# Tuesday, October 18

## Dedekind domains of interest

:::{.remark}
How to produce Dedekind domains:

- Start with $A\in \DD$, e.g. a PID, with $K = \ff(A)$.
- Take $L/K$ a finite degree field extension (not necessarily separable)
- Form the NT1 square, i.e. take the integral closure $B\da \intcl_L(A)$.
- Apply Krull-Akizuki: $B$ is always Dedekind.
- Check $B\in \amod^\fg$ when $L/K$ is separable.

Some examples:

- Starting with $A = \ZZ$ is classical: letting $K$ be a number field yields the ring of integers $B = \ZZ_K$.
- Starting with $A = k[t]$ for $k$ a field and $L/K$ any finite extension, then $B\in \amod^\fg$ even if $L/K$ is not separable -- the reason for this comes from geometry.

The latter is important geometrically: $L/K = k(C)/k(t)$ an extension of rational function fields where $C\slice k$ is a nice projective curve and $k(t)$ is the rational function field of $\PP^1$.
This yields a finite morphism of curves $\pi: C\to \PP^1$, which topologically is a branched cover where geometric ramification corresponds to algebraic ramification.
For $k=\CC$, this can be regarded as a branched covering of compact Riemann surfaces:

\begin{tikzpicture}
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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/NT/sections/figures}{2022-10-18_11-19.pdf_tex} };
\end{tikzpicture}

Taking $M = k(\tilde C)$ yields a cover $\tilde C\to C$ a further branched cover
How Dedekind domains show up: take $A$ with $k[t] = k[\AA^1]$ and take $B/A$ where $B = k[C^\circ]$, which yields a cover $C^\circ \mapsvia{\pi} \PP^1$ where $C^\circ = \pi\inv(\infty)$.
Note that the closed points of $\AA^1$ correspond to $\mspec k[t]$, and since $k[t]$ is a PID the points are monic irreducible polynomials.
Similarly the closed points of $C^\circ$ correspond to $\mspec k[C^\circ]$.
Over $\CC$ (or an algebraically closed field), these are of the form $\gens{t-\alpha}$ for $\alpha\in \CC$ and this corresponds to $\AA^1$.

Note that classically, Riemann-Hurwitz measures the extent of ramification of a branched cover.
For unbranched covers over $\CC$, one can pull back triangulations along a degree $d$ map to conclude that Euler characteristics multiply.
This generalizes to a formula involving the *genus* of a curve, which is proved using the *different ideal*.

> Note: the different can be used to compute discriminants of number fields.

Punchline: the point of building the theory of Dedekind domains (instead of just arbitrary number fields) is to encapsulate both theories, i.e. classical NT and the theory of function fields of algebraic curves.
:::

:::{.remark}
Three important finiteness theorems:

- Finiteness of the class number,
- Finiteness of $\rank \ZZ_K$,
- Hermite's finiteness theorem.

This may fail for arbitrary function fields, but have analogs when $k$ is finite.
:::

:::{.remark}
Take the NT1 square:

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

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJCIl0sWzAsMiwiQSJdLFsyLDIsIksiXSxbMiwwLCJMIl0sWzEsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDMsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMiwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=)

Recall that the trace form $\Tr_{L/K}$ is nondegenerate iff $L/K$ is separable.
For any $A\dash$lattice $\Lambda$ in $L$, take the dual lattice $\Lambda\dual$ with respect to the form $\inp x y \da \Trace_{L/K}(xy)$.
Hey: $B$ is a $K\dash$lattice in $L$ with $\Tr_{L/K}(B) \subseteq A$, and $B\dual = \ts{x\in L \st \Tr_{L/K}(xB) \subseteq A} \contains B$.

:::{.claim}
$B\dual$ is a fractional ideal.
:::

So for all $p\in \mspec B$, note $v_p(B\dual) \leq 0$, so invert it and define the **different ideal**
\[
D(B/A) \da (B\dual)\inv \normal B
.\]
We'll soon define the ideal norm map $N: \Frac(B) \to \Frac(A)$, and it will turn out that the discriminant can be written as $\delta(B/A) = N(D(B/A))$ -- note that $N$ is not injective in general.
:::

:::{.remark}
Consider $\mfp \normal A$ and $\mfp B\normal B$ its lift, which decomposes as $\mfp B = \prod p_i^{e_i}$.
A prime upstairs $p_i \divides \mfp$ is **ramified** iff $e_i \geq 2$ **or** the residual field $(B/p_1)/(A/\mfp)$ is an inseparable extension.
A prime downstairs is ramified iff any $p_i$ with $p_i \divides p$ is ramified.
:::

:::{.theorem title="Fundamental theorems"}
For $B/A$,

- $\mfp \normal A$ ramifies iff $\mfp \divides \delta(B/A)$,
- $p_i \normal B$ ramifies iff $p_i \divides D(B/A)$.

:::

:::{.remark}
Note that knowing the ramification points is not enough to know the discriminant or different, since this gives which primes divide it but not to which power.
For upper and lower bounds on valuations of differents, see Dedekind's different theorem.
However, extra information on the discriminant can improve the algorithm discussed previously 
:::

:::{.theorem title="?"}
Let $L/K$ be a finite degree extension.
TFAE:

1. The trace form $T: L\times L\to K$ where $T(x,y) = \Tr_{L/K}(xy)$ is nondegenerate.
2. There exists some $x\in L$ such that $\Tr_{L/K}(x) \neq 0$.
3. $\Tr_{L/K}: L\surjects K$ is surjective.[^clear_from_linear]
4. $L/K$ is separable.

[^clear_from_linear]: 
This is a $k\dash$linear map, so either surjective or zero.

:::

:::{.proof title="?"}
See Pete's "Field Theory" notes, uses Dedekind's theorem and is at most 1 page.
:::

:::{.corollary title="?"}
For $k$ a field and $A\in \kalg$, TFAE:

1. $A$ is étale,
2. The trace form $\inp\wait\wait: A\to k$ given by $\inp x y = \Tr_{A/k}(xy)$ is nondegenerate.

:::

:::{.proof title="?"}
$1\implies 2$: write $A = \prod _{1\leq i\leq r} L_i$ with $L_i$ finite degree separable extensions.
The trace form breaks into a direct sum: writing $\inp{\wait}{\wait}_i: L_i \times L_i \to k$ and $x = \tv{x_1,\cdots,x_r}, y = \tv{y_1,\cdots, y_r}$, then
\[
\inp x y = \sum_i \inp{x_i }{y_i }_i
.\]
In other words, the Gram matrix is block diagonal, which is invertible iff the blocks are invertible.
So nondegeneracy of the factors for all $i$ implies the overall trace form is nondegenerate.

$\not 1\implies\not 2$:
Suppose $A$ is not étale -- either there is a nonzero nilpotent, or if there are none then one can continue to write it as a product as above but one of the $L_i$ is inseparable.
In the first case, let $x$ be a nonzero nilpotent.
Then $xy$ is again nilpotent for any $y\in A$, and $\Tr(xy) = 0$ since any matrix representation of a nilpotent element has eigenvalues $\lambda_j = 0$ and thus $\sum_j \lambda_j = 0$.
In the second case, $A = \prod L_i$ with (say) $L_j$ inseparable, but by the previous argument some block in the Gram matrix is singular and thus the entire form is degenerate.
:::

:::{.remark}
We can now reconstruct the proof that $p$ ramifies iff $p\divides \delta(B/A)\normal A$ in this new setup:

\begin{tikzcd}
	B && L \\
	\\
	A && K
	\arrow[hook, from=3-1, to=1-1]
	\arrow["{\text{sep}}"', hook, from=3-3, to=1-3]
	\arrow[hook, from=1-1, to=1-3]
	\arrow[hook, from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJCIl0sWzAsMiwiQSJdLFsyLDIsIksiXSxbMiwwLCJMIl0sWzEsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCJcXHRleHR7c2VwfSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMywiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwyLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==)

Replace $A$ with $A_p$ and $B$ with $B_p \da (A\sm p)\inv B$.
Note that $A_p$ is a DVR and $\mspec B_p = \ts{q\in \mspec B \st q\divides p}$, so $B_p$ is not necessarily a DVR -- it is not necessarily local, but is semilocal.
Choose a basis $\ts{x_1,\cdots, x_n}$ for $B_p\in \mods{A_p}$ and set $a \da v_p(\Delta(x_1,\cdots, x_n))$, then $v_p(\delta(B/A)) = a$ since this exactly measures the power locally.
We then reduce to showing $p$ ramified iff $a\geq 1$ iff $\Delta(x_1,\cdots, x_n) \equiv 0 \mod p$.
Since the discriminant is compatible with base change, this happens iff considering $A_p/pA_p = A/p_A$ and $B_p/pB_p$ (which are algebras over the residue field) one has $\delta( (B_p/ pB_p) / (A/pA) ) = 0$.

Note that $A/pA$ is a field and $B_p/pB_p = B/p B_p$, so this is zero iff the trace form of $B/pB$ as an $A/pA\dash$algebra is degenerate, which happens iff $B/pB$ is not an étale $A/pA$ algebra. As before, this happens if (1) $B/pB$ contains a nonzero nilpotent, or (2) $B/pB$ has a nonseparable factor.
One can factor $pB = \prod _{1\leq i\leq g} p_i^{e_i}$ to write $B/pB \cong B/\prod p_i^{e_i} \cong \prod B/p_i^{e_i}$, and there is a nonzero nilpotent precisely when some $e_i \geq 2$.
If $e_i = 1$ for all $i$, then each factor is a field and is thus not etale iff some $B/p_i^{e_i}$ is not separable.
:::