Equivalence of definitions of tame ramification

Question

There are three different characterisations of a tamely ramified extension $L/K$ of a local field $K$ and I don't understand why they are equivalent.

$\bullet$ $p \nmid e$ where $p = \rm{char}(k)$, $k$ is the residue field of $K$ and $e$ is the ramification index.

$\bullet$ $G_1 = \{1\}$ where $G_1$ is the ramification group of $L/K$.

$\bullet$ $L/K$ has conductor $1$, where the conductor is the least integer $\frak f$ such that $1+\frak{m}^f \subseteq \rm{Ker}(\phi_{L/K})$. $\frak m$ is the unique maximal ideal of $K$ and $\phi_{L/K}: K^\times \longrightarrow \rm{Gal}(L/K)$ is the local Artin map.

The equivalence of the first two definitions has already been discussed on stackexchange but as far as I know, no proof has been provided.

Hints are very welcome too. Thank you.

Answer 1

$\newcommand{\mf}{\mathfrak}\newcommand{\Gal}{\operatorname{Gal}}\newcommand{\Ker}{\operatorname{Ker}}$ First, let $\mf M$ denote the maximal ideal of $L$, let $\pi_L$ be a uniformizer at $L$, and let $\pi_K=\pi_L^e$. Also, let $v_L$ denote the valuation on $L$, let $U_L^0=U_L=\{x\in L:v_L(x)=0\}$, let $U_L^k=1+\mf M^k$ for $k>0$, let $\ell$ denote $L$'s residue field, and make all the analagous defintions with $K$ in place of $L$. We'll need a couple lemmas that I won't bother proving.

Lemma. $U^0_L/U_L^1\simeq\ell^\times$ and $U_L^i/U_L^{i+1}\simeq\ell$ for $i>0$.

Lemma. There is an injection $\theta_i:G_i/G_{i+1}\to U_L^i/U_L^{i+1}$ induced by the map $s\in G_i\mapsto s(\pi_L)/\pi_L$.

Now we can show the equivalences.

Propostion. The second bullet implies the first.

We have that $G_0\simeq G_0/G_1$ is (isomorphic to) a finite subgroup of $U_L^0/U_L^1\simeq\ell^\times$. $\ell^\times$ contains no nontrivial $p$-torsion since $x^p-1=(x-1)^p$. Hence, $G_0$ also contains no nontrivial $p$-torsion, so $p\nmid|G_0|=e$. $\square$

Propostion. The first bullet implies the second.

We claim that $G_1$ is a $p$-group, from which the claim follows since $|G_1|\mid|G_0|$. Note that $G_i/G_{i+1}$ is (isomorphic to) a finite subgroup of $\ell$, which is a finite-dimensional vector space over $\mathbb F_p$. Hence, $G_i/G_{i+1}$ has order a $p$-power, and so $$|G_1|=\prod_{i=1}^\infty\frac{|G_i|}{|G_{i+1}|}$$ is a $p$-power as well. $\square$

Propostion. The first bullet implies the thrid.

Pick any $\alpha\in U_K^1$. Since $\alpha$ is a unit, it maps to the identity under the composition $$K^\times\xrightarrow{\phi_{L/K}}\Gal(L/K)\to\Gal(\ell/k),$$ so $\phi_{L/K}(\alpha)\in G_0$, the inertia group. Now, let $\mf f$ be the conductor of $L/K$. Then, $\alpha^{p^{\mf f}}\in U_K^{\mf f}$, so $\phi_{L/K}(\alpha)^{p^{\mf f}}=1$, meaning that the $\phi_{L/K}(\alpha)$ has order a $p$-power. At the same time, $|G_0|=e$ is coprime to $p$, so $\phi_{L/K}(\alpha)=1$. $\square$

For the last implication, there's the following fact (that I'm not sure how to prove) which should probably come in handy.

Lemma. The image of $U_K^n$ under the local Artin map is $G^n$.

I don't like thinking about the upper filtration, so I'll leave figuring out how to make use of this to someone else.

Answer 2

$L$ is a finite abelian extension of local field $K$. I will briefly describe the upper numbering to prove that conductor $1$ implies $G_1=\{1\} $.

Let $g_i=\# G_i $ so $g_0 \geq g_1 \geq g_2 \geq \cdots \geq1 $. Define a piecewise linear and continuous function $\phi : [0, \infty) \rightarrow [0, \infty) $ such that $\phi(0)=0 $ and it has slope $g_i/g_0 $ in $(i,i+1)$. You can see that $\phi$ is strictly increasing and thus invertible.

For $v\in \mathbb R_{\geq0 }$ define the upper numbering as $G^v=G_{\lceil \phi ^{-1} v \rceil}$ where $\lceil . \rceil $ is ceiling function.

The image of $U^{m}_K $ under local reciprocity map is $\theta_{L/K}(U^m_{K})=G^m $.

Thus $G^1=\{1\} $ when the conductor is 1. We say $u$ is a jump in uppper numbering if $G^u \neq G^{u+\epsilon} $ for all $\epsilon >0 $.

Theorem (Hasse - Arf): $L/K$ abelian extension of local fields. The jumps in upper numbering are at integers.

Thus $G^1 =G^v $ for all $v \in (0,1] $. If we let $v = g_1/g_0 $ then $$\{1\}=G^1 =G^v=G_{\lceil \phi ^{-1}(g_1/g_0) \rceil} = G_1$$ which is what we wanted.

The proofs of the theorems I have used above, I believe, are given in Serre's article on Local Class Field Theory in Cassels- Frohlich Algebraic Number Theory.