# Lecture 3 

<!--> [Video](https://www.youtube.com/watch?v=_4di81CzUt0&list=PLhsb6tmzSpiysoRR0bZozub-MM0k3mdFR&index=3)-->

:::{.remark}
Plans: group theoretic properties of $G(\bar K/K)$ for $K$ a \(p\dash \)adic field, tamely ramified extensions, and the theorems of local CFT.
:::

:::{.remark}
Setup: $K/\QQpadic$ finite with $L/K$ algebraic, normal, and separable (automatic if we assume $\characteristic k = 0$), so Galois.
The goal is to understand $G_K\da \Gal(\bar K/K)$, we'll first show it surjects onto an easier group and investigate the kernel:
:::

:::{.definition title="Inertia subgroup"}
The **inertia subgroup** is defined as the kernel of the reduction map to residue fields:

\begin{tikzcd}
	0 & \textcolor{rgb,255:red,214;green,92;blue,92}{I_{L/K} \da \ker \kappa} \\
	\\
	& {G(L/K)} \\
	\\
	0 & {G(\kappa_L, \kappa_K) \cong \Gal(L/K)/I_{L/K}}
	\arrow["\kappa", from=3-2, to=5-2]
	\arrow[from=1-2, to=3-2]
	\arrow[from=1-1, to=1-2]
	\arrow[from=5-2, to=5-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMSwyLCJcXEdhbChML0spIl0sWzEsNCwiXFxHYWwoXFxrYXBwYV9MLCBcXGthcHBhX0spIFxcY29uZyBcXEdhbChML0spL0lfe0wvS30iXSxbMSwwLCJJX3tML0t9IFxcZGEgXFxrZXIgXFxrYXBwYSIsWzAsNjAsNjAsMV1dLFswLDAsIjAiXSxbMCw0LCIwIl0sWzAsMSwiXFxrYXBwYSJdLFsyLDBdLFszLDJdLFsxLDRdXQ==)

:::

:::{.remark}
Note that $\kappa_L$ is finite, so $G(\kappa_L/\kappa_L)$ is an easier group to understand, and in fact is procyclic (topologically monogenic).
It's clear that $I_{L/K} \leq G_{L/K}$ is a closed subgroup, and thus corresponds to some $M$ with $K \subseteq M \subseteq L$ where $G(L/M) = I_{L/K}$ and $\Gal(M/K) \cong \Gal(\kappa_L/\kappa_K)$.
In fact, $M$ will be the union of all subfields of $L$ containing $K$ which are unramified over $K$.
:::


:::{.remark}
A special interesting case is when $L = \bar K$, then $M = K^\unram$ will be the maximal unramified extension:

\begin{tikzcd}
	{\bar{K}} \\
	& {} \\
	{K^\unram} \\
	& {} \\
	K
	\arrow[from=5-1, to=3-1]
	\arrow[from=3-1, to=1-1]
	\arrow["{I_{\bar L/K}}"', shift right=5, curve={height=30pt}, dashed, no head, from=3-1, to=1-1]
	\arrow["{\hat{\ZZ}}"', shift right=5, curve={height=30pt}, dashed, no head, from=5-1, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJcXGJhcntLfSJdLFswLDIsIkteXFx1bnJhbSJdLFswLDQsIksiXSxbMSwxXSxbMSwzXSxbMiwxXSxbMSwwXSxbMSwwLCJJX3tcXGJhciBML0t9IiwyLHsib2Zmc2V0Ijo1LCJjdXJ2ZSI6NSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn0sImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwxLCJcXGhhdHtcXFpafSIsMix7Im9mZnNldCI6NSwiY3VydmUiOjUsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9LCJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d)

Note that 
\[
G(K^\unram/K)\caniso \Gal(\bar \kappa_K/\kappa_K) \caniso \hat{\ZZ}
.\]
:::


:::{.question}
Two stories in CFT: what are the groups, and what are the corresponding fields?
:::

:::{.remark}
These are two genuinely different stories, e.g. for $K\in \NF$, for $K^\ab$ the maximal abelian extension it's precisely known what the Galois group is, but what is the actual extension (the Hilbert class field)?
If $K = \QQpadic$, then 
\[
K^\unram = \Union_{m\geq 1, p\notdivides m} \QQpadic(\zeta_m)
,\]
being careful because $\QQpadic(\zeta_p)$ is ramified.
This might also be true for other fields $K$.
:::

:::{.remark}
Setup: let $L/K$ be Galois, but now assume $I_{L/K}\in\Fin\Grp$, e.g. when $L/K$ is a finite extension.
We know $I_{L/K} \normal G(L/K)$ is normal since it is a kernel, and if $L/K$ is finite then the quotient is finite cyclic.
Put a filtration on the inertia group and we'll look at the filtered pieces: note that if $\sigma\in I_{L/K}$, then $\sigma: L\to L$ descends to a local morphism $\sigma: \OO_L\to\OO_L$ preserving $\mfp_L$.


:::{.question}
How much does $\sigma$ disrupt these local pieces?
:::


The claim is that since $I_{L/K}$ finite implies $\mfp_L = \gens{\pi_L}$ is principal, where the idea is that the uniformizer in the base extends all the way to $L$:

\begin{tikzcd}
	L && {\mfp_L} \\
	\\
	M && {\mfp_M = \gens{\pi_K} \in \spec \OO_M} \\
	\\
	K && {\mfp_K = \gens{\pi_K} \in \spec \OO_K}
	\arrow[from=5-1, to=3-1]
	\arrow[from=3-1, to=1-1]
	\arrow["{\text{unram}}", shift left=4, curve={height=-30pt}, dashed, no head, from=5-1, to=3-1]
	\arrow["{I_{L/K}, \text{finite}}", shift left=5, curve={height=-30pt}, dashed, no head, from=3-1, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

Moreover the discrete valuation $v_L: L\surjects \ZZ$ satisfies $v_L = \size I_{L/K} v_K$ on $K\units$.
So write $\mfp_L = \gens{\pi_L}$ for some uniformizer.
Define a filtration by
\[
I_{L/K, i} = \ts{\sigma\in I_{L/K} \st { \sigma(\pi_L)\over \pi_L}\in 1 + \mfp_L^i  }, \qquad I_{L/K, 0} \da I_{L/K}
,\]
where we note that $\sigma(\pi_L)$ will still be a uniformizer, so the quotient is a unit, and we are measuring how far it is from 1.
All of the subquotients turn out to be abelian.

> Setting $i=1$ should recover the Sylow subgroup.

One can check that this defines a series of normal subgroups of $G(L/K)$:
\[
I_{L/K} = I_{L/K, 0} \contains I_{L/K, 1} \contains I_{L/K, 2}\contains \cdots
.\]
Note that since $L = M\gens{\pi_L}$, so if $\sigma$ fixes $\pi_L$ it fixes all of $L$ and must be the identity.
Thus if $\sigma\neq \id$, then $\sigma$ moves $\pi_L$ and $\sigma(\pi_L)/\pi_L \neq 1$, so for $i\gg 1$ one has $\sigma(\pi_L)/\pi_L\neq 1\mod \mfp_L^i$.
The conclusion is that $I_{L/K, i} = 1$ for $i\gg 1$, making this a finite filtration.

We can write down an embedding
\[
I_{L/K}/ I_{L/K, 1} &\embeds \kappa_L\units \\
\sigma &\mapsto \sigma(\pi_L)/\pi_L
.\]

> Jackie is now happy about this.

In particular, the domain is cyclic of order prime to $p$.
So given the full Galois group, we factored out the cyclic unramified part and were left with inertia, and now we've factored something simple out of the remaining inertia part.
Bad news: $I_{L/K, 1}$ is complicated!
:::


:::{.definition title="Lower Numbering"}
For $i\geq 1$, define an embedding 
\[
I_{L/K, i} / I_{L/K, i+1} &\embeds \mfp_L^i/\mfp_L^{i+1} \isoin{\Ab\Grp} \GG_a(\kappa_L) \\
\sigma &\mapsto {\sigma(\pi_L) \over \pi_L} - 1
.\]
Note that the RHS could be an infinite field, but turns out to be a finite-dimension $\FF_p\dash$vector space and thus the domain is isomorphic as a group to $C_p^n$ for some $n$. 
We call this the filtration the **lower numbering**.
In particular, its order is a power of $p$.
:::

:::{.remark}
Upshot: $I_{L/K, 1}$ has $p\dash$power order, is a normal subgroup, and the quotient is cyclic of order prime to $p$, making it the unique Sylow $p\dash$subgroup of $I_{L/K}$.
In particular, this is a $p\dash$group, hence solvable.
:::

:::{.definition title="Tame and wild ramification"}
We say $L/K$ is **tamely ramified** iff $I_{L/K, 1} = 1$.
Otherwise, we say $L/K$ is **wildly ramified**.
:::

:::{.warnings}
This seems to mean "ramified but not too badly", but unramified extensions are also tamely ramified!
However, wildly ramified does imply ramified.
:::

:::{.remark}
We're interested in $L = \bar K$ where $I_{L/K}$ is not finite, so we'll need to glue using a limit.
Note that the lower numbering doesn't behave well with respect to extensions of $L$ since extensions $L'/L$ change the uniformizer.
More precisely, if $L'/L/K$ with $L/K$ and $L'/K$ Galois with $I_{L'/K}$ finite, there is a map $I_{L'/K}\surjects I_{L/K}$ but $I_{L'/K, i}$ does not get identified with $I_{L/K, i}$ -- the issue is that $\pi_L$ may have nothing to do with $\pi_{L'}$, and one is trying to control things $\mod \pi_L^i$ and $\mod \pi_{L'}^i$.

The fix: introduce a relabeling.

:::

:::{.definition title="Upper numbering"}
Set $g_i = \size I_{L/K, i}$ so $g_0 \geq g_1 \geq \cdots \geq g_M = 1$ for some $M\gg 0$.
Define a PL continuous function $\phi$, a scaling factor.
Some facts about it:

- $\phi$ will be linear on $(i, i+1)$ for all $i$ 
- $\phi(0) = 0$
- The slope of $\phi$ on $(i, i+1)$ will be ${g_{i+1} \over g_0}$, 
- $\phi: [0, \infty)\to [0,\infty)$ is a strictly increasing bijection.

Note that the slopes are decreasing, and limit to a constant slope $g_M/g_0$.
For $v\in \RR_{\geq 0}$, define 
\[
I_{L/K, v} \da I_{L/K, \ceiling{v}}
.\]
Then define the renormalized **upper numbering**
\[
I_{L/K}^u \da I_{L/K, \phi\inv(u)}
.\]
:::

:::{.remark}
The point: the upper numbering will now lift to infinite extensions.
:::


:::{.proposition title="?"}
If $L'/L/K$ are all Galois with $I_{L'/K}$ finite, then
\[
I_{L/K}^u = \im I_{L'/K}^u
.\]
:::

:::{.proof title="?"}
Omitted, see Serre's *Local Fields*, or Cassels-Froehlich.
:::

:::{.remark}
If one graphs $v\mapsto \size I_{L/K, v}$, there are jump discontinuities at random $\ZZ\dash$points.
After the rescaling, the graph of $u\mapsto \size I_{L/K, i}^u$ still jumps, but now at $\QQ\dash$points instead of on $\ZZ$.
The denominators of all of the jumps will all divide $g_0$.
:::

:::{.theorem title="Hasse-Arf"}
If $L/K$ is abelian, then the jumps in $I_{L/K}^u$ are in fact in $\ZZ$.
:::

:::{.remark}
If $L/K$ is any Galois extension, define $I_{L/K}^u$ by gluing $I_{M/K}^u$ for all $M/K$ algebraic and $I_{M/K}$ finite.
:::

:::{.remark}
It turns out that $L/K$ is tamely ramified $\iff I_{L/K, \eps} = 1$ for all $\eps> 0 \iff I_{L/K}^\delta = 1$ for all $\delta > 0$, and this last condition makes sense for any (possibly infinite) Galois extensions.

Note that compositing tame extensions is still tame, so $L/K$ contains a maximal tamely ramified extensions.
So we've split off some easier parts of the full extension:

\begin{tikzcd}
	L \\
	\\
	{K^2} \\
	& {} \\
	{K^1} \\
	& {} \\
	K
	\arrow[from=7-1, to=5-1]
	\arrow[from=5-1, to=3-1]
	\arrow["{\text{maximal unramified, cyclic}}"', shift right=5, curve={height=30pt}, dashed, no head, from=7-1, to=5-1]
	\arrow[from=3-1, to=1-1]
	\arrow["{\text{maximal tamely ramified, cyclic}}", shift left=5, curve={height=-30pt}, dashed, no head, from=7-1, to=3-1]
	\arrow["{\text{wildly ramified, pro-}p}"', shift right=5, curve={height=30pt}, dashed, no head, from=5-1, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)


Recall that $G(K_2/K_1) \cong I_{L/K} / I_{L/K, 1}$ was cyclic of order $m$ with $p\notdivides m$.
By Kummer theory, one can generally write down cyclic extensions of order $m$ over some field as long as the field contains $m$th roots of unity.

> See Birch's article in Cassels-Froehlich.

Since $p\divides m$, $K^\unram \contains \mu_m$ the $m$th roots of unity in $\bar K$.
So if $K^2/K^\unram$ is Galois with $G(K^2/K^\unram)\cong C_m$, then by Kummer theory, this extension arises by taking an $m$th root, so $K^2 \cong \K^\unram(a^{1\over m})$ for some $a\in K^\unram$.
E.g. since $\QQ$ contains square roots of unity $(\pm 1)$, every quadratic extension $K/\QQ$ is obtained as $K = \QQ(\sqrt a)$.
Here $K^\unram$ only has one prime, so $K^2 = K^\unram(\pi_K^{1\over m})$ since $m$th roots of units give unramified extensions and $a = \pi_K^\ell u$ for some unit.

Thus the tamely ramified extension $K^2 = K^t\da \Union_{m\geq 1, p\notdivides m} K^\unram(\pi_K^{1\over m})$.
Moreover there is a canonical isomorphism
\[
G(K^\unram(\sqrt[m]{\pi_K})/K^\unram ) &\caniso \mu_m \\
\sigma &\mapsto \sigma(\sqrt[m]{\pi_K})/\sqrt[m]{\pi_K}
.\]
Thus
\[
G(K^t/K^\unram) = \cocolim \mu_m \cong \cocolim \ts{ C_m \st {p\notdivides m}} = \prod_{\ell \neq p} \ZZelladic
.\]
:::

:::{.remark}
Our decomposition of the absolute Galois group is now:

\begin{tikzcd}
	{\bar{K}} \\
	\\
	{K^t} \\
	\\
	{K^\unram} \\
	\\
	K
	\arrow[from=7-1, to=5-1]
	\arrow[from=5-1, to=3-1]
	\arrow[from=3-1, to=1-1]
	\arrow["{=\hat{\ZZ} = \prod_{\ell} \ZZelladic}"', curve={height=30pt}, dashed, no head, from=7-1, to=5-1]
	\arrow["{\cong \prod_{\ell\neq p}\ZZelladic}"', curve={height=30pt}, dashed, no head, from=5-1, to=3-1]
	\arrow["{\text{pro\dash}p}"', curve={height=30pt}, dashed, no head, from=3-1, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

Note that if the top field is a finite extension, this is just factoring the full Galois group as 

- a cyclic group with a canonical generator, 
- a cyclic group with with a non-canonical generator, and
- a finite $p\dash$group.

The story thus far packages all of this together for all finite extensions at once by taking direct limits.
:::

:::{.remark}
Note that $G(K^t/K)$ is unknown at this point, and will in general be a semidirect product of the two (pro)cyclic group and its (pro)cyclic quotient.
Take a canonical generator to write 
\[
\gens{\Frob} = G(K^\unram/K) = \Gal(\kappa(K^\unram)/\kappa(K))
\]
which contains the map $x\mapsto x^q$ where $q \da \size \kappa(K)$, and $\Frob$ is defined by pulling this back to $\Gal(K^\unram/K)$ along the canonical isomorphisms.
We can lift $\Frob$ to $G(K^t/K)$ (which we want to understand), then it acts on the normal subgroup $\Gal(K^t/K^\unram)$ by conjugation, i.e. $\sigma \mapsto \Frob\circ \sigma\circ \Frob\inv$.
This is independent of choice of lift, since another lift is of the form $\psi \Frob$ where $\psi$ is in the abelian part and thus commutes with $\sigma$.
So we need to specify this action to say what $G(K^t/K)$ is.
We have a canonical isomorphism 
\[
G(K^t/K^\unram) \caniso \cocolim_m \mu_M(\bar K)
,\]
so it suffices to give an automorphism of this group.
The following exercise yields the glue:

:::{.exercise title="?"}
Note that Frobenius acts on the RHS, and $x\mapsto x^q$ makes sense here since $q$ is a power of $p$ and $m$ is prime to $p$.
Check that the map induced by $\Frob$ is $\zeta \mapsto \zeta^q$.
:::

:::

:::{.remark}
Next: statements of local CFT.
:::