% This file is a solution template for:
% - Giving a talk on some subject.
% - The talk is between 15min and 45min long.
% - Style is ornate.
% Copyright 2004 by Till Tantau <tantau@users.sourceforge.net>.
%
% In principle, this file can be redistributed and/or modified under
% the terms of the GNU Public License, version 2.
%
% However, this file is supposed to be a template to be modified
% for your own needs. For this reason, if you use this file as a
% template and not specifically distribute it as part of a another
% package/program, I grant the extra permission to freely copy and
% modify this file as you see fit and even to delete this copyright
% notice. 
\mode<presentation>
{
% \setbeamertemplate{background canvas}[vertical shading][bottom=red!10,
% top=blue!10]
  \usetheme{Warsaw}
  \usefonttheme[onlysmall]{structurebold}
}
% or whatever

\usepackage{amsmath,amssymb}
\usepackage[latin1]{inputenc}
\usepackage{colortbl}
\usepackage[english]{babel}
% Or whatever. Note that the encoding and the font should match. If T1
% does not look nice, try deleting the line with the fontenc.

% This file is a solution template for:
% - Giving a talk on some subject.
% - The talk is between 15min and 45min long.
% - Style is ornate.

\title[
\ \kern-190pt \blank{Jean-Pierre Demailly (Grenoble I), 26/02/2010\kern53pt
Geometric constructions \& algebraic numbers}]
% (optional, use only with long paper titles)
{Geometric constructions in relation with
algebraic and transcendental numbers}

%% \subtitle{Presentation Subtitle} % (optional)

\author[] % (optional, use only with lots of authors)
{Jean-Pierre Demailly}

\institute[]{Acad\'emie des Sciences de Paris, and\\
Institut Fourier, Universit\'e de Grenoble I, France}

% - Use the \inst command only if there are several affiliations.
% - Keep it simple, no one is interested in your street address.

\date[]% (optional)
{February 26, 2010 / Euromath 2010 / Bad Goisern, Austria}

%%\subject{Talks}
% This is only inserted into the PDF information catalog. Can be left
% out. 

% If you have a file called "university-logo-filename.xxx", where xxx
% is a graphic format that can be processed by latex or pdflatex,
% resp., then you can add a logo as follows:

\def\hexnbr#1{\ifnum#1<10 \number#1\else
 \ifnum#1=10 A\else\ifnum#1=11 B\else\ifnum#1=12 C\else
 \ifnum#1=13 D\else\ifnum#1=14 E\else\ifnum#1=15 F\fi\fi\fi\fi\fi\fi\fi}
\font\tenmathx=mathx10
\font\eightmathx=mathx8
\font\sevenmathx=mathxm7
\font\fivemathx=mathxm5
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  \textfont\mathxfam=\tenmathx
  \scriptfont\mathxfam=\sevenmathx
  \scriptscriptfont\mathxfam=\fivemathx
\def\mathx{\fam\mathxfam\tenmathx}
\def\mathxtype{\hexnbr\mathxfam}

\def\overacute{\mathaccent"0\mathxtype79}
\def\overobtuse{\mathaccent"0\mathxtype7D}

\def\arg{\mathop{\rm arg }}
\def\angle{\mathop{\rm angle}}


\definecolor{ColClaim}{rgb}{0,0,0.8}
\definecolor{Alert}{rgb}{0.8,0,0}
\definecolor{Blank}{rgb}{1,1,1}
\def\claim#1{{\color{ColClaim}#1}}
\def\alert#1{{\color{Alert}#1}}
\def\blank#1{{\color{Blank}#1}}

\def\srelbar{\vrule width0.6ex height0.65ex depth-0.55ex}
\def\merto{\mathrel{\srelbar\kern1.3pt\srelbar\kern1.3pt\srelbar
    \kern1.3pt\srelbar\kern-0.78ex\raise0.3ex\hbox{${\scriptscriptstyle>}$}}}

\newcommand{\Hom}{\operatorname{Hom}}
\newcommand{\Ker}{\operatorname{Ker}}
\newcommand{\tors}{\operatorname{torsion}}
\newcommand{\rk}{\operatorname{rk}}
\newcommand{\reg}{\operatorname{reg}}
\renewcommand{\div}{\operatorname{div}}

\newcommand{\bB}{{\mathbb B}}
\newcommand{\bC}{{\mathbb C}}
\newcommand{\bD}{{\mathbb D}}
\newcommand{\bF}{{\mathbb F}}
\newcommand{\bG}{{\mathbb G}}
\newcommand{\bK}{{\mathbb K}}
\newcommand{\bN}{{\mathbb N}}
\newcommand{\bP}{{\mathbb P}}
\newcommand{\bQ}{{\mathbb Q}}
\newcommand{\bR}{{\mathbb R}}
\newcommand{\bZ}{{\mathbb Z}}

\newcommand{\cA}{{\mathcal A}}
\newcommand{\cC}{{\mathcal C}}
\newcommand{\cD}{{\mathcal D}}
\newcommand{\cE}{{\mathcal E}}
\newcommand{\cF}{{\mathcal F}}
\newcommand{\cH}{{\mathcal H}}
\newcommand{\cI}{{\mathcal I}}
\newcommand{\cK}{{\mathcal K}}
\newcommand{\cM}{{\mathcal M}}
\newcommand{\cN}{{\mathcal N}}
\newcommand{\cO}{{\mathcal O}}
\newcommand{\cP}{{\mathcal P}}
\newcommand{\cX}{{\mathcal X}}

\newcommand{\dbar}{\overline\partial}
\newcommand{\ddbar}{\partial\overline\partial}
\newcommand{\ovl}{\overline}
\newcommand{\wt}{\widetilde}
\newcommand{\lra}{\longrightarrow}
\newcommand{\bul}{{\scriptscriptstyle\bullet}}

% mathematical operators
\renewcommand{\Re}{\mathop{\rm Re}\nolimits}
\renewcommand{\Im}{\mathop{\rm Im}\nolimits}
\newcommand{\Id}{\mathop{\rm Id}\nolimits}
\newcommand{\Supp}{\mathop{\rm Supp}\nolimits}

\newcommand{\Constr}{\mathop{\rm Constr}\nolimits}
\newcommand{\RC}{\mathop{\rm RC}\nolimits}
\newcommand{\Ori}{\mathop{\rm Ori}\nolimits}
\newcommand{\ssm}{\mathop{\Bbb r}}
\newcommand{\smallvee}{{\scriptscriptstyle\vee}}

% figures inserted as PostScript files
\special{header=/home/demailly/psinputs/mathdraw/grlib.ps}
\long\def\InsertFig#1 #2 #3 #4\EndFig{\par
\hbox{\hskip #1mm$\vbox to#2mm{\vfil\special{"
#3}}#4$}}
\long\def\LabelTeX#1 #2 #3\ELTX{\rlap{\kern#1mm\raise#2mm\hbox{#3}}}
\def\ovl{\overline}
\def\build#1^#2_#3{\mathrel{\mathop{\null#1}\limits^{#2}_{#3}}}
\def\bibitem[#1]#2#3{\medskip{\bf[#1]} #3}

\begin{document}

% Delete this, if you do not want the table of contents to pop up at
% the beginning of each subsection:
%%\AtBeginSubsection[]
%%{
%% \begin{frame}<beamer>
%%    \frametitle{Outline}
%%    \tableofcontents[currentsection,currentsubsection]
%%  \end{frame}
%%}


% If you wish to uncover everything in a step-wise fashion, uncomment
% the following command: 

%\beamerdefaultoverlayspecification{<+->}

\begin{frame}
  \pgfdeclareimage[height=1cm]{acad-logo}{academie}
  \pgfuseimage{acad-logo}
  \pgfdeclareimage[height=1cm]{ujf-logo}{logo_ujf}
  \pgfuseimage{ujf-logo}
  \titlepage
\end{frame}

%%\begin{frame}
%%  \frametitle{Outline}
%%  \tableofcontents
%% You might wish to add the option [pausesections]
%%\end{frame}


% Since this a solution template for a generic talk, very little can
% be said about how it should be structured. However, the talk length
% of between 15min and 45min and the theme suggest that you stick to
% the following rules:  

% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.

%% \section*{Basic concepts}
%%\def\pause{}

\catcode`\@=11
\def\openup{\afterassignment\@penup\dimen@=}
\def\@penup{\advance\lineskip\dimen@
  \advance\baselineskip\dimen@
  \advance\lineskiplimit\dimen@}
\newdimen\jot \jot=3pt
\newskip\centering \centering=0pt plus 1000pt minus 1000pt
\def\ialign{\everycr{}\tabskip\z@skip\halign}
\def\eqalign#1{\null\,\vcenter{\openup\jot\m@th
  \ialign{\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil
      \crcr#1\crcr}}\,}
\catcode`\@=12

\def\nthroot#1#2{\kern3pt{}^#1\kern-7pt\sqrt{#2}}

\begin{frame}
  \frametitle{Ruler and compasses vs. origamis}
  Ancient Greek mathematicians have greatly developed geometry
 (Euclid, Pythagoras, Thales, Eratosthenes...)
  \vskip6pt\pause
  They raised the question whether certain constructions can be
  made by \alert{ruler and compasses}
  \vskip6pt\pause
  \alert{Quadrature of the circle ?} This means: constructing a square
  whose perimeter is equal to the perimeter of a given circle.\\
  \claim{It was solved only in 1882 by Lindemann, after more than
  2000 years~: construction \alert{is not possible with ruler and 
  compasses~!}\\
  \vskip6pt\pause
  Neither is it possible to \alert{trisect an angle} (Wantzel 1837)}
  \vskip6pt\pause
  In Japan, on the other hand, there is a rich tradition of making 
  \alert{origamis}~:
  it is the art of \alert{folding paper} and maker nice geometric
  constructions out of such foldings.
\end{frame}

\begin{frame}
  \frametitle{Trisection of an angle with origamis}
 \pgfdeclareimage[height=6cm]{Trisection}{Trisection}
 \pgfuseimage{Trisection}
 \vskip1pt
 Folding along $(F3)$ so that $O$ is brought to $O'\in (F_1)$ and
 $I$ is brought to $I'\in D'$ constructs the 
 \alert{trisection of angle$(D,D')$}~!
\end{frame}

\begin{frame}
  \frametitle{Cube root of 2 with origamis}
 \pgfdeclareimage[height=6cm]{cube_root}{cube_root}
 \pgfuseimage{cube_root}
 \vskip1pt
 \claim{Exercise.} Show that this construction can be used to
 produce $\nthroot{3}{2}$ (side of the square is $3$).
\end{frame}

\begin{frame}
  \frametitle{``Axioms'' for ruler and compasses}
  \begin{itemize}
  \item One starts from a \alert{given set of points $S$}\\
  (quit often just two points $S=\{O, A\}$)\\
  Then enlarge $S$ into \alert{$S'\supset S$} by constructing lines and 
  circles according to the following rules:
  \vskip6pt\pause
\item \claim{Axiom (RC1).} Given two points $M$, $N$ already constructed
  in $S'$, one can construct \alert{the line $(MN)$ or the circle of center
  $M$ passing through $N$} (or vice versa).
  \vskip6pt\pause    
  \item \claim{Axiom (RC2).} Given 2 lines, 1 line and a circle, or 2 circles
  constructed from RC1, \alert{$S'$ contains all points of intersection 
  of these}.
  \vskip6pt\pause
  \item \claim{Question :} Describe the set of points $\Constr_{\RC}(S)$
  which \alert{can be constructed from $S$ in finitely many steps}.
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Complex numbers}
  \begin{itemize}
  \item One introduces the \alert{``imaginary number'' denoted 
  $i=\sqrt{-1}$} which does
  not exist among real numbers ($x^2=-1$ has no solution in $\bR$)
  \vskip6pt\pause
  \item 
  Complex numbers are \alert{combinations $x+iy$} where $x,y$ are real numbers,
  e.g. $2+3i$. Their set is \alert{denoted $\bC$}.
  \vskip6pt\pause
  \item \claim{Addition in $\bC$}
  \alert{
  $$(x+iy)+(x'+iy')=(x+x')+i(y+y'),$$}
  e.g.~~$(2+3i)+(-7+8i)=-5+11i$
  \vskip6pt\pause
  \item \claim{Multiplication in $\bC$}
  \alert{$$\eqalign{
  (x+iy)\times(x'+iy')&=xx'+ixy'+iyx'+i\times i\times yy'\cr
                      &=xx'+ixy'+iyx'+(-1)\times yy'\cr
                      &=(xx'-yy')+i(xy'+yx')\cr}$$}
  \end{itemize}
  \end{frame}

\begin{frame}
  \frametitle{Geometric interpretation of complex numbers}
  \begin{itemize}
  \item Complex numbers are identified with points of a euclidean plane, 
  \alert{once one chooses an origin $O$ and a point $A$ representing $1$}
  \vskip6pt\pause
  \item \claim{Interpretation of addition in $\bC$}
   Addition corresponds to adding vectors in the plane: use a
  \alert{parallelogram}
  \vskip6pt\pause
  \item \claim{Interpretation of multiplication in $\bC$}
  Introduce $|z|=\sqrt{x^2+y^2}$ and $\arg(z)=\angle(0x,Oz)$.
  Then
  \alert{$$|zz'|=|z|~|z'|,\qquad\arg(zz')=\arg(z)+\arg(z')~~
  \hbox{mod $2\pi$}$$}
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Square roots always exist in $\bC$~!}
  \begin{itemize}
  \item \alert{$\sqrt{-x}=\pm i\sqrt{x}$} if $x$ is a positive real number.
  \vskip6pt\pause
  \item \claim{Square root of a complex number:} if $z=x+iy$, then
  \alert{$$\sqrt{z}=\pm\Big(\sqrt{{x+\sqrt{x^2+y^2}}\over 2}+
  \varepsilon i\sqrt{{-x+\sqrt{x^2+y^2}}\over 2}\Big)$$}
  where $\varepsilon=+1$ if $y\ge 0$ and $\varepsilon=-1$ if $y<0$.
  \vskip6pt\pause
  \item \claim{Theorem (d'Alembert-Gauss)} Every polynomial of \hbox{degree~$d$
   \kern-9pt}\\
   $a_dz^d+a_{d-1}z^{d-1}+\ldots+a_1z+a_0$ with coefficients in $\bC$ has
   \alert{exactly $d$ roots} when counted with multiciplicities.
  \vskip6pt\pause
  \item \claim{Definition} One says that $z\in\bC$ is an \alert{algebraic
  number} if it is a solution of a polynomial with \alert{$a_j\in\bQ$} (or 
  \alert{$a_j\in\bZ$}), \alert{a~transcendental number otherwise}
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Example of algebraic and transcendental numbers}
  \begin{itemize}
  \item Although irrational, $z=\sqrt{2}$ is algebraic since $z^2-2=0$.
  \vskip6pt\pause
  \item $z=i\sqrt{2}$ is also algebraic since $z^2+2=0$.
  \vskip6pt\pause
  \item \claim{Hermite (1872):} $e=\exp(1)$ is transcendental.
  \vskip6pt\pause
  \item \claim{Lindemann (1882):} $\pi$ is transcendental.\\
  In fact if $\alpha$ is algebraic and non zero, then\\
  \alert{$e^\alpha$ is transcendental} (Lindemann-Weierstrass 1885).\\
  Now $\pi$ cannot be algebraic since $e^{i\pi}=-1$ is algebraic~!
  \vskip6pt\pause
  \item \claim{Gelfond / Schneider (1934):} if $\alpha$ and
  $\beta$ are algebraic, $\alpha\ne 0,1$ and $\beta\notin\bQ$, then
  \alert{$\alpha^\beta$ is transcendental}.\\
  For example, \alert{$2^{\sqrt{2}}$ is
  transcendental}, as well as \alert{$e^\pi=(e^{i\pi})^{-i}=(-1)^{-i}$}.
  \vskip6pt\pause
  \item Unknown whether \alert{$e/\pi$} is transcendental, not even known
  that \alert{$e/\pi\notin\bQ$}~!
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Subfields of the field of complex numbers}
  \begin{itemize}
  \item A subset $\bF\subset\bC$ is called a \alert{field} (but 
  there is a more general concept than just for numbers...) 
  if $\bF$ contains $0,1$, and is \alert{stable by
  addition, subtraction, multiplication and division},
  (i.e.\ for $z,w\in\bF$, we have $z+w\in\bF$, $z-w\in\bF$, 
  $zw\in\bF,~z/w\in\bF$ if $w\ne 0$)
  \vskip6pt\pause
  \item If $\bF$ contains $0,1,-1$, it is enough for $\bF$ to be stable by
  addition, multiplication and especially \alert{inverse
  ($z\in\bF$, $z\ne 0\Rightarrow 1/z\in\bF$)}.
  \vskip6pt\pause
  \item For example, $\bQ$, $\bR$, $\bC$ are fields but $\bZ$ is not
  ($2\in\bZ$ but $1/2\notin\bZ$), nor is the set $\bD$ of decimal numbers
  \vskip6pt\pause 
  \item The set denoted $\bQ[\sqrt{2}]$ of numbers of the form
  $x+y\sqrt{2}$, $x,y\in\bQ$ is a field~:
  $\displaystyle(x+y\sqrt{2})^{-1}={x-y\sqrt{2}\over x^2-2y^2}=
  {x\over x^2-2y^2}-{y\over x^2-2y^2}\sqrt{2}$
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Algebraic number fields}
  \begin{itemize}
  \item Similarly the set denoted $\bQ[\sqrt{-2}]$ of numbers of the form
  $x+y\sqrt{-2}=x+iy\sqrt{2}$, $x,y\in\bQ$ is a field \alert{(exercise~!)}\\
  These fields are called \alert{quadratic fields}
  \vskip6pt\pause 
  \item The set $\bQ[\nthroot{3}{2}]$ of numbers of the form
  $x+y\nthroot{3}{2}+z(\nthroot{3}{2})^2$, $x,y,z\in\bQ$ is a field
  \alert{(cubic field)}\\
  This is a bit harder to prove.\\
  \claim{Hint:} calculate $\omega^3$ where $\omega=-{1\over 2}+i{\sqrt{3}\over 2}$, and
  then show that the product
  $$
  (x+y\omega\nthroot{3}{2}+z(\omega\nthroot{3}{2})^2)
  (x+y\omega^2\nthroot{3}{2}+z(\omega^2\nthroot{3}{2})^2)$$
  no longer involves $\omega$ and yields a rational number when multiplied
  by $x+y\nthroot{3}{2}+z(\nthroot{3}{2})^2$.
  \end{itemize}
\end{frame}


\begin{frame}
  \frametitle{Degree of a number field}
  \begin{itemize}
  \item One can show (but this is yet harder) that if 
   $\alpha,\,\beta,\,\gamma,\ldots$ are algebraic numbers, then
   the sets $\bQ[\alpha]$, $\bQ[\alpha,\beta]$, 
   $\bQ[\alpha,\beta, \gamma]$ of polynomials $P(\alpha)$,
   $P(\alpha,\beta)$, $P(\alpha,\beta,\gamma)$ (...) with rational
   coefficients are \alert{fields}.
  \vskip6pt\pause 
  \item If $\bF\subset\bG$ are fields and every element
   $y\in\bG$ can be written in a \alert{unique way}
   $y=x_1\alpha_1+\ldots+x_p\alpha_p$ for $x_i\in\bF$ and
   certain (well chosen) elements \alert{$\alpha_i\in\bG$}, one
   says that $\bG$ has (finite) \alert{degree $p$ over $\bF$}, with
   \alert{basis $(\alpha_j)$ over $\bF$}, and 
   one writes \alert{$[\bG:\bF]=p$}
  \vskip6pt\pause 
  \item \claim{Example:} $[\bQ[\sqrt[2]:\bQ]=2$ and
  $[\bQ[\nthroot{3}{2}:\bQ]=3$.
  \item \claim{Exercise.} If $\bG=\bF[\alpha]$ where $\alpha\in\bG$,
  $\alpha\notin\bF$ and $\alpha$ satisfies an equation of 
  degree $2$ with coefficients in $\bF$, then $[\bG:\bF]=2$.
  Idem for degree $d$ if $\alpha$ does not satisfy any equation
  of lower order (take $\alpha_j=\alpha^j$, $0\le j\le d-1$).
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Successive extensions of fields}
  \begin{itemize}
  \item \claim{Theorem.} If $\bF\subset\bG\subset\bK$ are fields
  then
  \alert{$$[\bK:\bF]=[\bK:\bG]\times [\bG:\bF]$$}
  if the degrees are finite.
  \vskip6pt\pause
  \item \claim{Proof.} Write $p=[\bG:\bF]$ and 
  $q=[\bK:\bG]$.\\ 
  Every $z\in\bK$ can be written in a unique way\\
  $\displaystyle
  z=\sum_k y_k\beta_k,~~y_k\in\bG~~\hbox{for a basis
  $\beta_1,\ldots,\beta_q\in\bK$},$\\
  and each $y_k\in\bG$ can then be written in a unique way\\
  $\displaystyle
  y_k=\sum_j x_{jk}\alpha_j,~~x_{jk}\in\bF~~\hbox{for a basis
  $\alpha_1,\ldots,\alpha_p\in\bG$},$\\
  so, uniquely in terms of the  $\alpha_j\beta_k$
  \alert{(check it!)}
  $\displaystyle z=\sum_{j,k} x_{jk}\alpha_j\beta_k$.\\
  Thus $(\alpha_j\beta_k)$ is a basis of $\bK$ over $\bF$ and
  $[\bK:\bF]=pq$.
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Re-interpretation of constructions with ruler
   and compasses}
  \begin{itemize}
  \item We start from a set of points $S$ in the plane (of at least 
  two points) and interpret them
  as complex numbers in coordinates. By a rotation, change of
  origin and change of unit, we mant assume that two of these
  numbers are $s_1=0,\,s_2=1$, the other ones are complex numbers
  $s_3\ldots,s_n$, $n=\sharp S$. 
  \vskip6pt\pause
  \item \claim{Basic observation.} The set of points constructible
  from $S$ by ruler and compasses is stable by \alert{addition,
  multiplication, inverse}, and also by \alert{conjugation and 
  square root}.
  \vskip6pt\pause
  \item The set $\bQ(S)$ of all
  rational fractions $P(s_3,\ldots,s_n)/Q(s_3,\ldots,s_n)$
  \alert{is a field} (equal to $\bQ$ if we start from only 
  two points).
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Necessary and sufficient condition for 
  constructibility}
  \begin{itemize}
  \item When we construct a bigger set $S'\subset S$ with ruler
  and compasses, we only solve linear and quadratic equations
  (intersections of lines and/or circles) with coefficients
  in $\bQ(S)$ for the first step.
  \vskip6pt\pause
  \item In general, our construction consists of producing a
  \alert{``tower of quadratic extensions''}
  $$
  \bQ(S)=\bF_0\subset\bF_1\subset\ldots\subset\bF_k=\bQ(S')
  $$
  where each field $\bF_{j+1}=\bF_j[\alpha_j]$ is obtained by
  adjoining a point $\alpha_j$ satisfying at most a quadratic 
  equation.
  \vskip6pt\pause
  \item \claim{Remark.} The ``quadratic tower'' condition is
  necessary and sufficient: any such tower starting with $\bQ(S)$ 
  consists of points which are constructible step by step from~$S$.
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{The degree must be a power of $2$}
  \begin{itemize}
  \item \claim{Consequence:} $[\bQ(S'):\bQ(S)]$ must 
  be \alert{a power of~$2\,$!}
  \vskip6pt\pause
  \item \claim{Theorem (Gauss, just before 1800)} A regular
  $n$-agon (polygon with $n$-sides), is constructible if and
  only if the prime factorization of $n$ is of the form
  \alert{$n=2^kp_1\ldots p_m$} where the $p_j$ are Fermat
  primes, i.e.\ prime numbers of the form 
  \alert{$p_j=2^{2^{q_j}}+1$}.
  \vskip6pt\pause
  \item \claim{Proof.} -- We are using $n$-th rooths of $1$, i.e. the
  field $\bQ[\omega]$, $\omega^{n-1}+\ldots+\omega+1=0$, of degree 
  $d\le n-1$.\\
  -- Degree can be $d<n-1$ (example $d=2$ for $n=6$).\\
  -- Reduction to the case $n=p^r$ is a prime power\\
  -- When $n=p^r$, $\omega$ is of degree $d=(p-1)p^r$ exactly (this has to
  be proved!). Thus either $p=2$ or $r=1$ and $p-1$ has to be a pover of 
  $2$, i.e.\ $p=2^s+1$, and then $s$ itself has to be a power of~$2$.
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{Axioms of construction by origamis (1)} 
  One constructs lines by folding paper; points can be constructed
  by taking intersections of folding lines.
  \vskip4pt\pause
  \begin{itemize}
  \item \claim{Axiom O1.} Given two points $P$ ,$Q$, one can fold paper through
  line $(PQ)$
 \vskip7mm
 \pgfdeclareimage[height=2cm]{p11}{p11}
 \pgfuseimage{p11}
  \end{itemize}
\end{frame}


\begin{frame}
  \frametitle{Axioms of construction by origamis (2)} 
  \claim{Axiom O2.} Given two points $P$, $Q$, one can fold paper to
  bring $P$ to $Q$ (through the median line of segment $[P,Q]$.
 \vskip3mm
 \pgfdeclareimage[height=5cm]{p12}{p12}
 \pgfuseimage{p12}
\end{frame}

\begin{frame}
  \frametitle{Axioms of construction by origamis (3)} 
  \claim{Axiom O3.} Given two lines $(D_1)$, $(D_2)$ one can fold paper 
  to bring $(D_1)$ onto $(D_2)$ (through one of the bissecting lines)
 \vskip3mm
 \pgfdeclareimage[height=5cm]{p14}{p14}
 \pgfuseimage{p14}
\end{frame}

\begin{frame}
  \frametitle{Axioms of construction by origamis (4)} 
  \claim{Axiom O4.} Given one point $P$ and a line $(D)$, one can 
  fold through point $P$ in such a way that $(D)$ is brought to itself
  (thus perpendiculary to $(D)$ through $P$)
 \vskip3mm
 \pgfdeclareimage[height=5cm]{p16}{p16}
 \pgfuseimage{p16}
\end{frame}

\begin{frame}
  \frametitle{Axioms of construction by origamis (5)} 
  \claim{Axiom O5.} Given a line $(D)$ and two points $P,Q$, one can 
  (whenever possible) fold paper through $P$ in such a way that $Q$ is brought
  to a point of~$(D)$.
 \vskip3mm
 \pgfdeclareimage[height=5cm]{p19}{p19}
 \pgfuseimage{p19}
\end{frame}

\begin{frame}
  \frametitle{Axioms of construction by origamis (6)} 
  \claim{Axiom O6.} Given two lines $(D_1)$ and $(D_2)$ and two points 
  $P,Q$, one can (whenever possible) fold paper to bring $P$ to a point of 
  $(D_1)$ and $Q$ to a point of $(D_2)$
 \pgfdeclareimage[height=3.5cm]{p22}{p22}
 \pgfuseimage{p22}
  \vskip6pt\pause
  In fact, axiom O6 can be seen to imply all others. As in the case of
  compass and ruler, one can see that the axioms allow to take arbitrary
  integer multiples or quotients, as well as addition, multiplication or 
  division of complex numbers.
\end{frame}

\begin{frame}
  \frametitle{Origamis and cubic equations}
 \pgfdeclareimage[height=6cm]{equation}{equation}
 \pgfuseimage{equation}
 \vskip4pt
 \claim{Problem.} Bring $A(a,c)$ given onto $A'\in Ox$ and
 $B(b,d)$ given onto $B'\in Oy$ by folding.
\end{frame}

\begin{frame}
  \frametitle{Origamis and cubic equations (calculation)}
  One gets
  $t={}$slope$(AA')={}$slope$(BB')$ 
  $\displaystyle\Rightarrow t={d-y\over b}={c\over a-x}$
  \vskip6pt
  $\displaystyle
  I\Big({a+x\over 2},{c\over 2}\Big)~~
  J\Big({b\over 2},{d+y\over 2}\Big),$~~
  slope $\displaystyle(IJ)={-1\over t}=
  {d+y-c\over b-(a+x)}$\\
  Therefore
  $$
  x=a-{c\over t},~~~y=d-bt,~~~{-1\over t}=
  {2d-c-bt\over b-2a+{c\over t}}
  $$
  whence the equation
  $$
  bt^3+(c-2d)t^2+(2a-b)t-c=0
  $$
  which is equivalent to the most general cubic equation
  $t^3 + pt^2 + qt + r = 0$ by putting
  $$
  a ={q+1\over 2},~~b=1,~~c=-r,~~d=-{p+r\over 2}.
  $$
\end{frame}

\begin{frame}
  \frametitle{Necessary and sufficient condition for 
  constructibility by origamis}
  \begin{itemize}
  \item \claim{Theorem:} a set $S'$ can be constructed by origamis from
  $S=\{0,1,s_3,\ldots,s_n\}$ if and only if there is a
  tower of field extensions
  $$
  \bQ(S)=\bF_0\subset\bF_1\subset\ldots\subset\bF_k=\bQ(S')
  $$
  where each extension $\bF_{j+1}=\bF_j[\alpha_j]$ is a
  \alert{quadratic or cubic extension}.
  \vskip6pt\pause
  \item \claim{Corollary.} A polygon with $n$ sides can be constructed 
  with origamis if and only if $n=2^k3^\ell p_1\ldots p_m$ where each $p_j$
  is a prime number with the property that each $p_j-1=2^{a_j}3^{b_j}$.
  \end{itemize}
\end{frame}

\begin{frame}
  \frametitle{References}

  \claim{James King}, Origami constructible numbers, 2004
   http://citeseerx.ist.psu.edu/viewdoc/download?
   doi=10.1.1.123.6667\&rep=rep1\&type=pdf
  \vskip7pt

  \claim{Michael Filaseta} Lecture Notes, Section 6, the transcendence of $e$ 
  and $\pi$,
  http://www.math.sc.edu/\~{}filaseta/gradcourses/
  Math785/Math785Notes6.pdf
  \vskip7pt

  \claim{Serge Lang}, Algebra, Graduate Texts in Mathematics 211, Springer,
  3rd edition, 2002.
\end{frame}

\end{document}