# Thursday, October 20

:::{.remark}
Today: knocking out some standard NT, e.g. the degree inequality.
Let $A \in \Dedekind$ with $K = \ff(A)$ and $L/K$ a degree $n$ field extension, let $B = \intcl_L A$.
Suppose $B$ is finitely-generated over $A$, which is true e.g. if $L/K$ is separable.
Let $p\in \mspec A$, and recall there is a pushforward $i_*: \ff(A)\to \ff(B)$ where $I\mapsto BI = \gens{I}_B = I\tensor_A B$, taking the $B\dash$submodule generated by $I$.
Note that $i_*$ is a group morphism.

In our case, $\Frac(A) = \bigoplus _{\mspec A} \ZZ$ and so $i_*$ is determined by what happens to prime ideals: write $i_*(\mfp) = B\mfp = \prod p_i^{e_i}$ with $e_i \in \ZZ_{\geq 0}$.
Figuring out how this factors is a deep question of number theory, e.g. class field theory studies how primes split in abelian extensions of number fields.
For any $P_i \divides \mfp$, we define 

- $e(P_i \divides \mfp) \da e_i$ the **ramification index**,
- $f(P_i\divides \mfp) \da [B/P_i : A/\mfp]$ the **residual degree**.

:::

:::{.exercise title="Indices multiply in towers"}
Add a new third floor to our favorite edifice:

\begin{tikzcd}
	\mcp && C && M \\
	P && B && L \\
	\mfp && A && K
	\arrow[hook, from=3-3, to=2-3]
	\arrow[hook, from=2-3, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=2-3, to=2-5]
	\arrow[from=3-3, to=3-5]
	\arrow["{\text{finite}}"', hook, from=3-5, to=2-5]
	\arrow["{\text{finite}}"', hook, from=2-5, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwxLCJCIl0sWzIsMiwiQSJdLFs0LDEsIkwiXSxbNCwyLCJLIl0sWzIsMCwiQyJdLFs0LDAsIk0iXSxbMCwyLCJcXG1mcCJdLFswLDEsIlAiXSxbMCwwLCJcXG1jcCJdLFsxLDAsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsNCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCw1XSxbMCwyXSxbMSwzXSxbMywyLCJcXHRleHR7ZmluaXRlfSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsNSwiXFx0ZXh0e2Zpbml0ZX0iLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==)

Assume $B/A$ is finitely-generated and $C/B$ is finitely-generated.
As a consequence, $C/A$ is finitely-generated.
Show that
\[
e(\mcp \divides \mfp) = e(\mcp \divides P) \cdot e(P\cdot \mfp) \\
f(\mcp \divides \mfp) = f(\mcp \divides P) \cdot f(P\cdot \mfp) 
.\]

This shouldn't be that tough.
:::

:::{.theorem title="?"}
Take the setup as before: $A\in \Dedekind, K = \ff(A), [L: K] = n, B = \intcl_L(A)$.
Then
\[
[L:K] = \sum_{i=1}^r e(P_i \diides \mfp) f(P_i \cdots \mfp)
,\]
where $r$ is the number of distinct primes lying above $\mfp$, i.e. $\mfp B = \prod_{i=1}^r P_i^{e_i}$.
:::

:::{.proof title="?"}
Note that for $i\geq 1$, we have $\dim_{B/\mfp} \mfp^i/\mfp^{i+1} = 1$.
We showed earlier that $\mfp^i/\mfp^{i+1} = \mfp^i/(\mfp^i)\mfp \cong R/\mfp$ which is more patently dimension 1.
Replace $A$ with the DVR $A_\mfp$ and $B$ with the semilocal PID $B_\mfp = B\tensor_A A_\mfp = (A\sm\mfp)\inv B$.
Note that $\mspec B_\mfp$ is precisely the fiber over $\mfp$, i.e. $\ts{P_1,\cdots, P_r}$, which preserves the ramification index $e$ and residual degree $f$.
So these indices can be computed after passing to these localizations, and
we assume that $\mfp$ is the only prime of $A$ and $P_i$ are the only primes above $\mfp$.

- Step 1: Claim, $n\da [L:K] = \dim_{A/\mfp} B/\mfp B$.
  Proof: $B\in \amod^\free$, so if $B\isoas{\amod} A^n$ then $B/\mfp B = B\tensor_A A/\mfp \cong A^n\tensor_A A/\mfp \cong (A/\mfp)^n$.
- Step 2: Compute $B/\prod p_i^{e_i} \caniso \prod_{i=1}^r B/p_i^{e_i}$, so $\dim_{A/\mfp} B/\prod p_i^{e_i} = \sum_{i=1}^r \dim_{A/\mfp} B/p_i^{e_i}$.
  Consider $\dim_{A/\mfp} B/p_i^{e_i}$: take a $B/p_i$ composition series $B \contains p_i \contains p_i^2 \contains \cdots \contians p_i^{e_i}$.
  Each successive quotient is isomorphic to $B/p_i$ in $\mods{B}$, and is a vector space over $A/\mfp$ and thus isomorphic to $(A/p)^{f_i}$ in $\mods{A/\mfp}$ for some number $f_i$.
  Each step in the series breaks into $f_i$ steps when taken as a composition series over $A/\mfp$, so the total length is $e_i \cdot f_i$:

\begin{tikzcd}
	{p_i^{e_i}} & \cdots & {p_i^2} & {p_i}
	\arrow["{f_i}", hook, from=1-1, to=1-2]
	\arrow["{f_i}", hook, from=1-2, to=1-3]
	\arrow["{f_i}", hook, from=1-3, to=1-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJwX2lee2VfaX0iXSxbMiwwLCJwX2leMiJdLFszLDAsInBfaSJdLFsxLDAsIlxcY2RvdHMiXSxbMCwzLCJmX2kiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDEsImZfaSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMiwiZl9pIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=)

:::

:::{.remark}
In more general cases this is an inequality and one can study the *defect*.
:::

:::{.warnings}
Saying things happen for $L/K$ usually means they actually happen in the associated Dedekind domains $\OO_L$ and $\OO_K$.
For examples, "primes ramify in $L/K$" doesn't strictly make sense, since there are no nontrivial primes in $K$, and this always means primes ramify in $\OO_L$ over $\OO_K$.
:::

:::{.definition title="Ramification"}
\envlist

- $L/K$ is **totally ramified at $\mfp \in \mspec A$** iff there exists a unique $P\divides \mfp$ with $e(P\divides \mfp) = [L: K]$.
- $L/K$ is **ramified at $\mfp$** iff f$e(P\divides \mfp) = 1$ for all $P\divides \mfp$ **and** and $B/\mfp$ is a separable extension of $A/\mfp$.
- $L/K$ is **inert over $\mfp$** if $\mfp B$ is a prime ideal, i.e. $r=1$ and $e_1 = 1$.
- $L/K$ **splits completely** if $r=n$, i.e. $e_i = f_i = 1$ for all $i$.

:::

:::{.remark}
Interpretation in terms of branched covers of Riemann surfaces: removing the branch points upstairs and downstairs gives an actual unramified cover of curves in the topological sense.
A topological Galois cover $Y\to X$ is one such that $\pi_1 Y \normal \pi_1 X$ and $\Aut(Y/X)$ acts transitively on fibers -- similarly, for $L/ K$ Galois, $G = \Aut(L/K) = \Gal(L/K)$ acts on $p_i\divides \mfp$ transitively and $n=efr$.
:::

:::{.fact}
Consider the NT1 square: $L/K$ separable with $B/A$ as before.
Recall $D(B/A) \normal B$ and $\delta(B/A) \normal A$, the ideal norm $N: \Frac(B) \to \Frac(A)$, and the pushforward $i_*: \Frac(A) \to \Frac(B)$.
A fact:
\[
N(D(B/A)) = \delta(B/A)
.\]
:::

:::{.definition title="The ideal norm"}
Recall $\Frac(B) = \bigoplus _{\mspec B} \ZZ$, so it suffices to define $N(P)$ for $P\in \mspec B$.
Let $\mfp\in \mspec A$ be the unique prime with $P\divides \mfp$, then define
\[
N(P) \da \mfp^{f(P\divides \mfp)}
.\]
:::

:::{.example title="?"}
Take $K/\QQ$ with $\ZZ_K/\ZZ$ and $p\in \spec \ZZ$ with $P\in \spec \ZZ_K$.
Then $\ZZ_K/P$ is finite, and 
\[
\size \ZZ_K/P = p^{[\ZZ_K/p : \ZZ/p\ZZ]} \da p^{f(P \divides p)}
,\]
so $N(P) = \size \ZZ_K/p$.

> Note that the Frohlich invariant is sometimes called the "module index".

:::

:::{.remark}
Now if $I\normal B$ is nonzero, define
\[
N(I) \da \chi_{\amod}(B/I)
,\]
where we take the characteristic ideal not as a finite-length $B\dash$modules, but rather as a finite length $A\dash$module.

Claim: $B/I\in \amod$ is finite length iff $B/I\in \amod$ is torsion.
Write $a\da I \intersect A$, then this is nonzero since writing $I = \prod_{i=1}^r p_i^{e_i}$ shows $I \intersect A \contains \prod_{i=1}^r (p_i^{e_i} \intersect A)$, and if $P_i\divides p_i$ then $P_i \intersect A = p_i \neq 0$ and any nonzero element $x_i\in p_i$ yields an element $x^{e_i} \in p_i^{e_i} \intersect A$.
Therefore $B/I\in\mods{A/I \intersect A}$ and so $B/I \in \mods{A}^\tors$.
:::

:::{.remark}
One checks that if $P\in \mspec B$ then $N(P) = \chi_{\amod}(B/P) = p^f$ where $p\divides P$.
Note that $P \subseteq B$ is a $\mods{B}$ composition series where $B$ over $P$ is rank $f(P\divides p)$ in $\amod$. 
Refining this to $P \subseteq M_1 \subseteq \cdots \subseteq M_{f-1} \subseteq B$ where each quotient is isomorphic to $A/p$ requires $f(P\divides p)$ steps.
:::