%! TEX root = LF.tex % vim: tw=50 % 02/11/2024 12PM \begin{fcdefn}[Valuation on a Dedekind domain] \label{defn_9_6} \glssymboldefn{ddv}% % Definition 9.6 If $R$ is a \gls{dedd}, and $p \subseteq R$ a non-zero prime ideal, we write $v_p$ for the normalised \gls{valt} on $\Frac(R) = \Frac(R\localp{(p)})$ corresponding to the \gls{dvr} $R\localp{(p)}$. \end{fcdefn} \begin{example*} $R = \Zbb$, $p = (p)$, then $v_p$ is the $p$-adiv \gls{valt}. \end{example*} \begin{fcthm}[] \label{thm_9_7} % Theorem 9.7 Assuming: - $R$ is a \gls{dedd} - $I \subseteq R$ a non-zero ideal Then: $I$can be written uniquely as aproduct of prime ideals: \[ I = p_1^{e_1} \cdots p_r^{e_r} \] (with $p_i$ distinct). \end{fcthm} \begin{remark*} Clear for PIDs (PID implies UFD). \end{remark*} \begin{proof}[Proof (Sketch)] We quote the following properties of \gls{locon}: \begin{enumerate}[(i)] \item $I = J \iff I R\localp{(p)} = JR\localp{(p)}$ for all prime ideals $p$. \item If $R$ a \gls{dedd}, $p_1, p_2$ non-zero ideals, then \[ p_1 R\localp{(p_2)} = \begin{cases} p_2 R\localp{(p_2)} & p_1 = p_2 \\ R\localp{(p_2)} & p_1 \neq p_2 \end{cases} \] \end{enumerate} Let $I \subseteq R$ be a non-zero ideal. By \cref{lemma_9_3}, there are distinct prime ideals $p_1, \ldots, p_r$ such that $p_1^{\beta_1} \cdots p_r^{\beta_r} \subseteq I$, where $\beta_i > 0$. Let $0 \neq p$ be a prime ideal, $p \notin \{p_1, \ldots, p_r\}$. Then property (ii) gives that $p_i R\localp{(p)} = R\localp{(p)}$, and hence $I R\localp{(p)} = R\localp{(p)}$. \cref{coro_9_5} gives $IR\localp{(p_)} = (p_i R\localp{(p_i)})^{\alpha_i} = p_i^{\alpha_i} R\localp{(p_i)}$ for some $0 \le \alpha_i \le \beta_i$. Thus $I = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$ by property (i). For uniqueness, if $I = p_1^{\alpha_1} \cdots p_r^{\alpha_r} = p_1^{\gamma_1} \cdots p_r^{\gamma_r}$ then $p_i^{\alpha_i} R\localp{(p_i)} = p_i^{\gamma_i} R\localp{(p_i)}$ hence $a_i = \gamma_i$ by unique factorisation in \glspl{dvr}. \end{proof} \newpage \section{Dedekind domains and extensions} Let $L / K$ be a finite extension. For $x \in L$, we write $\Tr_{L / K}(x) \in K$ for the trace of the $K$-linear map $L \to L$, $y \mapsto xy$. If $L / K$ is separable of degree $n$ and $\sigma_1, \ldots, \sigma_n : L \to \ol{K}$denotes the set of embeddings of $L$ into an algebraic closure $\ol{K}$, then $\Tr_{L / K}(x) = \sum_{i = 1}^{n} \sigma_i(x) \in K$. \begin{fclemma}[] % Lemma 10.1 Assuming: - $L / K$ a finite separable extension of fields Then: the symmetric bilinear pairing \begin{align*} (\bullet, \bullet) &\to K \\ (x, y) &\mapsto \Tr_{L / K}(xy) \end{align*} is non-degenerate. \end{fclemma} \begin{proof} $L / K$ separable tells us that $L = K(\alpha)$ for some $\alpha \in L$. Consider the matrix $A$ for $(\bullet, \bullet)$ in the $K$-basis for $L$ given by $1, \alpha, \ldots \alpha^{n - 1}$. Then $A_{ij} = \Tr_{L / K}(\alpha^{i + j}) = [BB^\top]_{ij}$ where \[ B = \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \sigma_1(\alpha) & \sigma_2(\alpha) & \cdots \sigma_n(\alpha) \\ \vdots & \vdots & \ddots & \vdots \\ \sigma_1(\alpha^{n - 1}) & \sigma_2(\alpha^{n - 1}) & \cdots & \sigma_n(\alpha^{n - 1}) \end{pmatrix} \] So \[ \det A = \det(B)^2 = \left[ \prod_{1 \le i < j \le n} (\sigma_i(\alpha) - \sigma_j(\alpha)) \right]^2 \] (Vandermonde determinant), which is non-zero since $\sigma_i(\alpha) \neq \sigma_j(\alpha)$ for $i \neq j$ by separability. \end{proof} \textbf{Exercise:} On Example Sheet 3 we will show that a finite extension $L / K$ is separable if and only if the trace form is non-degenerate. \begin{fcthm}[] \label{thm_10_2} % Theorem 10.2 Assuming: - $\O_K$ a \gls{dedd} - $L$ a finite separable extension of $K = \Frac(\O_K)$ Then: the \gls{intc} $\O_L$ of $\O_K$ in $L$ is a \gls{dedd}. \end{fcthm} \begin{proof} $\O_L$ a subring of $L$, hence $\O_L$ is an integral domain. Need to show: \begin{enumerate}[(i)] \item $\O_L$ is Noetherian. \item $\O_L$ is \gls{intlyc} in $L$. \item Every $\neq 0$ prime ideal $P$ in $\O_L$ is maximal. \end{enumerate} Proofs: \begin{enumerate}[(i)] \item Let $e_1, \ldots, e_n \in L$ be a $K$-basis for $L$. Upon scaling by $K$, we may assume $e_i \in \O_L$ for all $i$. Let $f_i \in L$ be the dual basis with respect to the trace form $(\bullet, \bullet)$. Let $x \in \O_L$, and write $x = \sum_{i = 1}^n \lambda_i f_i, \lambda_i \in K$. Then $\lambda_i = \Tr_{L / K}(x e_i) \in \O_K$. (For any $z \in \O_L$, $\Tr_{L / K}(z)$ is a sum of elements in $\ol{K}$ which are \gls{int} over $\O_K$. Hence $\Tr_{L / K}(z) \in K$ is \gls{int} over $\O_K$, hence $\Tr_{L / K}(z) \in \O_K$.) Thus $\O_L \subseteq \O_K f_1 + \cdots + \O_K f_n \subseteq L$. Since $\O_K$ is Noetherian, $\O_L$ is finitely generated as an $\O_K$-module, hence $\O_L$ is Noetherian. \item Example Sheet 2. \item Let $P$ be a non-zero prime ideal of $\O_L$, and $p \defeq P \cap \O_K$ be a prime ideal of $\O_K$. Let $0 \neq x \in P$. Then $x$ satisfies an equation \[ x^n + a_{n - 1} x^{n - 1} + \cdots + a_0 = 0, \qquad a_i \in \O_K ,\] with $a_0 \neq 0$. Then $a_0 \in P \cap \O_K$ is a non-zero element of $p$, hence $p$ is non-zero, hence $p$ is maximal. We have $\O_K / p \injto \O_L / P$, and $\O_L / P$ is a finite dimensional vector space over $\O_K / p$. Since $\O_L / P$ is an integral domain and finite, it is a field. \qedhere \end{enumerate} \end{proof} \begin{remark*} \cref{thm_10_2} holds without the assumption that $L / K$ is separable. \end{remark*} \begin{corollary} % Corollary 10.3 The ring of integers of a number field is a \gls{dedd}. \end{corollary} \textbf{Convention:} $\O_K$ is the ring of integers of a number field -- $p \le \O_K$ a non-zero prime ideal. We normalise $|\bullet|_p$ (absolute value associated to $\ddv_p$, as defined in \cref{defn_9_6}) by $|x|_p = (N p)^{-\ddv_p(x)}$, where $N_p = |\O_K / p|$.