From e1c3718c0195b86676deb8a2c1b277032307c438 Mon Sep 17 00:00:00 2001
From: Ayushi <46610680+Ayushi141@users.noreply.github.com>
Date: Thu, 17 Oct 2024 13:04:51 +0200
Subject: [PATCH] Raw lecture 2 notes

---
 main.tex     | 95 ++++++++++++++++++++++++++++++++++++++++++++++++++++
 preamble.tex |  2 +-
 2 files changed, 96 insertions(+), 1 deletion(-)

diff --git a/main.tex b/main.tex
index cf3b073..bdb0d26 100644
--- a/main.tex
+++ b/main.tex
@@ -149,7 +149,102 @@ \subsection{The fixed point functor and exact sequences}
 
 \end{outline}
 
+\section{Preliminaries from algebraic number theory.}
 \marginnote{Lecture 2, 17.10.24}
+\marginnote{User: GRK, password: 2240.}
 
+\subsection{Number fields}
+
+\begin{outline}
+\begin{definition}
+    An algebraic number field is a finite field extension $k/\Q$. 
+\end{definition}
+This implies it is of characteristic 0 and primitiive element theorem is available $\implies k=\Q(a)$ for some $a\in k$ with a unique minimal polynomial $f\in \Q[X]$ of $\deg d = [k:\Q]$.
+    \1 The roots $(a_1,\ldots,a_d)$ must not lie in $\Q$ but rather in algebraic closure of $\Q$ inside $\bC$ and are called \textbf{the Galois conjugates} of $a$.
+    \1 Requiring $a\mapsto a_i$ defines isomorphism $\Q(a)\cong \Q(a_i)$ and any embedding $\Q \rightarrow \bC$ must send $a$ to some $a_i$, so there are exactly $d$ embeddings $k\rightarrow \bC$.
+    \1 Note that $(a_1,\ldots,a_d)=\overline{(a_1,\ldots,a_d)}$, so $\sigma_i(k)\subseteq \R$ iff $\overline{a}_i=a_i$.
+        \2 So we can split embeddings in real ones (real places of $k$ denoted $r_1$) and complex ones (complex places of $k$, denoted $2r_2$ because of complex conjugation). This implies $d=r_1+2r_2$.
+            \3 For $k=\Q(\sqrt[3]{2})$, $r_1=1, r_2=1$.
+            \3 For $k=\Q(\exp(2 \pi i/n)$, $r_1=0, r_2 = \phi(n)/2, n\geq 3$.
+\0 \begin{definition}
+    Associated to any $\alpha\in k$, we have two rational numbers: the \textbf{norm} and the \textbf{trace}.
+    $N_{k/\Q}(\alpha)=\sigma_1(\alpha)\ldots \sigma_d(\alpha)$ and $Tr_{k/\Q}(\alpha)=\sigma_1(\alpha)+\ldots+\sigma_d(\alpha)$
+\end{definition} 
+\marginnote{$N_{k/\Q}(\alpha)=\det (\alpha N \rightarrow N)$, similarly for trace.}
+Let $(\alpha_1,\ldots,\alpha_d)\in k$, $\lambda_1,\ldots,\lambda_d\in\Q$ such that $\sum \lambda_i\alpha_i = 0 \iff \sum \lambda_i \sigma(\alpha_i)=0$ for all $i$. Then $\{\alpha_i\}$ is a basis of $k$ $\iff \det (\sigma_i(\alpha_j))\neq 0$.
+
+\begin{definition}
+    The \textbf{discriminant} of a basis $\{\alpha_1,\ldots,\alpha_d\}$ is given by $\det^2(\sigma_i(\alpha_j))\in\Q$.
+\end{definition}
+\end{outline}
+
+\begin{exercise}
+    Show that discr$\{\alpha_i\}=\det Tr_{k/\Q}(\alpha_i,\alpha_j)$. Show that $k\in \Q(a) \implies discr\{1,a,a^2,\ldots,a^{d-1}\}=\prod_{i<j}(\sigma_i(a)-\sigma_j(a))^2$ 
+\end{exercise}
+
+Similarly we want to introduce relative versions for $l/k$. We define $discr_{l/k}$ using only those $\sigma_i:l\xhookrightarrow{} \bC$ which restrict to identity on $k$.
+
+\subsection{Integrality in number fields}
+
+\begin{outline}
+\marginnote{Algebraic number theory is not (algebraic) number theory but rather (algebraic number) theory.}
+\0 Algebraic number theory really is about algebraic integers. So now we define them. For the remainder let $k$ be an algebraic number field.
+\begin{definition}
+    The ring of integers in $k$ is given by $\cO_k=\{\alpha \in k : f(\alpha)=0 \text{ for some monic }f\in \Z[X]\}$.
+\end{definition}
+    \1 Example: $\cO_\Q = \Z$. $\Z$ is often denoted as \enquote{rational integers}.
+\0 Why do they form a ring?
+\begin{proposition}
+    Let $(\alpha_1,\ldots,\alpha_r)\in k$. Then $(\alpha_1,\ldots,\alpha_r)\in\cO_k \iff \Z[\alpha_1,\ldots,\alpha_r]$ is finitely generated.
+\end{proposition}
+The corolllary is that $\cO_k$ is a ring. (why?)
+
+\begin{lemma}
+    For $\alpha \in k$, there exist $\beta \in \cO_k, n\in\Z$ such that $\alpha=\frac{\beta}{n}$.
+\end{lemma}
+
+From now on we can assume that our algebraic number field is generated by a primitive element which is an algebraic integer.
+
+\begin{proposition}
+    We can sandwhich the ring $\cO_k$ between $\Z[a]$ and $\frac{1}{discr\{1,\ldots,a^{d-1}\}} \Z[a]$. (This 1/discr is in $\Z$ because it is in the intersection of algebraic integers in $k$ and $\Q$). Because it lies between two free abelian groups of the same rank, it has to be free abelian of the same rank.
+\end{proposition}
+
+\begin{corollary}
+    $\cO_k$ has a $\Z$-basis of rank $d$.Any such is called an integral basis.
+\end{corollary}
+(Z lattice in a Q vector space and you exhaust it by multiplying with the integers? What? Minkowski geometry of numbers (covolumes?))
+
+\begin{corollary}
+    $\cO_k$ is noetherian. 
+\end{corollary}
+
+\begin{definition}
+    The discriminant of $k$ is given by $discr_k=d_k=discr\{ \alpha_1, \ldots, \alpha_d \}$ for an integral basis. Well-defined because $\det(T...)=\pm 1$.
+\end{definition}
+
+More generally, we can also define relative discriminants $ fancy d_{l/k} = discr(\beta_i)\subseteq \cO_k$. This d is an ideal because in general we might be not in a PID anymore. 
+
+\begin{exercise}
+    $k=\Q(\sqrt{D})$, $D$ square-free integer. If $D \equiv 1 (4)$ or $D \equiv 2,3 (4)$  then the integral basis is ... and discriminant is $D$ or $4D$.
+\end{exercise}
+\end{outline}
+
+\subsection{The arithmetic of algebraic intgers}
+
+\begin{outline}
+\0 For $k=\Q(\sqrt{5})$, $\cO_k = \Z[\sqrt{-5}]$. In $\cO_k$, we have $21=3\cdot 7 = (1+2\sqrt{-5})\cdot (1-2\sqrt{-5})$ and these factors are irreducible. (something something norm of a number). So it is not a UFD. Kummer suggested: in an ideal world, there would be ideal numbers $p_1\cdot p_2 = 3$ and $p_3\cdot p_4 = 7$, with $p_1\cdot p_3 = 1+2\sqrt{-5}$ and $p_2 p_4 = 1-2\sqrt{-5}$, hence $21 = p_1p_2p_3p_4=p_1p_3p_2p_4$ so they would differ only by a permutation and factorization would be unique. Apparently: $p_1 \mid 3$ and $p_1 \mid 1+2\sqrt{-5} \implies p_1 \mid \lambda 3 + \mu (1+2\sqrt{5})$ and $p_1$ should be determined by the set of all $\alpha \in \cO_k$ that it divides.  So set $p_1=(3,1+2\sqrt{5})$ and $p_2=(3,1-2\sqrt{-5})$ and so on... So the idea is that one might get unique factorization in ideals instead. 
+
+\begin{theorem}
+    The ring $\cO_k$ is noetherian, integrally closed and of dimension $1$.
+\end{theorem}
+
+The hard thing is to single out that these three properties are key to a ring being a Dedekind domain. 
+
+\begin{definition}
+An integral domain satisfying these three properties is called a Dedekind domain. 
+\end{definition}
+\end{outline}
+
+\marginnote{Lecture 3, ...}
 \end{document}
  
\ No newline at end of file
diff --git a/preamble.tex b/preamble.tex
index 3fad7b8..59f1ce0 100644
--- a/preamble.tex
+++ b/preamble.tex
@@ -204,7 +204,7 @@
 ]{proposition}
 
 \declaretheorem[
-name=Korollar, 
+name=Corollary, 
 style=lightblue, 
 sibling=definition
 ]{corollary}