# Thursday, September 23

:::{.remark}
Let $k\in \Field$, we have a SES $1\to k\units \to \II_k \to C_k \to 1$.
An exercise from last time: for $\places{k}$ the places of $k$, prove that 
\[
H^2(L\slice k; \II_L) = \bigoplus_{v\in \places{k}} \Br(\Lv/\kv)
,\]
where $\Lv$ was obtained by choosing any place above $v$ in $L$ and completing.
:::

## Proof of Theorem

:::{.remark}
For $S \subseteq \places{k}$ a finite set of places containing all of the infinite places and $T$ a set of places of $L$ above $S$, we have
\[
\II_{L, T} = \prod_{w\in T}L_w\units \times \prod_{w\not\in T} \OO_{\Lv}\units
.\]
We can also write $H^2(L\slice k; \II_L) = \directlim_T H^2(L\slice k; \II_{L, T})$, so it's enough to show the following:
\[
H^2(L\slice k; \II_{L, T}) &= \bigoplus_{v\in S} \Br(\Lv / \kv) \\
H^2(L\slice k; \II_{\bar{k}, T}) &= \bigoplus_{v\in S} \Br(\kv)
.\]
We have
\[
H^2(L\slice k; \II_{L, T}) = \prod_{v\in S} H^2(L\slice k; \prod_{w\slice v} L_w\units ) \times \prod_{v\in S} H^2(L\slice k; \prod_{w\slice v} \OO_{L_w} \units) 
,\]
noting that we need to take the entire product to actually get a Galois module.
:::

:::{.claim}
\[
H^2(L\slice k; \prod_{w\slice v} \Lv \units) &= \Br(\Lv {}\slice{ \kv } ) \\
H^2(L\slice k; \prod_{w\slice v} \Lv \units) &= 0 
.\]

:::

:::{.proof title="of 1"}
\[
H^2(L\slice k; \prod_{w\slice v} L_w\units ) 
&=
H^2(L\slice k; \Ind_{\Gal(L/k)}^{\Gal(\Lv/\kv)} \Lv\units ) \\
&= H^2(\Lv \slice{\kv}; \Lv\units) \\
&\da \Br(\Lv \slice {\kv})
.\]
:::

:::{.proof title="of 2"}
Write
\[
H^2(L\slice k; \prod_{w\slice v} L_w\units) 
&= 
H^2(L\slice k; \Ind_{\Gal(L/k)}^{\Gal(\Lv/\kv)} \OO_{\Lv}\units ) \\
&= 
H^2(\Lv\slice {\kv}; \OO_{\Lv}\units)\\
&= 0
.\]

:::

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

\begin{tikzcd}
	{\Br(k)} && {\bigoplus_{v} \Br(\kv)} \\
	\\
	{H^2(k; \bar{k}\units)} && {H^2(k; \mathbb{I}_{\bar k}) }
	\arrow[from=1-1, to=1-3]
	\arrow[from=3-1, to=3-3]
	\arrow[Rightarrow, no head, from=1-1, to=3-1]
	\arrow[Rightarrow, no head, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXEJyKGspIl0sWzIsMCwiXFxiaWdvcGx1c197dn0gXFxCcihrX3YpIl0sWzAsMiwiSF4yKGs7IFxcYmFye2t9XFx1bml0cykiXSxbMiwyLCJIXjIoazsgXFxJSV97XFxiYXIga30pIl0sWzAsMV0sWzIsM10sWzAsMiwiIiwxLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMSwzLCIiLDEseyJsZXZlbCI6Miwic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==)


:::

## Injectivity 

:::{.theorem title="Injectivity"}
$\Br(k) \injects \bigoplus_v \Br(\kv)$ is injective, since $H^1(L\slice k; C_L) = 0$.
:::

:::{.theorem title="Actual IRL application of Sylow theorems"}
If $G$ is a finite group and $M\in \mods{G}$ then $H^i(G; M) = 0$ if $H^i(G_p; \ro{M}{G_p}) = 0$ for all $p$ where $G_p$ is a $p\dash$Sylow subgroup of $G$.
:::

:::{.proof title="?"}
There's a map 
\[
\qty{H^1(G; M) \mapsvia{\res}H^1(G_p; \ro{M}{G_p}) \mapsvia{\cores} H^1(G; M) } = \mult_d, d \da [G: G_p]
.\]
Since $d$ is prime to $p$, $\res$ is injective on $p\dash$power torsion, making $H^1(G; M)$ torsionfree.
Then since $G$ is finite, $H^i(G; M)$ is torsion, and the only torsion torsionfree group is zero.
:::

:::{.remark}
There will be multiple steps:

- It's enough to prove this for $\Gal(L\slice k)$ a $p\dash$group, using theorem on applications of Sylow.
  We know enough about the structure of $p\dash$groups to make induction arguments!

- It's enough to show that $H^i(L\slice k; C_L) = 0$ for $L\slice k$ cyclic.
  Letting $L\slice k$ be Galois with $G\da H^i(L\slice k)$ a $p\dash$group, then let $H \leq G$ be a nontrivial normal cyclic subgroup.
  Then the inflation-restriction exact sequence yields
  \[
  0 \to H^1(G/H; C_L^H) \to H^1(G; C_L) \to H^1(H; C_L) 
  ,\]
  using idele class groups and writing $C_L^H$ for the class group of the fixed field by $H$, and recalling that this comes from the Hochschild-Serre spectral sequence.
  By induction on the size of $G$, we'll know the right-hand side is 0, and the left-hand side is 0 by induction on $\# G$.
  However, note that we have to show this for *all* cyclic extensions!

- Prove the following theorem;

:::

:::{.remark}
Note that $C_L$ will not even be finitely generated!
:::

:::{.theorem title="?"}
If $L\slice k$ is cyclic, then $H^1(L\slice k; C_L) = 0$, and $\# H^2(L\slice k; C_L) = [L: k]$.
:::

:::{.warnings}
Note that $\# H^1 = 1$ in this case!
:::

:::{.definition title="Herbrand Quotient"}
If $G$ is finite cyclic and $M\in \mods{G}$, define the **Herbrand quotient** as
\[
q(M) \da { \# H^2(G; M) \over \# H^1(G; M) }
,\]
whenever this ratio is defined.
:::

:::{.remark}
Taking logs makes this look like an Euler characteristic.
:::

:::{.lemma title="Herbrand quotients are multiplicative"}
Suppose $0\to A\to B\to C\to 0$ is a SES of $G\dash$modules for $G$ cyclic.
Then 
\[
q(A) q(C) = q(B)
.\]
:::

:::{.exercise title="A fun one"}
Prove this!
It's the same proof that $\chi(A) + \chi(C) = \chi(B)$.
:::

:::{.lemma title="?"}
If $M$ is finite, then $q(M) = 1$, so this invariant for infinite modules.
:::

:::{.proof title="?"}
We first claim that $\# M^G = \# M_G$, recalling that $M_G = M/\gens{g-1} = M/IM$ for $I$ the augmentation ideal.
Note that in finite groups, for a SES $0\to A\to B\to C\to 0$ yields $\#B = (\# A)\cdot (\# C)$, or equivalently $(\#A)\cdot (\# B)\inv \cdot (\# C) = 1$ and this extends to longer exact sequences.

Now use the exact sequence 
\[
0 \to M^G \to M \mapsvia{g-1} M \to M_G \to 0
,\]
and so 
\[
(\# M^G) \cdot (\# M)\inv \cdot (\# M) \cdot (\# M_G)\inv = 1
.\]

Now to show that the sizes are equal,
Recall that
\[
H^*(G; M) = H^*\qty{ M \mapsvia{g-1} M \mapsvia{\sum g^i} M \to \cdots}
.\]
Thus we get
\[
0 \to H^1(G; M) \to \coker( M \mapsvia{g-1} M) \mapsvia{\sum g^i} \ker(M \mapsvia{g-1} M) \surjects H^2(G; M) \to 0 \\
\implies 0 \to H^1(G; M) \to M_G \mapsvia{\sum g^i} M^G \surjects H^2(G; M) \to 0
,\]
so 
\[
\# H^1(G; M) \cdot (\# M_G)\inv \cdot(\# M^G) \cdot (\# H^2(G; M))\inv = 1 = q(M)\inv
.\]
:::

:::{.lemma title="?"}
If $M, N$ are finitely generated in $\mods{G}$ and $M\tensor \RR \cong N\tensor \RR \in \mods{G}$, then $q(M) = q(N)$.
:::

:::{.remark}
Analogy: Reidemeister torsion!
Tensoring up to $\RR$ somehow doesn't lose all torsion information.
:::

:::{.proof title="of lemma"}
We'll first show $M\tensor \QQ \iso N\tensor \QQ$ implies $q(M) = q(N)$.

:::{.claim}
There is a map $M/\tors\to N/\tors$ with finite kernel and cokernel.
:::

\begin{tikzcd}[column sep=0.25in]
	0 && 0 \\
	\\
	{M_\tors} && {N_\tors} \\
	&&&& 0 & {M_\tors} & M & {N/N_\tors} & {\text{torsion!}} & 0 \\
	M && N \\
	\\
	{M\tensor \QQ} && {N\tensor \QQ}
	\arrow[from=1-1, to=3-1]
	\arrow[from=3-1, to=5-1]
	\arrow[from=5-1, to=7-1]
	\arrow["{\sim, \quad n\cdot p}", from=7-1, to=7-3]
	\arrow[from=3-3, to=5-3]
	\arrow[from=5-3, to=7-3]
	\arrow[from=1-3, to=3-3]
	\arrow[from=5-1, to=5-3]
	\arrow[from=3-1, to=3-3]
	\arrow[dashed, from=5-1, to=7-3]
	\arrow[from=4-5, to=4-6]
	\arrow[from=4-6, to=4-7]
	\arrow[from=4-7, to=4-8]
	\arrow[from=4-8, to=4-9]
	\arrow[from=4-9, to=4-10]
\end{tikzcd}

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


:::{.claim}
\[
M\tensor \RR \cong N\tensor \RR
\iff
M\tensor \QQ \cong N \tensor \QQ
.\]
:::

:::{.exercise title="?"}
Prove this!
:::

:::