From 8990e83e79f8ce1d7d2acfd0a508b079cce76cab Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 23 Sep 2022 13:03:54 +0200 Subject: [PATCH] Bloomer Anfang Alg Numbertheory --- inputs/algnumbertheory.tex | 153 +++++++++++++++++++++++++++++++++++++ 1 file changed, 153 insertions(+) create mode 100644 inputs/algnumbertheory.tex diff --git a/inputs/algnumbertheory.tex b/inputs/algnumbertheory.tex new file mode 100644 index 0000000..409d46c --- /dev/null +++ b/inputs/algnumbertheory.tex @@ -0,0 +1,153 @@ +% Alg 1 2022 Blomer + +\section{Introduction to algebraic number theory} + +\subsection{Number Fields and rings of integers} + +\begin{definition}[Number field] + A \vocab{number field} $K$ is a finite (hence algebraic) extension of $\Q$. + Its \vocab{ring of integers} $\cO_k$ is the integral closure of $\Z$ in $K$. +\end{definition} +\begin{remark} + \begin{enumerate}[(a)] + \item It is not obvious that this is a ring, but we have shown this in % TODO REF + \item For $K = \Q(\i)$ we have $\cO_K = \Z[\i]$ by % Blomer (3.5) TODO + \item Every $x \in K$ of the form $\frac{y}{n}$ with $y \in \cO_K$, $n \in \Z \setminus \{0\}$ since $\exists f \in \Z[X]: f(x) = 0$ with leading coefficient $0 \neq c \in \Z$. + Then $c^{n-1}f(x) = 0$ implies that $cx$ is integral over $\Z$. + \item An element $x \in K$ is in $\cO_k$ if and only if its minimal polynomial is in $\Z[X]$.% see Blomer 3.5 TODO + \item Since $\char(\Q) = 0$, every extension is separable. + \end{enumerate} + +\end{remark} + +\begin{definition} % and Lemma TODO + For $a \in K$consider the $\Q$-vectorspace-homomorphism $T_a : K \to K, x \mapsto ax$ and define the \vocab{norm} and \vocab{trace} + of $a$ by + \[ + \Tr(a) \coloneqq \Tr(T_a) + \] + and \[ + \Nr(a) = \det(T_a) +\] +\end{definition} +\begin{lemma} + $\Tr : k \to \Q$ and $\Nr : K^\ast \to \Q^\ast$ are homomorphisms. +\end{lemma} +\begin{proof} + $\Nr(T_a)$ and $\Tr(T_a)$ are coefficients of the characteristic polynomial of $T_a$, + hence in $\Q$. + Since $T_a + T_b = T_{a+b}$ and $T_{ab} = T_a T_b$, we see that $\Tr$ and $\Nr$ are homomorphisms. +\end{proof} + +\begin{theorem} + \begin{enumerate}[(a)] + \item Let $K$ be a number field of degree $n$ and denote by $\sigma$ the $n$ distinct homomorphisms + from $K$ to $\overline{\Q}$. + Then + \[ + \Tr(a) = \sum_{\sigma} \sigma a, \Nr(a) = \prod_{\sigma} \sigma a + \] + \item If $a \in \cO_K$ then $\Tr(a), \Nr(a) \in \Z$ + \end{enumerate} +\end{theorem} +\begin{proof} + \begin{enumerate}[(a)] + \item + If $a $ is of degree $m$ over $\Q$, then $(1, a, a^2, \ldots, a^{m-1})$ is a $\Q$- basis of $\Q(a) / \Q$. + Let $f_a(X) = X^n + c_1 X^{m-1} + \ldots+ c_m \in \Q[X]$ be the minimal polynomial of $a$. + Let $n = md$ and $(b_1, \ldots, b_d)$ a basis of $K / \Q(a)$. + Then $(b_1, b_1a, b_1a^2, \ldots, b_1a^{m-1}, \ldots, b_d, b_da, \ldots, b_da^{m-1})$ is a basis of $K / \Q$. + Then the matrix of $T_a$ with respect to basis has block structure with blocks: + \[ + \begin{pmatrix} & & & & -c_m \\ 1& & & & -c_{m-1} \\ & 1 & & -c_{m-2} \\ & & \ddots & \\ & & & 1 & -c_1 \end{pmatrix} + \] + The charactersitic polynomial is $f_a$ for every block, hence $f_a^d$ for the entire matrix. + On the set $\Hom(K, \overline{Q})$ we introduce an equivalence relation by $\sigma \sim \tau(a)$ if $\sigma(a) = \tau(a)$. + Then the set of equivalence classes is in bijection with $\Hom(\Q(a), \overline{\Q})$ via + \[ + [\sigma] \mapsto \sigma \defon{\Q(a)} + \] + Since $\{\sigma(a) | \simga \in \Hom(\Q(a), \overline{\Q})\}$ is the set of roots of $f_a$, by Vietas theorem we obtain + $\Tr_K(a) = d\Tr_{\Q(a)}(T_a) = d \sum_{\sigma \in \Hom(\Q(a), \overline{\Q}} \sigma(a) = \sum_{\delta \in \Hom(K, \overline{\Q}} \sigma(a)$. + + Basically the same proof works for $\Nr(a)$. + +\item If $a$ is integral over $\Z$, then all $\simga(a)$ are integral over $\Z$ (since they are roots of the same poynomial), so $\Nr(a)$ and $\Tr(a) \in \Q$ are integral over $\Z$ by % TODO Blomer 3.3 and a) + and so $\Nr(a), \Tr(a) \in \Z$. + + \end{enumerate} + +\end{proof} + + +\begin{definiton} + Let $K$ be a humber field with $\Q = \basis(\alpha_1, \ldots, \alpha_n)$. We define the \vocab{discriminant} + \[ +D(\alpha_1, \ldots, \alphan) = \det \left( \left( \delta_i \alpha_j \right)_{\substack{ \le j \le n \\ \delta_i \in \Hom(K, \overline{\Q})\} \right)^{2} + \] +\end{definition} + +\begin{lemma} + \begin{enumerate}[(a)] + \item We hae $D(\alpha_1, \ldots, \alpha_n) = \det(\Tr(\alpha_i \alpha_j)_{i,j}) \in \Q$ + \item The map $(x,y) \mapsto \Tr(xy)$ is a non-degenerate bilinear form + \item We have $D(\alpha_1, \ldots, \alpha_n) \neq 0$ + \end{enumerate} +\end{lemma} +\begin{proof} + \begin{enumerate}[(a)] + \item We have $\Tr(\alpha_i \alpha_j) = \sum_\sigma \sigma(\alpha_i \alpha_j ) = \sum_\sigma \sigma(\alpha_i) \sigma(\alpha_j)$, + so $\left( \Tr(\alpha_i \alpha_j) \right) = (\sigma \alpha_i)^T_{\sigma, i} \cdot (\sigma \alpha_j)_{\sigma, j}$. + \item By additivity of the trace, it's clear that this is a bilinear form. + The associated symmetric matrix is $M = \Tr(\alpha_i \alpha_j)_{i,j}$ where $\{\alpha_i\}$ is a basis. + We can assume that $K = \Q(\theta)$, so $(1, \theta, \theta^2, .., \theta^{n-1})$ is a basis. + So $(\sigma \alpha_i)_{\sigma, i}$ is a Vandermonde matrix % TODO? + , hence it has non-vanishing determinant and so $\det M \neq 0$. + \item follows from (a) and (b). + \end{enumerate} +\end{proof} + +\begin{theorem} + Every ideal $\{0\} \neq \mathcal a \subseteq \cO_K$ (in particular $\cO_k$ itself) is a free $\Z$-module of rank $n = [K : \Q]$. + In particular, every ideal of $\cO_K$ is finitely generated as a $\Z$-module, in particular as an $\cO_K$-module. + Hence $\cO_K$ is noetherian. +\end{theorem} +\begin{proof} + Let $\alpha_1, \ldots, \alpha_n$ be a basis of $K / \Q$. W.l.o.g.~ $\alpha_j \in \cO_K$, with discriminant $D \in \Z$. + Let $x = \sum a_j \alpha_j \in \cO_K$ for some $a_j \in \Q$. + Then $\alpha_i x = \sum \alpha_i \alpha_j \cdot a_j$. + This implies $\Z \ni \Tr(\alpha_i x) = \sum_j \Tr(\alpha_i \alpha_j) a_j$ for all $i$. + We have $\Tr(\alpha_i \alpha_j) \in \Z$. This is an $n \times n$ linear system of equations in $a_j$. + By Cramer's rule, $a_j \in \frac{1}{D} \Z$. Hence $Dx \in \sum \Z \alpha_j$, so $D \cO_K \subseteq \sum \Z \alpha_j$ is a free module as a submodule of a ree $\Z$-module (by % TODO Blomer 0.9 + ), obviously of rank $n$. + Now let $\{0\} \neq \mathcal{a} \subseteq \cO_k$ be an ideal. + This is again a $\Z$-submodule of $\cO_k$, hence a free module. If $0 \neq a \in \mathcal{a}$, then $a \cO_K \subseteq \mathcal{a}$, + hence the $\Z$-rank of $\mathcal{a}$ is $n$. +\end{proof} + +\begin{example} + A $\Z$-basis of $\Z[\i]$ is $\{1,\i\}$ and the discriminant is + \[ + \det \begin{pmatrix} 1 & \i \\ 1 & -\i \end{pmatrix} ^{2} = (-2\i)^{2} = -4 + \] +\end{example} + +\begin{corollary} + \begin{enumerate}[(a)] + \item A $\Z$-module basis of $\cO_K$ is called an \vocab{integral basis} of $K$. The discriminant with respect to this basis is \vocab{the discriminant} $D_K$ of $K$. + It is indpendent of the basis. +\item Every ideal $\{0\} \neq \mathcal a \subseteq \cO_K$ has finite index. + We denote by $N \mathcla a = (\cO_k : \mathcal a) = \# \cO_K / \mathcal a$, \vocab{the norm} of $\mathcal a$. + If $(\apha_1, \ldots, \alpha_n)$ is a $\Z$-basis of $\mathcal a$, denn $D(\alpha_1, \ldots, \alpha_n) = (N \mathcal a)^2 D_K$ + \end{enumerate} +\end{corollary} +\begin{proof} + \begin{enumerate}[(a)] + \item A change of basis in $\Z$ corresponds to multiplying with a matrix of determinant $\pm 1$. + + \item Since $\mathcal a$ is a submodule of $\cO_K$ of full rank, we can choose compatible bases + $(\alha_1, \ldots, \alpha_n) \subseteq \cO_K$, $(d_1 \alpha_1, \ldots, d_n \alpha_n) \subseteq \mathcal a$. + + $(\cO_k, \mathcal a) = d_1 \cdot \ldots \cdot d_n$. + \end{enumerate} +\end{proof}