# Tuesday, September 28

> See fppf cohomology.
> Note: statements of the form $A\tensor C \cong B\tensor C \implies A\cong B$ aren't quite descent!
> There's no descent data or e.g. Galois equivariance, and the downstairs maps may not be related to the original map at all.

:::{.theorem title="?"}
For $L\slice k$ cyclic of degree $n$, 
\[
q(C_L)=n
.\]

:::

:::{.remark}
Recall that $q$ is multiplicative in exact sequences, equals 1 for finite \(G\dash\)modules, and if $M\tensor R \cong N\tensor R$ then $q(M)= q(N)$.
:::

:::{.proof title="of 3rd property"}
It's enough to show this for $M, N$ torsionfree, since $q(M) = q(M/M_\tors)$.
The claim is that for $R$ sufficiently divisible, letting $M\tensor \QQ \mapsvia{\phi} N \tensor \QQ$, $\ro{\phi}{M}$ factors through $N$ with torsion kernel.
Use that $M\tensor \RR \iso N\tensor \RR$ implies $M\tensor \QQ \iso N\tensor \QQ$
Now we claim that if $G \in \Grp$ and $V_1, V_2\in \mods{G}$ over a field $k$ and $L\slice k$ is any extension, then $V_1\tensor L \iso V_2\tensor L$ implies $V_1 \iso V_2$.
:::

:::{.proof title="of claim"}
1: Use that $\Hom$ commutes with tensor products in the following way:
\[
\Hom_G(V_1\tensor L, V_2\tensor L) = \Hom_G(V_1, V_2)\tensor L
.\]
We can write the LHS as $(V_1\dual \tensor V_2 \tensor_k L)^G$, and the right-hand side as $(V_1\dual \tensor V_2)^G \tensor_k L$.
It's enough to show that for any $G\dash$representation $V$, since $V^G \tensor L \cong (V\tensor L)^G$ where $V^G \da \ker( V\sumpower{q} \mapsvia{q} \bigoplus _{g\in G} V )$.
But now we're done since $L\slice k$ is flat.

2: If both $V_i$ are irreducible over $L$, this follows from Schur.
For $V_i$ irreducible over $k$ an infinite field, then being an isomorphism is a Zariski open condition, and any Zariski open subset of $\AA^{n}\slice k$ has infinitely many rational points.
:::

:::{.theorem title="?"}
If $L\slice k$ is cyclic and $S$ is a set of primes of $K$ including all infinite primes, all primes that ramify, and all primues  under a set of generators of the class group of $L$, letting $T$ be the set of primes of $L$ over $S$, we have

- $q(\II_{L, T}) = \prod_{v\in S} [\Lv : \kv]$
- $[L: k] q(\OO_{L, T}\units) = \prod_{v} [\Lv: \kv]$
- $q(C_L) = [L : k]$

:::

:::{.proof title="1 and 2 imply 3"}
There is a SES
\[
0 \to \OO_{L, T}\units \to \II_{L, T} \to C_L \to 1
,\]
where $\OO_{L, T}\units$ allows denominators in $T$.
Then using (1) and (2),
\[
q(C_L) = q(\II_{L, T}) / q(\OO_{L, T}\units) = [L: k]
.\]
:::

:::{.proof title="of 1"}
Write $\II_{L, T} = \prod_{v\in T} \Lv\units \times \prod_{v \not\in T} \OO_{\Lv}\units$, so
\[
q(\II_{L, T})
&= \prod_{v\in S} q( \prod_{w\in \places{/L} } L_w\units) \\
&= \prod_{v\in S} { \# H^2(\Lv \slice{\kv}; \Lv\units) \over \# H^2(\Lv \slice {\kv}; \Lv\units ) }\\
&= \prod_{v\in S} \# \Br(\Lv \slice {\kv})\\
&= \prod_{v\in S} [\Lv: \kv]
.\]

:::

:::{.proof title="of 2"}
Write $L_1 \da \Hom_\Set(T, \ZZ)$ and
\[
L_2 \da m(\lambda: \OO_{L, T}\units &\to L_1\tensor \RR )
\alpha &\mapsto ( \log\abs{ \alpha}_w )_{w\in T} 
.\]
Dirichlet's unit theorem implies $L_2 \injects L_1^0\tensor \RR \da \ts{\vector x \st \sum x_i = 0}$ is a lattice.
We can write
\[
L_1 
&= \bigoplus _{v\in S} \bigoplus _{w\slice v} \ZZ \\
&= \bigoplus _{v\in S} \Ind_{\Gal(\Lv \slice {\kv})}^{ \Gal(L\slice k)} \ZZ
,\]
Thus
\[
q(L_1)
&=
\prod _{v\in S} q\qty{ \Ind_{\Gal(\Lv \slice {\kv})}^{ \Gal(L\slice k)} \ZZ } \\
&=
\prod_{v\in S} q(\Lv \slice {\kv}, \ZZ) \\
&= \prod_{v\in S} [\Lv: \kv]
.\]

To compute the other side, use that there is a SES $0\to L_1^0 \to L_1 \mapsvia{\Sigma} \ZZ \to 0$.
So
\[
q(L_1^0) = q(L_1) / q(L\slice k; \ZZ) = {\prod [\Lv: \kv] \over [L: k] } 
.\]
Now note $q(L_k) = q(\OO_{L, T}\units)$ and there is a SES
\[
0 \to \mu(L) \to \OO_{L, T}\units \to L_k \to 0 \implies q(\OO_{L, T}\units ) = q(L_k)
,\]
where $\mu(L)$ are the roots of unity in $L$, which form a finite group.
Then
\[
q(\OO_{L, T}\units) = q(L_1^0) = {\prod [\Lv: \kv] \over [L:k] }
.\]

:::

:::{.fact title="from class field theory"}
\[
\#\qty{\II_K \over  k\units\Norm_{L\slice k} \II_L } = [L: k]
.\]
How to prove: reduce to Kummer extensions, adjoin $p$th roots of unity, etc.
:::

:::{.remark}
This fact implies $H^1(L\slice k; C_L) = 1$.
The proof is that $\#\qty{H^2/H^1} = [L: k]$, which implies $\# H^1 = 1$.
:::

:::{.theorem title="?"}
Severi-Brauer varieties over $k$ satisfy the Hasse principle, i.e. the following sequence is exact:
\[
0 \to \Br(k) \to \bigoplus _{v\in \places{k}} \Br(\kv)
.\]
:::

## Proof

:::{.theorem title="Hasse-Minkowski"}
Let $q$ be a quadratic form over a number field $k$, then the projective quadric $X\da \ts{q=0} \subseteq \PP^n\slice k$ satisfies the Hasse principle: $X$ has rational points over $k$ iff $X$ has rational points over $\kv$ for all $v\in\places{k}$.
:::

:::{.definition title="Quadratic forms representing elements"}
Given $q$ a quadratic form over $k$ a field (e.g. a number field or a local field), then for $a\in k$, we say $q$ **represents** $a$ if there exist elements $\vector x\in k^n\smz$ such that $q(\vector x) = 0$.
:::

:::{.theorem title="a stronger one"}
Given $a\in k$, $q$ represents $a$ iff over $k$ iff $q$ represents $a$ over $\kv$ for all $v\in \places{k}$.
Moreover, rational points are Zariski dense on $q(x) = a$.
:::

:::{.remark}
That this implies the first theorem is easy, setting $a=0$.
Conversely, consider $q'(\vector x, z) \da q(\vector x) - az^2$.
Then $q$ represents $a$ iff $q'$ represents 0 -- however, this can go wrong if $z=0$!
Exercise: find a good proof.
:::

:::{.proof title="?"}
Let $n$ be the number of variables.

- For $n=1$, we saw that $x^2=a$ satisfies the Hasse principle in the first class.
  Moreover rational points are Zariski dense on the projective variety $x^2 = ay^2$.
- For $n=2$, consider $q(x_1, x_2) = a$.
  We'll pick this up next time!

:::