# 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.

• 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. – user14972 Jun 10 '12 at 14:29
• 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 $K$. – KCd Jun 10 '12 at 15:56
• @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! – Spyam Jun 10 '12 at 16:41
• 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,...). – KCd Jun 13 '12 at 12:20
• This is particular case of the question I answered here: math.stackexchange.com/questions/176532/… – Keenan Kidwell Aug 3 '12 at 22:57

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.

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.

• I found a simpler proof. I may refer the lemma of my previous answer elsewhere, so I leave it. – Makoto Kato Aug 3 '12 at 23:32