# Thursday, October 06

:::{.theorem title="Normalization"}
Let $R$ be an integrally closed domain with $K = \ff(R)$ and $L/K$ a degree $n$ **separable** extension.
Form the integral closure / normalization $\intcl_B(A)$:

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

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJCIl0sWzIsMCwiTCJdLFswLDFdLFswLDIsIkEiXSxbMiwyLCJLIl0sWzMsNCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDAsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzQsMSwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=)

Then $B \in \amod^\fg$ and hence a Noetherian ring.
:::

:::{.remark}
Note 

- There is an equality $\krulldim B = \krulldim A$, 
- If $A$ is Dedekind then so is $B$ (integrally closed dimension 1!),
- If $A$ is a PID, then $B \cong_A A^{[L: K]}$ is a free $A\dash$module.
  Note that this is checked by tensoring to the fraction field and checking dimension.

:::

:::{.proof title="?"}
For $x,y\in B$, define $\inp xy \da \Trace(xy)$ where $\Trace: L\to k$ is a $k\dash$linear map.
We'll use two facts:

1. Since $L$ is separable (and finite degree), $\inp\wait\wait: L\tensorpowerk{2}$ is nodegenerate.[^degen]
2. $\inp B B \subseteq A$, i.e. $\Trace(B) \subseteq A$.

Choose a $k\dash$basis $\ts{e_1, \cdots, e_n} \subseteq L$ that is contained in $B$ (by clearing denominators) and let $\Lambda \da \gens{e_1,\cdots, e_n}_A$ be the corresponding free $A\dash$module.
Note that $\inp \Lambda \Lambda \subseteq A$ and \( \Lambda \subseteq B \).
Define \( \Lambda\dual = \ts{x\in L \st \inp x \Lambda \subseteq A} \), then \( \Lambda \subseteq \Lambda\dual \) and we have
\[
\Lambda \subseteq B \subseteq B\dual \subseteq \Lambda\dual
.\]
Moreover $\Lambda\dual$ is free with dual basis $\ts{e^1,\cdots, e^n}$, so $\Lambda\dual \in \amod^\fg$ and $B\leq \Lambda\dual$.
Since $A$ is Noetherian, $B\in \amod^\fg$.

[^degen]: 
If $L$ were purely inseparable, the trace form would be identically zero.
Later we'll see how to compute traces in terms of minimal polynomials.

:::

:::{.remark}
Moral: in Dedekind domains, any lattice $B$ can be sandwiched between two free lattices.
:::

:::{.example title="?"}
For $A = \ZZ$ and $K/\QQ$ a degree $n$ number field,

\begin{tikzcd}
	{\ZZ_K} && K \\
	\\
	\ZZ && \QQ
	\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=WzAsNCxbMCwwLCJcXFpaX0siXSxbMiwwLCJLIl0sWzAsMiwiXFxaWiJdLFsyLDIsIlxcUVEiXSxbMiwwLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMSwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==)

Recall that $\ZZ_K$ is the ring of algebraic integers in $K$, is a Dedekind domain, and there exists a $\ZZ\dash$basis $\ts{\alpha_1, \cdots \alpha_n} \subseteq \ZZ_K$ such that $\ZZ_K = \sum \ZZ \alpha_i$.
:::

:::{.remark}
If $A\in \Dedekind$ and $L/K$ is inseparable, then $B$ is not necessarily in $\amod^\fg$, but we always have $B\in \Dedekind$ -- this is the **Krull-Akizuki** theorem.
One can consider examples like $K = \FF_q\functionfield{t}$ and $L = \FF_q\functionfield{t^{1\over p}}$, so $L/K$ is a degree $p$ inseparable extension.

Another example:

\begin{tikzcd}
	{C^0 \text{ a curve}} & B && L & {C\slice k} \\
	\\
	{\AA^1\slice k = \PP^1\slice k\smts{\infty}} & {k[t]} && {k(t)} & {\PP^1\slice k}
	\arrow[from=3-2, to=3-4]
	\arrow[from=1-2, to=1-4]
	\arrow[from=1-2, to=3-2]
	\arrow[from=1-4, to=3-4]
	\arrow[dashed, from=1-1, to=3-1]
	\arrow[dashed, from=1-5, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMSwyLCJrW3RdIl0sWzAsMiwiXFxBQV4xXFxzbGljZSBrID0gXFxQUF4xXFxzbGljZSBrXFxzbXRze1xcaW5mdHl9Il0sWzEsMCwiQiJdLFswLDAsIkNeMCBcXHRleHR7IGEgY3VydmV9Il0sWzMsMiwiayh0KSJdLFs0LDIsIlxcUFBeMVxcc2xpY2UgayJdLFszLDAsIkwiXSxbNCwwLCJDXFxzbGljZSBrIl0sWzAsNF0sWzIsNl0sWzIsMF0sWzYsNF0sWzMsMSwiIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzcsNSwiIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d)

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/NT/sections/figures}{2022-10-06_11-37.pdf_tex} };
\end{tikzpicture}

:::

## Algorithm for finding $\ZZ_K$

:::{.remark}
Algorithm for computing $\ZZ_K$ for $K\in \NF$: let $K = \QQ[t]/\gens{f(t)}$ with $f\in \ZZ[t]$ monic irreducible.
Let $\alpha$ be a root of $f$ and consider the order $\OO \da \ZZ[ \alpha ] \cong \ZZ[t]/\gens{f}$.
First compute $\Delta(\OO) \subseteq \ZZ\smz$ the discriminant of the $\ZZ\dash$lattice $\OO$ with respect to the trace form, so $\OO = \bigoplus_{k=1}^n \ZZ \alpha^k$.
Compute the Gram matrix $G(i, j) = \Trace{K\slice \QQ}( \alpha^i \alpha^j)$.
Note that $\OO$ might not be the maximal order, but writing $\Delta_K = \Delta(\ZZ_K)$ we have 
\[
\Delta(\OO) = [\ZZ_K : \OO]^2 \Delta_K
.\]
If $\Delta(\OO)$ is squarefree then $\OO = \ZZ_K$.
Otherwise, record the distinct primes $p_1,\cdots, p_n$ dividing $\Delta(\OO)$.
If $v_{p_i}(\Delta(\OO)) = 1$ then $v_{p_i}(\Delta(K)) = 1$, otherwise the difference is by an even power due to the square.

If $v_{p_i}(\Delta(\OO)) = 0, 1$ then locally $\OO_{ (p) } = (\ZZ_K)_{(p)}$ and tensoring $(\wait)\tensor_\ZZ \ZZ_{(p)}$.
If $v_{p_i}(\Delta(\OO)) \geq 2$ then decide if $\OO_{(p)} = (\ZZ_K)_{(p)}$ or not.
If not maximal at $p$ there is some $\beta\in \ZZ_K\smz$ such that $p\beta \in \OO$.

Thus we are trying to find elements in $\ZZ_K \intersect {1\over p} \OO/\OO$.
Note that since $\OO \subseteq \ZZ_K$, if $\beta \in {1\over p} \OO$ then $\beta\in \ZZ_K \iff \beta + c\in \ZZ_K$ for all $c\in \OO$.
Note 
\[
{1\over p} \OO / \OO \cong \OO/p\OO \cong \ZZ^n/p\ZZ^n \implies \size{1\over p} \OO/\OO = p^n
.\]
One can check that that if $\beta = \sum_{i=0}^{n-1} x_i \alpha^i$ then each $x_i \in \ts{j /p \st 0\leq j\leq p-1}$ and there are $p^n$ choices.
To see whether this lies in $\ZZ_K$, one can compute minimal polynomials.
If any element is not integral, this yields an element in $\ZZ_K\sm \OO$, so one can enlarge the order $\OO$ -- given such a $\beta$, replace $\OO$ with $\OO' \da \OO\adjoin{ \beta}$.
Since $\OO'\contains \OO$, the index $[\ZZ_K: \OO']$ divides $[\ZZ_K : \OO]$ and is smaller by at least a factor of $p$ (since the lattice has been dilated by a factor of $p$) and one can start the entire process again with $\OO'$.

If one does not find such a $\beta$, then $\OO$ maximal at $p$ and $v_p(\Delta(\OO)) = v_p(\Delta(\ZZ_K))$ is the correct valuation and we can cross the prime $p$ off.

In any case, either the index decreases or the number of primes decreases, so this terminates.
:::

:::{.example title="?"}
Let $K = \QQ(\sqrt{5}) = \QQ[t]/\gens{f}$ where $f(t) = t^2-5$.
Then $\alpha = \sqrt 5$ and $\OO = \ZZ[\sqrt 5]$, and
\[
\Delta(\OO) = \matt{ \Trace(1 \cdot 1) }{ \Trace(1 \cdot \sqrt 5)}{\Trace(\sqrt 5 \cdot 1 ) }{ \Trace(\sqrt 5 \cdot \sqrt 5)}
.\]
One can check by summing over Galois conjugates that
\[
\Trace(x\cdot 1 + y\sqrt 5) = (x\cdot 1 + y\sqrt 5) + \sigma(x\cdot 1 + y\sqrt 5) = 2x
\]
This yields $\Delta(\OO) = \det \matt 2 0 0 {10} = 20 = 2^2\cdot 5$.
We know $\ZZ[\sqrt 5]$ is maximal except possibly at $p=2$.
So test 4 elements to see if they're in $\ZZ_K$: $\ts{\beta_i} = \ts{{1\over 2}, {\sqrt 5\over 2}, {1\over 2}(1+\sqrt 5)}$.

- Since $\ZZ$ is integrally closed, $\beta_0 \not\in \ZZ_K$.
- Checking $(2\beta_1)^2=5$ yields minimal polynomial $4t^2 - 5$ which is not monic, so $\beta_1\not\in \ZZ_K$.
- Check $(2\beta_2-1)^2 = 5\implies 4\beta_2^2 - 4\beta_2 + 1 = 5\implies \beta_2^2 - \beta_2 - 1 = 0$ so $\beta_2 \in \ZZ_K$.

The previous steps showed $[\ZZ_K: \ZZ[\sqrt 5]] = 1,2$, and finding the element $\beta_2$ shows the index is 2 and $\ZZ_K = \ZZ[\beta_2]$.
:::

:::{.exercise title="?"}
Let $K/\QQ$ be a quadratic number field, write $K = \QQ(\sqrt D)$ with $D \in \ZZ\sm\ts{0, 1}$ squarefree, check when $\ZZ_K = \ZZ[\sqrt D]$.
Set $\alpha = \sqrt D$, which has minimal polynomial $t^2-D$.
Compute traces $\Trace(1), \Trace( \alpha)$ from the matrix definition: write $K = \QQ\cdot 1 \oplus \QQ \cdot \alpha$.
One can check 
- $\Trace(\cdot 1) = \Trace(I) = 2$, 
- $\Trace\qty{\matt 0 D 1 0 } = 0$ by showing $\alpha.e_1 = \alpha.1 = \alpha = e_2$ and $\alpha.e_2 = \alpha.\alpha = D = De_1$,
- $\Trace(\cdot D) = \Trace\qty{\matt D 0 0 D} = 2D$.

Thus
\[
\Disc(\ZZ[\sqrt D]) = \det \matt{\Trace(1.1)}{\Trace(1.\alpha)}{\Trace( \alpha.1)}{\Trace(\alpha.\alpha)} = \det\matt200{2D} = 4D
,\]
thus for all primes $p > 2$ we have $v_p(\Delta(\OO)) = 0, 1$ since $D$ is squarefree, and $\OO$ is maximal except possibly at $p=2$.
We have
\[
v_2 ( \Delta(\OO) ) = 
\begin{cases}
2 & D \text{ odd} 
\\
3 & D\text{ even}.
\end{cases}
\]
and so $[\ZZ_K: \ZZ[\sqrt D]] = 1,2$.

Now test for integrality:

- $\beta_0 \da {1\over 2}\not\in \ZZ_K$.
- $\beta_1 \da {\sqrt D\over 2}$: minimal polynomial of $4t^2-D = 0 \implies t^2 - {D \over 4} = 0$, and $4\not\divides D$ since $D$ is squarefree.
- $\beta_2 \da {1+ \sqrt D\over 2}$: check $(2\beta_2 - 1)^2 = = 0\implies 4\beta_2^2 - 4\beta_2 + 1 = 0$, so $\beta_2^2 - \beta_2 + {1-D\over 4} = 0$. 

So if $D\equiv 1\mod 4$ then $\ZZ_K = \ZZ[\beta_3]$ with $\Delta_K = D$ and $\ZZ_K = \ZZ[\sqrt D]$ with $\Delta_K = 4D$.
:::