# Thursday, September 16

## Computing Brauer Groups

:::{.remark}
Let $k$ be a \(p\dash \)adic field, our goal is to show $\Br(k) = \QQ/\ZZ$.
We were trying to show

1. $H^2(k^\unram\slice k; (k^\unram)\units) = \QQ/\ZZ$,
2. Any Brauer class is split by an *unramified* extension

This says that we can split the computation of $\Br(k)$ into an interesting part (the ramified case) and a trivial part (the unramified case).

\todo[inline]{Check 2!}

To prove 1, we used
\[
1 \to U_{k^\unram} \to (k^\unram)\units\to \ZZ\to 0
,\]

and 

a. $H^2(k^\unram\slice k; \ZZ) = \QQ/\ZZ$,
b. $H^{>0}(k^\unram\slice k; U_{k^\unram}) = 0$,

where we used a filtration 
\[
\Fil_r U_{k^\unram} 
=
\begin{cases}
U_{k^\unram} &  r=0
\\
 \ts{x\st x\equiv 1\mod \pi^r}& r\geq 1.
\end{cases}
\]
and
\[
\gr^r(\Fil_\bullet U_{k^\unram}) 
=
\begin{cases}
\kappa\units & r=0 
\\
\kappa & r\geq 1.
\end{cases}
.\]


We now want to show

- $H^{>0}(k^\unram\slice k; \bar \kappa\units) = H^{>0}(\kappa; \bar\kappa\units)$
- $H^*( k^\unram\slice k; \bar \kappa\units) = H^*(\kappa; \bar \kappa\units) = 0$, and we were working on $*=2$.

:::

:::{.proposition title="?"}
For $k$ any field, $H^1(k; (k\sep)^+) = 0$, where $k^+$ denotes taking the additive group.
:::

:::{.proof title="?"}
$H^1$ here classifies forms of SESs
\[
0\to k \to V\to k \to 0
,\]
since automorphisms of this SES correspond to matrices $\ts{ \matt 1 * 0 1 \st *\in k^+} \cong k^+$.
But any form of this splits, since any SES of vector spaces splits.
:::

:::{.theorem title="?"}
For $k$ any field, $H^{>0}(k; (k\sep)^+) = 0$.
:::

:::{.proof title="?"}
It's enough to show this for finite extensions, so consider $H^{>0}(L\slice k; L^+) = 0$.
The normal basis theorem implies that $L^+ \cong k[G]$ as a \(G\dash\)module, since this is the regular representation.
We'll use the following common lemma:


:::{.lemma title="Shapiro's Lemma"}
If $H\leq G$ are finite groups and $M \in \mods{H}$ then 
\[
H^*(G; \Ind_H^G M)\cong H^*(H; M), &&
\Ind_H^G M = M\tensor_{\ZZ[H]} \ZZ[G]
.\]
:::

Now use that
\[
H^*( \Gal(L\slice k); k[\Gal(L\slice k)] ) = H^*(1; k) = 0 && *>0
.\]

:::

:::{.proof title="of Shapiro's lemma"}
Let $P^\bullet \projresolve \ZZ \in \mods{\ZZ[G]}$ be a free resolution and use Frobenius reciprocity to write
\[
H^*(G; \Ind_H^G M) 
&= H^*(\Hom(P^\bullet, \Ind_H^G M)) \\
&= H^*(\Hom( \Res^G_H P^\bullet, M)) \\
&= H^*(H; M)
,\]
where $\Res_H^G P^\bullet \projresolve \ZZ \in \mods{\ZZ[H]}$ is a free resolution, since $P^\bullet = \ZZ[G]\sumpower{I}$ (using that it's free) and thus $\Res_H^G P^\bullet = \ZZ[H]\sumpower{I'}$.
:::

:::{.proof title="of theorem, part b"}
We now want to prove (3),
\[
H^*(k^\unram \slice k; U_{k^\unram} / \Fil^r) = 0 && *>0
.\]
By induction on $r$, since we have a SES
\[
0 \to\Fil^{r-1}/\Fil^r \to U_{k^\unram}/\Fil^r \to U_{k^\unram}/ \Fil_{r-1} \to 1
,\]
where $H^*$ of the two outer terms vanishes and thus so does $H^*$ of the middle by the LES in cohomology.

For (4), we want to show $H^*(k^\unram\slice k; U_{k^\unram}) \to \lim_r H^*(k^\unram\slice k; U_{k^\unram} / \Fil_r)$.
We can move an inverse limit in:
\[
\invlim_r \invlim_n H^*({k_n}\slice k; U_{k^\unram} / \Fil^r)
&= \invlim_r \invlim_n H^*(\Hom(P^\bullet, U_{k^\unram} / \Fil^r)) \\
&= H^*(k^\unram \slice k; \invlim_r U_{k^\unram} / \Fil^r)
.\]


This uses the Mittag-Leffler condition to show that $\lim_1$ vanishes, which applies because we actually have surjectivity.
:::

:::{.theorem title="Hasse"}
If $D\slice k$ is a division algebra over $k$ a \(p\dash \)adic field (or any local field) with $\dim_k D = n^2$ (using that it's a form of a matrix algebra), then $D$ is split by the unique **unramified** extension of $k$ of degree $n$.
:::

:::{.remark}
That there is a unique such extensions follows from the fact that $\hat \ZZ$ has a unique subgroup of every index.
:::

## Proof of theorem


:::{.remark}
Write $k_n$ for the unique unramified extension of degree $n$.
We'll want to show

1. Show that it's enough to show $K_n \subseteq D$,
2. Actually show $k_n \subseteq D$.

:::

:::{.lemma title="?"}
For $k$ any field and $D\slice k$ any division algebra of $\dim_k D = n^2$, then if $L \slice k \subseteq D$ is a Galois extension of degree $n$, then $D$ splits over $L$.

> This is true without the extension being Galois.

:::

:::{.proof title="of lemma, using $\Tw$"}
Write $D = \Endo(V) \in \Tw\slice k$ for some $V\in \Tw$ of dimension $n$, then 
\[
D \basechange{L} = \Endo(V\basechange{L}) \in \Tw\slice L
.\]
Then since $L \subseteq D$, we have $L\actson V$ so $L\tensor_k L \actson V_L$, then use that $L\tensor_k L \iso L^{n}$ for $n\da \# \Gal(L\slice k)$.

> Why: write $L\tensor_k L = k[x] / I$ and use the Chinese remainder theorem!

We can write $L^n = \oplus Le_i$, so $V_L = \oplus e_i V_L$ which is dimension 1 and thus its Brauer class is trivial.
:::

:::{.remark}
Other proofs of this seem much more difficult!

So now let's show $k_n$ splits $D$.
We'll need to develop some valuation theory for division algebras.
:::

:::{.definition title="Valuations on division algebras"}
Define a valuation $v: D\to \ZZ \union\ts{\infty}$ extending the valuation on $K \subseteq D$ given by $1/n \val(\Norm_{D\slice k}(x))$.
Equivalently, for $x\in D$, use that $k(x) \subseteq D$ is a finitely generated $k\dash$algebra in which every nonzero element is a unit, so it's a field and carries a natural valuation.
:::

:::{.definition title="Valuation ring"}

Define
\[
\OO_D &\da \ts{x\st v(x) \geq 0} \subseteq D\\
\mfm_D &\da \ts{x\st v(x) > 0} \subseteq D \\
\mci &\da \OO_D/\mfm_D
,\]
where $\mfm_D \in \mspec D$, and set 
\[
f &\da [\mci : k] && \text{Degree of field extensions} \\
e &\da [\val(k) : \val(D)] && \text{Ramification index}
.\]
:::

:::{.remark}
Note that $\mci$ is a field, since all division algebras over finite fields are field extensions (using our computation of the Brauer groups of fields).
:::

:::{.fact}
$ef=n^2$, where the same proof for extensions of \(p\dash \)adic fields goes through.
:::

:::{.claim}
\[
e=f=n
.\]
:::

:::{.remark}
We'll show

1. $e\leq n$,

2. $f\leq n$,

Then since $ef=n^2$ this forces $e=f=n$.
:::

:::{.lemma title="?"}
Any commutative $L \in \Alg\slice k$ with $L \subseteq D$ satisfies $\dim_k L \leq n$.
:::

:::{.proof title="?"}

It's enough to prove this for $\Mat(n\times n; k)$, since the dimension won't change after passing to a finite extension, and proving here is classical.

:::{.exercise title="?"}
Prove this!
:::

:::

:::{.proof title="of claim"}
For (1): 
chose $\pi\in \OO_D$ with $v(\pi) = 1/e$, i.e. something with minimal positive valuation.
Then $k(\pi) \subseteq D$ is an extension over $k$ of degree at most $n$, by the lemma.

For (2):
Write $\mci = \kappa( \alpha)$ for $\alpha$ a primitive element, and let $\tilde \alpha \in D$ be a lift.
Then $k( \tilde \alpha) \subseteq D$ is a field extension of degree $\leq n$ by the lemma, and its residue field is $\mci$.
:::

:::{.corollary title="?"}
We have an exact equality
\[
[k( \tilde \alpha): k] = n
,\]
so $k(\tilde \alpha)\slice k$ is unramified, and there's a unique such extension, and since $\kappa(\tilde \alpha) \subseteq D$.
:::

:::{.remark}
A proof of this theorem using $\Tw$ or $\SB$ would be clarifying.
:::

:::{.claim}
The following map is an isomorphism:
\[
\QQ/\ZZ \cong H^2(k^\unram\slice k; (k^\unram)\units ) \iso H^2(k; \bar k\units)
.\]
:::

:::{.proof title="of claim"}
Use that the LHS is isomorphic to $H^2(k^\unram\slice k; H^0(k^\unram; \bar k \units ))$, and consider the Hochschild-Serre spectral sequence
\[
H^p(k^\unram\slice k; H^q(k^\unram; \bar k \units)) \abuts H^{p+q}(k; \bar k \units)
.\]

The spectral sequence reads:

\begin{tikzcd}[column sep=0.05in]
	& {} \\
	\\
	\\
	\\
	&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^2(k^\unram\slice k; H^0(k^\unram; \bar k \units)) = \QQ/\ZZ} \\
	&& \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(k^\unram\slice k; H^0(k^\unram; \bar k \units)) =_{90} 0} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(k^\unram\slice k; H^1(k^\unram; \bar k \units)) = 0} \\
	&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^0(k^\unram\slice k; H^0(k^\unram; \bar k \units)) = k\units} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^0(k^\unram\slice k; H^1(k^\unram; \bar k \units)) =_{90}0} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^0(k^\unram\slice k; H^2(k^\unram; \bar k \units)) = 0} \\
	{} &&&&&& {} \\
	& {}
	\arrow[from=1-2, to=9-2]
	\arrow[from=8-1, to=8-7]
\end{tikzcd}


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

Then for degree reasons, there are no nontrivial differentials to kill the two nonzero terms.
One can alternatively use the SES
\[
0 \to \Br(k^\unram\slice k) \to \Br(k) \to \Br(k^\unram)
.\]
:::