# Tuesday, November 02

:::{.remark}
Setup: let $k\in \Global\Field$, $S \subseteq \places{k}$ a finite set of places, $\mcg\in \smooth\Grp\Sch\slice{\OO_{k, S}}$ yielding a SES
\[
0\to \mcg^0 \to \mcg\to\mcf\to 0
,\]
where $\mcf\in \Grp\Sch$ is finite étale.
After base-changing to $k$, this yields a SES of groups
\[
0\to G^0\to G\to F\to 0
.\]
We defined $H^1_S(k; G) \da \ts{\tau \in H^1(k; G) \st \tau_{\kv} \in \im \phi \forall v\not\in S}$ where $\phi: H^1(\OO_{\kv}; \mcg)\to H^1(k;\mcg)$.
:::

:::{.slogan}
General principal we'll use: there should only be finitely many things over $\OO_{k_S}$, see Shafarevich's conjecture.
:::

:::{.theorem title="?"}
\envlist

a. $H_S^1(k; G)\to H^!_S(k; F) \to \prod_{v\in S} H^1(\kv; F)$ has finite fibers, and
b. If $k$ is a number field, then $H_S^1(k; G)$ is finite.

:::

## Proof

### Proof of a

:::{.claim title="Step 1"}
For $v\not\in S$, $H^1(\OO_v; \mcf) \to H^1(\kv; F)$ is injective.
:::

:::{.proof title="Sketch"}
First consider a fiber that is a trivial torsor over $\ts{\pt}$.
$\mcf\slice{\OO_{\kv}}\in\torsors{F}$. 
Why?
Finite things are proper (finite iff proper and affine), so by the valuative criterion of properness $\mcf(\kv)$ nonempty implies $\mcf(\OO_{\kv})$ nonempty.

Now one can use a twisting argument to show that every fiber is finite.
:::

:::{.claim title="Step 2"}
$H^1(\OO_{\kv}; \mcg^0)$.
:::

:::{.proof title="of step 2"}
Let $\kappa(v)$ be the residue field.

1. We know $H^1(\kappa(v); \ro{ \mcg^0 }{\kappa(v)} ) = 1$ by Lang's theorem. 
2. Let $\tau\in H^1(\OO_{\kv}; \mcg^0)$ be a torsor.
  Now by the infinitesimal lifting criterion for smoothness, i.e. Hensel's lemma, if $\tau(\kappa(v))$ is nonempty then $\tau_{\OO_{\kv}}$ is nonempty.

3. $H^1_S(k; G)\to H^1_S(k; F)$ has finite fibers.
  This uses input from a hard theorem which we'll black box here.

  The proof is that $H^1(\OO_{\kv}; \mcg)\to H^1(\OO_k; \mcf) \injects H^1(k; F)$, where the first map is injective by (2) above, and the second by (1).
  Let $\Sha$ denote the Tate-Shafarevich group, then $\ker h \subseteq \Sha^1_S(k; G) \da \ker \qty{H^1(k; C) \to \prod_{v\in S} H^1(k; G_{\kv}) }$
  By Borel-Serre, for $\mcg$ affine[^equiv_bsd] the latter is finite.
  
  Now a twisting argument shows this for every fiber.

:::{.claim}
$H^1_S(k; \mcf) \to \prod_{v\in S} H^1(\kv; F_{\kv})$ has finite fibers.
:::

:::{.proof title="?"}
Now use the source of all finiteness in arithmetic geometry: Hermite-Minkowski's theorem, using that the previous information determines the degree and discriminant.

- Choose $(\tau_v)\in \prod_{v} H^1(\kv; F_{\kv})$, and suppose $\tau \to (\tau_v)_{v\in S}$.
  As a scheme, $\tau = \spec R$ for $R\in \kalg$ finite étale, so $R = \prod_i L_i$ with $L_i$ unramified away from $S$.
  This bounds the discriminant, so there are finitely many possibilities for $L_i$.

- For each $R$ there are finitely many $F$ actions making $\spec R$ into a torsor.
  Using representability, then $F\to \ul{\Aut}_k(R)$ is a finite étale group scheme.
  Since both are now finite, there are only finitely many such maps.

:::

[^equiv_bsd]: 
(conjecturally for $\mcg$ arbitrary, thought to be true but equivalent to BSD!), 

:::

### Proof of b

:::{.remark}
For the proof of (b), it's enough to show that $\prod_v H^1(\kv; \ro{F}{\kv} )$ is finite.
Krasner's lemma shows that are only finitely many degree $d$ extensions over a \(p\dash \)adic field (false for $\fp((t))$! ), so $\kv$ has finitely many extensions.
Then there are only finitely many extensions of $k$ of a given degree, so there are only finitely many possibility for $F\dash$torsors.
:::

:::{.exercise title="?"}
Produce infinitely many extensions of $\fp((t))$.
:::

:::{.remark}
Note: étale is stronger than smooth?
In the following proof, we could take everything to be étale to simplify things.
:::

:::{.remark}
Reminder of setup: let $X\in \Var\slice k$ be proper, $G\in\smooth\Grp\Sch\slice k$ finite , $T\in \torsors{G}\slice X$.
We're trying to show that $\Sel_{T\slice X}(k; G)$ is finite.

**Step 1**: spread out.
There are finitely many denominators and thus finitely many places and $\mcg$ finite smooth and $\OO_{k, S'}$ a group scheme with $\mcx$ proper over $\OO_{k, S'}$.
Then $\mct\slice X$ is a $\mcg\dash$torsor with $\mcg_k = G, \mct_k = T, \mcx_k = X$.
Let $\tau\in \Sel_{T\slice X}(k; G)$, then $\tau \in H^1_{S'}(k; G)$ since $\tau_{\kv}$ is in the image of $X(\kv) \to H^1(\kv; G_{\kv})$, but $X(\kv) = \mcx(\OO_{\kv})$ by the valuative criterion of properness.
So for a number field, the set we want is contained in a finite set and we're done in this case.

For function fields, consider $\tau \in \Sel_{T\slice X} (k; G) \subseteq \im\qty{ H^1_{S'}(k; G) \to \prod_{v\in S'} H^1(\kv; F) }$, and let $\pi$ be the inclusion.
The claim is that $\pi$ has finite image.
The proof is by the local constancy lemma applied to $X(\kv) \to H^1(\kv; G/G^0)$, which is essentially Krasner's lemma plus compactness -- here we used that $X(\kv)$ is compact.
Finite image plus finite fibers implies finite, so we're done.
:::

## Misc

:::{.theorem title="Minchev"}
Let $k \in \Number\Field$, $X,Y \in \Var\slice k$ geometrically integral, $F:Y\to X$ finite étale and not an isomorphism, and let $S \subseteq \places{k}$ be a finite set of places and suppose $X(\AA^S) \neq \emptyset$.
Then $X$ does not satisfy strong approximation, i.e. $X(k) \embeds X(\AA^S)$ is not dense.
:::

:::{.corollary title="?"}
If $X$ is geometrically integral and $\pi_1^\et(X_{\kbar}) \neq 1$, then $X(\AA^S)$ nonempty implies $X$ does not satisfy strong approximation (since $X$ admits no interesting étale covers)
:::

:::{.remark}
Goal: find $v\not\in S$ and $U \in X(\kv)$ open where $U$ does not contain any $k\dash$points.
:::