image of $\alpha$ by the canonical homomorphism $B \rightarrow F'$.
Lemma
There exists $\theta \in B$ such that $L = K(\theta)$ and $F' = F(\bar \theta)$.
Proof:
There exists $\gamma \in B$ such that $L = K(\gamma)$.
Since $F'$ is a finite field, there exists $\alpha \in B$ such that $F' = F(\bar \alpha)$.
Let $r$ be the number of elements of $B/\mathfrak{P}$.
Let $\theta = \alpha + rt\gamma$, where $t$ is a rational integer.
Since $r \in \mathfrak{P}$, $\theta \equiv \alpha$ (mod $\mathfrak{P}$).
We can choose $t$ such that all the conjugate of $\theta$ over $K$ is distinct.
Then $\theta$ satisfies the desired properties.
QED
Let $\theta$ be as in the lemma.
Let $f(X)$ be the minimal polynomial of $\theta$ over $K$.
Then $f(X) \in A[X]$.
Let $\bar f(X)$ be the reduction of $f(X)$ mod $\mathfrak{p}$.
Since $f(\theta) = 0$, $\bar f(\bar \theta) = 0$.
Let $n = [L : K]$.
Then the degree of $f(X)$ is $n$.
Since $L/K$ is unramified, $n = [F' : F]$.
Hence $\bar f(X)$ is the minimal polynomial of $\bar \theta$ over $F$.
Since $F'/F$ is Galois, $\bar f(X)$ splits in $F'$.
Hence $f(X)$ splits in $L$ by Hensel's lemma and we are done.
A: Let $A$ and $B$ be the rings of integers of $K$ and $L$ respectively.
Let $\mathfrak{p}$ and $\mathfrak{P}$ the unique maximal ideals of $A$ and $B$ respectively.
Let $F = A/\mathfrak{p}$, $F' = B/\mathfrak{P}$.
For any $\alpha \in B$, we denote by $\bar \alpha$ the image of $\alpha$ by the canonical homomorphism $B \rightarrow F'$.
There exists $\theta \in B$ such that $L = K(\theta)$.
Let $f(X)$ be the minimal polynomial of $\theta$ over $K$.
Then $f(X) \in A[X]$.
Let $\bar f(X)$ be the reduction of $f(X)$ mod $\mathfrak{p}$.
Since $f(\theta) = 0$, $\bar f(\bar \theta) = 0$.
Let $n = [L : K]$.
Then the degree of $f(X)$ is $n$, hence the degree of $\bar f(X)$ is also $n$.
Since $L/K$ is unramified, $n = [F' : F]$.
Hence $\bar f(X)$ is the minimal polynomial of $\bar \theta$ over $F$.
Since $F'/F$ is Galois, $\bar f(X)$ splits in $F'$.
Hence $f(X)$ splits in $L$ by Hensel's lemma and we are done.
