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%\title{What is a ring?}
\author{David Pierce}
\date{Istanbul, December, 2009}

\theoremstyle{definition}
\newtheorem*{theorem}{Theorem}
\newtheorem*{corollary}{Corollary}
\newtheorem*{example}{Example}
\newtheorem*{examples}{Examples}
\newtheorem*{definition}{Definition}
\newtheorem*{Qu}{Question}
%\theoremstyle{definition}
\newtheorem*{A}{Answer}

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\newcommand{\Ham}{\stnd H}
\newcommand{\Oct}{\stnd O}
\newcommand{\Sed}{\stnd S}
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\raggedright

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\begin{document}

%\layout
%\newpage
\huge

%\maketitle

%\fbox{
\parbox[t][390pt][b]{355pt}{
  \includegraphics[width=355pt]{mondrian-lozenge-2.eps}

\Large
Piet Mondrian, 
\emph{Tableau No.\ IV; Lozenge Composition with Red, Gray, Blue,
  Yellow, and Black}
% c. 1924/1925, National Gallery of Art, Washington
%  (Gift of Herbert and Nannette Rothschild)
} 
%}
\hfill
\parbox[t][400pt]{170pt}{
  \begin{center}
  \textsc{Interacting Rings}
\vfill\vfill\vfill
{David Pierce}
\vfill\vfill
December, 2009
\vfill
Istanbul
\vfill
  \end{center}
  %  \begin{tabular}[t]{c}D\\A\\V\\I\\D\\ \\P\\I\\E\\R\\C\\E\end{tabular}
}

\pagebreak

A \textbf{derivation} of a field is an operation $D$ on the field satisfying
\begin{align*}
D(x+y)&=Dx+Dy,&D(x\cdot y)=Dx\cdot y+x\cdot Dy.
\end{align*}
\begin{example}
``Taking the derivative,''
\begin{equation*}
f\mapsto f', 
\end{equation*}
on $\R(x)$ or the field of meromorphic functions on $\C$.
\end{example}

The derivations of a field $K$ compose a \textbf{vector space} over $K$, 
\begin{equation*}
\Der K,
\end{equation*}
where the vector-space operations are given by
\begin{align*}
(D_0+D_1)x&=D_0x+D_1x,&(a\cdot D)x=a\cdot(Dx).
\end{align*}
Then $\Der K$ also has a \textbf{multiplication,} given by
  \begin{equation*}
    [D_0,D_1]=D_0\circ D_1-D_1\circ D_0;
  \end{equation*}
  this is the \textbf{Lie bracket} operation, which I may denote by
  \begin{equation*}
\brkt.
\end{equation*}


\pagebreak

In this context, a \textbf{multiplication} is an operation $\cdot$ on an
abelian group that distributes over addition: 
\begin{align*}
x\cdot(y+z)&=x\cdot y+x\cdot z,&(x+y)\cdot z&=x\cdot z+y\cdot z.
\end{align*}
A \textbf{ring} in the most general sense is an abelian group with a
multiplication. 

\begin{examples}\mbox{}
\begin{enumerate}
	\item 
	$\Z$ and $\Q$;
	\item
	the \textbf{Cayley--Dickson} algebras $\R$, $\C$, $\Ham$, $\Oct$, $\Sed$, \dots;
	\item
	the ring $\operatorname M_n(R)$ of $n\times n$ matrices over a ring $R$;
	\item
	$(\R^3,\times)$;
	\item
	$(\Der K,\brkt)$.
\end{enumerate}
\end{examples}

\pagebreak

A \textbf{group operation} is another kind of multiplication.

The \textbf{permutations} of a set $A$ compose a group,
\begin{equation*}
(\Sym A,\circ),
\end{equation*}
the operation being \textbf{composition.}

If there is a homomorphism from a group $(G,\cdot)$ to $(\Sym A,\circ)$,

then $(G,\cdot)$ \textbf{acts} on $A$.

The action is \textbf{faithful} if the homomorphism is one-to-one.

\begin{theorem}[Cayley]
A group acts faithfully on its underlying set.

Indeed, if $(G,\cdot)$ is a group, and $g,x\in G$, define
\begin{equation*}
\lambda_g(x)=g\cdot x.
\end{equation*}
Then
\begin{equation*}
g\mapsto\lambda_g\colon(G,\cdot)\to(\Sym G,\circ).
\end{equation*}
\end{theorem}

\pagebreak

Now let $V$ be an \emph{abelian} group.

The \textbf{endomorphisms} of $V$ compose an abelian group,
\begin{equation*}
\End V.
\end{equation*}

\begin{examples} 
$\phi\mapsto\phi(1)\colon\End{\Z}\overset{\cong}{\to}\Z$,\hfill $\End{\Z\oplus\Z}\cong\operatorname M_2(\Z)$.
\end{examples}

Then $(\End V,\circ)$ is an \textbf{associative ring:} a ring $(R,\cdot)$ satisfying
\begin{equation*}
x\cdot(y\cdot z)=(x\cdot y)\cdot z.
\end{equation*}
If there is a homomorphism from a field $K$ to $(\End V,\circ)$,\\
then $V$ is a \textbf{vector space} over $K$.

We may say then $K$ \textbf{acts} on $V$.

\begin{example} $K$ acts on $\Der K$. \end{example}

But also $(\Der K,\brkt)$ may be said to \textbf{act} on $K$.

So $K$ and $(\Der K,\brkt)$ are \textbf{interacting rings.}


\pagebreak

The multiplications of %the abelian group 
$V$ compose an abelian group,
\begin{equation*}
\Mult V.
\end{equation*}
This has an involutory automorphism,
\begin{math}
\mult\mapsto\invo{\mult}
\end{math},
where 
%$\invo{\mult}$ is the \textbf{opposite} of $\mult$, so
\begin{equation*}
\invo{\mult}(x,y)=\mult(y,x).
\end{equation*}

\begin{example}
$\mult\mapsto\mult(1,1)\colon\Mult{\Z}\overset{\cong}{\to}\Z$, but $\invo{\mult}=\mult$.
%\hfill$\Mult{\Z\oplus\Z}\cong\operatorname M_2(\Z)\oplus\operatorname M_2(\Z)$.
\end{example}

%Then $\Mult{\End V}$ contains all integral combinations of $\circ$ and $\invo{\circ}$.

%So $\End V$ is a ring in many ways:

%\pagebreak

\begin{examples}
In place of $V$, consider $\End V$: %\mbox{}
\begin{enumerate}
	\item 
	$(\End V,\circ)$ is an associative ring, as above.
	\item 
	$(\End V,\circ-\invo{\circ})$ is a \textbf{Lie ring,} namely, a ring $(R,\cdot)$
	in which
	\begin{align*}
(x\cdot y)\cdot z&=x\cdot(y\cdot z)-y\cdot(x\cdot z),& x\cdot x&=0.
\end{align*}
In particular, $(\Der K,\brkt)$ is a Lie ring.
	\item 
	$(\End V,\circ+\invo{\circ})$ is a \emph{Jordan ring,} %namely, a ring 
	in which
	\begin{align*}
(x\cdot y)\cdot (x\cdot x)&=x\cdot(y\cdot(x\cdot x)),&x\cdot y&=y\cdot x.
\end{align*}
\end{enumerate}
\end{examples}

\pagebreak

If $(R,\cdot)$ is a ring, $p,q\in\Z$, and
\begin{equation*}
x\mapsto \lambda_x\colon(R,\cdot)\to(\End R,\pq)
\end{equation*}
(where again $\lambda_x(y)=x\cdot y$),
let $(R,\cdot)$ be called a \textbf{$(p,q)$-ring.}


\begin{theorem}\mbox{}
\begin{enumerate}
\item
All associative rings are $(1,0)$-rings.  
\item
All Lie rings are $(1,1)$-rings.  
\end{enumerate}
\end{theorem}

\begin{corollary}
If 
\begin{equation*}
(p,q)\in\{(0,0),(1,0),(1,1)\},
\end{equation*}
then 
$(\End V,\pq)$
is a $(p,q)$-ring.
\end{corollary}

\begin{theorem}[P]
The converse holds.
\end{theorem}

\emph{Proof.}
%\begin{proof}%[Proof of theorem.]
$
x\mapsto\lambda_x\colon(\End V,p{\circ}-q\invo{\circ})\to(\End{\End V},p{\circ}-q\invo{\circ})$
\begin{align*}
{}\iff&
\lambda_{x\cdot y}=\lambda_x\cdot\lambda_y\\
{}\iff&
  \lambda_{px\circ y-qy\circ x}(z)
=(p\lambda_x\circ\lambda_y-q\lambda_y\circ\lambda_x)(z)\\
{}\iff&p(px\circ y-qy\circ x)\circ z-qz\circ(px\circ y-qy\circ x)=\dots
\end{align*}
%\qedhere\end{proof}


\pagebreak

\begin{comment}

\begin{proof}%[Proof of theorem.]
We have 
\begin{equation*}
x\mapsto\lambda_x:(\End
E,p{\circ}-q\invo{\circ})\to(\End{\End V},p{\circ}-q\invo{\circ})
\end{equation*}
if and only if 
\begin{equation*}
\lambda_{x\cdot y}=\lambda_x\cdot\lambda_y, 
\end{equation*}
that is,
\begin{equation*}
  \lambda_{px\circ y-qy\circ x}(z)
=(p\lambda_x\circ\lambda_y-q\lambda_y\circ\lambda_x)(z),
\end{equation*}
that is,
\begin{multline*}
p(px\circ y-qy\circ x)\circ z-qz\circ(px\circ y-qy\circ x)\\
=  p\bigl(px\circ(py\circ z-qz\circ y)
-q(py\circ z-qz\circ y)\circ x\bigr)\\
-q\bigl(py\circ(px\circ z-qz\circ x)
-q(px\circ z-qz\circ x)\circ y\bigr),
\end{multline*}
that is,
\begin{align*}
  p^2&=p^3,&
pq&=p^2q,&
qp&=q^3,&
p^2q&=pq^2,&
pq&=pq^2
\end{align*}
---assuming the 6 compositions $x\circ y\circ z$ \emph{etc.}\ are
independent in some example; and they are when $E=\Z^4$.
%, and $x$, $y$, and $z$ are appropriate transpositions of coordinates.
\end{proof}

\pagebreak

\end{comment}

A \textbf{differential field} is a pair
\begin{equation*}
(K,V),
\end{equation*}
where
\begin{enumerate}
\item 
$K$ is a field, %---in particular, a \textbf{commutative ring;}
\item
$V$ is both a subspace and a sub-ring of $\Der K$.
%\item
%$V$ is a sub-ring of $\Der K$.
\end{enumerate}

\begin{theorem}
If $(K,V)$ is a differential field, and $\dim_K(V)=m$, then $V$ has a basis 
\begin{equation*}
\{\partial_0,\dots,\partial_{m-1}\}, 
\end{equation*}
where in each case
\begin{equation*}
[\partial_i,\partial_j]=0.
\end{equation*}
\end{theorem}

The \emph{structures}
\begin{math}
(K,\partial_0,\dots,\partial_{m-1})
\end{math}
have a \textbf{theory,} which I  denote by
\begin{equation*}
\DF^m.
\end{equation*}

\begin{example}
%$\Bigl(\C(x_0,\dots,x_{m-1}),\displaystyle\frac{\partial}{\partial
%x_0},\dots,\frac{\partial}{\partial x_{m-1}}\Bigr)\models\DF^m$.
$\Bigl(\C(x_0,\dots,x_{m-1}),{\partial}/{\partial
x_0},\dots,{\partial}/{\partial x_{m-1}}\Bigr)\models\DF^m$.
\end{example}

\pagebreak

Let $\str A$ be a \textbf{structure} with underlying set $A$. 

(So $\str A$ might be a group, a differential field, an ordered set, \dots)

By introducing \emph{names} for all elements of $A$, we get the structure
\begin{equation*}
\str A_A.
\end{equation*}
The \textbf{diagram} of $\str A$ is the quantifier-free theory of $\str A_A$.

\begin{example}
The diagram of the field $\F$ is axiomatized by
\begin{gather*}
\begin{gathered}
0+0=0,\\0\cdot0=0,
\end{gathered}
\qquad\begin{gathered}
1+0=1,\\1\cdot0=0,
\end{gathered}
\qquad\begin{gathered}
0+1=1,\\0\cdot1=0,
\end{gathered}
\qquad\begin{gathered}
1+1=0,\\1\cdot1=1,
\end{gathered}\\
0\neq1.
\end{gather*}
This does \emph{not} entail field-theory.  

For example, it does not entail 
\begin{equation*}
\Forall x\Forall yx\cdot y=y\cdot x.
\end{equation*}
Neither does field-theory entail $1+1=0$.
\end{example}

\pagebreak

Let $\ACF$ be the theory of \textbf{algebraically closed fields,} such as $\C$.  That is, $\ACF$ has the field axioms, along with, for each positive integer $n$, the axiom
\begin{equation*}
\Forall{u_0}\dots\Forall{u_{n-1}}\Exists xu_0+u_1\cdot x+\cdots+u_{n-1}\cdot x^{n-1}+x^n=0.
\end{equation*}


\begin{theorem}
If $K$ is a field, then the theory
\begin{equation*}
\ACF\cup\diag K
\end{equation*}
is \textbf{complete}  (it entails either $\sigma$ or $\lnot\sigma$ for each $\sigma$\dots).
\end{theorem}

\begin{proof}
Use the \emph{\L o\'s--Vaught Test.} 

(This relies on \emph{G\"odel's Completeness Theorem.})
\begin{enumerate}
	\item 
	The theory $\ACF\cup\diag K$ has no finite models.
	\item
	by Steinitz, all algebraically closed fields that include $K$, but are of cardinality $(\lvert K\rvert+\aleph_0)^+$, are isomorphic over $K$.\qedhere
\end{enumerate}
\end{proof}

(G\"odel's \emph{Incompleteness} Theorem: a \emph{particular} theory\\
---namely $\Th{\N,+,\cdot,<}$---has no complete axiomatization.)

\pagebreak

\begin{definition}[A. Robinson]
A theory $T$ is
\textbf{model~complete} if, for all models $\str A$ of $T$, the theory
\begin{equation*}
T\cup\diag{\str A}
\end{equation*}
is complete, that is,
\begin{equation*}
T\cup\diag{\str A}\entails\Th{\str A_A}.
\end{equation*}
\end{definition}

\begin{examples}[A. Robinson]
\mbox{}
\begin{enumerate}
\item 
Torsion-free divisible abelian groups (\emph{i.e.}\ vector spaces over
$\Q$), 
\item
algebraically closed fields, such as $\C$ (by the last slide),
\item
real-closed fields, such as $\R$.
\end{enumerate}
\end{examples}

\begin{theorem}[A. Robinson]
  A theory $T$ is model complete if, for all models $\str A$ of $T$,
  \begin{equation*}
T\cup\diag{\str A}\entails\Th{\str A_A}_{\forall},
  \end{equation*}
  that is, if $\str A\included\str B$, and $\str B\models T$, then:
  
  every \textbf{system} over $\str A$ soluble in $\str B$ is soluble in $\str A$. 
\end{theorem}

\pagebreak

Let 
\begin{equation*}
\DF_0^m=\DF^m\cup\{%\overbrace{1+\cdots+1}^
p\neq0\colon p\text{ prime}\}.
\end{equation*}


\begin{theorem}[McGrail, 2000]
$\DF_0^m$ has a
\textbf{model companion,} $\DCF_0^m$: that is,
\begin{equation*}
  (\DF_0^m)_{\forall}=(\DCF_0^m)_{\forall}
\end{equation*}
%---a model of either theory embeds in a model of the other---, 
and $\DCF_0^m$ is model complete.
\end{theorem}

\begin{theorem}[Yaffe, 2001]
  The theory of fields of characteristic $0$ with $m$ derivations
  $D_i$, where
  \begin{equation*}
  [D_i,D_j]=\sum a^k_{i\,j}D_k,
  \end{equation*}
has a model companion.
\end{theorem}

\begin{theorem}[P, 2003; Singer, 2007]
  The latter follows readily from the former.
\end{theorem}

\begin{theorem}[P, submitted March, 2008]
$\DF^m$ has a
model companion, $\DCF^m$, given in terms of varieties.
\end{theorem}

\pagebreak

\parbox[t][390pt][t]{170pt}{\raggedright

If $(K,V)$ is a differential field,
what is the model theory of $V$?
}
\hfill
%\fbox{
\parbox[t][390pt][b]{370pt}{
  \includegraphics[width=370pt]
{mondrian-broadway-boogie-woogie.eps}

\Large Piet Mondrian, \emph{Broadway Boogie Woogie}}
%}

\pagebreak

\begin{theorem}
Let $(V,\cdot)$ be a Lie ring, and
\begin{equation*}
R=(\End V,\circ).
\end{equation*}
Then $(V,\cdot)$ acts on $R$ as a Lie ring of derivations.  

The action takes $D$ to the derivation
\begin{equation*}
f\mapsto Df
\end{equation*}
of $R$, where
\begin{equation*}
Df(x)=D\cdot(f(x))-f(D\cdot x).
\end{equation*}
That is,
\begin{equation*}
Df=[\lambda_D,f].
\end{equation*}
In short,
\begin{equation*}
D\mapsto\lambda_{\lambda_D}%[\lambda_D,{}\cdot{}]
\colon(V,\cdot)\to(\Der R,\brkt).
\end{equation*}
\end{theorem}

%Why does this work?

\pagebreak

\begin{comment}

\begin{proof}
We have already observed
\begin{equation*}
x\mapsto\lambda_x\colon(V,\cdot)\to(\End V,\brkt).
\end{equation*}
If $R=(\End V,\circ)$, we have also
\begin{equation*}
x\mapsto\lambda_x\colon(\End V,\brkt)\to(\Der R,\brkt).
\end{equation*}
Indeed, $\lambda_{[x,y]}=[\lambda_x,\lambda_y]$
%$x\mapsto\lambda_x$ is a homomorphism
by the \textbf{Jacobi identity,}
\begin{equation*}
[[x,y],f]=[x,[y,f]]-[y,[x,f]];
\end{equation*}
and $\lambda_x$ is a derivation of $R$ by the computation
\begin{align*}
  [x,f\circ g]
&=x\circ f\circ g\phantom{{}-f\circ x\circ g+f\circ x\circ g}-f\circ g\circ x\\
&=x\circ f\circ g-f\circ x\circ g+f\circ x\circ g-f\circ g\circ x\\
&=[x,f]\circ g\phantom{{}-f\circ x\circ g}+f\circ[x,g].\qedhere
\end{align*}
\end{proof}

\pagebreak

\end{comment}

Again $(V,\cdot)$ is a Lie ring, so it acts on $R$, namely $(\End V,\circ)$.

Let $t\in\End V$.  It may happen that $(\{Dt\colon D\in V\},\circ)$\\
---is a well-defined sub-ring of $R$, \\
---is closed under the action of $(V,\cdot)$, and \\
---is a field.

Then $V$ is a vector space over $K$, \\
and $(V,\cdot)$ acts on $K$ as a ring of derivations.

It may happen further that $V$ acts on $K$ as a \emph{space} of derivations:

That is, if $a,f\in K$ and $D\in V$, it may happen that
\begin{equation*}
a(D)f=a\circ(Df).
\end{equation*}

Then let $(V,\cdot,t)$ be called a \textbf{vector Lie ring.}

\begin{example}
If $(K,V)$ is a differential field, $t\in K$, and $Dt\neq0$ for some $D$ in~$V$, then $(V,\brkt,t)$ is a vector Lie ring, and 
\begin{equation*}
(\{Dt\colon D\in V\},\circ)=K.
\end{equation*}
\end{example}

\pagebreak

\begin{theorem}[P]
The class of $m$-dimensional vector Lie rings is elementary, with $\forall\exists$ axioms.  Its theory
has a model companion, whose models are those
 $(V,\cdot,t)$ such that, when we let
  \begin{equation*}
  K=(\{Dt\colon D\in V\},\circ), 
  \end{equation*}
then $V$ has a commuting basis
$(\partial_i\colon i<m)$ over $K$, and
  \begin{equation*}
    (K,\partial_0,\dots,\partial_{m-1})\models\DCF^m.
  \end{equation*}
Here $\dim_C(V)=\infty$, where $C$ is the constant field.
\end{theorem}


However, for an infinite field $K$, the theory of Lie algebras over
$K$ apparently has no model-companion (Macintyre, announced 1973).

Is there a model-complete theory of infinite-dimensional Lie algebras
with no extra structure? 


\pagebreak

%\fbox{
\parbox[t][400pt][b]{255pt}{
  \includegraphics[width=255pt]{gottlieb-centrifugal.eps}

{\Large Adolph Gottlieb, \emph{Centrifugal}}}
%}
\hfill
\parbox[t][400pt]{280pt}{
\mbox{}\vfill
We can also consider $(V,K)$ as a two-sorted structure.
\vfill
}

\pagebreak

A {vector space} can be understood model-theoretically as a triple
\begin{equation*}
(V,K,*), 
\end{equation*}
where
\begin{enumerate}
\item 
$V$ is an abelian group;
\item
$K$ is a field;
\item
$*$ is the \textbf{action} of $K$ on $V$, that is,
\begin{equation*}
(x,\vctr v)\mapsto\sca x{\vctr v}\colon K\times V\to V, 
\end{equation*}
and $\sca x{\vctr v}=\lambda_x(\vctr v)$, where
\begin{math}
x\mapsto\lambda_x\colon K\to(\End V,\circ)
\end{math}.
\end{enumerate}

Let the theory of vector spaces of dimension $n$ be
\begin{equation*}
  T_n,
\end{equation*}
where
$n\in\{1,2,3,\dots,\infty\}$.

\begin{theorem}[Kuzichev, 1992]
  $T_n$ admits elimination of quantified vector-variables.
\end{theorem}

\pagebreak

A theory is \textbf{inductive} if unions of chains of models are models.

\begin{theorem}[\L o\'s \&\ Suszko 1957, Chang 1959]
A theory $T$ is inductive if and only if
  \begin{equation*}
  T=T_{\forall\exists}.
  \end{equation*}
\end{theorem}

Hence all model complete theories have $\forall\exists$ axioms.

Of an arbitrary $T$, a model $\str A$ is \textbf{existentially closed}
if
\begin{equation*}
T\cup\diag{\str A}\entails\Th{\str A_A}_{\forall}.
 % \str A\included\str B\implies\str A\elsub_1\str B
\end{equation*}
%for all models $\str B$ of $T$.

\begin{theorem}[Eklof \&\ Sabbagh, 1970]
  Suppose $T$ is inductive. 
  \begin{enumerate}
  \item
   $T$ has a model companion if and
  only if the class of its existentially closed models is elementary.
  \item
  In this case, the theory of this class is the model companion.
\end{enumerate}
\end{theorem}

\pagebreak

Again, $T_n$ is the theory of vector spaces of dimension $n$.

If $n>1$, then no completion $T_n{}^*$ of $T_n$ can be model complete,
because it cannot be $\forall\exists$ axiomatizable.

For example, let 
\begin{gather*}
a_0=\vctr v^0=2,\qquad a_{s+1}=\vctr v^{s+1}=\sqrt{a_s}%=2^{1/2^{s+1}}
,\\
K_s=%\Q(a_0,\dots,a_s)=
\Q(a_s),\\
V_s=\lspan[K_s]{\vctr v^s,\dots,\vctr v^{s+n-1}}.
\end{gather*}
Then
\begin{equation*}
\sca{a_{s+1}}{\vctr v^{s+1}}=\vctr v^s,
\end{equation*}
so we have a chain
\begin{equation*}
(V_0,K_0)\included(V_1,K_1)\included\cdots
\end{equation*}
of models of $T_n$ whose union has dimension $1$.

\begin{comment}

There is a
chain 
\begin{equation*}
  (V_0,K_0)\included(V_1,K_1)\included\dotsb
  \included(V_s,K_s)\included\dotsb  
\end{equation*}
of models of $T_n{}^*$,
where 
\begin{enumerate}
\item 
$(V^{(s)},K^{(s)})$ has basis $(\vctr v_s,\dots,\vctr v_{s+n-1})$, but
\item
  $\vctr v_{s}=\sca{x_s}{\vctr v_{s+1}}$ for some $x_s$ in $K^{(s+1)}\setminus
    K^{(s)}$, so 
\item
the union of the chain has dimension $1$.
\end{enumerate}


\end{comment}
The situation changes if there are \emph{predicates} for linear dependence.
\pagebreak

Let $\VS_n$ (where $n$ is a positive integer) be the theory of
vector spaces with a new $n$-ary 
predicate $P^n$ for linear dependence.  So $P^n$ is defined by
\begin{equation*}
  \Exists{x^0}\dotsb\Exists{x^{n-1}}
  \Bigl(\sum_{i<n}\sca{x^i}{\vctr v_i}=0\land \bigvee_{i<n}x^i\neq0\Bigr).
\end{equation*}
Let $\VS_{\infty}$ be the union of the $\VS_n$.

\begin{theorem}[P]\mbox{}
  \begin{enumerate}
  \item 
  $\VS_n$ has a model companion, the theory of $n$-dimensional spaces
  over algebraically closed fields.
\item
$\VS_{\infty}$ has a model companion,
  the theory of infinite-dimensional spaces over algebraically closed
  fields. 
  \end{enumerate}
\end{theorem}

\pagebreak

\begin{proof}
Given a field-extension $L/K$, where
where 
\begin{equation*}
[L:K]\geq n+1, 
\end{equation*}
we can embed $(K^{n+1},K)$ in $(L^n,L)$,
\emph{as models of $\VS_n$,}  
under
\begin{equation*}
  \begin{pmatrix}
    x^0\\\vdots\\x^{n-1}\\x^n
  \end{pmatrix}
\mapsto
  \begin{pmatrix}
    x^0\\\vdots\\x^{n-1}
  \end{pmatrix}-x^n
   \begin{pmatrix}
    a^0\\\vdots\\a^{n-1}
  \end{pmatrix},
  \end{equation*}
where the $a^i$ are chosen from $L$ so that the tuple
\begin{equation*}
(a^0,\dots,a^{n-1},1)
\end{equation*}
is linearly independent over $K$.
\end{proof}

%The key is lowering dimension to $n$.

\begin{comment}

Given a field-extension $L/K$, where
where 
\begin{equation*}
[L:K]\geq n+1, 
\end{equation*}
we can embed $(K^{n+1},K)$ in $(L^n,L)$,
\emph{as models of $\VS_n$,}  
under
\begin{equation*}
  \begin{pmatrix}
    x^0\\\vdots\\x^{n-1}\\x^n
  \end{pmatrix}
\mapsto
\begin{pmatrix}
   1&      &0&-a^0\\
    &\ddots& &\vdots\\
   0&      &1&-a^{n-1}
\end{pmatrix}
  \begin{pmatrix}
    x^0\\\vdots\\x^{n-1}\\x^n
  \end{pmatrix},
\end{equation*}
that is,
\begin{equation*}
  \tuple x\mapsto\left(
  \begin{array}{c|c}
    I_n&-\tuple a
  \end{array}
  \right)
\tuple x,
\end{equation*}
where the $a^i$ are chosen from $L$ so that the tuple
\begin{equation*}
(a^0,\dots,a^{n-1},1)
\end{equation*}
is linearly independent over $K$.

\end{comment}

%from $K^{n+1}$ to $L^n$, along with the inclusion of $K$ in $L$.


\pagebreak

\begin{comment}

Why?  Given an $(n+1)\times n$ matrix $U$ over $K$, we want to show
\begin{equation*}
  \rank U=n\iff\det\left(\left(
  \begin{array}{c|c}
    I&-\tuple a
  \end{array}
  \right)
U\right)\neq0.
\end{equation*}
Write $U$ as
$\left(\begin{array}{c}
  X\\\hline\tuple y\transp
\end{array}\right)$. 
Then
\begin{equation*}
  \rank U=n\iff
\det\left(\begin{array}{c|c}
  X&\tuple a\\\hline
\tuple y\transp&1
\end{array}\right)\neq0.
\end{equation*}
Moreover,
\begin{align*}
\det\left(\begin{array}{c|c}
  X&\tuple a\\\hline
\tuple y\transp&1
\end{array}\right)
=
\det(&X-\tuple a\tuple y\transp),\\
&X-\tuple a\tuple y\transp
=
\left(
  \begin{array}{c|c}
    I&-\tuple a
  \end{array}
  \right)
\left(\begin{array}{c}
  X\\\hline\tuple y\transp
\end{array}\right)
=
\left(
  \begin{array}{c|c}
    I&-\tuple a
  \end{array}
  \right)
U.
\end{align*}
That does it.

\pagebreak

\end{comment}

Compare:  

Let $T$ be the theory of fields with an algebraically
closed subfield.

The existentially closed models of $T$ have
transcendence-degree~$1$, because of

\begin{theorem}[A. Robinson]
We have an inclusion
\begin{equation*}
  K(x,y)\included L(y)
\end{equation*}
of pure transcendental extensions, where
\begin{equation*}
  K(x,y)\cap L=K,
\end{equation*}
provided
\begin{equation*}
  L=K(\alpha,\beta),
\end{equation*}
where 
\begin{align*}
\alpha&\notin K(x,y)\alg,&
\beta&=  \alpha x+y.
\end{align*}
\end{theorem}

(Hence $T$ has no model companion.)

\pagebreak

A \textbf{Lie--Rinehart pair} is a quadruple $(V,K,\der{}{},\sca{}{})$, where
\begin{asparaenum}
\item 
$V$ and $K$ are abelian groups, 
\item
$D$ is an action of $V$ on $K$; and $*$, of $K$ on $V$; so
%\begin{align*}
%(\vctr v,x)\mapsto\der{\vctr v}x&\colon V\times K\to K,&
%(x,\vctr v)\mapsto\sca x{\vctr v}&\colon K\times V\to V,
%\end{align*}
%and
\begin{equation*}
\begin{gathered}
\der{(\vctr u+\vctr v)}x=\der{\vctr u}x+\der{\vctr v}x,\\
\der{\vctr v}{(x+y)}=\der{\vctr v}x+\der{\vctr v}y,
\end{gathered}
\qquad
\begin{gathered}
\sca{(x+y)}{\vctr v}=\sca x{\vctr v}+\sca y{\vctr v},\\
\sca x{(\vctr u+\vctr v)}=\sca x{\vctr u}+\sca x{\vctr v};
\end{gathered}
\end{equation*}
\item
The actions are faithful:
\begin{align*}
  \Exists x(\der{\vctr v}x=0&\lto\vctr v=0),&
  \Exists{\vctr v}(\sca x{\vctr v}=0&\lto x=0);
\end{align*}
\item
if $\vctr u,\vctr v\in V$, there is a unique element $[\vctr u,\vctr v]$ of $V$ such that
%for all $x$ in $K$,
\begin{equation*}
  \der{[\vctr u,\vctr v]}x=\der{\vctr u}{(\der{\vctr v}x)}-\der{\vctr v}{(\der{\vctr u}x)},
  \end{equation*}
%  and then
  \begin{equation*}
  {\sca{(\der{\vctr u}x)}{\vctr v}}
={[\vctr u,\sca x{\vctr v}]}-{\sca x{[\vctr u,\vctr v]}};
  \end{equation*}
  \item
  if $x,y\in K$, there is a unique element $x\cdot y$ of $K$ such that
  %for all $\vctr v$ in $V$,
  \begin{equation*}
\sca{(x\cdot y)}{\vctr v}=\sca x{(\sca y{\vctr v})},
\end{equation*}
%and then
\begin{gather*}
\der{(\sca x{\vctr v})}y=x\cdot(\der{\vctr v}y).
\end{gather*}
\end{asparaenum}

\pagebreak
Assuming $(V,K,\der{}{},\sca{}{})$ is a Lie--Rinehart pair, one shows
that $V$ does act on $K$ as a Lie ring \textbf{of
  derivations:}
\begin{equation*}
  \der{\vctr v}{(x\cdot y)}=(\der{\vctr v}x)\cdot y+x\cdot(\der{\vctr v}y).
\end{equation*}
\begin{comment}



Indeed,
  \begin{align*}
&\phantom{{}={}}  \sca{(\der {\vctr v}{(x\cdot y)})}{\vctr u}\\
&=[{\vctr v},\sca{(x\cdot y)}{\vctr u}]-\sca{(x\cdot y)}{[{\vctr v},{\vctr u}]}\\
&=  [{\vctr v},\sca x{(\sca y{\vctr u})}]-\sca x{(\sca y{[{\vctr v},{\vctr u}]})}\\
&  \begin{aligned}=
\sca{(\der {\vctr v}x)}{(\sca y{\vctr u})}  &+\sca x{[{\vctr v},\sca y{\vctr u}]}\\
                            &-\sca x{[{\vctr v},\sca y{\vctr u}]}+\sca x{(\sca{(\der
                              {\vctr v}y)}{\vctr u})}
  \end{aligned}\\
&=\sca{(\der {\vctr v}x)}{(\sca y{\vctr u})}+\sca x{(\sca{(\der {\vctr v}y)}{\vctr u})}\\
&=\sca{((\der {\vctr v}x)\cdot y)}{\vctr u}+\sca{(x\cdot(\der {\vctr v}y))}{\vctr u}\\
&=\sca{((\der {\vctr v}x)\cdot y+x\cdot(\der {\vctr v}y))}{\vctr u}.
  \end{align*}
  
  
\end{comment}
Let the theory of those Lie--Rinehart pairs
$(V,K,\der{}{},\sca{}{})$ in which $(K,\cdot)$ is a field be denoted by
\begin{equation*}
\LR{}{}.
\end{equation*}
In this case, $(K,V)$ is a differential field.

The theory $\LR{}{}$ is not inductive.
However, let the theory of those models $(V,K,\der{},\sca{}{})$ of $\LR{}{}$ in which
\begin{equation*}
\dim_K(V)\leq m
\end{equation*}
be denoted by
\begin{equation*}
\LR m{}.
\end{equation*}
Then $\LR m{}$ is inductive and companionable.


\pagebreak

Let $T$ be the theory of pairs $(V,K)$, where $K$ is a field, $\Char K=0$, and $V$ acts on $K$ as a space of derivations.

Let $\DCF_0^{(m)}$ be the model-companion of the theory of fields of
characteristic $0$ with
$m$ derivations \emph{with no required interaction.}

\begin{theorem}[\"Ozcan Kasal]
The
existentially closed models of $T$ are just those models $(V,K)$ such that
\begin{enumerate}
\item 
$\trdeg{K/\Q}=\infty$;
\item
$(K,\vctr v_0,\dots,\vctr v_{m-1})\models\DCF^{(m)}_0$ whenever
  $(\vctr v_0,\dots,\vctr v_{m-1})$ is linearly independent over $K$;
\item
if $(x^0,\dots,x^{n-1})$ is algebraically independent, and
$(y^0,\dots,y^{n-1})$ is arbitrary, then 
for some $\vctr v$ in $V$,
\begin{equation*}
  \bigwedge_{i<n}\der{\vctr v}{x^i}=y^i.
\end{equation*}
\end{enumerate}
\end{theorem}

These are not first-order conditions: they require the constant field
to be $\Q\alg$.

\pagebreak

The picture changes when (for each $n$) a predicate $Q_n$ is
introduced for the $n$-ary relation on scalars defined by
\begin{equation*}
%Q^n(x^0,\dots,x^{n-1})\iff
\bigvee_{i<n}\Forall{\vctr v}\Bigl(\bigwedge_{j\neq i}\der{\vctr v}{x^j}=0\lto \der
{\vctr v}{x^i}=0\Bigr). 
\end{equation*}
Let the new theory be
\begin{equation*}
  T',
\end{equation*}
so this entails
\begin{equation*}
  %\Forall{(x^0,\dots,x^{n-1})}
  \lnot Q_nx^0\cdots x^{n-1}
  \Leftrightarrow\Exists{(\vctr v_0,\dots,\vctr v_{n-1})}
   \bigwedge_{\substack{i<n\\j<n}}\der{\vctr v_i}{x^j}=\delta_i^j.
\end{equation*}
Say $(a^0,\dots,a^{n-1})$ from $K$ is
\textbf{$D$-dependent} if
\begin{equation*}
(V,K)\models Q_na^0\cdots a^{n-1}.
\end{equation*}
So algebraic dependence implies
$D$-dependence. 

Also, $D$-dependence also makes $K$ a pregeometry.

\pagebreak

\begin{theorem}[\"Ozcan Kasal]
The existentially closed models of $T'$ are those $(V,K)$ such that
$\Ddim K=\infty$ and
whenever 
\begin{enumerate}
\item
$U$ is quasi-affine over $\Q(a^0,\dots,a^{k-1},\vec b)$ with a generic
  point
  \begin{equation*}
    (x^0,\dots,x^{\ell+m-1},\vec y),
  \end{equation*}
where $\vec x$ is algebraically independent over
$\Q(\vec a,\vec b)$,
\item
$(\vctr v_0,\dots,\vctr v_{k+\ell-1})$ is linearly independent,
\item 
$\left(\begin{array}{c|c}I_k&0\end{array}\right)=(\der{\vctr v_j}{a^i})^{i<k}_{j<k+\ell}$,
\item
$\left(\begin{array}{c|c}
F&I_{\ell}\\\hline G&H
\end{array}\right)$ 
is
$(\ell+m)\times(k+\ell)$ with entries from $\Q(\vec a,\vec b)[U]$,
\end{enumerate}
then $U$ contains $(\vec c,\vec d)$ such that
\begin{enumerate}
\item
$\left(\begin{array}{c|c}
F&I_{\ell}\\\hline G&H
\end{array}\right)(\vec c,\vec d)=(\der{\vctr v_j}{c^i})^{i<\ell+m}_{j<k+\ell}$,
\item
$\Ddim{c^{\ell},\dots,c^{\ell+m-1},\vec d/\vec a,c^0,\dots,c^{\ell-1}}=0$.
\end{enumerate}
\end{theorem}

\pagebreak

\begin{comment}

\begin{theorem}[\"Ozcan Kasal]
The existentially closed models of $T'$ are those $(V,K)$ such that
$\Ddim K=\infty$ and
whenever 
\begin{enumerate}
\item
$U$ is quasi-affine over $\Q(a^0,\dots,a^{k-1},\vec b)$ with a generic
  point
  \begin{equation*}
    (x^0,\dots,x^{\ell+m-1},\tuple z),
  \end{equation*}
where $\vec x$ is algebraically independent over
$\Q(\vec a,\vec b)$,
\item
$g_i^j\in\Q(\vec a,\vec b)[U]$, where $i<k+\ell$ and $j<\ell+m$,
\item
$(\vctr v_0,\dots,\vctr v_{k+\ell-1})$ is linearly independent,
\item
$\displaystyle
\bigwedge_{\substack{i<k+\ell\\j<k}}\der{\vctr v_i}{a^j}
=\delta_i^j$ 
and
$\displaystyle\bigwedge_{\substack{i<\ell\\j<\ell}}g_{k+i}^j=\delta_i^j$;
\end{enumerate}
then $U$ contains $(\vec c,\vec d)$ such that
\begin{enumerate}
\item
$\text{$D$-dim}(c^{\ell},\dots,c^{\ell+m-1},\vec d/\vec a,c^0,\dots,c^{\ell-1})=0$,
\item
$\displaystyle
\bigwedge_{\substack{i<k+\ell\\j<\ell+m}}\der{v_i}{a^j}
=g_i^j(\vec c,\vec d)$.
\end{enumerate}
\end{theorem}

\pagebreak

\end{comment}

\begin{comment}

\begin{theorem}[\"Ozcan Kasal]
The existentially closed models of $T'$ are those $(V,K)$ such that
$\Ddim K=\infty$ and
whenever 
\begin{enumerate}
\item
$(v_0,\dots,v_{k+\ell-1})$ is linearly independent,
and
\begin{equation*}
\displaystyle
\bigwedge_{\substack{i<k+\ell\\j<k}}\der{v_i}{a^j}
=\delta_i^j,
\end{equation*}
\item
$U$ is a quasi-affine variety over $\Q(\tuple a,\tuple b)$ with a generic
  point
  \begin{equation*}
    (x^0,\dots,x^{\ell-1},y^0,\dots,y^{m-1},\tuple z),
  \end{equation*}
where $(\tuple x,\tuple y)$ is algebraically independent over
$\Q(\tuple a,\tuple b)$,
\item
$g_i^j\in\Q(\tuple a,\tuple b)[U]$, where $i<k+\ell$ and $j<m$;
\end{enumerate}
then $U$ contains $(a^k,\dots,a^{k+\ell-1},\tuple c,\tuple d)$ such that
\begin{enumerate}
\item
each $c^j$ and $d^j$ is $D$-dependent on $(a^0,\dots,a^{k+\ell-1})$,
\item
$\displaystyle
\bigwedge_{\substack{i<k+\ell\\j<k+\ell}}\der{v_i}{a^j}
=\delta_i^j\land
\bigwedge_{\substack{i<k+\ell\\j<m}}\der{v_i}{c^j}
=g_i^j(a^k,\dots,a^{k+\ell-1},\tuple c,\tuple d)$.
\end{enumerate}
\end{theorem}


\pagebreak

\end{comment}

\renewcommand{\thepage}{{\huge\textsc{fin}}}
\includegraphics[height=375pt]{kline-palladio.eps}

{\Large Franz Kline, \emph{Palladio}}

%\layout
\end{document}
