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.
Best Answer
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.