# Tuesday, August 24


:::{.remark}
Last time: if $K$ is a number field and $X\slice K$ is geometrically irreducible, then $X(K_v) \neq \emptyset$ for almost all $v$.
:::

:::{.proof title="?"}
Choose $X\slice{\OO_K\invert{N}}$ such that $X$ has geometrically integral fibers.
It's enough to show that $X(K(v)) \neq \emptyset$ for almost all $v$, where $K(v)$ is the residue field at finite places $v$.

Now use the following theorem:
:::

:::{.theorem title="Lang-Weil Estimates"}
If $X$ over $\OO_K\invert{N}$ is geometrically integral, then
\[
\# X(\FF_{q^k}) = ( 1 + O(q^{1\over 2}) ) q^{k\dim X}
.\]

:::

:::{.claim}
If $X\slice{\OO_{K_v}}$ is smooth then 
\[
X(K(v)) \neq \emptyset
\implies
X(K_v) \neq \emptyset
.\]
:::

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

Use 

- Slice and Hensel, or the formal smoothness criterion, i.e. 

\begin{tikzcd}
	{\spec R} && X \\
	\\
	{\spec R'} && Y
	\arrow["\smooth", from=1-3, to=3-3]
	\arrow["\cl", from=1-1, to=3-1]
	\arrow[from=1-1, to=1-3]
	\arrow["\exists"{description}, dashed, from=3-1, to=1-3]
	\arrow[from=3-1, to=3-3]
\end{tikzcd}

Taking $R \da R'/I$ with $I$ nilpotent.

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgUiJdLFswLDIsIlxcc3BlYyBSJyJdLFsyLDIsIlkiXSxbMiwwLCJYIl0sWzMsMiwiXFxzbW9vdGgiXSxbMCwxLCJcXGNsIl0sWzAsM10sWzEsMywiXFxleGlzdHMiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMSwyXV0=)

See Hartshorne chapter 3, in the exercises!
:::

:::{.remark}
As a black box, we'll use that this is true for $\dim_{{\OO_{K_v}}} X = 1$, i.e. for curves.
This follows from the Weil conjectures for curves, see Severi/Bombieri.
If $X$ is genus $g$, then in fact we have a finer estimate:
\[
\abs{ \# X(\FF_{q^k}) - q^n } \leq q^{1\over 2} + 1
.\]
:::

:::{.proof title="?"}
We'll show this for $\dim_{\OO_K\invert{n}} = 2$.
Idea: try to fiber with curves.

- Suppose $\reldim X = 1$ for $X \to S$ over $\OO_K\invert{n}$ where $S$ is a curve with geometrically integral fibers.

- Without loss of generality, $X\to S$ where

  - $S$ is smooth of genus $g'$,
  - $X/S$ is smooth with fibers of genus $g$.

  - Now take the count

  \[
  X(\FF_{q^k}) 
  &= (1 + O_{g'}(q^{- {k \over 2} }) )q\cdot (1 + O_g(q^{- {k\over 2} }))q \\
  &= (1 + O_{g, g'}(q^{- {k\over 2} }))q^2
  .\]

- Such an $X\to S$ after replacing $X$ by an open subvariety.
  The proof of this follows from [Bertini](https://www.wikiwand.com/en/Theorem_of_Bertini): for $X \subseteq \PP^n$, take geometric projections and delete the singular locus.
  The fibers are slices by hyperplanes, and thus the fibers are geometrically integral.

:::

## Brauer Groups

:::{.remark}
Some upcoming topics:

- Severi-Brauer varieties (so $X\slice{K}$ where $X\slice{\bar{K}} \cong \PP^n$) satisfy the Hasse principle.
  Implies Hasse-Minkowski!

- The Brauer-Manin obstruction to the Hasse principle.

:::

### The Brauer-Manin Obstruction

:::{.remark}
Setup:

- $X$ is a variety,
- $\Br(X)$ is an abelian group
- Given $X \mapsvia{f} Y$, there is an induced map $f^*: \Br(Y)\to \Br(X)$.

For $K$ a number field (which we can view as a variety with a single point), we have
\[
\Br(K_v)
= 
\begin{cases}
\QQ/\ZZ &  v \text{ finite}
\\
\ZZ/2 & v \text{ real}
\\
0 & v \text{ complex},
\end{cases}
\]
which fits into a SES
\[
0 \to \Br(K) \to \bigoplus_v \Br(K_v) \mapsvia{\Sigma} \QQ/\ZZ \to 0
.\]
Note that most of the terms in the middle sum are $\QQ/\ZZ$, making $\Br(K)$ a large group.
:::

:::{.remark}
The yoga of the Hasse principle says we should try to solve things in adelic points first.
Write 
\[
\AA_K = \prod_v' (K_v, \OO_v) \subseteq \prod_v K_v
\]
where we take the restricted product.
There is a map $X(K)\to X(\AA_K)$, and taking $\alpha\in \Br(X)$ one gets a map $\alpha^*: X(K) \to \Br(K)$.
This yields a diagram

\begin{tikzcd}
	{X(K)} && {X(\AA_K)} \\
	\\
	{\Br(X)} && {\Br(\AA_k) \cong \bigoplus_v \Br(K_v)}
	\arrow["{\alpha^*}", from=1-1, to=3-1]
	\arrow["{\tilde\alpha^*}", from=1-3, to=3-3]
	\arrow[from=1-1, to=1-3]
	\arrow[from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJYKEspIl0sWzAsMiwiXFxCcihYKSJdLFsyLDAsIlgoXFxBQV9LKSJdLFsyLDIsIlxcQnIoXFxBQV9rKSBcXGNvbmcgXFxiaWdvcGx1c192IFxcQnIoS192KSJdLFswLDEsIlxcYWxwaGFeKiJdLFsyLDMsIlxcdGlsZGVcXGFscGhhXioiXSxbMCwyXSxbMSwzXV0=)

Using that $\Sigma: \Br(\AA_K) \to \QQ/\ZZ$, for a fixed $\alpha\in \Br(K)$, 
\[
X(K) \subseteq (\Sigma \circ \tilde \alpha)\inv (0) \subseteq X(\AA_K)
,\]
and $(\Sigma \circ \tilde\alpha)\inv(0) = X(\AA_K)^\alpha$.
Thus the Hasse principle is violated if $X(\AA_K)$ is nonempty but $X(\AA_K)^\alpha$ is empty.
More generally, it's violated if 
\[
X(\AA_K)^{\Br}
\da
\Intersect_{\alpha\in \Br(X)} X(\AA_K)^{\alpha} = \emptyset
.\]

:::

### The Hasse Principle for Severi-Brauers

:::{.remark}
Let $X\slice{K}$ be a Severi-Brauer, then $[X] \in \Br(K)$ and $X\cong \PP^n\slice{K} \iff [X] = 0$.
Using that 
\[
\oplus \iota_v: \Br(K) \embeds \bigoplus_v \Br(K_v)
,\]
we have
\[
[X] = 0 \iff \iota_v(X) = 0\,\, \forall v
&& \text{since } \iota_v(X) = [X_{K_v}] \in \Br(K_v)
.\]
:::

:::{.fact}
It turns out that $X\cong \PP^n \iff X(K) \neq \emptyset$.
:::

## Brauer Groups and Galois Cohomology

:::{.definition title="Brauer Groups"}
Let $K \in \Field$, then
\[
\Br(K) \da H^2_\Gal(K, \bar{K}\units ) = H^2_{\Grp}(\Gal(K^s/K), (K^s)\units )
.\]
:::

:::{.remark}
Let $G \in \Grp$ be discrete, so we're not considering any topology on it.
Let $M \in \mods{G}$, or equivalently $M\in \modsleft{\ZZ[G]}$.

We can take invariants and coinvariants:
\[
M^G &\da \ts{m\in M \st gm=m \,\, \forall g\in G } = \Hom_{\ZZ[G]}(\ZZ, M) \\
M_G &\da M /\gens{\ts{gm-m \st g\in G}} = \ZZ \tensor_{\ZZ[G]} M
.\]
These are the largest submodules/quotient modules respectively on which $G$ acts trivially.
:::

:::{.exercise title="?"}
Why are these equal to homs and tensors respectively?
:::

:::{.definition title="Group cohomology"}
\[
H^i(G; M) &\da \Ext^i_{\ZZ[G]}(\ZZ; M)\\
H_i(G; M) &\da \Tor_i^{\ZZ[G]}(\ZZ; M)
.\]
:::

:::{.example title="Cyclic groups"}
For $G\da \ZZ$, we have $\ZZ[G] = \ZZ\adjoin{x, x\inv}$.
Take a projective resolution
\[
0 \to \ZZ[G] \mapsvia{\cdot (x-1)} \ZZ[G] \mapsvia{x\mapsto 1} \ZZ \to 0
.\]

Deleting the augmentation and applying $\Hom_{\ZZ[G]}(\wait, \ZZ)$ yields $0\to \ZZ \mapsvia{f: \cdot (x-1)}  \ZZ\to 0$, and noting that $x$ acts by 1, $f$ is the zero map.
This yields
\[
H^*(G; \ZZ) &=
\begin{cases}
\ZZ &  * = 0, 1
\\
0 & \text{else}.
\end{cases}
\\
H_*(G; \ZZ) &=
\begin{cases}
\ZZ &  * = 0, 1
\\
0 & \text{else}.
\end{cases}
\]
:::