# Conductor $0$ implies unramified

Suppose that $L/K$ is a finite abelian extension of $K$ a local field. Suppose that $U_K$ is contained in $N(L^*)$. Then $L/K$ is unramified.

What I've tried:

I know that $L/K$ is unramified is equivalent to saying that the inertia subgroup is trivial. So take an $s \in G$ such that $s$ acts trivially on $\mathcal{O}_L / p_L$, i.e. $s(x) - x \in p_L$ for every $x \in \mathcal{O}_L$. However, I don't see how to relate this with using that $U_K = \mathcal{O}_K - p_K \subseteq N(L^*)$.

I try to take a uniformizers $\pi_L$ and $\pi_K$. Then I know that $|\pi_L|_L = |\pi_K|_K ^ e$ and try showing that $e = 1$ here. Once again though, I struggle to relate units in $\mathcal{O}_K$.

I know that units of $\mathcal{O}_K$ should be elements with magnitude $1$ or equivalently outside the maximal ideal. I don't understand how to relate this with the norm of $L^*$.

As the problem is presented (a finite abelian extension $L/K$ + an additional condition on norms), the use of CFT seems inevitable. But let us try to keep it minimal. First suppose that $L/K$ is only galois, of degree $n$ and residual degree $f$. Consider the exact sequence of valuation at level $L$, say $0 \to U_L \to L^* \to Z \to 0$ and norm it down to the analogous exact sequence at level $K$. (Ideally, one should write down a commutative diagram, but I don't know how to). By definition of the inertia index, the norm just induces multiplication by $f$ on $Z$. By hypothesis, the norm from $U_L$ to $U_K$ is surjective, hence the snake lemma applied to our commutative diagram immediately shows that $K^*/NL^*\cong Z/fZ$ . Then suppose $L/K$ abelian, so that by local CFT, $Gal(L/K) \cong K^*/NL^* \cong Z/fZ$. In particular $[L:K]=f$, i.e. $L/K$ is unramified.
$N(L^{\times})$ will also contain some power of $\pi_K$ since it has finite index in $K^{\times}$. (By local reciprocity, the index is $[L:K]$)
If $U_K \times \pi_K^n$ is contained in $N(L^{\times})$, then $N(L^{\times})$ contains the norm group of the unramified extension of $K$ of degree $n$, so $L$ is contained in the unramified extension of degree $n$, so it is also unramified.
• Am I understanding this correctly: $N(L^*)$ contains the units and also the $\pi_K$ after some level $n$. These together form the unramified extension, say $F$, where $\pi_F = \pi_K^n$? Commented Jan 2, 2017 at 1:58
• @Future Yes, the norm group of the unramified extension $F$ is not hard to work out since the norm symbol is explicit: $(a,F/K)$ is the Frobenius taken to the power of $v(a)$ (the valuation) so $a \in N(F^{\times})$ if and only if $(a,F/K) = 1$, if and only if $v(a)$ is a multiple of $[F:K]$ Commented Jan 2, 2017 at 2:05
• How do you guarantee unramified extension of degree n? (Suppose char of $K$ is not 0) Commented Sep 23, 2020 at 17:47