%%%%
\documentclass[a4paper,landscape]{article}
\special{landscape} % apparently needed for gv to display file properly
\usepackage{amsmath,amsthm,amssymb,url,amscd}
%\usepackage{mathdots}
\usepackage{/usr/TeX/texmf-local/tex/generic/mathdots}
%\usepackage{layout}
\usepackage[polutonikogreek,english]{babel}
%\usepackage[oxonia]{psgreek}

\usepackage{bm}

%\usepackage{eco} This would prevent my getting Greek letters

\usepackage{parskip}

\usepackage[nohead]{geometry}
\addtolength{\textheight}{30pt}
\usepackage{pstricks}
\usepackage{graphics}
\usepackage{wrapfig}

\usepackage{verbatim}

\newcommand{\Z}{\mathbb Z}
\newcommand{\VS}[1]{\operatorname{VS}_{#1}}
\newtheorem*{theorem}{Theorem}
\newcommand{\alg}[1]{{#1}^{\mathrm{alg}}}
\begin{document}
%\thispagestyle{empty}

\huge

%David 
D.
Pierce, \textbf{Vector-spaces over unspecified fields,}
Nijmegen, 2006 
\vfill
From Euclid of Alexandria,
\emph{The Elements,}
Proposition II.5:
\vfill
\begin{wrapfigure}{r}{0.6\textwidth}\centering
  \begin{pspicture}[0.5]
(-0.75,-0.75)(9.75,5.25)
\psset{unit=15mm}
\psline(0,3)(6,3)(6,1)(0,1)(0,3)
\psline(6,1)(6,0)(3,0)(3,3)
\psline(4,0)(4,3)
\psline(3,0)(6,3)
\uput[ul](0,3){$A$}
\uput[ur](6,3){$B$}
\uput[u](3,3){$\mathnormal\Gamma$}
\uput[u](4,3){$\mathnormal\Delta$}
\uput[dl](3,0){$E$}
\uput[dr](6,0){$Z$}
\uput[d](4,0){$H$}
\uput[dr](4,1){$\mathnormal\Theta$}
\uput[dl](0,1){$K$}
\uput[dl](3,1){$\mathnormal\Lambda$}
\uput[r](6,1){$M$}
    \end{pspicture}
\end{wrapfigure}

\selectlanguage{polutonikogreek}
\sf % without this, I don't get Greek letters
E>uje~ia g'ar tic <h AB tetm'hsjw e>ic m`en >'isa kat`a t`o G, e>ic
d`e >'anisa kat`a t`o D; l'egw, <'oti t`o <up`o t~wn AD, DB
perieq'omenon >orjog'wnion met`a to~u >ap`o t~hc GD tetrag'wnou >'ison
>esti t~w| >ap`o t~hs GB tetrag'wnw|.\rm
\vfill
\selectlanguage{english}
`For, let a straight [line] $AB$ be cut into equal [segments] at
$\mathnormal\Gamma$, and unequal at $\mathnormal\Delta$; I say that
the rectangle bounded by $\mathnormal{A\Delta,\Delta B}$%
% [namely $\mathnormal{A\Delta\Theta K}$]
, with the square on $\mathnormal{\Gamma\Delta}$%
% [namely $\mathnormal{\Lambda\Theta HE}$]
, is equal to the square on $\mathnormal\Gamma B$%
% [namely $\mathnormal\Gamma BZE$]
.'
\vfill
We might say
\begin{equation*}
  (x+y)(x-y)+y^2=x^2,
\end{equation*}
where $x$, $y$ are the lengths of $A\mathnormal\Gamma$,
$\mathnormal{\Gamma\Delta}$ respectively; maybe Euclid too
(Descartes, \emph{Rules for the Direction of the Mind,} 4).

\newpage

From Ren\'e Descartes, second page of \emph{The Geometry:}
\vfill
\begin{wrapfigure}{l}{0.6\textwidth}
\centering
%\begin{center}
      \begin{pspicture}
%      (0,0)(2.4,1.6)
(0,-0.5)(6,4.75)
%\psset{unit=16mm}
\psline(0,0)(6,0)(1,4)
\psline(1,0)(1.5,3.6)
\psline(3,0)(3.3,2.16)
\uput[d](1,0){$D$}
\uput[d](3,0){$A$}
\uput[dr](6,0){$B$}
\uput[ur](1.5,3.6){$E$}
\uput[ur](3.3,2.16){$C$}
    \end{pspicture}
%\end{center}
\end{wrapfigure}

%\begin{itshape}
\hfill\emph{%
\guillemotleft Soit par exemple $AB$ l'vnit\'e, \& qu'il faille
multiplier $BD$ par $BC$, ie n'ay qu'a ioindre les poins $A$ \& $C$,
puis tirer $DE$ parallele a $CA$, \& $BE$ est le produit de cete
Multiplication.\guillemotright
}
%\end{itshape}
\vfill
`For example, suppose $AB$ is unity, and that one must multiply $BD$ by
$BC$; I need only join points $A$ and $C$, then draw $DE$ parallel to
$CA$, and $BE$ is the product of this multiplication.'
\vfill
Take $B$ as an origin of \textbf{vectors.}  Then $D$ is the multiple of $A$
by a \textbf{scalar,} say $[D:A]$.  Because  $E-D\parallel C-A$, we
have 
\begin{equation*}
  [D:A]*C=E\qquad \&\qquad [D:A]=[E:C].
\end{equation*}
%So we can do field-arithmetic in a vector-space with parallelism.
\newpage

Precisely, a \textbf{vector-space} is a two-sorted structure
$(V,K,*)$, where:
\begin{center}
$V$ is an abelian group;
\quad
$K$ is a field;
\quad
$*:K\times V\to V$;
\end{center}
\begin{equation*}
  (t\mapsto(\bm x\mapsto(t*\bm x)))\in
\operatorname{Hom}(K,(\operatorname{End}(V),\circ)). 
\end{equation*}
So the signature of $(V,K,*)$ is
$\{+,-,0\}\amalg\{+,-,\cdot,0,1\}\amalg\{*\}$; there are variables
$\bm x$ for vectors, and $t$ for scalars.
\vfill
Abraham Robinson (\emph{Complete Theories,} Amsterdam, 1956) showed
model-completeness of the theory of non-trivial vector-spaces over
a \emph{fixed} scalar field: `partial' model-completeness.
\vfill
If $n\in\{1,2,3,\dots,\infty\}$, let $T_n$ be the theory of
$n$-dimensional vector-spaces.  Andrey Kuzichev (\emph{Z. Math.\ Logik
  Grundlag.\ Math.,} 1992) showed elimination of quantified
\emph{vector-}variables in $T_n$.  Hence, if $U$ is a complete
field-theory, then $T_n\cup U$ is complete.
\vfill
$T_n$ has $\exists\forall$ axioms for raising dimension.  `Hence:'
\vfill
$T_n\cup U$ is not inductive
($\forall\exists$);
much less model-complete (unless $n=1$).
\vfill
We can remedy this `problem' with a predicate for parallelism.

\newpage

The vector-space $(V,K,*)$ expands to $(V,K,*,\parallel)$, where
\begin{equation*}
    \bm x\parallel\bm y\iff\exists s\;\exists
  t\;(s*\bm x+t*\bm y=\bm 0\land(s\neq0\lor t\neq0)).
\end{equation*}
If $\dim_KV>1$, then $(V,K,*,\parallel)$ is interpretable uniformly in
the reduct $(V,\parallel)$, and 
\begin{equation*}
  (V,K,*,\parallel)\subseteq(W,L,*,\parallel)\iff
(V,\parallel)\subseteq(W,\parallel).
\end{equation*}
\vfill
(By contrast, three-sorted structures $(G,N,G/N)$, where
%$G$ is a group, and 
$N\triangleleft G$, are interpretable in the
reducts $(G,N)$; but
\begin{equation*}
  (\Z,4\Z,\Z_4)\nsubseteq(\Z,2\Z,\Z_2)\quad\&\quad(\Z,4\Z)\subseteq(\Z,2\Z).)
\end{equation*}
\vfill
The theory $\VS 2$ of the expansions $(V,K,*,\parallel)$ is inductive.
\vfill
Hence the theory of the reducts $(V,\parallel)$ is inductive.
\vfill
The model-companion of $\VS 2$ is the theory $\VS 2{}\!^*$ of
\textbf{two-dimensional} vector-spaces over \textbf{algebraically
  closed} fields.  In particular, $\VS 2{}\!^*$ is model-complete: 

\newpage

Replace the binary $\parallel$ with an $n$-ary predicate $\parallel^n$
for linear dependence; 
in the new signature, let $\VS n$ be the theory of vector-spaces.
\vfill
\begin{theorem}
The existentially closed models of $\VS n$ are $n$-dimensional (over
algebraically closed fields; since these models compose an elementary
class, their theory is the model-companion of $\VS n$).
\end{theorem}
\vfill
\emph{Proof.}  Say $K\subset L$ (both fields), and $\dim_KL\geqslant
n+1$.  
\vfill
I say that $(K^{n+1},K,\parallel^n)$ embeds in $(L^n,L,\parallel^n)$:
\vfill
Suppose $\{a_0,\dots,a_{n-1},1\}$ from  % correction
$L$ is linearly independent over $K$.  Let
 \begin{equation*}
A=
\begin{pmatrix}
   1&   0&      &      0\\
   0&   1&      &      0\\
    &    &\ddots&       \\
   0&   0&      &      1\\
-a_0&-a_1&\cdots&-a_{n-1}
\end{pmatrix}=
\left(
   \begin{array}{c}
     I_n\\\hline-\bm a
   \end{array}
\right).
 \end{equation*}
The embedding is $\bm x
\mapsto\bm x\cdot A$, taking the rows of $I_{n+1}$ to the rows of
$A$:

\newpage

Indeed,
  write $n$ elements of $K^{n+1}$ as the rows of %
%the $n\times(n+1)$-matrix 
$(\begin{array}{c|c}
   U&\bm v
  \end{array})$.
These rows are dependent if and only if 
\begin{equation*}
0=
  \det\left(
\begin{array}{c|c}
  U&\bm v\\\hline \bm a&1
\end{array}\right)=
\det(U-\bm v\cdot\bm a).
\end{equation*}
But $U-\bm v\cdot\bm a=(\begin{array}{c|c}
    U&\bm v
  \end{array})\cdot
\left(
   \begin{array}{c}
     I_n\\\hline-\bm a
   \end{array}
\right)$, whose rows are the images in $L^n$ of the rows of
$(\begin{array}{c|c}
   U&\bm v
  \end{array})$.\hfill$\qed$
\vfill
Compare with structures $(L,K,P^n)$, where 
%$K\subset L$, both algebraically closed fields, and 
$P^n$ is $n$-ary algebraic
dependence.  From a standard counterexample:
% showing that the pre-geometry $(L,X\mapsto{K(X)})$ is not
% modular gives us also 
\begin{equation*}
  {(K(a_0,\dots,a_n)},K,P^n)\subset
({K(a_0,\dots,a_n,b,c)},{K(b,c)},P^n), 
\end{equation*}
where
$\dim(a_0\cdots a_n/K)=n+1$, and
$(b,c)$ is a generic solution to $\sum_{i=0}^na_ix^i=y$, so that
$\dim(a_0\cdots a_n/Kbc)=n$.
\vfill
\begin{center}
  \begin{pspicture}
    (-5,-1)(5,1)
\psellipse(-2,0)(3,1)
\rput(0,0){$K$}
\rput(-3,0){$a_0\cdots a_n$}
\psellipse(1.5,0)(2.5,1)
\rput(2,0){$b$}
\rput(3,0){$c$}
  \end{pspicture}
\end{center}

\end{document}

\newpage

\layout

\end{document}
\begin{align*}
  \det(U-\bm v\cdot\bm a)
&=
\det    \left(
    \begin{array}{c|c}
      U-\bm v\cdot\bm a & \bm v\\\hline\bm 0 & 1
    \end{array}\right)\\
&=
  \det  \left(\left(
    \begin{array}{c|c}
      U & \bm v\\\hline\bm a & 1
    \end{array}\right)
\cdot
    \left(
    \begin{array}{c|c}
      I_n & \bm 0\\\hline-\bm a & 1
    \end{array}\right)\right)
=\det\left(
    \begin{array}{c|c}
      U & \bm v\\\hline\bm a & 1
    \end{array}\right),
\end{align*}

\begin{comment}
  
 >E`an e>uje~ia gramm`h tmhj~h| e>ic >'isa ka`i >'anisa, t`o <up`o t~wn
 >an'iswn t~hc <'olhc tmhm'atwn perieq'omenon >orjog'wnion met`a to~u
 >ap`o t~hc metax`u t~wn tom~wn tetrag'wnou >'ison >est`i t~w| >ap`o
 t~hc <hmise'iac tetrag'wnw|.

\end{comment}
\begin{comment}
  
`If a straight line be cut into equal and unequal [segments], the rectangle
bounded by the unequal segments of the whole, with the square on the
[segment] between the cuts, is equal to the square on the half.

\end{comment}