[top] [TitleIndex] [WordIndex]

ant07/homework/hw7

Homework 7

pdf

\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: 



2013-05-11 18:33