Degree of cyclotomic extension of local field

Question

Let $K$ be a local field and let $L$ be a degree $n$ separable extension. Let $\kappa_K$ and $\kappa_L$ be the associated residue fields, which satisfy $|\kappa_K| = q$ and $[\kappa_L:\kappa_K]=f$. By Hensel's lemma, we know that $L$ contains a primitive $q^f$-$1$st root of unity $\zeta$, and that $\kappa_K(\bar{\zeta}) = \kappa_L$. Therefore, $\kappa_L=\kappa_{K(\zeta)}$, so $[\kappa_{K(\zeta)}:\kappa] = f$. I'd like to prove that $K(\zeta)$ is unramified, which boils down to proving $[K(\zeta):K]=f$. I feel like Hensel's lemma should explain this part too, but I'm sort of stuck. Any help would be appreciated.

Answer 1

I’m not sure what the problem is. You have $K\subset L$, and between them you have the maximal unramified extension of $K$ in $L$, call it $K^{\text u}$. You know that the residue field extensions $f^L_K=f^{K^{\text u}}_K=[\kappa_L:\kappa_K]$, while for the ramification degrees, $e^L_K=e^L_{K^{\text u}}$, i.e. the top layer is totally ramified, the bottom is totally unramified. You have to use multiplicativity of both the residue field degrees $f^?_?$ and the ramification indices $e^?_?$.

If this isn’t enough for you, get back to me in a comment, and I’ll try to fill in any missing details.

EDIT — Addendum:
You have asked why, when adjoining a primitive $q^f$-th root of unity to $K$ you get the maximal unramified extension of $K$ in $L$. Let’s see how this works:

I’ll change your notation slightly, letting the residue-class fields be $\kappa$ and $\lambda$, so that my $\kappa$ is your $\kappa_K$, and my $\lambda$ is your $\kappa_L$. Now, by definition, the residue-field extension degree of $L$ over $K$, which we’re denoting $f$, is $[\lambda:\kappa]$. Let $\zeta_0$ be a primitive ($q^f-1$)-th root of unity in $\lambda$, say with minimal $\kappa$-polynomial $\varphi(X)\in\kappa[X]$. I’m going to call $q^f=Q$, for convenience.

Now lift $\varphi$ to any monic polynomial $\Phi(X)\in K[X]$; this is still $K$-irreducible. Since $\varphi$ splits into linear factors over $\lambda$, one of these being $X-\zeta_0$, Hensel tells us that $\Phi(X)$ also splits into linear factors over $L$, one of these being $X-\xi$ with $\xi\mapsto\zeta_0\in\lambda$, but of course $\xi$ need not be a root of unity at all.

Nonetheless, $\xi^{q^f}\equiv\xi\mod{\mathfrak M}$, where $\mathfrak M$ is the maximal ideal in the integers of $L$. I’ll ask you to consider the sequence $$\xi, \xi^Q, (\xi^Q)^Q=\xi^{Q^2},\xi^{Q^3},\cdots$$ which you easily show is $\mathfrak M$-adically convergent, all of the terms being congruent to $\xi$ modulo $\mathfrak M$. And of course the limit is a ($Q-1$)-th root of unity reducing to $\zeta_0$ in $\lambda$. This limit is what I’ll call $\zeta$; it’s a primitive such root. Its minimal polynomial is congruent to $\Phi$ modulo $\mathfrak M$, that is, it reduces to $\varphi(X)$, and it’s still irreducible.

Putting it all together, what do we have? $K(\zeta)=K(\xi)$ is of degree $f$ over $K$, unramified since its residue-field degree is equal to its vector-space degree. And certainly there is no larger unramified extension of $K$ in $L$. So I think that you have your maximal unramified extension, generated by a root of unity of the right type.