# Tuesday, November 29

## Maximal unramified extensions

:::{.remark}
Setup: the NT1 square, where now the extension is Galois.

\begin{tikzcd}
	P & B && L \\
	\\
	{p } & A && K
	\arrow[from=3-2, to=1-2]
	\arrow["{\text{Galois}}"', from=3-4, to=1-4]
	\arrow[from=3-2, to=3-4]
	\arrow[from=1-2, to=1-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwwLCJCIl0sWzMsMCwiTCJdLFsxLDIsIkEiXSxbMywyLCJLIl0sWzAsMCwiUCJdLFswLDIsInAgIl0sWzIsMF0sWzMsMSwiXFx0ZXh0e0dhbG9pc30iLDJdLFsyLDNdLFswLDFdXQ==)

Define $D \da D(P/p) \da \ts{ \sigma\in G \st \sigma P = P }$, then there is an isomorphism:

\begin{tikzcd}
	{B/P} && {B/\sigma(P)} \\
	\\
	& {A/p}
	\arrow["\sim", from=1-1, to=1-3]
	\arrow[from=3-2, to=1-3]
	\arrow[from=3-2, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJCL1AiXSxbMiwwLCJCL1xcc2lnbWEoUCkiXSxbMSwyLCJBL3AiXSxbMCwxLCJcXHNpbSJdLFsyLDFdLFsyLDBdXQ==)

Set $\Aut(B/P, A/p) = \Aut(\ell(P) / \kappa(p))$ where $\kappa(p) = A/p$ and $\ell(P) = B/P$.
If $\sigma \in D$ then it defines a reduction map $r: D(P/p) \to \Aut(\ell(P) / \kappa(p))$.
We define the **inertia group** $I(P/p) = \her r = \ts{\sigma\in G \st \forall x\in B, \sigma(x) - x\in P}$ as the kernel of the reduction map.
Note that $I(P/p) \normal D(P/p) \leq G$.

Consider a tower:

\begin{tikzcd}
	P & {B \subseteq} & L \\
	{p_D} & {A_D \subseteq} & {L^D} \\
	p & {A \subseteq} & K
	\arrow[from=3-3, to=2-3]
	\arrow[from=2-3, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwwLCJMIl0sWzIsMSwiTF5EIl0sWzIsMiwiSyJdLFsxLDAsIkIgXFxzdWJzZXRlcSJdLFsxLDEsIkFfRCBcXHN1YnNldGVxIl0sWzEsMiwiQSBcXHN1YnNldGVxIl0sWzAsMiwicCJdLFswLDEsInBfRCJdLFswLDAsIlAiXSxbMiwxXSxbMSwwXV0=)

:::

:::{.claim}
\[
e(P/p_D) &= e(P/p) \da e \\
f(P/p_D) &= f(P/p) \da f \\
r(P/p_D) &= 1
.\]
:::

:::{.remark}
Note $r=1$ since $\Aut(L/L^D)= D$.
Using multiplication in towers,
\[
e(P/p_d) f(P/p_D) =
e(P/p_d) f(P/p_D) r(P/p_D) 
= \size D = ef
,\]
using that the extension is Galois, and so $e(P/p_D)\divides e$ and $f(P/p_d)\divides f$.
This forces $e(P/p_d) = e$ and $f(P/p_D) = f$.
Again using multiplication in towers, $e(P/p_D) = 1$ and $f(P_D/p) = 1$, so there a lot of splitting.
If $L^D/K$ were Galois (if $D$ is normal, e.g. if the extension is abelian), we would know that $p$ splits completely in $A_D$.
:::

:::{.theorem title="?"}
Suppose that $\kappa(p) = A/p$ is perfect. Then

a. $\ell(P)/\kappa(p)$ is Galois.
b. $r: D(P/p) \surjects \Aut(\ell(P) / \kappa(p) )$.
c. $\size I(P/p) = e \da e(P/p)$.

:::

:::{.proof title="Part a"}
We have $e(p_D/p) = f(p_D/p) = 1$, so $A_D/p_D = A/p$.
Then $\ell(P) / \kappa(p)$ is a finite separable extension, thus monogenic, so write $\ell(P) = \kappa(p)[\bar\alpha]$ for some $\bar\alpha$.
Lift $\bar\alpha$ to any $\alpha\in B$, and let $f\in L^D[t]$ be its minimal polynomial.
Note $\alpha$ is integral over $A$ and thus over $A_D$, so in fact $f\in A_D[t]$.
Since $L/L^D$ is Galois, $f$ splits and so every root is of the form $\sigma(\alpha)$ for some $\sigma \in D$.

Let $\bar f$ be the image of $f$ in ${A_D\over P_D}[t] = {A\over p][t]$, then the roots of $\bar f$ are of the form $r(\sigma)(\bar \alpha)$ for some $\sigma\in D$.
Since the other roots are conjugate, all the roots lie in $\ell(P)$.
So $\ell(P)$ is a normal extension on $\kappa(p)$, and since it was assumed separable, this makes it the splitting field of $\bar f$.
:::

:::{.proof title="Parts b and c"}
Since every conjugate of $\bar\alpha$ over $\ell(P)$ is of the form $r(\sigma)(\bar\alpha)$, every element of $\Aut(\ell(P) / \kappa(p))$ is of the form $r(\sigma)$ for some $\sigma\in D$.
So $r$ is surjective, and part c follows immediately.
:::

:::{.remark}
The only piece that may not be Galois is $L^D/K$, so assume it is.
We can extend the tower:

\begin{tikzcd}
	P & B & L \\
	{p_I} & {A_I} & {L^I} \\
	{p_D} & {A_D} & {L^D} \\
	p & A & K
	\arrow["{\deg = r}"', from=4-3, to=3-3]
	\arrow["{\deg = f}"', from=3-3, to=2-3]
	\arrow["{\deg =e}"', from=2-3, to=1-3]
	\arrow[from=4-2, to=3-2]
	\arrow[from=3-2, to=2-2]
	\arrow[from=2-2, to=1-2]
	\arrow[from=4-2, to=4-3]
	\arrow[from=3-2, to=3-3]
	\arrow[from=2-2, to=2-3]
	\arrow[from=1-2, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTIsWzIsMywiSyJdLFsyLDIsIkxeRCJdLFsyLDEsIkxeSSJdLFsyLDAsIkwiXSxbMSwwLCJCIl0sWzEsMSwiQV9JIl0sWzEsMiwiQV9EIl0sWzEsMywiQSJdLFswLDMsInAiXSxbMCwyLCJwX0QiXSxbMCwxLCJwX0kiXSxbMCwwLCJQIl0sWzAsMSwiXFxkZWcgPSByIiwyXSxbMSwyLCJcXGRlZyA9IGYiLDJdLFsyLDMsIlxcZGVnID1lIiwyXSxbNyw2XSxbNiw1XSxbNSw0XSxbNywwXSxbNiwxXSxbNSwyXSxbNCwzXV0=)

- The first extension: $p$ splits into $r$ different primes.
- The second extension: each $p_i$ above $p$ remains inert (no ramification or splitting, only larger residue fields).
- The third extension: totally ramified.

:::

:::{.proposition title="?"}
Let $L/K$ be a finite separable extension, not necessarily Galois:

\begin{tikzcd}
	& B & K \\
	p & A & L
	\arrow[from=2-3, to=1-3]
	\arrow[from=2-2, to=1-2]
	\arrow[from=2-2, to=2-3]
	\arrow[from=1-2, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMSwwLCJCIl0sWzEsMSwiQSJdLFsyLDEsIkwiXSxbMiwwLCJLIl0sWzAsMSwicCJdLFsyLDNdLFsxLDBdLFsxLDJdLFswLDNdXQ==)

a. There exists a unique subextension $L^s$ of $L/K$ such that for any subextension $F$ of $L/K$, $p$ splits complete in $F$ iff $F \subseteq L^s$.
  So $L^s$ is the largest extension in which $p$ splits completely.
  Moreover, if $M$ is the Galois closure of $L/K$ and $P_1,\cdots, P_r$ are the primes of $M$ lying over $p$, then 
\[
L^s = \intersect_{i=1}^r M^{D(P_i/p)} \intersect L
.\]

b. If $L/K$ is Galois, so is $L^s$.

:::

:::{.theorem title="?"}
Let $A\subseteq K$ and let $K\sep$ be a separable closure, and let $K_1,\cdots, K_r$ be finite degree extensions of $K\sep/K$.
Let $L \da K_1\cdots K_r$ be their compositum, then if $p\in \mspec A$, then $p$ splits completely in each of $K_1,\cdots, K_r$ iff $K$ splits completely in $L$.
:::

:::{.remark}
Note that $\impliedby$ is clear by multiplication in towers, and $\implies$ follows immediately from the proposition.
:::

:::{.corollary title="?"}
If $L/K$ is separable and $p$ splits completely in $L$ then $p$ splits completely in the Galois closure $M$ of $L/K$.
:::

:::{.proof title="?"}
This follows from the fact that $M$ is the compositum of all Galois conjugates of $L$.
:::

:::{.proposition title="?"}
Consider a tower where $L/K$ is separable:

\begin{tikzcd}
	& B & K \\
	p & A & L
	\arrow[from=2-3, to=1-3]
	\arrow[from=2-2, to=1-2]
	\arrow[from=2-2, to=2-3]
	\arrow[from=1-2, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMSwwLCJCIl0sWzEsMSwiQSJdLFsyLDEsIkwiXSxbMiwwLCJLIl0sWzAsMSwicCJdLFsyLDNdLFsxLDBdLFsxLDJdLFswLDNdXQ==)

There exists a unique subextension $L^i$ of $L/K$ such that for any subextension $F$ of $L/K$, $p$ is unramified in $F$ iff $F \subseteq L^i$.
If $L/K$ is Galois, so is $L^i$, and can be construct as $L^i = \intersect_{i=1}^r M^{I(P_i/p)} \intersect L$.
Moreover, a theorem and corollary identical to the one above holds with "splits completely" replaced by "unramified".
:::

:::{.remark}
We've already used this to prove Stickelberger's theorem.
:::

## Toward Chebotarev Density

:::{.definition title="Frobenius elements"}
Take an NT1 square with Galois group $G$

\begin{tikzcd}
	P & B & K \\
	p & A & L
	\arrow[from=2-3, to=1-3]
	\arrow[from=2-2, to=1-2]
	\arrow[from=2-2, to=2-3]
	\arrow[from=1-2, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMSwwLCJCIl0sWzEsMSwiQSJdLFsyLDEsIkwiXSxbMiwwLCJLIl0sWzAsMSwicCJdLFsyLDNdLFsxLDBdLFsxLDJdLFswLDNdXQ==)

Assume $\kappa(p) \cong \FF_q$ and $\ell(P)/\kappa(p)$ of degree $f = f(P/p)$.
Then $\ell(P) \cong \FF_{q^f}$, and $\Aut(\FF_{q^f}/\FF_q)$ is cyclic of order $f$ with canonical generator $F_q: x\mapsto x^q$.
A **Frobenius element** $\tau_{P/p}$ is any $\sigma(D)$ such that $\pi(\sigma) = F_q$, which is well-defined up to an element of $I(P/p)$ and thus defines a coset.
If $e(P/p) = 1$ then $\tau_{P/p}$ is well-defined on the nose.
:::

:::{.remark}
Assume $e=1$ (unramified).
If $P'/p$ is another prime over $p$, then $\tau_{P'/p}$ is conjugate to $\tau_{P/p}$.
Conversely, every conjugate of $\tau_{P/p}$ is of this form.
Thus there is a *conjugacy class* $C(p) \subseteq G$ which is an invariant of $p$.
Thus we get a map $\mspec A\to \ts{\text{Conjugacy classes of } G}$, and when $G$ is abelian this reduces to a map $\mspec A\to G$.
:::

:::{.question}
What are the fibers of this map? Is every class in $G$ the Frobenius element of some prime?
Are the number of primes with a fixed Frobenius equidistributed? (Yes: Chebotarev density!)
:::