# Tuesday, October 05

:::{.remark}
Goal: prove the Hasse-Minkowski theorem.
We looked at $n\leq 3$, so today we'll look at $n=4$.
:::

:::{.theorem title="?"}
Let $Q \subseteq \PP^3\slice k$ be a smooth quadric, then
\[
Q(k) \neq \emptyset \iff Q(\kv) \neq \emptyset
\]
for all places $v\in \places{k}$.
:::

:::{.proof title="$n=4$ case"}
Let $X \subseteq \Gr_1(\PP^3)$ be the variety of lines in $Q$, and consider $\mci \to X$ the universal family.
Then $X_{\kbar} = \PP^1 \union \PP^1$ since $Q_{\kbar} = \PP^1 \times \PP^1 \injectsvia{\OO(1, 1)} \PP^3$.
Consider the case when $X$ is not connected.
Then

- $\mcu_c \to Q$ is an isomorphism, which can be checked over $\kbar$.
- $Q(\kv) \neq \emptyset$ implies $\mcu_c(\kv) \neq \emptyset$ and thus $C(\kv) \neq \emptyset$ for all $v$.
- By the Hasse principle for Severi-Brauers, if $C=\PP^1$ implies $C(k) \neq \emptyset$.
- Then $\mcu_c \to C$ is Zariski trivial, so $\mcu_c \cong Q$ has rational points.

\begin{tikzcd}
	{\mcu_c} && \mcu && Q \\
	\\
	C && X
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=3-3]
	\arrow[hook, from=3-1, to=3-3]
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-3, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJcXG1jdV9jIl0sWzAsMiwiQyJdLFsyLDIsIlgiXSxbMiwwLCJcXG1jdSJdLFs0LDAsIlEiXSxbMCwzXSxbMywyXSxbMSwyLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDFdLFszLDRdXQ==)

Now consider the case when $X$ is connected.
The claim is that there exists a quadratic extensions $k'/k$ where $X_{k'}$ is not connected:

- $k' = \globsec{X; \OO_X}$, which is a rank 2 vector bundle (which can be checked over $\kbar$).
So
\[
\globsec{X_{k'}; \OO_{X_{k'}}} = \globsec{X; \OO_X}_{k'} = k' \tensor_k k' \cong k' \oplus k'
,\]
so $X_{k'}$ is disconnected.

- We can take $[Q] \in H^1(k; \Orth_n) \mapsvia{\det} H^1(k; \mu_2)$, and it maps to $[k']$.
- $\Gal(\kbar\slice k)$ acts on $\Pic(Q) \cong \ZZ\cartpower{2}$, and this action factors through $\ts{\pm 1}$.
  Here $\Orth_n = \Aut(\sum x_i^2)$.

  - Why: this action preserves the *effective cone* in $\Pic(Q_{\kbar})$ spanned by $\pi_1^* \OO(1)$ and $\pi_2^* \OO(1)$, which are those bundles with global sections (which is preserved by Galois).


:::{.warnings}
Even if $[L] \in \Pic Q$ is Galois-invariant, this does not imply that $L$ is defined over $k$!
This can be a common source of errors.
:::

By case 1, $X(k')\neq \emptyset$, so there exists a rational line $L \subseteq Q$ contained in one connected component of $X_{k'}$.
There is an action $\Gal(k'\slice k)\actson L$, so take $\sigma \in \Gal(k'\slice k)$.
Then $\sigma(L)$ is in the other component $X_{k'}$, since Galois interchanges its components pointwise.
Then considering the two rulings of the quadric yields the following picture, where they intersect at a point:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/RationalPoints/sections/figures}{2021-10-05_10-08.pdf_tex} };
\end{tikzpicture}

But then $\sigma(p) = p$ is Galois fixed, and is thus a $k\dash$rational point.
:::



:::{.theorem title="?"}
Let $n\geq 5$ and $Q = \sum_{l\leq i \leq n} a_i x_i^2$ be a nondegenerate quadratic form over a number field $k$.
Then $Q$ satisfies the Hasse principle.
:::

:::{.proof title="General case"}
We'll proceed by induction on $n\geq 5$.
Write $Q = a_1 x_1^2 + a_2 x_2^2 + G(x_3,\cdots, x_n)$.

:::{.claim}
$G$ represents $\kv$ for almost all $v\in \places{k}$.
:::

It's enough to show that $G'(x_3,\cdots, x_{n+1}) \da G(x_3, \cdots, x_n) + ax_{n+1}^2$ represents 0 for all $a\in \kv$ and almost all $v$.
Without loss of generality, we can assume $G$ is nondegenerate over the residue field $\kappa(v)$, by throwing out finitely many things.
Then $G'$ has rank at least $n-2$ over $\kappa(v)$ for almost all $v$.


:::{.claim}
$G'$ has a smooth rational point for for all 
:::

Using the Lang-Weil estimates (using absolute irreducibility), $G'(\kappa(v))$ has about $(\# \kappa(v))^{n-3}$ rational points, where the error term is uniform in $v$.
The singular locus is a dimension smaller, so about $(\# \kappa(v))^{n-4}$, and for $n$ large enough for this to hold, the former is larger.

Now use Hensel's lemma, any smooth rational point on the special fiber lifts to the generic fiber (i.e. the infinitesimal smoothness criteria).
This proves the first claim that $G$ represents $\kv$ for almost all $v$.


:::{.fact}
For almost all $v$, $G(x_3,\cdots, x_n)$ represents *every* element of $k$.
:::

Let $U \subseteq (\prod \kv) \cartpower{n-2} k[v]$ be the set $\ts{ (x_3, v), \cdots, (x_n, v)}$ such that there exists an $(x_1, v), (x_2, v)$ with $Q(x_1, \cdots, x_n) = 0$.
Some claims:


- $U$ is open, which follows from the fact above,
- $U$ is nonempty since $Q$ represents 0 locally by hypothesis,
- The set $U' \subseteq (\prod \kv)\cartpower{2}$ of pairs $(x_1, v), (x_2, v)$ such that there exist $(x_3,v),\cdots,(x_n, v)$ with $Q(x_1,\cdots, x_n)=0$ is also open.

Then by weak approximation, there exist $x_1, x_2\in k$ such that $(x_1, x_2)\in U$.
So write $c = a_1 x_1^2 + a_2 x_2^2\in k$ and define $Q'(z, x_3,\cdots, x_n) = -cz^2 + G(x_3,\cdots, x_n)$.
This is a quadratic form in $n-1$ variables that represents 0 locally.
Now by induction, $Q'$ represents zero globally.
:::