Merge overleaf-2024-10-17-1255 into main

main.tex

@@ -19,6 +19,7 @@
 \newcommand{\acts}{\curvearrowright}
 \newcommand\Gal[1]{\operatorname{Gal}({#1})}
 \renewcommand{\phi}{\varphi}
+\newcommand\discr[1]{\operatorname{discr}({#1})}
 
 \begin{document}
 
@@ -154,79 +155,90 @@ \section{Preliminaries from algebraic number theory.}
 \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}
+    \begin{definition}
+        An algebraic number field is a finite field extension $k/\Q$. 
+    \end{definition}
+        \1 This definition implies the following properties:
+            \2 The field $k$ has characteristic 0.
+            \2 By the Primitive Element Theorem, $k = \mathbb{Q}(a)$ for some $a \in K$.
+            \2 There exists a unique minimal polynomial $f \in \mathbb{Q}[X]$ for $a$, with $\deg(f) = d = [k:\mathbb{Q}]$.
+        \1 Let $(a_1, \ldots, a_d)$ be the roots of $f$ in the algebraic closure of $\mathbb{Q}$ within $\mathbb{C}$. These roots are called the \textbf{Galois conjugates} of $a$. Note that these roots do not lie in $\mathbb{Q}$.
+        \1 Properties of embeddings:
+            \2 For each $i$, the map $a \mapsto a_i$ defines an isomorphism $\mathbb{Q}(a) \cong \mathbb{Q}(a_i)$.
+            \2 Any embedding $k\rightarrow \mathbb{C}$ must send $a$ to some $a_i$.
+            \2 There are exactly $d$ embeddings $k \rightarrow \mathbb{C}$, denoted $\sigma_1, \ldots, \sigma_d$.
+        \1 Classification of embeddings:        
+            \2 Note that $(a_1, \ldots, a_d) = \overline{(a_1, \ldots, a_d)}$, so $\sigma_i(k) \subseteq \mathbb{R}$ if and only if $\overline{a_i} = a_i$.
+            \2 We can thus classify the embeddings as:
+                \3 Real embeddings (real places of $K$): $r_1$
+                \3 Complex embeddings (complex places of $K$): $2r_2$ (counted in pairs due to complex conjugation)
+            \2 This classification implies $d = r_1 + 2r_2$
+        \1 Examples:
+            \2 For $k = \mathbb{Q}(\sqrt[3]{2})$: $r_1 = 1, r_2 = 1$
+            \2 For $k = \mathbb{Q}(\exp(2\pi i/n))$, $n \geq 3$: $r_1 = 0, r_2 = \phi(n)/2$ (odd $n$)
+    \0 \begin{definition}
+        For any $\alpha \in K$, we define two rational numbers:\\
+        1. The norm: $N_{K/\mathbb{Q}}(\alpha) = \prod_{i=1}^d \sigma_i(\alpha)$\\
+        2. The trace: $Tr_{K/\mathbb{Q}}(\alpha) = \sum_{i=1}^d \sigma_i(\alpha)$
+    \end{definition}
+    \marginnote{Note: $N_{K/\mathbb{Q}}(\alpha) = \det(\alpha: K \rightarrow K)$, and similarly for the trace.}
+        \1 Basis criterion: Let $(\alpha_1,\ldots,\alpha_d)\in k$ and $\lambda_1,\ldots,\lambda_d\in\Q$. Then $\sum_{i=1}^d \lambda_i\alpha_i = 0 \iff \sum_{i=1}^d \lambda_i \sigma_j(\alpha_i)=0$ for all $j$. 
+        Moreover, $\{\alpha_i\}_{i=1}^d$ is a basis of $k$ if and only if $\det(\sigma_i(\alpha_j))\neq 0$.
+    \0 \begin{definition}
+        The \textbf{discriminant} of a basis $\{\alpha_1,\ldots,\alpha_d\}$ of a number field $k$ of degree $d$ over $\Q$ is defined as: $\discr{\{\alpha_1,\ldots,\alpha_d\}}=\det^2(\sigma_i(\alpha_j))\in\Q$, where $\sigma_1,\ldots,\sigma_d$ are the $d$ distinct embeddings of $k$ into $\bC$.
+    \end{definition}
+    \begin{exercise}
+        Prove that $\discr{\alpha_i} = \det(Tr_{k/\Q}(\alpha_i\alpha_j))_{1 \leq i,j \leq d}$. Show that if $k = \Q(a)$ for some $a \in k$, then $\discr{\{1,a,a^2,\ldots,a^{d-1}\}} = \prod_{1 \leq i < j \leq d}(\sigma_i(a)-\sigma_j(a))^2$.
+    \end{exercise}
+    \0 To introduce relative versions for an extension $l/k$, we define the relative discriminant $\discr{}_{l/k}$ using only those embeddings $\sigma_i : l \hookrightarrow \mathbb{C}$ which restrict to the identity on $k$.
 \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}
+    \marginnote{Algebraic number theory is not (algebraic) number theory but rather (algebraic number) theory.}
+    \0 Let $k$ be an algebraic number field for the following discussion.
+    \begin{definition}
+        The ring of integers in $k$ is defined as:
+        $$\cO_k=\{\alpha \in k : f(\alpha)=0 \text{ for some monic }f\in \Z[X]\}.$$
+    \end{definition}
+        \1 Example: $\cO_\Q = \Z$. It is often referred to as the ring of \enquote{rational integers}.
+    \begin{proposition}
+        For $(\alpha_1,\ldots,\alpha_r)\in k$, the following are equivalent:\\
+        1. $(\alpha_1,\ldots,\alpha_r)\in\cO_k$\\
+        2. $\Z[\alpha_1,\ldots,\alpha_r]$ is finitely generated as a $\Z$-module.
+    \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}