# Thursday, August 18

:::{.remark}
Course goal: treat algebraic number theory using standard tools of field theory and commutative algebra.
Very much necessary for those interested in arithmetic geometry and/or algebraic geometry.
We'll be most interested in Dedekind rings, generalizing rings of integers (and their localizations/completions) and coordinate rings of affine algebraic curves.
:::

:::{.remark}
Upcoming plan:

- 2 lectures of commutative algebra "review".
- Dedekind domains (see Pete's CA notes).
- ANT in Dedekind domains (see "NT1 square" from earlier).
- Specialize to number fields, "geometry of numbers" methods, with complete proofs.

Note that we'll only lightly discuss proofs for the first few items, since the statements are relatively easy to use.
:::

:::{.theorem title="?"}
Let $K\in \NF$ with discriminant[^disc_imp] $\delta_K$, a prime $p$ **ramifies** in $K$ $\iff$ $p\divides \delta_K$.

[^disc_imp]: 
A very important integer!

:::

## Proof

:::{.remark}
Pete proved each direction separately in a reading course in the 90s with Raghavan Narasimhan in a reading course.
The book he used involved two separate arguments, each taking a few pages.
This is an if and only if though, so the dream would be to understand each side well enough to simply *see* both directions at once.
We'll aim for an argument that proves both at once.
Toward a proof: we first define the discriminant $\delta_K$ for $\ZZ_K$ the ring of integers of a number field $K$.
:::

:::{.definition title="Trace and discriminant"}
There exists a $\ZZ\dash$basis $\ts{\alpha_1, \cdots, \alpha_d}$ with $d = [K: \QQ]$.
Note that multiplication by $\alpha$ for $\alpha\in K$ is a $K\dash$linear and thus a $\QQ\dash$linear map, so expanding in the above basis and taking the matrix trace of $\cdot \alpha$ yields $\Tr_{K\slice \QQ}( \alpha)$.
The **trace form** is a (nondegenerate) symmetric bilinear form
\[
T: \ZZ_K \times \ZZ_K &\to \ZZ \\
T(\alpha, \beta) &\mapsto \Tr_{K\slice \QQ}(\alpha \beta)
.\]
One can form its Gram matrix $G_{ij} = T( \alpha_i, \alpha_j) \in \Mat_{d\times d}(\ZZ)$, and the **discriminant** is defined as $\delta_K \da \det G \in \ZZ\smz$.
:::

:::{.remark}
Reduce everything $\mod p$ in the definition of the trace form (or apply $\wait\tensor_\ZZ \ZZ/p\ZZ$) to get reduced trace form on the quotient $\ZZ_K/p\ZZ_K$, an extension of $\ZZ/p\ZZ$.
Note that the $\ZZ\dash$basis reduces to $\ts{\bar \alpha_1, \cdots, \bar \alpha_d}$, a $\ZZ/p\ZZ\dash$basis for $\ZZ_K/p\ZZ_K$.
Then $p\divides \delta_K \iff$ the discriminant of the $\mod p$ trace form is zero $\iff$ the reduced trace form is degenerate.
Note that $\ZZ_K/p\ZZ_K$ is a finitely generated free $\ZZ/p\ZZ\dash$module, and we now have a clean question: 
:::

:::{.question}
Let $k\in \Field$ and $A\in \Alg^{\fd}\slice k$, when is the trace form nondegenerate?
:::

:::{.answer}
This happens iff $A$ is an etale $k\dash$algebra, i.e. $A \cong \prod_{i=1}^r \ell_{i}$ wit each $\ell_i{}\slice k$ a finite degree separable field extensions.
:::

:::{.remark}
One has to prove this, but this readily reduces to the case where $r=1$, so $A\slice k$ is a finite degree field extension.
This is a standard result in field theory!
Note that since $\ZZ/p\ZZ$ is perfect, separability is automatic.
Thus $p\divides\delta_K$ exactly when $\ZZ_K/p\ZZ_K$ splits into a product of fields.
This is clearly related to ramification, which we'll now describe.
:::

:::{.definition title="Ramification"}
Write the pushforward of $p$ as $p\ZZ_K = \prod p_i^{e_i}$ with the $p_i$ distinct, then $p$ is **unramified** in $K$ iff $e_i =1$ for all $i$.
:::

:::{.remark}
Now write 
\[
\ZZ_K/p\ZZ_K = \ZZ_K/\prod p_i^{e_i} \cong \prod \ZZ_K /p_i^{e_i}
\]
using comaximality of these ideals and apply the CRT, since primes are maximal in Dedekind domains.
Since rings modulo maximal ideals are fields, if $p$ is unramified then $\ZZ_K/p\ZZ_K \cong \prod \ZZ_K/p_i$ is a product of fields.
On the other hand, if $p$ is ramified then some $e_i \geq 2$, note that products of fields are reduced.
For $R/p_i^{e_i}$ with $e_i \geq 2$, take $\pi \in p_i\sm p_i^2$ nonzero to get $\pi^{e_i} = 0$ a nonzero nilpotent, concluding the proof.
:::

## Generalizing

:::{.fact}
In fact this proves more:
if $A \in \Dedekind$ and $K = \ff(R)$ and $L\slice K$ is a finite degree separable field extension with $B = \intcl_{L}(A)$, we get a standard square:

\begin{tikzcd}
	B && L &&& \textcolor{rgb,255:red,92;green,92;blue,214}{\ZZ_K} && \textcolor{rgb,255:red,92;green,92;blue,214}{K} \\
	\\
	A && K &&& \textcolor{rgb,255:red,92;green,92;blue,214}{\ZZ} && \textcolor{rgb,255:red,92;green,92;blue,214}{\QQ}
	\arrow[no head, from=1-1, to=3-1]
	\arrow[no head, from=1-3, to=3-3]
	\arrow[no head, from=1-6, to=3-6]
	\arrow[no head, from=1-8, to=3-8]
	\arrow["\subseteq", hook, from=3-1, to=3-3]
	\arrow["\subseteq", hook, from=1-1, to=1-3]
	\arrow["\subseteq", hook, from=3-6, to=3-8]
	\arrow["\subseteq", hook, from=1-6, to=1-8]
\end{tikzcd}

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

We can define a notion of discriminant:
:::

:::{.definition title="Discriminants for ideals in integral closures"}
Define $\delta_{B\slice A}$ the discriminant.
Let $p\in \Spec A\smz$ and write $pB = \prod p_i^{e_i}$.
Then $p$ **ramifies in $B$** iff

- Some $e_i > 1$, or
- For some $i$, the residual extension $B/p_i{}\slice{A/p}$ is not separable.
:::

:::{.theorem title="?"}
$p\divides \delta_{B\slice A} \iff p$ ramifies in $B$.
:::

:::{.remark}
Note that for covers of Riemann surfaces, one can discuss ramification upstairs or downstairs.
We also want to discuss in the course the upstairs analog of $\Delta_{B\slice A}$ using the **different ideal** $\Delta_{B\slice A} \normal B$.
This has the property that $p$ ramifies iff $p\divides \Delta_{B\slice A}$.
Note that later we'll be able to push ideals down using norms.
:::

:::{.example title="?"}
The simplest number fields: quadratic fields ($[K:\QQ] = 2$).
Write $K = \QQ(\sqrt d)$ with $d\in \ZZ$ squarefree.
An important result for number theory exams: 
\[
\delta_{\QQ(\sqrt d)} = 
\begin{cases}
d & d \equiv 1\mod 4 
\\
4d & d \equiv 2,3\mod 4.
\end{cases}
.\]
For $p>2$, take the \(p\dash \)adic valuation to see $\Ord_p(\delta_{\QQ(\sqrt d)}) \leq 1$.
However, $\Ord_2(\delta_{\QQ(\sqrt d)}) \in \ts{0,2,3}$.
So odd primes can only divide the discriminant once, but 2 can divide it more!
This is explained by **Dedekind's different theorem**.
:::

## Commutative Algebra "Review"

:::{.remark}
All rings are commutative and unital.

- Ideals $I, J\normal R$ are comaximal (or coprime) iff $I+J = R$, so $\gens{I, J} = I+J$.
  Note that for $\ZZ$, this recovers $p, q$ being relatively prime.

  - Collections $\ts{I_i}$ are pairwise comaximal if the obvious thing holds: $I_i + I_j = R$ for all $i\neq j$.
  - Distinct maximal ideals are comaximal.

- Radicals: $\sqrt{I} = \ts{x\in R\st x^n\in I \text{ for some } n\in \ZZ_{\geq 0}}$.
  - For $R=\ZZ$, $\sqrt{\gens{24}} = \sqrt{\gens{2^3\cdot 3}} = \gens{2\cdot 3}$.
  More generally, take the prime factorization and reduce all exponents to 1.

- For $\ts{p_i}$ distinct maximal ideals, $\sqrt{p_i^{e_i}} = p_i$, the ideals $\ts{p_i^{e_i}}$ are pairwise comaximal.

- CRT: if $\ts{I_i}$ are pairwise comaximal, then
  - $\prod I_i = \Intersect I_i$
  - There is a natural map $R/ \intersect I_i \to \prod R/I_i$ which is an isomorphism.

:::

:::{.exercise title="?"}
For $I, J\normal R$, show that if $\sqrt{I}, \sqrt{J}$ are comaximal then $I, J$ are comaximal.
:::