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%                                    Springer Heidelberg 2004/12/07
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%%%%% General:

\newcommand{\Gk}[1]{\selectlanguage{polutonikogreek}#1\selectlanguage{english}}
\newcommand{\defn}[2]{\textbf{#1#2}}   % for words being defined (the
				% second argument is for a plural
				% ending, punctuation, etc.)

\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}
\renewcommand{\setminus}{\smallsetminus}
\renewcommand{\phi}{\varphi}

\newcommand{\eqcl}[1]{[\,#1\,]}  % equivalence-class
\newcommand{\tuple}[1]{\bm{#1}}         % tuples

\newcommand{\alg}[1]{#1^{\mathrm{alg}}}    % algebraic closure
\newcommand{\included}{\subseteq}      % [the name suggests the meaning here]
\newcommand{\pincluded}{\subset}       % proper inclusion    

\newcommand{\gpgen}[1]{\langle#1\rangle}  % group (or space) generated by #1
\newcommand{\size}[1]{\lvert#1\rvert}  % cardinality
\newcommand{\radix}[2][1]{\sqrt{\vphantom{#1}}#2}   % square root
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\DeclareMathOperator{\trdeg}{tr-deg} % transcendence-degree
\newcommand{\inv}{^{-1}}                % multiplicative inverse
\newcommand{\End}[1]{\operatorname{End}(#1)}  % Endomorphism group


%%%%%  Quantifiers:

\newcommand{\qsep}{\;}                 % follows a quantified variable
\newcommand{\Forall}[1]{\forall{#1}\qsep }
\newcommand{\Exists}[1]{\exists{#1}\qsep }
\newcommand{\Existsunique}[1]{\exists!\,{#1}\qsep }

%%%%%  Other logical:

\newcommand{\proves}{\vdash}
\newcommand{\amp}{\;{}\mathbin{\&}{}\;}
\newcommand{\Iff}{\iff}

%%%%%  General model-theory:

\newcommand{\lang}{\mathcal{L}}        % a language or signature
\DeclareMathOperator{\modelclass}{Mod}
\newcommand{\Mod}[1]{\modelclass^{\included}(#1)}  % the category of
				% models with embeddings
\newcommand{\Model}[1]{\modelclass^{\elsub}(#1)}  % the category of
				% models with elementary embeddings
\DeclareMathOperator{\diag}{diag}      % (Robinson) diagram
\newcommand{\str}[1]{\mathfrak{#1}}    % structure
\newcommand{\elsub}{\preccurlyeq}      % is an elementary substructure of

%%%%%  Interpretations

\newcommand{\intp}[2]{{#1}_{#2}}      % interpretation of form. #1 w.r.t. #2
\newcommand{\iuni}[1]{\delta_{#1}{}}  % universe for interpretation #1
\newcommand{\ifunc}[2]{f_{#1}^{#2}}   % surj. map for interpretation
				      % #1 in structure #2

%%%%%  Specific theories:

\newcommand{\VSp}[1]{\operatorname{VS}_{#1}} % vector-spaces with an
				% n-ary predicate for linear dependence
\newcommand{\VSpm}[1]{\operatorname{VS}_{#1}^{\mathrm m}} % vector-spaces 
%of minumum dimension n with an n-ary predicate for linear dependence  
\newcommand{\VSpr}[1]{\operatorname{VS}_{#1}^{\mathrm r}} % reducts of
				% models of this
\newcommand{\VSpst}[1]{{\VSp{#1}}\!^*}      % model-companion for these

%%%%% Specific structures:

\newcommand{\stnd}[1]{\mathbb{#1}}
\newcommand{\C}{\stnd{C}}         % complex numbers
\newcommand{\R}{\stnd{R}}         % real numbers
\newcommand{\Q}{\stnd{Q}}         % rational numbers
\newcommand{\Z}{\stnd{Z}}         % integers
\newcommand{\F}[1]{\stnd{F}_{#1}} % prime field of char. #1
\newcommand{\vnn}{\upomega}       % von-Neumann natural numbers

%%%%% Vectors:

\newcommand{\vsp}{vector-space}   % I like to hyphenate; others might not
\newcommand{\vsps}{\vsp s}
\newcommand{\Vsp}{Vector-space}
\newcommand{\Vsps}{\Vsp s}
\newcommand{\vect}[1]{\bm{#1}}         % vectors 
\newcommand{\vx}{\vect x}
\newcommand{\vy}{\vect y}
\newcommand{\vz}{\vect z}
\newcommand{\vw}{\vect w}
\newcommand{\va}{\vect a}
\newcommand{\vb}{\vect b}
\newcommand{\svmult}{\ast}  % multiplication of scalar and vector
\newcommand{\crossprod}{\bm{\times}}  % cross-product
\newcommand{\ldep}[1]{P^{#1}}    % predicate for linear dependence
\newcommand{\transp}[1]{{#1}^{\mathrm t}} % transpose
\newcommand{\Id}[1][n]{\mathrm{I}_{#1}}  % identity matrix

%
\journalname{Archive for Mathematical Logic}
%
\begin{document}

\title{Model-theory of \vsps\\ over
unspecified fields
}
%\subtitle{Do you have a subtitle?\\ If so, write it here}

\titlerunning{\Vsps}        % if too long for running head

\author{David Pierce
}

%\authorrunning{Short form of author list} % if too long for running head

\institute{David Pierce \at
              Mathematics Department, Middle East Technical University, Ankara, Turkey, 06531 \\
              \email{dpierce@metu.edu.tr}
}

\date{Received: date / Revised: date}
% The correct dates will be entered by the editor


\maketitle

\begin{abstract}
\Vsps{} over unspecified fields can be axiomatized as
one-sorted 
structures, namely, abelian groups with the
relation of parallelism.  Parallelism is binary linear dependence.
When equipped with the $n$-ary relation of linear dependence for some
 positive integer $n$, a \vsp{} is existentially closed if and
 only if it is
$n$-dimensional over an algebraically closed field.  In the signature
with an $n$-ary predicate for linear dependence for \emph{each}
positive integer $n$, the theory of infinite-dimensional \vsps{}
over algebraically closed fields is the model-completion of the theory
of \vsps. 
%Insert your abstract here. Include up to five keywords.
\keywords{\vsp{} \and model-completeness \and sort}
\subclass{03C10 \and 03C60}
\end{abstract}









\setcounter{section}{-1}

\section{Introduction}\label{sect:intro}

In Abraham Robinson's original definition \cite[\S2.2]{MR0472504}, a
(first-order) theory $T$ is \defn{model-complete}{} if the theory
$T\cup\diag{\str M}$ is
complete whenever $\str M\models T$.  Two equivalent formulations are
as follows,
in current notation,
where $\str M$ and $\str N$ are models of $T$:
  \begin{enumerate}
\item
$\str M\included\str N\implies\str M\elsub_1\str N$ (\defn{Robinson's
  Test}{}) \cite[2.3.1]{MR0472504}; \item
$\str M\included\str N\implies\str
  M\elsub\str N$
  \cite[2.4.1]{MR0472504}.  \end{enumerate}
Model-completeness is of use in the discovery of complete
  theories, since a
model-com\-plete theory with a model that embeds in every other model is
complete \cite[4.1.6]{MR0472504}.  Discovery of
\emph{model}-complete theories---complete or not---is also of
interest in itself.

Robinson gives some examples of model-complete theories, including the
theories of non-trivial
torsion-free divisible abelian groups \cite[\S3.1]{MR0472504},
algebraically closed fields \cite[\S3.2]{MR0472504}, and
real-closed fields \cite[\S3.3]{MR0472504}; these examples are
now standard \cite[ch.~3]{MR1924282}.  Robinson notes also, in effect,
that the
theory of non-trivial \vsps{} \emph{over a given field} is
model-complete \cite[\S3.6]{MR0472504}.  However, although his
notation would seem to allow it, he does not appear to consider the
theory of \vsps{} \emph{tout court} (over an unspecified field).
%; this would have been a theory of \emph{two}-sorted structures.

In the most usual formulation, a
\vsp\ is a two-sorted structure.
About early versions of model-theory, Angus Macintyre
notes \cite[\S2.4, p.~198]{MR1988966}:
\begin{quote}
  For no good reason, natural structures were forced into a one-sorted
  formulation (and by now it is a considerable nuisance that there is
  no basic model-theory text in a many-sorted setting).
\end{quote}
Nonetheless, the possibility of treating \vsps{} as two-sorted
structures has
been acknowledged in one textbook  \cite[\S5.4]{MR1800596}; also, modules
in general have been treated as two-sorted structures
\cite[ch.~9]{MR1057608}.
A natural signature for this treatment of vector-spaces includes:
\begin{enumerate}
  \item
the signature of
abelian groups, for the vectors;
\item
the signature of rings, for the scalars; \item
a symbol for multiplication of vectors by scalars  (in the present
paper, this symbol is $\svmult$).
\end{enumerate}
In this signature, let $T$ be the theory of \vsps, and if
$n\in\{1,2,3,\dots\}\cup\{\infty\}$, let $T_n$ be the theory of
\vsps{} of dimension $n$.  Andrey
Kuzichev \cite{MR1255099} gives an
explicit elimination of quantified
vector-variables in $T_n$.  Indeed, suppose $C$ is a formula whose
vector-variables are from the tuple $(x_1,\dots,x_m,y)$.  (Kuzichev
says $m\geq1$, but this is not necessary if summations $\sum_{i=1}^0f_i$
are understood to be $0$.)  Let $C'$ be the
result of replacing each $y$ in $C$ with $\sum_{i=1}^m\alpha_i\svmult x_i$,
and let $C^*$ be the result of replacing each atomic sub-formula
$\lambda\svmult y=\sum_{i=1}^m\mu_i\svmult x_i$ of $C$ with
$\lambda=0\land\sum_{i=1}^m\mu_i\svmult x_i=0$.  Then
\begin{equation*}
  T\proves\Exists yC\leftrightarrow
(\Exists{\tuple{\alpha}}C'\lor
(C^*\land\Exists y\Forall{\tuple{\alpha}}y\neq\sum_{i=1}^m\alpha_i\svmult x_i)).
\end{equation*}
In each of the theories $T_n$, we can express the formula $\Exists
y\Forall{\tuple{\alpha}}y\neq\sum_{i=1}^m\alpha_i\svmult x_i$ without
quantified vector-variables.  Kuzichev gives the corollary that, if
$U$ is a complete theory
of the scalar-field, then $T_n\cup U$ is complete.

However, these complete theories $T_n\cup U$ are never model-complete,
unless $n=1$.  They are not even inductive: the union of a chain of
models need not be a model (no matter what $U$ is, unless $n=1$; see
\S\ref{sect:n} below).
The problem is the lack of a way to express linear independence with
an existential formula.  The picture
changes if predicates for linear independence or dependence are added.

%\begin{wrapfigure}{r}{0.45\textwidth}
  \begin{figure}
\centering
    \begin{pspicture}
      (0,0)(4,1.7)
\psset{unit=5mm}
\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[u](1.5,3){$x$}
\uput[u](3.5,3){$y$}
\uput[r](6,2){$x-y$}
    \end{pspicture}
\caption{Euclid's II.5}\label{fig:euclid}
   \end{figure}
%\end{wrapfigure}
I arrive, myself, at the example of \vsps{} by noting first that,
in the
\emph{Elements} \cite{MR17:814b}, Euclid gives a
geometric formulation of certain
field-theoretic identities.  For example, the identity
$(x+y)(x-y)+y^2=x^2$
is Euclid's Proposition~II.5, but Euclid expresses it in terms of
the squares and rectangles bounded by certain straight lines
(\Gk{e>uje~iai gramma'i,} what we might call \emph{line-segments;}
see Figure~\ref{fig:euclid}). 

At the beginning of the \emph{Geometry} \cite{Descartes-Geometry},
Ren\'e{} Descartes shows
how straight lines (\emph{lignes droites}) can be multiplied to
produce \emph{lines} rather than
rectangles, if one line is chosen as unit:

%\begin{wrapfigure}{l}{0.6\textwidth}
\begin{figure}
\centering
    \begin{pspicture}
      (0,0)(2.4,1.6)
\psset{unit=4mm}
\psline(0,0)(6,0)(1,4)
\psline(1,0)(1.5,3.6)
\psline(3,0)(3.3,2.16)
\uput[ul](1,0){$D$}
\uput[ul](3,0){$A$}
\uput[ur](6,0){$B$}
\uput[ur](1.5,3.6){$E$}
\uput[ur](3.3,2.16){$C$}
    \end{pspicture}
\caption{Descartes's geometric multiplication}\label{fig:descartes}
 \end{figure}
%\end{wrapfigure}

\begin{quote}
  \begin{itshape}
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%
\footnote{`For example, let $AB$ be taken as unity, and
  let it be required to multiply $BD$ by $BC$.  I have only to join
  the points $A$ and $C$, and draw $DE$ parallel to $CA$; then $BE$ is
  the product of $BD$ and $BC$' \cite[p.~5]{Descartes-Geometry}.}
[Fig.~\ref{fig:descartes}].
\end{itshape}
\end{quote}

By means of Descartes's observation, we can interpret the scalar-field
in a \vsp{} of dimension at least two by means of the binary relation of
\emph{parallelism,} that is, binary linear dependence
(\S\ref{sect:parallel}).
So \vsps{} of dimension at least two are axiomatizable, as one-sorted
structures, in the signature of abelian groups
with a binary predicate.  Also, the axioms can be chosen to be
$\forall\exists$.

When the predicate for parallelism is used,
then
the existentially closed \vsps{} are
two-dimensional.  (It does not matter whether a separate sort for
the scalars is retained or not.)  If, more generally, an $n$-ary
predicate for linear dependence is used, then the existentially
closed \vsps{} are $n$-dimensional over
algebraically closed fields (\S\ref{sect:n}).  The class of such
models being elementary, its theory is model-complete, although it
does not admit full quantifier-elimination.

On the value of looking at basic structures, I quote again Descartes,
this time from near the end of the
fourth of his \emph{Rules for the Direction of the Mind}
  \cite{Descartes-Rules}:

  \begin{quote}
    \begin{itshape}
  Quant \`a moi, conscient de ma
  faiblesse, j'ai  d\'ecid\'e d'observer
  opini\^a\-trement, dans ma qu\^ete de connaissances, un ordre tel
  qu'en partant toujours des choses les plus simples et les plus
  faciles, je m'interdise de passer \`a d'autres, avant que dans les
  premi\`eres il ne m'apparaisse qu'il ne reste plus rien \`a
  d\'esirer; c'est pourquoi j'ai pouss\'e jusqu'\`a ce jour, aussi
  loin que j'ai pu, l'\'etude de cette math\'ematique
  universelle\dots%
\footnote{`Aware how slender my powers are, I have resolved in my
  search for knowledge of things to adhere unswervingly to a definite
  order, always starting with the simplest and easiest things and
  never going beyond them till there seems to be nothing further which
  is worth achieving where they are concerned.  Up to now, therefore,
  I have devoted all my energies to this universal mathematics\dots'
  \cite[p.~20]{Descartes-Eng}.}
 \end{itshape}
  \end{quote}

The result about existentially closed \vsps\ means, in part, that
a \vsp{} of $n+1$ dimensions can be embedded in an
$n$-dimensional space
so that linear independence of $n$-element sets (more precisely,
$n$-tuples) is preserved.  
A similar result arises in the model-theory of
fields:  Let $(a_0,\dots,a_n)$ be algebraically
independent over a field $K$, and let $(b,c)$ be a generic solution to
$\sum_{i=0}^na_ix^i=y$.
(This generalizes \cite[Example 8.1.2, p.~291]{MR1924282}; see also
\cite[Example 2.1.8, p.~77]{MR1429864}.) 
  Then $\trdeg(K(a_0,\dots,a_n)/K)=n+1$, while
$\trdeg(K(a_0,\dots,a_n,b,c)/K(b,c))=n$, although every $n$-tuple of
elements of $K(a_0,\dots,a_n)$ that is algebraically independent over
$K$ is still independent over $K(b,c)$.

Recently and independently, Moshe Kamensky \cite{Kamensky} has worked
with (two-sorted) vector-spaces and has given theories of them that
have elimination of quantifiers; this elimination is
achieved by means of additional sorts, as for the Grassmannians of the
finite powers of the scalar-field.  


\section{Interpretation of the field in the
  \vsp{}}\label{sect:parallel}

In a chapter of his own textbook on the subject, Robin Hartshorne
\cite[ch.~3]{MR1761093}
presents Descartes's approach to geometry.  It is perhaps a Cartesian
spirit that
allows Alfred Tarski \cite{MR0106185,MR1791303} to axiomatize
Euclidean geometry by means of the ternary relation of between-ness and
the quaternary relation of equidistance; these relations allow the
underlying (real-closed) field to be recovered from the space.  

In the present
section, I specify an origin in space
and work with the group-structure determined by it; then only a binary
relation is needed to recover the underlying (arbitrary) field.  This
observation was useful to me in showing that a field equipped with a
space and Lie-ring of derivations can be described as a one-sorted
structure in two ways: as a field with certain operators, or as a
Lie-ring with certain operators.

A \defn{\vsp}{} can be defined as a triple $(V,K,\svmult)$,
where $V$ is an abelian group (of \defn{vector}{s}), $K$ is a field
(of \defn{scalar}{s}), 
and~$\svmult$ is a 
function $(x,\vect v)\mapsto x\svmult \vect v$ from $K\times V$ to
$V$ so that:
\begin{enumerate}
  \item
the functions $\vect v\mapsto a\svmult \vect v$ (where $a\in K$) are
endomorphisms of $V$, and 
\item
the function $x\mapsto (\vect v\mapsto
x\svmult \vect
v)$ is a
ring-homomorphism from $K$ into $(\End V,\circ)$.  
\end{enumerate}
Since $K$ is a
field, the homomorphism into $(\End V,\circ)$ is an embedding,
unless $V$ is trivial;
excluding this case, we may treat the homomorphism as an inclusion, so that
multiplication in $K$ is literally composition.  Even in the trivial
case, let us retain the symbol for composition to denote
multiplication of scalars.

Model-theoretically then, a \vsp{} is a
two-sorted structure, in the signature
$\{+,-,\vect0,\circ,0,1,\svmult\}$; here the elements of $\{+,-\}$ do
double duty, taking arguments from the sort of vectors \emph{or} the
sort of scalars.  I shall use normal-face variables for scalars (as $0$), 
and boldface for vectors (as $\vect0$) or, in more general contexts,
for tuples of variables. 

In every vector-space, the binary relation $\parallel$ on the sort of
vectors is defined by the sentence
\begin{equation}\label{eqn:pardef}
  \vx\parallel\vy\leftrightarrow\Exists u\Exists
  v(u\svmult\vx+v\svmult\vy=\vect0\land(u\neq0\lor v\neq0)).
\end{equation}
(Here and throughout the paper, outer universal quantifiers are
suppressed.)  The sentence
\eqref{eqn:pardef} has the form \begin{equation}\label{eqn:form}
  \alpha\leftrightarrow\exists{z_0}\cdots\Exists{z_{n-1}}\beta, \end{equation}
where $\alpha$
and $\beta$ are quantifier-free, and none of the
$z_i$ appears in $\alpha$; such a sentence is equivalent to the
conjunction of the $\forall\exists$-sentence
$\exists{z_0}\cdots\Exists{z_{n-1}}(\alpha\to\beta)$ and the universal
sentence $\beta\to\alpha$.

Let $\VSp 2$ be the theory of \vsps{} that have been \emph{expanded}
to the signature $\{+,-,\vect0,\circ,0,1,\svmult,\parallel\}$
so as to satisfy~\eqref{eqn:pardef}.  
(The subscript $2$ alludes to the number of arguments that $\parallel$
takes.) 
Models of the theory $\VSp 2$
can be \emph{extended} so as to satisfy also
\begin{equation}\label{eqn:2}
  \Exists{\vx}\Exists{\vy}\vx\nparallel\vy.
\end{equation}
Let $\VSpm 2$ be the theory of the resulting structures: the
\vsps{}, with parallelism, of dimension at least two.  (The
superscript `m' refers to a \emph{m}inimum dimension.)  This
theory is inductive.

\defn{Reduction}{} of a one-sorted structure is the discarding of some
named operations or relations; we may understand reduction of a
many-sorted structure to involve also the discarding of some sorts.
In this sense, every model  $(V,K,\svmult,\parallel)$ of $\VSpm 2$ has
a one-sorted reduct 
$(V,\parallel)$ in the signature $\{+,-,\vect0,\parallel\}$.
This reduct
is an abelian group with a certain binary relation; we might just call
the reduct an \defn{abelian group with parallelism}.
Let $\VSpr 2$ be
the theory of these structures.  
The theories $\VSpm 2$ and $\VSpr 2$ will be related in
Theorem~\ref{thm:V}.

In general, the models of a theory $T$ in a signature $\lang$ are the
objects of a category whose 
morphisms are either embeddings or elementary embeddings.  Let the two
possibilities for the category be $\Mod T$ and $\Model T$
respectively.  
If $\lang\included\lang'$, and $T'$ is a theory of $\lang'$, and
reducts to $\lang$ of models of $T'$ are models of $T$, then we may
speak of reduction from $T'$ to $T$: it is a functor $R$ from
$\Model{T'}$ to $\Model{T}$, and from $\Mod{T'}$ to $\Mod{T}$ (this is
a trivial case 
of Lemma~\ref{lem:int} below).    

Possibly, along with the reduction-functor $R$  from $\Model{T'}$ to
$\Model{T}$, there is an
\defn{expansion}-functor $E$ from $\Model{T}$ to $\Model{T'}$ such that
$R\circ E$ is the identity, and for each model $\str A$ of $T'$, there
is a `reasonable' isomorphism $\sigma_{ER}^{\str A}$ from $\str A$ to
$E(R(\str A))$---reasonable in the sense that, for all models $\str A$
and $\str A^*$ of $T'$, for every elementary embedding $h$ of $\str A$ in
$\str A^*$, the following diagram commutes:
\begin{equation*}
  \begin{CD}
    \str A @>{h}>> \str A^*\\
@V{\sigma^{\str A}_{EF}}VV  @VV{\sigma^{\str A^*}_{ER}}V\\
E(R(\str A)) @>>{E(R(h))}> E(R(\str A^*))
  \end{CD}
\end{equation*}
In a word then, $R$ is an \defn{equivalence}{} of categories;
we might say also that $R$ is a \defn{weakly conservative}{} reduction. 
For example, $R$ is weakly conservative in case $T'$ is $\VSp 2$, and
$\lang$ is 
$\{+,-,\vect0,\circ,0,1,\svmult\}$ (so that $T$ is just the theory of
vector-spaces).
  However, in this
case, $E$ is not
functorial on $\Mod{T}$.  
For example, $(\C,\R,\svmult)\included(\C,\C,\svmult)$, but
$(\C,\R,\svmult,\parallel)\not\included(\C,\C,\svmult,\parallel)$. 
This is a point to be
developed in \S\ref{sect:n}.  Briefly, $E$ is not functorial here,
because non-parallelism does not have an existential definition.  In
cases where $E$ \emph{is} functorial on $\Mod{T}$, 
and the diagram above commutes whenever $h$ embeds $\str A$ in $\str A^*$,
so that $R$ is an
equivalence of $\Mod{T'}$ and $\Mod{T}$,---in such cases, 
we may say that
$R$ is simply a \defn{conservative}{} reduction.

Now the statement of Theorem~\ref{thm:V} makes sense.  For the proof,
I first note that
reduction and expansion are instances of
\emph{interpretation.}
In the account of Wilfrid Hodges \cite[\S5.3,
  p.~212]{MR94e:03002},
an \defn{interpretation}{} of a (one-sorted) structure $\str A$ whose
signature is
$\lang$ in a structure $\str B$ whose signature is $\lang'$ consists, for
some positive integer
$n$, of:
\begin{enumerate}
\item\label{item:form}
a function $\phi\mapsto\intp{\phi}I$ converting each $k$-ary unnested
atomic formula 
of $\lang$, for each $k$ in $\vnn$, into an $nk$-ary formula
of $\lang'$,
  \item\label{item:univ}
an $n$-ary formula $\iuni I$ of $\lang'$, and
\item\label{item:func}
a surjective function $\ifunc I{\str B}$ from $\iuni I^{\str B}$
to $A$, such that
\begin{equation}\label{eqn:interpr}
\str B\models\intp{\phi}I(\tuple b) 
\Iff 
  \str A\models\phi(\ifunc I{\str B}(\tuple b))
\end{equation}
for all $\tuple b$ from $\iuni I^{\str B}$.  
\end{enumerate}
We may assume that the universe $A$ of $\str A$ is literally
the set $\iuni I^{\str B}$, \emph{modulo}
the equivalence-relation defined in $\str B$ by $\intp{(x=y)}I$.
Then $\ifunc I{\str B}$ is the quotient-map, and $\str A$ is
uniquely determined by 
$\str B$ and $\phi\mapsto\intp{\phi}I$ and $\iuni I$
\cite[5.3.3, p.~216]{MR94e:03002}: we may
write $\str A=I(\str B)$.  Thus we have a function $I$ converting
structures of $\lang'$ into structures of $\lang$.

In case $\str A$ has several sorts, composing a tuple
$(A_{\alpha}\colon \alpha\in S)$ perhaps, then the
definition must be modified:  For each $\alpha$ in $S$, for some
positive integer $n_{\alpha}$, there will be
an formula $\iuni{I,\alpha}$ in some $n_{\alpha}$-tuple of variables
(each variable being assigned to a sort), 
along with a function $\ifunc{I,\alpha}{\str B}$ 
from $\iuni{I,\alpha}^{\str B}$ onto $A_{\alpha}$, so
that a variant of~\eqref{eqn:interpr} holds, namely,
\begin{equation}\label{eqn:interpr2}
  \str B\models\intp{\phi}I(\tuple b_{\alpha(0)},\dots,\tuple b_{\alpha(k-1)})
\Iff
\str A\models\phi(\ifunc{I,\alpha(0)}{\str B}(\tuple b_{\alpha(0)}),\dots,
\ifunc{I,\alpha(k-1)}{\str B}(\tuple b_{\alpha(k-1)}))
\end{equation}
where $\tuple b_{\alpha(j)}\in\iuni{I,\alpha(j)}^{\str B}$.

The function
$\phi\mapsto\intp{\phi}I$ extends so that its domain comprises 
all unnested formulas of $\lang$, and \eqref{eqn:interpr}
or~\eqref{eqn:interpr2} still holds 
\cite[5.3.2, p.~214]{MR94e:03002}:
the extension takes
$\phi\land\psi$ to $\intp{\phi}I\land\intp{\psi}I$, and
$\lnot\phi$ to $\lnot(\intp{\phi}I)$, and
$\Exists x\phi$ to
$\Exists{(x^0,\dots,x^{n-1})}(\intp{\phi}I\land\iuni
I(x^0,\dots,x^{n-1}))$ (with appropriate modifications in the
many-sorted case).

If there are theories $T$ of $\lang$ and $T'$ of $\lang'$ such that
$I(\str B)\models T$ whenever $\str B\models T'$, 
then let us say that $I$ is an interpretation from
$T'$ to $T$.  In this case, we can understand $I$ as
a functor from
$\Model{T'}$ to $\Model T$, where, if $\str B$ and $\str
B^{*}$ are models of $T'$, and $h$ is an elementary embedding of $\str B$
in $\str B^{*}$, then $I(h)$ is the well-defined elementary embedding of
$I(\str B)$ in $I(\str B^{*})$ given by
\begin{equation}\label{eqn:functor}
  I(h)(\ifunc I{\str B}(\tuple b))=\ifunc I{\str B^{*}}(h(\tuple b))
\end{equation}
\cite[5.3.4(a), p.~217]{MR94e:03002}.
If the formulas involved in $I$ are simple enough, then more follows: 

\begin{lemma}\label{lem:int}
If $I$ is an interpretation from $T'$ to $T$ (and therefore a functor
from $\Model{T'}$ to $\Model T$), then $I$ is also
a functor from $\Mod {T'}$ to $\Mod T$, \emph{provided} both that
the formula $\iuni I$
is existential, and also that the formula
$\intp{\phi}I$ is existential whenever $\phi$ is
an unnested atomic formula of $\lang$, 
or is $\lnot(x=y)$, 
or is $\lnot Rx^0\dotsb x^{k-1}$ for some 
predicate $R$ in $\lang$.
\end{lemma}

\begin{proof}
The claim is a variant of \cite[5.3.4(b), p.~217]{MR94e:03002}
and a refinement of \cite[\S5.4, exercise~3, p.~225]{MR94e:03002}.
\end{proof}

An interpretation $I$ from $T'$ to $T$, paired with an
interpretation~$J$ from $T$ to $T'$, is a
\defn{bi-interpretation}{} of $T'$ and $T$ if there are formulas
$\chi_{IJ}$ of
$\lang$, and $\chi_{JI}$ of $\lang'$, such that, whenever $\str
A\models T$, then the set
\begin{equation}\label{eqn:isomorphism}
\{(a,\ifunc{I}{J(\str A)}(\ifunc{J}{\str A}(\tuple b_0),\dots,
\ifunc{J}{\str A}(\tuple b_{n-1})))\colon \str
A\models\chi_{IJ}(a,\tuple b_0,\dots,\tuple b_{n-1})\}
\end{equation}
is (the graph of) an isomorphism $\sigma^{\str A}_{IJ}$ from $\str A$
to $I(J(\str A))$, and
similarly for all models of $T'$, with the places of $I$ and $J$
reversed (Hodges \cite[\S5.4(c),
p.~222]{MR94e:03002}, generalizing Ahlbrandt and Ziegler
\cite[p.~67]{MR831437}; in the case of several sorts, there will be
several formulas $\chi_{IJ,\alpha}$, indexed with the sorts, and the
graph of $\sigma_{IJ}^{\str A}$ will have components of the form 
$\{(a,\ifunc{I,\alpha}{J(\str A)}
(\ifunc{J,\beta(0)}{\str A}(\tuple b_0),\dots,
\ifunc{J,\beta(n_{\alpha}-1)}{\str A}(\tuple b_{n_{\alpha}-1})))\colon 
\str A\models\chi_{IJ,\alpha}(a,\tuple b_0,\dots,\tuple
b_{n_{\alpha}-1})\}$).  
A special case was alluded to above, where 
reduction can be `undone' by an expansion.

\begin{lemma}\label{lem:bi}
If $(I,J)$ is a bi-interpretation of $T'$ and $T$, then $I$ is an
equivalence of the
categories $\Model{T'}$ and $\Model T$; if $I$ and $J$ are
also functorial between $\Mod{T'}$ and $\Mod T$, then these categories
too are
equivalent, provided that the formulas $\chi_{IJ}$ and $\chi_{JI}$
are existential. 
\end{lemma}

\begin{proof}
To prove the first claim, if
$\str A$ and $\str A^*$ are models of $T$, and $h$ is an elementary
embedding of $\str A$ in $\str A^*$, then we show that
the following diagram commutes (and likewise for models of $T'$):
\begin{equation*}
  \begin{CD}
    \str A @>{h}>> \str A^*\\
@V{\sigma^{\str A}_{IJ}}VV  @VV{\sigma^{\str A^*}_{IJ}}V\\
I(J(\str A)) @>>{I(J(h))}> I(J(\str A^*))
  \end{CD}
\end{equation*}
Suppose $a\in A$ (and all models are one-sorted; extra sorts would be
  a challenge only to the notation.)  Then
(abbreviating the notation in~\eqref{eqn:isomorphism}) we have
  $\sigma_{IJ}^{\str A}(a)=\ifunc I{J(\str 
  A)}(\ifunc J{\str A}(\tuple b))$ for some $\tuple b$ such that
$\str A\models\chi_{IJ}(a,\tuple b)$.  Then
$\str A^*\models\chi_{IJ}(h(a),h(\tuple b))$ since $h$ is elementary,
  so by~\eqref{eqn:functor}
  \begin{multline*}
    \sigma_{IJ}^{\str A^*}(h(a))
=\ifunc I{J(\str A^*)}(\ifunc J{\str A^*}(h(\tuple b)))
=\ifunc I{J(\str A^*)}(J(h)(\ifunc J{\str A}(\tuple b)))\\
=I(J(h))(\ifunc I{J(\str A)}(\ifunc J{\str A}(\tuple b))
=I(J(h))(\sigma_{IJ}^{\str A}(a)).
  \end{multline*}
So the diagram does commute, and the first
claim is proved.  If $h$ is merely an embedding, but $I(h)$ is still
well-defined as an embedding by~\eqref{eqn:functor}, then the same
argument works, provided also $\chi_{IJ}$ is
existential. 
\qed
\end{proof}

Now let us return to the special case of reductions and expansions.
The following lemma singles out some of the basic properties of
\vsps{} to be
used in proving Theorem~\ref{thm:V}.  

\begin{lemma}\label{lem:basic}
  Suppose $(V,K,\svmult,\parallel)\models\VSpm 2$.   \begin{enumerate}
\item\label{item:p1}
Say $\vx,\vy\in V$, and $\vx\neq0$.  Then
\begin{equation*}
  \vx\parallel\vy\Iff\Exists uu\svmult\vx=\vy\Iff \Existsunique
  uu\svmult\vx=\vy.
\end{equation*}
\item\label{item:p2}
The relation
$\parallel$, restricted to $V\setminus\{\vect0\}$, is an
equivalence-relation. 
\item\label{item:p3}
The set
$\{\vy\in V\colon \vx\parallel\vy\}$ is a subgroup of $V$, for all $\vx$ in
  $V$.    \item\label{item:p4}
  Let $\vx$, $\vz$, and
  $\vw$ be vectors in $V$ such that $\vx\nparallel\vz$; let $u\in K$;
  and let $\vy=u\svmult\vx$.  Then
\begin{equation*}%\label{eqn:-}
  u\svmult\vz=\vw\Iff \vz\parallel\vw\amp \vx-\vz\parallel\vect
  y-\vw.
\end{equation*}
  \end{enumerate}
\end{lemma}

\begin{proof}
I write a proof for item~\ref{item:p4} only, assuming
items~\ref{item:p1},~\ref{item:p2}, and~\ref{item:p3}. 
%See Fig.~\ref{fig:parallel}.
Assume $\vx\nparallel\vz$ and $\vy=u\svmult\vx$.  Part of
item~\ref{item:p3} is that $\vect0$ is parallel to every vector; so
$\vz\neq\vect0$.  Similarly, but with the help of item~\ref{item:p2}, we have
$\vz\nparallel\vx-\vz$, so $\vx-\vz\neq\vect0$. Hence, if
$u\svmult\vz=\vw$, then $\vz\parallel\vw$ by item~\ref{item:p1}, and also
$\vy-\vw=u\svmult(\vx-\vz)$, which is parallel to $\vx-\vz$ for the
same reason.
Conversely, say
$\vz\parallel\vw$.  Then $\vw=t\svmult\vz$ for some $t$ in $K$, so that
$\vy-\vw=u\svmult\vx-t\svmult\vz=u\svmult(\vx-\vz)+(u-t)\svmult\vz$.
If also $\vx-\vz\parallel\vy-\vw$, then
$(u-t)\svmult\vz\parallel\vx-\vz$ by item~\ref{item:p3}, so $u=t$.
\qed\end{proof}

\begin{theorem}\label{thm:V}
Reduction from $\VSpm 2$ to $\VSpr 2$
is conservative.
\end{theorem}

\begin{proof}
We have $R$ from $\Mod{\VSpm 2}$ to $\Mod{\VSpr 2}$.  To go the other
way, we want to interpret an arbitrary model
$(V,K,\svmult,\parallel)$ of $\VSpm 2$ in its reduct, $(V,\parallel)$,
obtaining an expansion-functor $E$.
(It will follow that the reducts $(V,\parallel)$ compose an elementary
class.) 
In the notation for interpretations given earlier, we shall need formulas
$\iuni{E,\mathrm{vec}}$ and 
$\iuni{E,\mathrm{sca}}$; these will be $\vx=\vx$ and
\begin{equation*}
  \vx_0\neq\vect 0\land\vx_0\parallel\vx_1,
\end{equation*}
respectively.  Then $\ifunc{E,\mathrm{vec}}{(V,\parallel)}$ will be
the identity
on $V$, while $\ifunc{E,\mathrm{sca}}{(V,\parallel)}$ will be given by
\begin{equation}\label{eqn:function}
\ifunc{E,\mathrm{sca}}{(V,\parallel)}(\vect a_0,\vect a_1)=b 
\Iff
  (V,K,\svmult,\parallel)\models b\svmult\vect a_0=\vect a_1.
\end{equation}
Indeed, the function $\ifunc{E,\mathrm{sca}}{(V,\parallel)}$ is well
defined on $\iuni{E,\mathrm{sca}}^{(V,\parallel)}$, and is surjective onto $K$,
by Lemma~\ref{lem:basic} (item \ref{item:p1}):
\begin{enumerate}
  \item
If $(V,\parallel)\models\iuni{E,\mathrm{sca}}(\vect a_0,\vect a_1)$, 
then there is a unique $b$ in $K$ such that 
$(V,K,\svmult,\parallel)\models b\svmult\vect a_0=\vect a_1$.
\item
If
$b\in K$, and $\vect a\neq\vect0$,
then $(V,\parallel)\models\iuni{E,\mathrm{sca}}(\vect a,b\svmult\vect a)$.
\end{enumerate}
Let $\ifunc{E,\mathrm{sca}}{(V,\parallel)}$ be denoted by
\begin{equation*}%\label{eqn:f}
  (\vx,\vy)\mapsto\eqcl{\vx:\vy}.  
\end{equation*}
We have to define the function $\phi\mapsto\intp{\phi}E$.  This
will be the identity at all formulas with no
scalar-variables.  It is now
enough to suppose that $\phi$
is one of
\begin{equation*}
  x=y,\quad x+y=z,\quad x\circ y=z,\quad x\svmult\vy=\vz.
\end{equation*}
The formula $\intp{(x=y)}E$ will be the
equation
\begin{equation}\label{eqn:classes}
  \eqcl{\vx_0:\vx_1}=\eqcl{\vy_0:\vy_1},
\end{equation}
expressed in $\{+,-,0,\parallel\}$.
To obtain this expression, first note that
\begin{equation}\label{eqn:mult-int}
\vy_0\neq\vect0\amp  \eqcl{\vx_0:\vx_1}\svmult\vy_0=\vy_1\Iff
\eqcl{\vx_0:\vx_1}=\eqcl{\vy_0:\vy_1}.
\end{equation}
%The converse of \eqref{eqn:mult-int} also holds by definition.
Now let $\psi(\vx_0,\vx_1,\vy_0,\vy_1)$ denote the formula
\begin{equation*}
  \vx_0\nparallel\vy_0\land\vx_0\parallel\vx_1\land\vy_0\parallel\vy_1
  \land\vx_0-\vy_0\parallel\vx_1-\vy_1. 
\end{equation*}
(The situation can be illustrated by Fig.~\ref{fig:descartes} of
\S\ref{sect:intro}, 
with $(A,B,C,D,E)$ replaced with
$(\vx_0,\vect 0, \vy_0,\vx_1,\vy_1)$.)
This formula $\psi$ defines a subset of
$\iuni{E,\mathrm{sca}}^{(V,\parallel)}\times
\iuni{E,\mathrm{sca}}^{(V,\parallel)}$; that is, it 
defines a
binary relation on $\iuni{E,\mathrm{sca}}^{(V,\parallel)}$.  This
relation is symmetric, though not
reflexive, and not quite transitive.
However, by Lemma~\ref{lem:basic} (item~\ref{item:p4}) and
\eqref{eqn:mult-int}, and by the assumption that $\dim_KV\geq2$, we have
\begin{multline*}
\eqcl{\vx_0:\vx_1}=\eqcl{\vy_0:\vy_1}\amp
\vx_0\nparallel\vy_0\Iff
\psi(\vx_0,\vx_1,\vy_0,\vy_1);\\
\shoveleft{\eqcl{\vx_0:\vx_1}=\eqcl{\vy_0:\vy_1}\amp
\vx_0\parallel\vy_0\Iff
\vx_0\parallel\vy_0\amp\Exists{\vz_0}\Exists{\vz_1}}\\
(\psi(\vx_0,\vx_1,\vz_0,\vz_1)\amp
\psi(\vz_0,\vz_1,\vy_0,\vy_1)).
\end{multline*}
 From these observations, we obtain~\eqref{eqn:classes} as a
 formula of $\{+,-,0,\parallel\}$.
A slight simplification is possible: $\intp{(x=y)}E$ can be the
 existential formula
\begin{equation}\label{eqn:ker}
\Exists{\vz_0}\Exists{\vz_1}
(\psi(\vx_0,\vx_1,\vy_0,\vy_1)\lor
(\psi(\vx_0,\vx_1,\vz_0,\vz_1)\land\psi(\vz_0,\vz_1,\vy_0,\vy_1))).
\end{equation}
We obtain the remaining formulas $\intp{\phi}E$ by means of
\eqref{eqn:mult-int}
and the following identities:
\begin{gather*}
  \eqcl{\vx:\vy}+\eqcl{\vx:\vz}=\eqcl{\vx:\vy+\vz};\\
\eqcl{\vy:\vz}\circ\eqcl{\vx:\vy}=\eqcl{\vx:\vz}.
\end{gather*}
In particular, writing \eqref{eqn:classes} as an abbreviation for
\eqref{eqn:ker}, we can tabulate the results:
\begin{equation*}
  \begin{array}{c|c}
    \phi & \intp{\phi}E\\\hline
x+y=z &
\Exists{\vw}(\eqcl{\vx_0:\vw}=\eqcl{\vy_0:\vy_1}\land
\eqcl{\vx_0:\vx_1+\vw}=\eqcl{\vz_0:\vz_1})\\
x\circ y=z &
\Exists{\vw}(\eqcl{\vy_1:\vw}=\eqcl{\vx_0:\vx_1}\land
\eqcl{\vy_0:\vw}=\eqcl{\vz_0:\vz_1})\\
x\svmult\vy=\vz &
\eqcl{\vx_0:\vx_1}=\eqcl{\vy:\vz}\lor(\vy=\vect0\land\vz=\vect0). 
  \end{array}
\end{equation*}
These formulas $\intp{\phi}E$ are existential.
Also, $\lnot(\intp{(x=y)}E)$ is equivalent to the
existential formula
\begin{equation*}
  \Exists{\vw}(\eqcl{\vx_0:\vx_1}=\eqcl{\vy_0:\vw}\land\vw\neq\vy_1). 
\end{equation*}

Now we have $E$ as a functor from $\Mod{\VSpr 2}$ to $\Mod{\VSpm 2}$,
by Lemma~\ref{lem:int}.  The composition $R\circ E$ is the identity,
and $\sigma_{RE}^{(V,\parallel)}$ is the identity, and
$\chi_{RE}$ is $\vx=\vy$.  For the other side,
$\sigma_{ER}^{(V,K,\svmult,\parallel)}$ is the identity on $V$, while
on $K$ it takes $y$ to the equivalence-class $\{(\vx_0,\vx_1)\colon
\vx_0\neq\vect0\amp 
y\svmult\vx_0=\vx_1\}$; correspondingly, $\chi_{ER,\mathrm{vec}}$ is
$\vx=\vy$, while
$\chi_{ER,\mathrm{sca}}$ is $\vx_0\neq\vect0\amp
y\svmult\vx_0=\vx_1$.  By Lemma~\ref{lem:bi}, the proof is complete. 
\qed
\end{proof}

\section{Existentially closed \vsps{}}\label{sect:n}

It was said in \S\ref{sect:intro} that, if $U$ is a \emph{complete}
theory of fields, then the theory, in $\{+,-,\vect0,\circ,0,1,\svmult\}$, of
$n$-dimensional \vsps{} whose
scalar-fields are models of $U$ is complete, but not model-complete or
even inductive, if $n>1$.  Indeed, suppose $V$ is a \vsp{} with basis
$(\vect b_0,\dots,\vect b_{n-1})$ over a model $K$ of $U$, and $n>1$.  If
$K\pincluded K'$, then the $\vect b_i$ span a \vsp{} $W$
over $K'$, but we can have $\vect b_0=a\svmult \vect b_1$ for some
$a$ in $K'\setminus K$, so that $\dim_{K'}W\leq n-1$.  Still,
$(V,K,\svmult)$ is a {substructure} of $(W,K',\svmult)$.  We can have
$K'\models U$, and we can
extend $W$ to an $n$-dimensional space $V'$ over $K'$.  So $(V,K,\svmult)$
is a substructure of $(V',K',\svmult)$, and both are models of
$T_n\cup U$.  Continuing, we form a chain
\begin{equation}\label{eqn:chain}
(V,K,\svmult)\pincluded(V',K',\svmult)\pincluded\dotsb\pincluded
  (V^{(k)},K^{(k)},\svmult)\pincluded\dotsb 
\end{equation}
of models of
$T_n\cup U$; the union of this chain has dimension {at most} $n-1$.
The chain \emph{can}
have dimension one, even if $n=\infty$:  
for each $k$, let $V^{(k)}$ have basis
$(\vect b_k,\vect b_{k+1},\dots)$, and assume $\vect b_k=a_k\svmult\vect
b_{k+1}$ for some $a_k$ in $K^{(k+1)}\setminus K^{(k)}$.

However, when the structures in \eqref{eqn:chain} expand to models of
$\VSpm 2$, they no longer form a chain: non-parallelism is not
preserved in super-structures.
Nonetheless, we shall see that there are chains where non-parallelism
\emph{is} preserved in going up, but the union has dimension two.  The
conclusion will be that
the existentially closed models of $\VSpm 2$ are two-dimensional.
There is a generalization involving chains that preserve $n$-ary linear
independence.

If $n>0$, let $\ldep n$ be an $n$-ary predicate for linear
dependence.  (So $\ldep 2\vx\vy$ means $\vx\parallel\vy$; but $\ldep 1$
is superfluous, since $\ldep 1\vx$ is equivalent to $\vx=\vect0$.)
For each positive
$n$, there is a theory $\VSp n$ of \vsps{} in the signature
$\{+,-,\vect0,\circ,0,1,\svmult,\ldep n\}$, axiomatized
by the usual vector-space axioms, along with
\begin{gather}
\label{eqn:P}
\ldep n\vx_0\dotsb\vx_{n-1}\leftrightarrow
\exists{y^0}\cdots\Exists{y^{n-1}}
\Big(
\sum_{i<n}y^i\svmult
\vx_i=\vect0\land\bigvee_{i<n}y^i\neq0\Big).
\end{gather}
Having the form of \eqref{eqn:form}, the axiom~\eqref{eqn:P} is
equivalent to an $\forall\exists$-sentence.
So $\VSp n$ is inductive.  Models of $\VSp n$ have two sorts.  But let
$\VSpm n$ be the theory of models of $\VSp n$ of dimension at least
$n$; so $\VSpm n$ has the additional axiom
\begin{equation*}
  \exists{\vx_0}\cdots\Exists{\vx_{n-1}}\lnot\ldep
n\vx_0\dotsb\vx_{n-1}.
\end{equation*}
In this theory, when $n\geq2$, we can define 
non-parallelism and parallelism in $\VSpm n$ by means of the
existential formulas 
\begin{gather*}
\Exists{(\vx_2,\dots,\vx_{n-1})}\lnot\ldep n\vx_0\dotsb\vx_{n-1},\\
  \Exists{(\vx_2,\dots,\vx_n)}
\Big(\vx_1=0
\lor
\big(
\lnot\ldep n\vx_1\dotsb\vx_n\land
\bigwedge_{j=2}^n\ldep n\vx_0\dotsb\vx_{j-1}
\vx_{j+1}\dotsb\vx_n\big)\Big).
\end{gather*}
Indeed, if $(\vect a_1,\dots,\vect a_n)$ is linearly independent, but
$(\vect a_0,\dots,\vect a_{i-1},\vect a_{i+1},\dots,\vect a_n)$ is
not,
then $\vect a_0$ is a unique linear combination of 
$(\vect a_1,\dots,\vect a_n)$, and this combination does not use
$\vect a_j$.  If that is so, whenever $2\leq j\leq n$, then $\vect
a_0$ must be a multiple of $\vect a_1$.
If we reduce a model of $\VSpm n$ to the sort of vectors, in the
signature $\{+,-,\vect0,\ldep n\}$, the reduct might be termed an
\defn{abelian group with $n$-ary linear dependence}.  As a porism from
  Theorem~\ref{thm:V}, we have that the reduction from $\VSpm n$ to
  the theory of
  abelian groups with $n$-ary linear dependence is conservative.
By Theorem~\ref{thm:ldep} below, the
model-companion of $\VSp n$---hence, of $\VSpm n$---is the theory
$\VSpst n$ of
$n$-dimensional \vsps{} over algebraically closed fields.  Hence the
theory of abelian groups with $n$-ary linear dependence is likewise
companionable.

To warm up for the proof of Theorem~\ref{thm:ldep}, I consider first
$\VSp 1$, the theory of non-trivial \vsps.
I shall generally write models as $(V,K)$, rather than the
full $(V,K,\svmult)$:

\begin{theorem}\label{thm:1}
  The theory $\VSp 1$
is companionable; its model-companion is the theory $\VSpst 1$.
The model-companion is \emph{not} a model-\emph{completion;}
much less does it admit quantifier-elimination; but it is the
model-completion of the theory of \vsps{} of dimension at most
$1$. 
\end{theorem}

\begin{proof}
Suppose $(V,K)\models\VSp 1$ and has basis $B$. We may assume that $B$
is a subset of a field extending $K$ (while
still being linearly independent over $K$).  Then we
can form $(\alg{K(B)},\alg {K(B)})$, a model of $\VSpst 1$ in which
$(V,K)$ embeds.

Suppose $(V,K)\models\VSpst 1$ and is a substructure of the models
$(W,L)$ and $(W',L')$, which have equal cardinality greater than $\size
K$.  Then $\size L=\size{L'}$, so $L\cong L'$ over $K$; hence
\begin{equation*}
  (W,L)\cong(L\otimes_KV,L)\cong(L'\otimes_KV,L')\cong(W',L')
\end{equation*}
over $(V,K)$.  So $\VSpst 1\cup\diag{(V,K)}$ is categorical in
powers greater than 
$\size K$; having also no finite models, the theory is complete.
Therefore $\VSpst 1$ is
model-complete, so it is the model-companion of $\VSp 1$.

In the foregoing argument, it is enough to suppose merely that $(V,K)$ is
one-dimensional; $K$ need not be algebraically closed.  Even if
$V=\{\vect 0\}$,
the theory $\VSpst 1\cup\diag{(V,K)}$ is
complete.  This shows that
$\VSpst 1$ is the model-completion of the theory of \vsps{} of
dimension at most $1$.

The model $(\alg{\Q},\alg{\Q})$ of $\VSpst 1$ has the two
  substructures
  $(\gpgen{1,\radix 2},\Q)$ and $(\gpgen{1,\radix 3},\Q)$, which are
  two-dimensional isomorphic models of $\VSp 1$, although the
formula \begin{equation*}
  \Exists z(z\circ z=1+1\land z\svmult \vx_0=\vx_1)
\end{equation*}
is satisfied in $(\alg{\Q},\alg{\Q})$ by $(1,\radix 2)$, but not by
  $(1,\radix 3)$.  Therefore the theory
$\VSpst 1\cup\diag{(\gpgen{1,\radix 2},\Q)}$ is the same as
$\VSpst 1\cup\diag{(\gpgen{1,\radix 3},\Q)}$ (as long as the same
  symbol denotes $\radix 2$ and $\radix 3$ respectively), but this
  theory is not
  complete.  Thus $\VSpst 1$ is not a model-completion of $\VSp 1$.
\qed\end{proof}

Proving the more general Theorem~\ref{thm:ldep} below requires some
preliminary work
on matrices. I shall understand an $m\times n$ matrix ($m$ rows, $n$
columns) as a
function on $m\times n$, that is, on $\{(i,j)\in\vnn^2\colon i<m\amp j<n\}$.
Such a function
can be denoted by
\begin{equation*}
  (u^i_j)^{i<m}_{j<n}\;;
\end{equation*}
its transpose is
$(u^i_j)_{i<m}^{j<n}$.  I shall consider elements of a Cartesian
  power like $K^n$ as row-vectors, that is, $1\times n$ matrices.

\begin{lemma}\label{lem:dets}
  For all positive integers $n$, if $U$ is an $n\times n$ matrix (over
  some field), and
  $\vect u$ is $n\times 1$, and $\vect a$ is $1\times n$, then
  \begin{equation}\label{eqn:dets}
    \det\left(
    \begin{array}{c|c}
      U & \vect u\\\hline\vect a & 1
    \end{array}\right)
=\det(U-\vect u\cdot\vect a).
  \end{equation}
\end{lemma}

\begin{proof}
Start with the identity
\begin{equation*}
    \left(
    \begin{array}{c|c}
      U & \vect u\\\hline\vect a & 1
    \end{array}\right)
\cdot
    \left(
    \begin{array}{c|c}
      \Id & \vect 0\\\hline-\vect a & 1
    \end{array}\right)
=
    \left(
    \begin{array}{c|c}
      U-\vect u\cdot\vect a & \vect u\\\hline\vect 0 & 1
    \end{array}\right)
\end{equation*}
and take the determinant of either side.
\qed\end{proof}

The lemma can be interpreted as follows.  Let $K$ be a field, and let
$V$ be $K^{n+1}$.  The space of alternating $n$-forms on $V$ has
dimension $\binom{n+1}n$, or $n+1$.  Now understand
$V^n$ as comprising the
$n\times(n+1)$ matrices over $K$.  The function
$\left(
\begin{array}{c|c}
  U & \vect u
\end{array}
\right)
\mapsto\det\left(
    \begin{array}{c|c}
      U & \vect u\\\hline\vect a & 1
    \end{array}\right)$ from $V^n$ to $K$ is an alternating
$n$-form on $V$ that takes $\left(
\begin{array}{c|c}
  \Id & \vect 0
\end{array}
\right)$ to $1$, and that takes $\left(
\begin{array}{c|c}
  U & \vect u
\end{array}
\right)$ to $0$ if one of its rows is
$\left(
\begin{array}{c|c}
  \vect a & 1
\end{array}
\right)$.  There is only one such function; the function
$\left(
\begin{array}{c|c}
  U & \vect u
\end{array}
\right)
\mapsto\det(U-\vect u\cdot\vect a)$ has the same properties;
\eqref{eqn:dets} follows.

Suppose further $\pi_i$ is $\vx\mapsto x_i$ on $V$, so $\pi_i\in V^*$.  If
$B\in V^n$, let $\Phi B$ be the function
$\det\left(
    \begin{array}{c}
      B\\\hline
      \begin{matrix}
	\pi_0 & \cdots & \pi_n
      \end{matrix}
    \end{array}\right)$: this is an element
    $\sum_{i=0}^nc^i\cdot\pi_i$ of $V^*$.  Then $(c^0,\dots,c^n)$ is
    the \defn{cross-product}{} \cite[pp.~83 f.]{Spivak:CM} of (the rows
    of) $B$: it is the element $\times B$ of $V$ such that
$\vx\cdot(\times B)=\Phi B\vx$
for all $\vx$ in $V$.

\begin{lemma}\label{lem:cross}
  Let $K$ be a field, $V=K^{n+1}$, and $B\in V^n$.  Say $K\pincluded
  L$, and $\vect a\in L^{n+1}$, and $\vect a$ is
  linearly independent over $K$.  The following are equivalent:
  \begin{enumerate}
    \item\label{item:cross}
$\times B=0$;
\item\label{item:det}
$\det\left(
    \begin{array}{c}
      B\\\hline
\vect a
    \end{array}\right)=0$;
\item\label{item:rows}
The rows of $B$ are linearly dependent over $K$;
\item\label{item:rows-L}
The rows of $B$ are linearly dependent over $L$.
  \end{enumerate}
\end{lemma}

\begin{proof}
The equivalence of \eqref{item:cross} and \eqref{item:det} follows
from the definition of $\times B$ and the linear independence of
$\vect a$ over $K$.
  The equivalence of
    \eqref{item:rows} and \eqref{item:rows-L} follows from the
    observation that if the equation $\vect x\cdot B=\vect 0$ has a
    non-trivial solution at all, it has one in $K^n$.  Finally,
$\vect a$ is not in the span over $L$ of the rows of $B$: if it
    were, then $\vect c\cdot B=\vect a$ for some $\vect c$ in $L^n$,
    which would mean that the $n+1$ entries
    of $\vect a$ belonged to the span over $K$ of the $n$ entries of
    $\vect c$.  Hence \eqref{item:det} and \eqref{item:rows-L} are
    equivalent. \qed\end{proof}

\begin{theorem}\label{thm:cross}
Suppose $L/K$ is a field-extension, and $[L:K]\geq n+1$.  Then the \vsp{}
$(K^{n+1},K)$ embeds in $(L^n,L)$ so as to preserve linear
independence of $n$-tuples: that is, the embedding is of the
structure $(K^{n+1},K,\ldep n)$ in $(L^n,L,\ldep n)$.  
%If we consider $K^{n+1}$ and $L^n$ as consisting of row-vectors,
%then,  
One such
embedding is
\begin{equation}\label{eqn:transf}
\vx\longmapsto\vx\cdot
  \left(
  \begin{array}{c}
    \Id\\\hline -\vect a
  \end{array}\right),
\end{equation}
where $\vect a$ in $L^n$ is such that $(a_0,\dots,a_{n-1},1)$ is
linearly independent over $K$.
\end{theorem}

\begin{proof}
Assume that  $(a_0,\dots,a_{n-1},1)$ is
linearly independent over $K$.  Treating $n$
vectors from $K^{n+1}$ as the rows of an $n\times(n+1)$
matrix, we can write the result as $\left(
  \begin{array}{c|c}
    U&\vect u
  \end{array}\right)$.  By Lemma~\ref{lem:cross}, the rows of $\left(
  \begin{array}{c|c}
    U&\vect u
  \end{array}\right)$ are linearly dependent over $K$ if and only if
  \begin{equation}\label{eqn:n+1}
    \det\left(
    \begin{array}{c|c}
      U & \vect u\\\hline
        \vect a & 1
    \end{array}\right)    =0.
  \end{equation}
The images of those rows under the
  transformation in \eqref{eqn:transf} are the rows of the product
  $\left(
  \begin{array}{c|c}
    U&\vect u
  \end{array}\right)\cdot  \left(
  \begin{array}{c}
    \Id\\\hline -\vect a
  \end{array}\right)$.  This product is $U-\vect u\cdot\vect a$.  So the
  images are linearly dependent over $L$ if and only if
  \begin{equation}\label{eqn:n}
    \det(U-\vect u\cdot\vect a)=0.
  \end{equation}
By Lemma~\ref{lem:dets}, equations~\eqref{eqn:n+1} and~\eqref{eqn:n}
are equivalent.  Hence the transformation in~\eqref{eqn:transf} is an
embedding that preserves independence of $n$-tuples.
\qed\end{proof}

Not every embedding of $(K^{n+1},K)$ in
$(L^n,K)$ preserves independence of $n$-tuples:  Just use an
embedding as in \eqref{eqn:transf}, but replace some---not
all---entries of $\tuple a$ with $0$.  

\begin{theorem}\label{thm:ldep}
For each positive $n$, the theory $\VSpst n$ is the
  model-companion of $\VSp n$; it is not a model-completion of $\VSp
  n$ or even of $\VSpm n$,
  but is the model-completion of the theory of $n$-dimensional \vsps.
\end{theorem}

\begin{proof}
The proof is, in part, as for Theorem~\ref{thm:1}.  Since $\VSp n$ is
inductive, every model embeds in an existentially closed model.  Such
models are models of $\VSpst n$.  Indeed, every model $(V,K,\ldep n)$
of $\VSp n$ embeds in $(\alg K\otimes_KV,\alg K,\ldep n)$; so
existentially closed
models must have algebraically closed scalar-fields.  Also, if
$\dim_KV=m<n$, then $(V,K,\ldep n)$ fails to have a solution to
\begin{equation}\label{eqn:Pn}
  \lnot \ldep n\vx_0\dotsb \vx_{n-1};
\end{equation}
but $(V,K,\ldep n)$ embeds in $(V\oplus K^{n-m},K,\ldep n)$, where
\eqref{eqn:Pn} does have a solution; so $(V,K,\ldep n)$ was not
existentially closed.  Finally,
say $\dim_KV>n$, so that there is a linearly independent $(n+1)$-tuple
$(\vect a^0,\dots,\vect a^n)$ of vectors in $V$.  Then $(V,K,\ldep n)$
has no solution to 
\begin{equation}\label{eqn:dep}
  \sum_{i=0}^nx^i\svmult\vect a_i=\vect0\land\bigvee_{i=0}^nx^i\neq0.
\end{equation}
But analyse $V$ as $V_0\oplus V_1$, where $V_0$ is spanned by the
vectors $\vect a^i$.  By Theorem~\ref{thm:cross}, there is a model
$(W,L,\ldep n)$ of $\VSp n$ in which $(V_0,K,\ldep n)$ embeds, but
which has a solution to \eqref{eqn:dep}.  Then $(V,K,\ldep n)$ embeds
in $(W\oplus L\otimes_KV_1,L,\ldep n)$, and the latter has a solution to
\eqref{eqn:dep}.  So again $(V,K,\ldep n)$ was not existentially closed.

So the existentially closed models of $\VSp n$ are models of $\VSpst
n$; in particular, every model of $\VSp n$ embeds in a model of
$\VSpst n$.  If
$(V,K,\ldep n)$ is an $n$-dimensional model of $\VSp n$, then $\VSpst
n\cup\diag{(V,K,\ldep n)}$ is complete, by a categoricity argument as in
the proof of
Theorem~\ref{thm:1}.  Therefore $\VSpst n$ is the model-completion of
the theory of $n$-dimensional models of $\VSp n$.

However, $\VSpst n$ is not the model-completion of $\VSpm n$ (much
less of $\VSp n$).
Indeed, let $(a_0,\dots,a_n)$ be an $(n+1)$-tuple of algebraic
numbers such that the $(n+2)$-tuple $(a_0,\dots,a_n,1)$ is linearly
independent over
$\Q$.  By Theorem~\ref{thm:cross}, 
the row-spaces of the matrices 
\begin{equation*}
  \left(
  \begin{array}{c}
    \Id\\\hline
    \begin{matrix}
      a_0 & \cdots & a_{n-1}
    \end{matrix}
  \end{array}\right)
\quad\text{ and }\quad
  \left(
  \begin{array}{c}
    \Id\\\hline
    \begin{matrix}
       a_1 & \cdots & a_n
    \end{matrix}
  \end{array}\right)
\end{equation*}
are substructures $\str A$ and $\str B$ of
$((\alg{\Q})^n,\alg{\Q},\ldep n)$.  Being $(n+1)$-dimensional, they
are isomorphic models of $\VSpm n$; but no automorphism of 
$((\alg{\Q})^n,\alg{\Q},\ldep n)$ takes $\str A$ to
$\str B$.
\qed\end{proof}

We can let $\VSp{\infty}$ be the union of the theories $\VSp n$; so it
is the theory of \vsps{} in the signature
$\{+,-,\vect0,\circ,0,1,\svmult,\ldep 1,\ldep 2,\ldep 3,\dots\}$.  Let
$\VSpst{\infty}$ be the
theory, in the same signature, of infinite-dimensional \vsps{}
over algebraically closed fields.

\begin{theorem}\label{thm:infty}
  The theory $\VSpst{\infty}$ is the model-completion of
  $\VSp{\infty}$, but does not admit quantifier-elimination.
\end{theorem}

\begin{proof}
Every model $(V,K,\ldep 1,\dots)$ of $\VSp{\infty}$ embeds in
the model
\begin{equation*}
  (\alg K\otimes_KV\oplus(\alg K)^{\vnn},\alg K,\ldep 1,\dots)
\end{equation*}
of $\VSpst{\infty}$.  Also, let $(V,K,\ldep
1,\dots)\models\VSp{\infty}$, and $\kappa=\aleph_0+\size V+\size K$.
Then $\VSpst{\infty}\cup\diag(V,K,\ldep 
1,\dots)$ is complete, since all of its
models $(W,L,\ldep 1)$ are isomorphic, provided
$\trdeg(L/K)=\kappa$ and
$\dim_L(W/(L\otimes_KV))=\kappa$; but these are just the models of size
$\kappa$ that realize certain types.  So $\VSpst{\infty}$ is the
model-completion of $\VSp{\infty}$.  

Finally, the example at the end
of the proof of Theorem~\ref{thm:1} shows that $\VSpst{\infty}$ does
not admit elimination of quantifiers.  Indeed,   $(\gpgen{1,\radix
  2},\Q)$ expands to a
substructure of a model of $\VSpst{\infty}$ in which $1\parallel\radix
2$, but $\VSpst{\infty}\cup\diag(\gpgen{1,\radix 2},\ldep 1,\ldep
2,\dots)$ is not complete, since it does not specify which scalars
ensure that $1\parallel\radix 2$.
\qed\end{proof}

\begin{theorem}
  The completions of the model-complete theories $\VSpst n$ are
  obtained by specifying a
  characteristic for the scalar field.  The completions are
  $\vnn$-stable. 
\end{theorem}

\begin{proof}
  Let $\VSpst{n,p}$ be the theory of models of $\VSpst n$ whose scalar
  fields have characteristic $p$ (positive or zero).  If $n$ is finite, then
  $\VSpst{n,p}$ is uncountably categorical, hence complete and
  $\vnn$-stable. 

All models of $\VSpst{\infty,p}$ have a
  substructure isomorphic to $(\{\vect0\},\F p)$, where $\F p$ is
  a prime field of characteristic $p$.  Hence
  $\VSpst{\infty,p}\proves\diag(\{\vect 0\},\F p)$.  But here
  $(\{\vect0\},\F p)\models\VSp{\infty}$, so the theory
  $\VSpst{\infty}\cup\diag(\{\vect 0\},\F p)$ is complete by
  Theorem~\ref{thm:infty}.  Therefore $\VSpst{\infty,p}$ is complete.

Finally, let $(V,K)$ be a countable, \emph{definably closed}
substructure of a big model of $\VSpst{\infty}$.  This implies that,
if some vectors in $V$
are linearly dependent, then scalars witnessing this can be found in
$K$.  A complete type of $\vx$ over $(V,K)$ says either that $\vx$ is
linearly independent from $V$, or---for some $\vect a_i$ in $V$---that
$\vx=\sum_{i<n}t^i\svmult\vect a_i$ for some scalars $t^i$.  In the
latter case, the type also specifies a variety over $K$ of which
$(t^0,\dots,t^{n-1})$ is a generic point.  If $(u^0,\dots,u^{n-1})$ is
another generic point
of the same variety, there is an automorphism of the big model, over
$(V,K)$, 
taking 
$\sum_{i<n}t^i\svmult\vect a_i$ to
$\sum_{i<n}u^i\svmult\vect a_i$.
Thus there are
just countably many types of $\vx$ over $(V,K)$.  Similarly for types
of a scalar variable $t$:  If $V=\{\vect0\}$, then these types
correspond to types over $K$, of which there are countably many.  If
$V$ contains a non-zero vector $\vect a$, then for every type $p(t)$ over
$(V,K)$, there is a type of $\vx$ that includes $\{\Exists
t(\vx=t\svmult\vect a\land\phi(t)\colon \phi\in p\}$; but the union of any
two such sets is inconsistent.
\qed\end{proof}
  
\begin{acknowledgements}
%If you'd like to thank anyone, place your comments here
%and remove the percent signs.
I read Euclid and
Descartes as a student at St John's College, Annapolis and Santa Fe.  
At Logic Colloquium 2006 in Nijmegen,
Ehud Hrushovski pointed out an
error in an earlier version of this work (and informed me of
Kamensky's work \cite{Kamensky}). 
The anonymous referee offered some helpful suggestions and
corrections. 
\end{acknowledgements}


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