# Homework 7

\documentclass{article} \include{macros} \voffset=-0.05\textheight \textheight=1.1\textheight \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \begin{document} \begin{center} \Large\bf Homework 7 for Math 581F\\ Due FRIDAY November 16, 2007 \end{center} Each problem has equal weight, and parts of problems are worth the same amount as each other. \begin{enumerate} \item For each of the following three fields, determining if there is an order of discriminant $20$ contained in its ring of integers: $$ K = \Q(\sqrt{5}), \quad K=\Q(\sqrt[3]{2}), \quad\text{and}\ldots $$ $K$ any extension of $\Q$ of degree $2005$. [Hint: for the last one, apply the exact form of our theorem about finiteness of class groups to the unit ideal to show that the discriminant of a degree $2005$ field must be large.] \item Compute the class group of $\QQ(\sqrt{-15})$ following a similar approach to the computation of the class group of $\QQ(\sqrt{10})$ in the book. (Do not do this by typing 1 or 2 lines into Sage, but instead compute the Minkowski bound, etc.) \item Prove that the quantity $C_{r,s}$ in our theorem about finiteness of the class group can be taken to be $\left(\frac{4}{\pi}\right)^{s} \frac{n!}{n^n}$, as follows (adapted from \cite[pg.~19]{sd:brief}): Let $S$ be the set of elements $(x_1,\ldots, x_{n})\in\R^n$ such that $$ |x_1| + \cdots |x_{r}| + 2 \sum_{v=r+1}^{r+s} \sqrt{x_v^2 + x_{v+s}^2} \leq 1. $$ \begin{enumerate} \item Prove that $S$ is convex and that $M=n^{-n}$, where $$ M = \max\{ |x_1\cdots x_r\cdot (x_{r+1}^2 + x_{(r+1)+s}^2)\cdots (x_{r+s}^2 + x_n^2)| : (x_1,\ldots, x_n) \in S\}. $$ [Hint: For convexity, use the triangle inequality and that for $0\leq \lambda \leq 1$, we have \begin{align*} \lambda\sqrt{x_1^2 + y_1^2} &+ (1-\lambda)\sqrt{x_2^2+y_2^2}\\ &\geq\sqrt{(\lambda x_1 + (1-\lambda)x_2)^2 + (\lambda y_1 + (1-\lambda)y_2)^2} \end{align*} for $0\leq \lambda \leq 1$. In polar coordinates this last inequality is $$ \lambda r_1 + (1-\lambda)r_2 \geq \sqrt{\lambda^2 r_1^2 + 2\lambda(1-\lambda) r_1 r_2 \cos(\theta_1 - \theta_2) + (1-\lambda)^2 r_2^2}, $$ which is trivial. That $M\leq n^{-n}$ follows from the inequality between the arithmetic and geometric means. \item Transforming pairs $x_v, x_{v+s}$ from Cartesian to polar coordinates, show also that $v=2^{r}(2\pi)^s D_{r,s}(1)$, where $$ D_{\ell,m}(t) = \int \cdots \int_{\mathcal{R}_{\ell,m}(t)} y_1 \cdots y_m dx_1 \cdots dx_{\ell} dy_1 \cdots dy_m $$ and $\mathcal{R_{\ell,m}}(t)$ is given by $x_{\rho}\geq 0$ ($1\leq \rho\leq \ell$), $y_{\rho}\geq 0$ ($1\leq \rho\leq m$) and $$ x_1 + \cdots + x_{\ell} + 2(y_1+\cdots +y_m) \leq t. $$ \item Prove that $$ D_{\ell,m}(t) = \int_{0}^t D_{\ell-1,m}(t-x)dx =\int_{0}^{t/2} D_{\ell,m-1}(t-2y)y dy $$ and deduce by induction that $$ D_{\ell,m}(t) = \frac{4^{-m}t^{\ell+2m}}{(\ell+2m)!} $$ \end{enumerate} \end{enumerate} \end{document} %%% Local Variables: %%% mode: latex %%% TeX-master: t %%% End: