\documentclass[12pt]{article}
\usepackage{amsfonts, amsthm, amsmath}
\usepackage[all]{xy}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}
\setlength{\topmargin}{0in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\parskip}{0pt}
\setlength{\parindent}{20pt}

\def\kbar{\overline{k}}
\def\AA{\mathbb{A}}
\def\CC{\mathbb{C}}
\def\FF{\mathbb{F}}
\def\PP{\mathbb{P}}
\def\QQ{\mathbb{Q}}
\def\RR{\mathbb{R}}
\def\ZZ{\mathbb{Z}}
\def\gotha{\mathfrak{a}}
\def\gothb{\mathfrak{b}}
\def\gothm{\mathfrak{m}}
\def\gotho{\mathfrak{o}}
\def\gothp{\mathfrak{p}}
\def\gothq{\mathfrak{q}}
\def\gothr{\mathfrak{r}}
\DeclareMathOperator{\ab}{ab}
\DeclareMathOperator{\coker}{coker}
\DeclareMathOperator{\disc}{Disc}
\DeclareMathOperator{\Frob}{Frob}
\DeclareMathOperator{\Gal}{Gal}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\Ind}{Ind}
\DeclareMathOperator{\Inf}{Inf}
\DeclareMathOperator{\inv}{inv}
\DeclareMathOperator{\Norm}{Norm}
\DeclareMathOperator{\Res}{Res}
\DeclareMathOperator{\Trace}{Trace}
\DeclareMathOperator{\unr}{unr}
\DeclareMathOperator{\Cl}{Cl}

\def\head#1{\medskip \noindent \textbf{#1}.}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{prop}[theorem]{Proposition}

\begin{document}

\begin{center}
\bf
Math 254B, UC Berkeley, Spring 2002 (Kedlaya) \\
Cohomology of local fields: some computations
\end{center}

\head{Reference} Milne, III.2 and III.3; Neukirch, V.1.

\head{Notation convention} If you catch me writing $H^i(L/K)$
for $L/K$ a Galois extension of fields,
that's shorthand for $H^i(\Gal(L/K), L^*)$. Likewise for $H_i$ or
$H^i_T$.

We now make some computations of $H^i_T(L/K)$ for $L/K$
a finite Galois
extension of local fields. The first one is one we already made back
during Kummer theory.
\begin{prop}[Hilbert-Noether Theorem 90]
For any finite extension $L/K$ of fields, $H^1(L/K) = 0$.
\end{prop}
Our ultimate goal today will be to prove the following.
\begin{prop}
For any finite Galois extension $L/K$ of local fields, $H^2(L/K)$
is cyclic of order $[L:K]$. Moreover, this group can be canonically
identified with $\frac{1}{[L:K]}\ZZ/\ZZ$, such that
if $M/L$ is another finite extension such that
$M/K$ is also Galois, the inflation homomorphism
$H^2(L/K) \to H^2(M/K)$ corresponds to the inclusion
$\frac{1}{[M:K]}\ZZ/\ZZ \subseteq \frac{1}{[L:K]}\ZZ/\ZZ$.
\end{prop}

Before continuing, it is worth keeping in a safe place the exact sequence
\[
1 \to \gotho_L^* \to L^* \to L^*/\gotho_L^* = \pi_L^\ZZ \to 1,
\]
In this exact sequence of $G
= \Gal(L/K)$-modules, the action on $\pi_L^\ZZ$ is always trivial
(since the valuation on $L$ is Galois-invariant). For convenience, we
write $U_L$ for the unit group $\gotho_L^*$.

\head{The unramified case}
Recall that unramified extensions are cyclic, since their Galois groups are
also the Galois groups of extensions of finite fields. 

\begin{prop}
For any finite extension $L/K$ of \emph{finite} fields, the map
$\Norm_{L/K}: L^* \to K^*$ is surjective.
\end{prop}
\begin{proof}
One can certainly give an elementary proof of this using the fact that
$L^*$ is cyclic (exercise).
But one can also see it using the machinery we have at hand.
Because $L^*$ is a finite module, its Herbrand quotient is 1. Also,
we know $H^1_T(L/K)$ is trivial by Hilbert Theorem 90. 
Thus $H^0_T(L/K)$ is trivial too, that is,
$\Norm_{L/K}: L^* \to K^*$ is surjective.
\end{proof}

\begin{prop}
For any finite unramified extension $L/K$ of local fields, the map
$\Norm_{L/K}: U_L \to U_K$ is surjective.
\end{prop}
\begin{proof}
Say $u \in U_K$ is a unit. Pick $v_0 \in L/K$ such that in the residue fields,
the norm of $v_0$ coincides with $u$. Thus $u/\Norm(v_0) \equiv 1 \pmod{\pi}$,
where $\pi$ is a uniformizer of $K$. Now we construct units
$v_i \equiv 1 \pmod{\pi^i}$ such that $u_i = u/\Norm(v_0\cdots v_i) \equiv 1 
\pmod{\pi^{i+1}}$: simply take $v_i$ so that 
$\Trace((1-v_i)/\pi^i) \equiv (1-u_i)/\pi^i \pmod{\pi}$. (That's possible
because the trace map on residue fields is surjective, since the residue
field extension is separable, or if you like by the normal basis theorem.)
Then the product $v_0v_1\cdots$ convergest to a unit $v$ with norm $u$.
\end{proof}
\begin{cor}
If $L/K$ is an unramified extension of local fields, then
$H^i_T(\Gal(L/K), U_L) = 1$ for all $i \in \ZZ$.
\end{cor}
\begin{proof}
For $i$ odd, this is Theorem 90; for $i$ even, it is the previous
proposition.  
\end{proof}

Using the Herbrand quotient, we get
$h(L^*) = h(U_L) h(L^*/U_L)$. The previous corollary says $h(U_L) = 1$.
And
\begin{align*}
h(L^*/U_L) &= h(\ZZ) \\
&= \#H^0_T(\Gal(L/K), \ZZ)/\#H^1_T(\Gal(L/K), \ZZ) \\
&= \#\Gal(L/K)^{\ab} / \#\Hom(\Gal(L/K), \ZZ) \\
&= [L:K].
\end{align*}
Since $H^1_T(\Gal(L/K), L^*)$ is trivial, we conclude $H^0_T(\Gal(L/K), L^*)$
has order $[L:K]$. In fact, it is cyclic: the long exact sequence of Tate
groups gives
\[
1 = H^0_T(\Gal(L/K), U_K) \to
H^0_T(\Gal(L/K), L^*) \to H^0_T(\Gal(L/K), \ZZ) \to H^1_T(\Gal(L/K), U_K) = 1
\]
and the middle map is an isomorphism.

Consider the short exact sequence
\[
0 \to \ZZ \to \QQ \to \QQ/\ZZ \to 0
\]
of modules with trivial Galois action. Since $\QQ$ is injective as an abelian
group, it is also injective as a $G$-module for any group $G$ (exercise).
Thus we get an isomorphism $H^0_T(\Gal(L/K), \ZZ) \to H^{-1}_T(\Gal(L/K),
\QQ/\ZZ)$. But the latter is
\[
H^1(\Gal(L/K), \QQ/\ZZ) = \Hom(\Gal(L/K),
\QQ/\ZZ);
\]
since $\Gal(L/K)$ has a canonical generator (Frobenius), we
can evaluate there and get a canonical map $\Hom(\Gal(L/K), \QQ/\ZZ)
\to \ZZ/[L:K]\ZZ \subset \QQ/\ZZ$. Putting it all together, we get a 
canonical map
\[
H^2(\Gal(L/K), L^*) \cong H^0_T(\Gal(L/K), L^*)
\cong H^1(\Gal(L/K), \QQ/\ZZ) \hookrightarrow \QQ/\ZZ.
\]
In this special
case, this is none other than the local invariant map! In fact, by
taking direct limits, we get a canonical isomorphism
\[
H^2(K^{\unr}/K) \cong \QQ/\ZZ.
\]

What's really going on here is that $H^0_T(\Gal(L/K), L^*)$ is a cyclic
group generated by a uniformizer $\pi$ (since every unit is a norm).
Under the map $H^0_T(\Gal(L/K), L^*) \to \QQ/\ZZ$, that uniformizer
is being mapped to $1/[L:K]$.

\head{The cyclic case}

Let $L/K$ be a cyclic but possibly ramified extension of local fields.
Again, $H^1_T(L/K)$ is trivial by Theorem 90, so all there is
to compute is $H^0_T(L/K)$. We are going to show again that it
has order $n$. (It's actually cyclic again, but we won't prove this just
yet.)

\begin{lemma}
Let $L/K$ be a finite Galois extension of local fields. Then there
is an open, Galois-stable subgroup $V$
of $\gotho_L$ such that $H^i(\Gal(L/K), V) = 0$ for all $i>0$
(i.e., $V$ is acyclic for cohomology).
\end{lemma}
\begin{proof}
By the normal basis theorem, there exists $\alpha \in L$ such that
$\{\alpha^g: g \in \Gal(L/K)\}$ is a basis for $L$ over $K$. Without loss
of generality, we may rescale to get $\alpha \in \gotho_L$; then put
$V = \sum \gotho_K \alpha^g$. As in the proof of additive Theorem 90,
$V$ is induced: $V = \Ind^G \gotho_K$, so is acyclic.
\end{proof}

The following proof uses that we are in characteristic 0, but you can
patch it up for the function field case.
\begin{lemma}
Let $L/K$ be a finite Galois extension of local fields. Then there
is an open, Galois-stable subgroup $W$
of $U_L = \gotho_L^*$ such that $H^i(\Gal(L/K), W) = 0$ for all $i>0$.
\end{lemma}
\begin{proof}
Take $V$ as in the previous lemma. If we chose $\alpha$ sufficiently divisible,
then $V$ lies in the radius of convergence of the exponential series
\[
\exp(x) = \sum_{i=0}^\infty \frac{x^i}{i!}
\]
(you need $v_p(x) > 1/(p-1)$, to be precise), and we may take $W = \exp(V)$.
\end{proof}

Since the quotient $U_L/V$ is finite, its Herbrand quotient is 1, so
$h(U_L) = h(V) = 1$. So again we may conclude that
$h(L^*) = h(U_L) h(\ZZ) = [L:K]$, and so $H^0_T(\Gal(L/K), L^*) = [L:K]$.
But we don't yet know that $H^0_T(\Gal(L/K), L^*)$ is cyclic, since
the groups $H^i_T(\Gal(L/K), U_L)$ don't necessarily vanish.

\head{Note} This is all that we need for ``abstract''
local class field theory. We'll revisit this point later.

\head{The general case}

For those in the know, there is a spectral sequence underlying this next
result; see Milne, Remark II.1.35.
\begin{prop}[Inflation-Restriction Exact Sequence]
Let $G$ be a finite group, $H$ a normal subgroup, and $M$ a $G$-module.
If $H^i(H, M) = 0$ for $i=1, \dots, r-1$, then 
\[
0 \to H^r(G/H, M) \stackrel{\Inf}{\to} H^r(G,M) \stackrel{\Res}{\to}
H^r(H,M)
\]
is exact.
\end{prop}
\begin{proof}
For $r=1$, the condition on $H^i$ is empty. In this case, $H^1(G,M)$
classifies crossed homomorphisms $\phi:G \to M$. If one of these
factors through $G/H$, then clearly it becomes trivial when restricted to
$H$. Conversely, if $\phi(h) = m^h - m$ for all $h \in H$, then
$g \mapsto \phi(g) - m^g + m$ is a crossed homomorphism which factors
through $G/H$. Thus the sequence is exact at $H^1(G,M)$; exactness at
$H^i(G/H,M)$ is similar but easier.

If $r>1$, we induct on $r$ by dimension shifting. Let $N$ be the $G$-module
which makes the sequence
\[
0 \to M \to \Ind^G (M) \to N \to 0.
\]
By the long exact sequence in homology, $H^i(N) \cong H^{i+1}(M)$. Thus the
case of $r$ for $M$ reduces to the case of $r-1$ for $N$.
\end{proof}
By Theorem 90, we thus have
\begin{cor}
If $M/L/K$ is a tower of fields with $M/K$ and $L/K$ finite and Galois,
the sequence
\[
0 \to H^2(L/K) \to H^2(M/K) \to H^2(M/L)
\]
is exact.
\end{cor}

We now prove the following.
\begin{prop}
For any finite Galois extension $L/K$ of local fields, $H^2(\Gal(L/K), L^*)$
has order at most $[L:K]$.
\end{prop}
A key fact we need to recall is that any finite extension of local fields
is \emph{solvable}: the maximal unramified extension is cyclic, the
maximal tamely ramified extension is cyclic over that, and the rest is
an extension of order a power of $p$, so its Galois group is automatically
solvable. This lets us induct on $[L:K]$.
\begin{proof}
We've checked the case $L/K$ cyclic, so we may use it as the basis for an
induction. If $L/K$ is not cyclic, since it is solvable, we can find a
Galois subextension $M/K$. Now the exact sequence
\[
0 \to H^2(M/K) \to H^2(L/K) \to H^2(L/M)
\]
implies that $\#H^2(L/K) \leq \#H^2(M/K)
\#H^2(L/M) = [M:K][L:M] = [L:K]$.
\end{proof}

To complete the proof that $H^2(L/K)$ is cyclic of order
$[L:K]$, it now suffices to produce a cyclic subgroup of order $n$.
Let $M/K$ be an unramified extension of degree $n$. Then we have a diagram
\[
\xymatrix{
& & H^2(M/K) \ar[r]^{\Res} \ar[d]^{\Res}
& H^2(ML/L) \ar[d]^{\Res} \\
0 \ar[r] & H^2(L/K) \ar[r] & H^2(ML/K) \ar[r]
& H^2(ML/L)
}
\]
in which the bottom row is exact and the vertical arrows are injective,
both by the inflation-restriction sequence. It suffices to show that
$H^2(M/K) \to H^2(ML/L)$ is the zero map; then we can push a generator
of $H^2(M/K)$ down to $H^2(ML/K)$, and pull it back to $H^2(L/K)$ by exactness
to get an element of order $n$.

Let $e = e(L/K)$ and $f = f(L/K)$ be the ramification index, so that
$[ML:L] = e$. Since unramified extensions are cyclic, we have
$H^2(M/K) \cong H^0_T(M/K)$ and $H^2(ML/L) \cong H^0_T(ML/L)$,
so the map $\Res: H^2(M/K) \to H^2(ML/L)$ can be computed as
$\Res: H^0_T(M/K) \to H^0_T(ML/L)$,
which is just the canonical map
$K^*/\Norm_{M/K}(M^*) \to L^*/\Norm_{ML/L}((ML)^*)$.
Now $K^*/\Norm_{M/K}(M^*)$ is a cyclic group of order $ef$
generated by $\pi_K$, a uniformizer
of $K$, and $L^*/\Norm_{ML/L}((ML)^*)$ is a cyclic group of order
$e$ generated by $\pi_L$, a uniformizer of $L$. But
$\pi^K$ is a unit of $\gotho_L$ times $\pi_L^e$. So the Res map
is indeed zero.

\head{Note}
If $L/K$ is a finite extension of degree $n$, then the map
$\Res: H^2(K^{\unr}/K) \to H^2(L^{\unr}/L)$ translates by the local
reciprocity map into a map from $\QQ/\ZZ$ to itself. This map turns out
to be multiplication by $n$ (see Milne, Proposition II.2.7).

\head{The local invariant map}

By staring again at the above argument, we can in fact prove that
$H^2(\overline{K}/K) \cong \QQ/\ZZ$.
First of all, we have
an injection $H^2(K^{\unr}/K) \to H^2(\overline{K}/K)$ by the
inflation-restriction exact sequence, and the former is canonically
isomorphic to $\QQ/\ZZ$; so we have to prove that this injection is
actually also surjective. Remember that $H^2(\overline{K}/K)$ is the
direct limit of $H^2(M/K)$ running over all finite extensions $M$ of $K$.
What we just showed above is that if $[M:K] = n$ and $L$ is the unramified
extension of $K$ of degree $n$, then the images of $H^2(M/K)$ and
$H^2(L/K)$ in $H^2(ML/K)$ are the same. In particular, that means that
$H^2(M/K)$ is in the image of the map $H^2(K^{\unr}/K) \to 
H^2(\overline{K}/K)$. Since that's true for any $M$, we get that the
map is indeed surjective, hence an isomorphism.

Next time, we'll use this map to obtain the local reciprocity map.

\head{Exercises}

\begin{enumerate}
\item
Give an elementary proof (without cohomology)
that the norm map from one finite field to another
is always surjective.
\item
Give an example of a cyclic ramified extension $L/K$ of local fields
in which the groups $H^i_T(\Gal(L/K), U_L)$ are nontrivial.
\end{enumerate}

\end{document}