# Tuesday, September 21

:::{.remark}
Last time: for $k$ a \(p\dash \)adic field, we have $\Br(k) = \QQ/\ZZ$.
The plan for today:

- Examples
- A SES for $L$ a number field:
\[
0\to \Br(L) \to \bigoplus_{v\in \places{k} } \Lv \to \QQ/\ZZ\to 0
.\]

- Possibly the Hasse-Minkowski theorem
- The Brauer-Manin obstruction.

:::

## Construction of Brauer classes over $K$

:::{.remark}
Fix a character to a cyclic group
\[
\chi: \Gal(k\sep \slice k) \to C_n = \gens{\sigma}
\]
and set $k_\chi$ to be the fixed field.
:::

:::{.definition title="Cyclic Algebra"}
For $a\in k\units/(k\units)\powers{n}$,
write
\[
(\chi, a) = k_\chi \gens{\sigma} / \gens{ \sigma s = s^{\sigma} \sigma, \sigma^n - a} && s\in k_\chi
.\]

:::

:::{.remark}
We have 
\[
[(\chi, a)] \da [X] \cupprod [a] \in H^1(K; C_n)\cup H^1(K; \mu_n) = \Br(k)
.\]
There are cases where it's not known if these types of algebras are generators of certain Brauer groups.
:::

:::{.remark}
For $k$ a \(p\dash \)adic field and $k_n$ the unique unramified degree $n$ extension, we can construct a character
\[
\chi_n: \Gal(k\sep\slice{k}) \to \Gal(k_n{}\slice k) \caniso C_n
,\]
where the isomorphism is canonical, sending the Galois group to the Frobenius.
:::

:::{.theorem title="?"}
Let $\pi$ be a uniformizer of $\OO_K$.
Every CSA is equivalent to one of the form
\[
(\chi_n, \pi^m) \to {m\over n} \in \QQ/\ZZ = \Br(k)
.\]

:::

:::{.remark}
If $m, n$ are coprime one gets a division algebra.
:::

:::{.proof title="Sketch"}
This is mostly a computation that involves unwinding the isomorphism $\Br(k) \to \QQ/\ZZ$.
A sketch:

- The class $[(\chi_n, \pi)]$ has order $n$,
- The class $[(\chi_n, \pi)]^m = [(\chi_n, \pi^m)]$, which is given by a cup product.

:::

:::{.remark title="An algorithm to compute"}
Let $D\slice k$ be a division algebra.

- Find a copy of $k_n$ in $D$, which can be done since this is a division algebra of dimension $n^2$.
- There exists a $\sigma \in D$ such that $\sigma\actson K_n$ by conjugation is the canonical generator of $\Gal(k_n {}\slice k) \caniso C_n$ (where we take $\Frob$ as the canonical generator).
- Then $[D] \mapsto {v(\sigma) \over n} \in \QQ/\ZZ = \Br(k)$, where $v$ is the normalized valuation on $D$ we constructed previously.
  Note that this is well defined since changing $D$ changes the output by an integer.
:::

:::{.example title="The simplest case: $n=2$"}
Using that there is in fact a canonical isomorphism $\mu_2 \cong C_2$ since there's only one nontrivial element in each group, we have
\[
H^1(k; C_2) = H^1(k; \mu_2) = k\units/(k\units)\powers{2}
.\]
Hence any character 
\[
\chi: \Gal(k\sep\slice k) \to C_2 = \mu_2
\]
is represented by some $b_\chi \in k\units/(k\units)\powers{2}$.
So we have an identification
\[
(\chi, a) \leadsto (b_\chi, a)_2 = (a, b_\chi)_2
=
\begin{cases}
0 & v(a) \equiv v(b) \mod 2 
\\
{1\over 2} & \text{else}.
\end{cases}
\]
For the corresponding extension to be unramified, one needs the valuation to be zero.
So for example taking $k(\pi)\slice k$ yields a ramified extension since $v(\pi ) = 1$.

> Note that here $(\wait, \wait)_n$ is generally a Hilbert or norm-residue symbol.

:::

:::{.exercise title="?"}
Prove that these cyclic algebras are CSAs.
:::

## The SES

:::{.remark}
Our goal for today: for $k$ a number field, show the following sequence is exact
\[
0\to \Br(k) \to \bigoplus_{v\in \places{k} } \kv \mapsvia{\sum} \QQ/\ZZ\to 0
.\]

:::

:::{.proposition title="?"}
For \( \alpha \in \Br(k) \), using the pullback of $i_v$, 
\[
\Br(K) \mapsvia{\prod_{i_v^*}} \prod_v \Br(\kv)
\]
factors through \( \bigoplus_v \Br(\kv) \), i.e. $i_v^*( \alpha) = 0$ for almost all $v$.
:::

:::{.proof title="of prop, proof 1"}
First represent $\alpha$ by $X\in \SB$, so $X(\kv) \neq \emptyset$ for almost all $v$.
This implies $X_{\kv} \cong \PP^n\slice k$ for almost all $v$.
:::

:::{.definition title="Ideles"}
\[
\II_k \da \prod_v' (\kv\units, \OO_{\kv}\units) = \ts{ (x_v) \in \prod_v \kv\units \st x_v \in \OO_{\kv}\units \, \text{ for almost all } v}
.\]
A basis of open sets is given by $(x) \cdot \prod_v \OO_{\kv}\units$.
:::

:::{.warnings}
There is a map
\[
\II_k &\embeds \AA_k^2 \\
x &\mapsto (x, x\inv)
,\]
and there is a subspace topology -- but this is not equivalent to the topology above, and is in fact a source of an infamous error!
:::

:::{.definition title="S-ideles"}
If $S$ is a finite set of places of $K$ containing all infinite places, then define
\[
\II_{k, S} = \prod_{v\in S} \kv\units \times \prod_{v\not\in S} \OO_{\kv}\units \subseteq \II_K
.\]
:::

:::{.fact}
\[
\II_k = \colim_{S} \II_{k, s}
.\]
:::

:::{.remark}
The idea will be to study the following SES of Galois modules:
\[
1 \to L\units \to \II_L \to C_L \to 1
,\]
where $C_L$ is the idele class group.
:::

:::{.proposition title="?"}
\[
H^2(L\slice k; \II_L) &= \bigoplus_{v\in\places{k}}\Br(\Lv {}\slice {\kv}) \\
H^2(k; \II_{k\sep}) &= \bigoplus_{v\in\places{k}} \Br(\kv)
,\]
:::

:::{.theorem title="?"}
\[
H^1(L\slice k; C_L) &= 0 \\
H^2(L\slice k; C_L) &= [d] \in \QQ/\ZZ, \quad d\da {1\over [L:k]}
.\]

This will imply
\[
H^1(k; C_{k\sep}) &= 0 \\
H^2(k; C_{k\sep}) &= \QQ/\ZZ
.\]

:::

:::{.proof title="sketch"}
We can write
\[
H^2(L\slice k; \II_L) 
&= H^2(L\slice k; \directlim_T \II_{L, T}) \\
&= \directlim_T H^2(L\slice k; \II_{L, T})
,\]
so it's enough to show that for $S$ a finite set of places of $K$ and $T$ a set of places over $S$ that we have
\[
H^2(L\slice k; \II_{L, T}) = \bigoplus_{v\in S} \Br(\Lv \slice {\kv})
.\]
:::

:::{.exercise title="?"}
Try to prove this, it uses Shapiro's lemma and isn't too difficult.
:::