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

\theoremstyle{definition}
\newtheorem*{theorem}{Theorem}
\newtheorem*{corollary}{Corollary}
\newtheorem*{Qu}{Question}
\theoremstyle{definition}
\newtheorem*{A}{Answer}

\newcommand{\Sym}[1]{\operatorname{Sym}(#1)}
\newcommand{\id}[1]{\operatorname{id}_{#1}}
\newcommand{\inv}{^{-1}}
\newcommand{\pow}[1]{\mathscr{P}(#1)}
\renewcommand{\emptyset}{\varnothing}
\newcommand{\symdiff}{\vartriangle}
\newcommand{\stnd}[1]{\mathbb{#1}}
\newcommand{\Z}{\stnd Z}
\newcommand{\Q}{\stnd Q}
\newcommand{\R}{\stnd R}
\newcommand{\C}{\stnd C}
\newcommand{\Ham}{\stnd H}
\newcommand{\Oct}{\stnd O}
\newcommand{\Sed}{\stnd S}
\newcommand{\vnn}{\upomega}
\newcommand{\Aff}{\stnd A}

\newcommand{\Hom}[1]{\operatorname{Hom}(#1)}
\newcommand{\End}[1]{\operatorname{End}(#1)}
\newcommand{\Mult}[1]{\operatorname{Mult}(#1)}
\newcommand{\mult}{\mathsf m}
\newcommand{\cmpt}{\mathsf c}
%\newcommand{\invo}[1]{\breve{#1}}
\newcommand{\invo}[1]{\accentset{\bullet}{#1}}
\newcommand{\Der}[1]{\operatorname{Der}(#1)}
\newcommand{\der}[2]{#1\mathop D#2}
\newcommand{\sca}[2]{#2*#1}
%\newcommand{\pdual}[1]{#1\sphat{}} % needs amsxtra
%\newcommand{\pdual}[1]{\widehat{#1}}

\newcommand{\class}[1]{\mathbf{#1}}

\newcommand{\DF}{\operatorname{DF}}
\newcommand{\DCF}{\operatorname{DCF}}
\newcommand{\VL}{\operatorname{VL}}
\newcommand{\VS}{\operatorname{VS}}

\newcommand{\lspan}[2]{\operatorname{span}_{#1}(#2)}
\newcommand{\Th}[1]{\operatorname{Th}(#1)}
%\newcommand{\conj}[1]{\overset{\ast}{#1}}
%\newcommand{\conj}[1]{\bar{#1}}
\newcommand{\conj}[1]{\accentset{\rule[1pt]{7pt}{1pt}}{#1}}
%\newcommand{\conj}[1]{#1^*}
%\newcommand{\conjlong}[1]{\overline{#1}}
\newcommand{\Mat}[2]{\operatorname{M}_{#1}(#2)}
\newcommand{\included}{\subseteq}
\newcommand{\pincluded}{\subset}
\renewcommand{\setminus}{\smallsetminus}
\newcommand{\Exists}[1]{\exists{#1}\;}
\newcommand{\Forall}[1]{\forall{#1}\;}
\newcommand{\lto}{\Rightarrow}
\newcommand{\trdeg}[1]{\operatorname{tr-deg}(#1)}
\newcommand{\alg}{^{\operatorname{alg}}}
\newcommand{\Ddim}[1]{D\operatorname{-dim}(#1)}
\newcommand{\tuple}[1]{\bm{#1}}
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\newcommand{\str}[1]{\mathfrak{#1}}
\newcommand{\diag}[1]{\operatorname{diag}(#1)}
\newcommand{\elsub}{\preccurlyeq}
\newcommand{\transp}{^{\operatorname{t}}}
\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}
\newcommand{\Stone}[1]{\operatorname{S}(#1)}
\newcommand{\rank}[1]{\operatorname{rank}(#1)}
\newcommand{\entails}{\vdash}

\raggedright

\usepackage{graphicx}
\usepackage{layout}

\setcounter{page}{0}
\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
July, 2009
\vfill
Lyon
\vfill
  \end{center}
  %  \begin{tabular}[t]{c}D\\A\\V\\I\\D\\ \\P\\I\\E\\R\\C\\E\end{tabular}
}

\pagebreak

The interacting rings in question arise from
\textbf{differential fields:} %structures
\begin{equation*}
  (K,\partial_0,\dots,\partial_{m-1}),
\end{equation*}
where
\begin{enumerate}
\item 
$K$ is a field---in particular, a \textbf{commutative ring;}
\item
each $\partial_i$ is a \textbf{derivation} of $K$: an endomorphism $D$ of
the abelian group of $K$ that obeys the \textbf{Leibniz rule,}
\begin{equation*}
D(x\cdot y)=D(x)\cdot y+x\cdot D(y);
\end{equation*}
\item
$[\partial_i,\partial_j]=0$ in each case, where $[{}\cdot{},{}\cdot{}]$ is the
  \textbf{Lie bracket,} so 
  \begin{equation*}
    [x,y]=x\circ y-y\circ x.
  \end{equation*}
%(the \textbf{Lie bracket} of $x$ and $y$).
\end{enumerate}
A standard example is 
\begin{math}
(\C(x_0,\dots,x_{m-1}),\frac{\partial}{\partial
x_0},\dots,\frac{\partial}{\partial x_{m-1}})
\end{math}.

In general, let
\begin{equation*}
  V=\lspan K{\partial_i\colon i<m}\included\Der K;
\end{equation*}
then $V$ is also a \textbf{Lie ring.}  

\pagebreak

Recall some notions due to Abraham Robinson:

The \emph{quantifier-free} theory of $\str A_A$ is denoted by
\begin{equation*}
  \diag{\str A}.
\end{equation*}
A theory $T$ is
\textbf{model~complete} under any of three equivalent conditions:
\begin{enumerate}
\item 
whenever $\str A$ is a \textbf{model} of $T$, the theory 
\begin{equation*}
T\cup\diag{\str A}
\end{equation*}
is \textbf{complete;}  
\item
whenever $\str A\models T$,
\begin{equation*}
T\cup\diag{\str A}\entails\Th{\str A_A};
\end{equation*}
\item
whenever $\str A,\str B\models T$,
\begin{equation*}
  \str A\included\str B\implies\str A\elsub\str B.
\end{equation*}
\end{enumerate}
Then $T$ is complete if all models have a common submodel.

\pagebreak


Robinson's examples of model complete theories include the theories of
\begin{enumerate}
\item 
torsion-free divisible abelian groups (\emph{i.e.}\ vector spaces over
$\Q$), 
\item
algebraically closed fields,
\item
real-closed fields.
\end{enumerate}

\begin{theorem}[Robinson]
  $T$ is model complete, provided
  \begin{equation*}
T\cup\diag{\str A}\entails\Th{\str A_A}_{\forall}
  \end{equation*}
whenever $\str A\models T$,
that is,
\begin{equation*}
\str A\included\str B\implies\str A\elsub_1\str B
\end{equation*}
whenever $\str A,\str B\models T$.
\end{theorem}

\begin{proof}
If $\str A\elsub_1\str B$, then
$\str A\elsub\str C$ for some $\str C$, where $\str B\included\str C$;
then $\str B\elsub_1\str C$, so continue:
  \begin{math}
    \xymatrix@!{
\str A\ar[rr]^{\elsub}\ar[dr]|{\elsub_1}&&\str
C\ar[dr]|{\elsub_1}\ar[rr]^{\elsub}&&\str
E\ar[dr]|{\elsub_1}\ar[rr]^{\elsub}&&\\ 
&\str B\ar[ur]|{\elsub_1}\ar[rr]_{\elsub}&&\str
D\ar[ur]|{\elsub_1}\ar[rr]_{\elsub}&&\str F\ar[r]_{\elsub}&
}
  \end{math}
\end{proof}



\pagebreak

Let 
\begin{align*}
\DF^m&=\Th{\{\text{fields with $m$ commuting derivations}\}},\\
\DF_0^m&=\DF^m\cup\{p\neq0\colon p\text{ prime}\}.
\end{align*}


\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}

\begin{comment}

\begin{theorem}[P, submitted March, 2008]
The theory $\DF^m$ of fields (of unspecified characteristic) with $m$
commuting derivations has a
\textbf{model companion,} $\DCF^m$: that is,
\begin{equation*}
  \DF^m{}_{\forall}=\DCF^m{}_{\forall}
\end{equation*}
---a model of either theory embeds in a model of the other---, and
$\DCF^m$ is model complete.
\end{theorem}

McGrail~(2000), Yaffe~(2001): the characteristic $0$ case.

If $(K,\partial_0,\dots,\partial_{m-1})\models\DF^m$, and
\begin{equation*}
\lspan K{\partial_i\colon i<m}=V=\lspan K{D_i\colon i<m},
\end{equation*}
then (in characteristic $0$) Yaffe gives a model companion of
\begin{equation*}
\Th{K,D_0,\dots,D_{m-1}}.
\end{equation*}
P~(2003), Singer~(2007):
this can be got from $\DCF^m$.

%What if $V$ is treated as a \textbf{sort,} like $K$?

\end{comment}

\pagebreak

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


What is the model theory of $V$?
\vfill
First consider rings in general.
}
\hfill
%\fbox{
\parbox[t][390pt][b]{370pt}{
  \includegraphics[width=370pt]
{mondrian-broadway-boogie-woogie.eps}

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


\pagebreak

In the most general sense, a \textbf{ring} is a structure
\begin{equation*}
  (E,\cdot),
\end{equation*}
where 
\begin{enumerate}
\item 
$E$ is an abelian group in $\{0,-,+\}$, and 
\item
the binary operation
$\cdot$ distributes over $+$ in both senses: it is a
\textbf{multiplication.} 
\end{enumerate}

Beyond this, there are axioms for:
\begin{align*}
  &\begin{gathered}
    \text{\textbf{commutative rings}}\\
xy-yx=0\\
(xy)z=x(yz)
  \end{gathered}
&&
  \begin{gathered}
\text{\textbf{Lie rings}}\\
    x^2=0\\
(xy)z=x(yz)-y(xz)
  \end{gathered}
\end{align*}

By itself, $(xy)z=x(yz)$ defines \textbf{associative rings;} 

and 
$(xy)z=x(yz)-y(xz)$ is the \textbf{Jacobi identity.}


\pagebreak

For rings, are there \textbf{representation theorems} like the following?

\begin{theorem}[Cayley]
  Every abstract group $(G,1,{}\inv,{}\cdot{})$ embeds in the symmetry group
  \begin{equation*}
  (\Sym{G},\id{G},{}\inv,{}\circ{})
  \end{equation*}
under $x\mapsto\lambda_x$, where
\begin{equation*}
  \lambda_g(y)=g\cdot y.
\end{equation*}
\end{theorem}

A ring is \textbf{Boolean} if it satisfies $x^2=x$.

\begin{theorem}[Stone]
Every abstract Boolean ring $(R,0,+,\cdot)$ or $\str R$ embeds in a
Boolean ring of sets
  \begin{equation*}
(\pow{\Omega},\emptyset,\symdiff,\cap).
  \end{equation*}
(Here $\Omega=\{\text{prime ideals of $\str R$}\}$, and the
  embedding is $x\mapsto\{\mathfrak p\colon x\notin\mathfrak p\}$.)
\end{theorem}

For associative rings and Lie rings \emph{only,} there are such
theorems. 

\pagebreak

I know no representation theorem for \textbf{abelian groups.}
There are just \textbf{`prototypical'} abelian groups, like $\Z$.
One might mention \emph{Pontryagin duality:}  Every (topological) abelian group
$G$ embeds in 
%$\pdual{\pdual G{}}$,
$G^*{}^*$,
where %$\pdual G=\Hom{G,\R/\Z}$.
$G^*=\Hom{G,\R/\Z}$.



%\pagebreak

\begin{comment}
  

\begin{Qu}
  What is a \textbf{ring?}
\end{Qu}


\end{comment}

Prototypical \textbf{associative rings} include

\begin{enumerate}
\item 
 $\Z$, $\Q$, $\R$, $\C$, and $\Ham$;
\item
matrix rings.
\end{enumerate}
But there are \textbf{non-associative rings:}

\begin{enumerate}
\item
$(\R^3,\times)$ is a \textbf{Lie ring} (in fact, the \emph{Lie algebra} of
  $\operatorname{SO}(3,\R)$);

\item 
the \textbf{Cayley--Dickson algebras} $\R$, $\R'$, \dots
become non-associative after $\R''$ (which is $\Ham$):

\end{enumerate}


\pagebreak

\begin{comment}
  

In the Cayley--Dickson construction,
\begin{equation*}
(\R,\C,\Ham,\Oct,\Sed,\dots)=(\R,\R',\R'',\R''',\R'''',\dots):
\end{equation*}

\end{comment}
Let $(E,\cdot)$ be a ring with
 an \emph{involutive
anti-automorphism} or \textbf{conjugation} $x\mapsto\conj x$.
The abelian group $\Mat2E$ is a ring under
  \begin{equation*}
  \begin{pmatrix}
    a&b\\
    c&d
  \end{pmatrix}
  \begin{pmatrix}
    x&y\\
    z&w
  \end{pmatrix}
=
\begin{pmatrix}
  ax+zb&ya+bw\\
  xc+dz&cy+wd
\end{pmatrix},
\end{equation*}
with conjugation
\begin{equation*}
    \begin{pmatrix}
  x&y\\z&w
    \end{pmatrix}
\mapsto
  \begin{pmatrix}
    \conj x&\conj z\\\conj y&\conj w
  \end{pmatrix}.
\end{equation*}
Let $E'$ comprise the matrices
  \begin{equation*}
    \begin{pmatrix}
      x&y\\-\conj y&\conj x
    \end{pmatrix}.
  \end{equation*}
Then $E'$ is closed under the operations, and $E$ embeds under
\begin{equation*}
  x\mapsto
  \begin{pmatrix}
    x&0\\0&\conj x
  \end{pmatrix}.
%\colon(E,\cdot)\rightarrowtail(\Mat2E,\cdot)
\end{equation*}

\begin{comment}

\pagebreak

If $(E,\cdot)$ is a ring, let
\begin{enumerate}
\item 
$\End E$ be the \emph{abelian group} of endomorphisms of the abelian
  group $E$,
\item
$x\mapsto\lambda_x\colon E\to\End E$, where 
  \begin{equation*}
  \lambda_x(y)=xy.
  \end{equation*}
\end{enumerate}


\begin{Qu}
  What is an \textbf{associative ring?}
\end{Qu}

\begin{A}
  A ring $(E,\cdot)$ is associative if and only if $x\mapsto\lambda_x$
  is a ring homomorphism from $(E,\cdot)$ to $(\End E,\circ)$, that
  is,
  \begin{equation*}
    (xy)z=\lambda_{xy}(z)
=\lambda_x\circ\lambda_y(z)=
x(yz).
  \end{equation*}
But $x\mapsto\lambda_x$ might not be an embedding, unless $(E,\cdot)$
has a unit.
\end{A}

\end{comment}

\pagebreak

If $E$ is an abelian group, then its multiplications compose an
abelian group
%\begin{equation*}
%  \Mult E,
%\end{equation*}
that has an involutory automorphism,
\begin{equation*}
  \mult\mapsto\invo{\mult},
\end{equation*}
where $\invo{\mult}$ is the \textbf{opposite} of $\mult$:
\begin{equation*}
  \invo{\mult}(x,y)=\mult(y,x).
\end{equation*}
Let $\End E$ be the \emph{abelian group} of endomorphisms of $E$.
Then
\begin{enumerate}
\item 
$(\End E,\circ)$ is an associative ring;
\item
$(\End E,\circ-\invo{\circ})$ is a Lie ring;
\item
$(\End E,\circ+\invo{\circ})$ is a \textbf{Jordan
  ring:}\footnote{\Large Pascual Jordan, 1902--80.} a ring satisfying
  \begin{align*}
  xy&=yx,&(xy)x^2&=x(yx^2).
  \end{align*}
\end{enumerate}

\pagebreak

If $(E,\cdot)$ is a ring, let 
\begin{equation*}
x\mapsto\lambda_x\colon E\rightarrow\End E,
\end{equation*}
where (as in the Cayley Theorem)
\begin{equation*}
  \lambda_a(y)=a\cdot y.
\end{equation*}
If $p$ and $q$ are in $\Z$, let $(E,\cdot)$ be called a
\textbf{$(p,q)$-ring} if
\begin{equation*}
x\mapsto\lambda_x\colon
(E,\cdot)\rightarrow(\End
E,p{\circ}-q\invo{\circ}).
\end{equation*}

\begin{theorem}
All associative rings are $(1,0)$-rings; all Lie rings are
$(1,1)$-rings.  In particular,
$(\End E,  p{\circ}-q\invo{\circ})$ is a $(p,q)$-ring if
\begin{equation*}
  (p,q)\in\{(0,0),(1,0),(1,1)\}.
\end{equation*}
\end{theorem}

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

\pagebreak

\begin{proof}%[Proof of theorem.]
We have 
\begin{equation*}
x\mapsto\lambda_x:(\End
E,p{\circ}-q\invo{\circ})\to(\End{\End E},p{\circ}-q\invo{\circ})
\end{equation*}
if and only if 
\begin{equation*}
\lambda_{xy}=\lambda_x\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

If $(V,\cdot)$ is a Lie ring, then each $\lambda_x$ is a
\textbf{derivation} of it:  Write the Jacobi identity
%  $(xy)z=x(yz)-y(xz)$,
as
\begin{equation*}
  x(yz)=(xy)z+y(xz);
\end{equation*}
this means
\begin{equation*}
  \lambda_x(yz)=\lambda_x(y)\cdot z+y\cdot\lambda_x(z). 
\end{equation*}
Thus $\lambda$ factors:% through $(\Der{V,\cdot},\circ-\invo{\circ})$.

  \begin{equation*}
    \xymatrix@%
%=5ex{
!{
(V,\cdot)\ar[r]^(.4){\lambda} \ar[dr]_(.4){\lambda}& 
(\Der{V,\cdot},\circ-\invo{\circ})\ar[d]^{\included}\\
&(\End V,\circ-\invo{\circ})
}
  \end{equation*}

\pagebreak
For any abelian group $V$, the Lie ring $(\End V,\circ-\invo{\circ})$
acts as a ring of derivations of the \textbf{associative} ring $(\End
V,\circ)$: 
\begin{align*}
  [z,x\circ y]
&=z\circ x\circ y\phantom{{}-x\circ z\circ y+x\circ z\circ y}-x\circ y\circ z\\
&=z\circ x\circ y-x\circ z\circ y+x\circ z\circ y-x\circ y\circ z\\
&=[z,x]\circ y\phantom{{}-x\circ z\circ y}+x\circ[z,y].
\end{align*}

  \begin{equation*}
\xymatrix@%
%=5ex{
R=6cm{
(\End V,\circ-\invo{\circ})\ar[r]^(.4){\lambda}
  \ar[dr]_(.4){\lambda}&  
(\Der{\End V,\circ},\circ-\invo{\circ})\ar[d]^{\included}\\
&(\End{\End V},\circ-\invo{\circ})
}
  \end{equation*}

\pagebreak

Combine the diagrams---again, $(V,\cdot)$ is a Lie ring:

  \begin{equation*}
\xymatrix@R=1cm@C=2cm{
(V,\cdot)\ar[r]^(.4){\lambda} \ar[dr]_(.4){\lambda}& 
(\Der{V,\cdot},\circ-\invo{\circ})\ar[d]^{\included}&\\
&(\End V,\circ-\invo{\circ})\ar[r]^(.4){\lambda}
  \ar[dr]_(.4){\lambda}&  
(\Der{\End V,\circ},\circ-\invo{\circ})\ar[d]^{\included}\\
&&(\End{\End V},\circ-\invo{\circ})
}
  \end{equation*}

%Thus $(V,\cdot)$ acts as a ring of derivations of $(\End V,\circ)$:  
Each $D$ in $V$ determines the derivation 
\begin{equation*}
f\mapsto Df
\end{equation*}
of
$(\End V,\circ)$, where 
\begin{equation*}
Df=\lambda_{\lambda_D}(f)=[\lambda_D,f], 
\end{equation*}
so that
\begin{equation*}
Df(x)=D\cdot(f(x))-f(D\cdot x).
\end{equation*}

\pagebreak

If $(K,\partial_0,\dots,\partial_{m-1})\models\DF^m$, and $V=\lspan
K{\partial_i\colon i<m}$, and $t$ in $K$ is not constant, then
\begin{equation*}
  K=\{Dt\colon D\in V\}.
\end{equation*}
Indeed, if $Dt=a\neq0$, then 
\begin{equation*}
x=\displaystyle\frac xa(Dt)
=\left(\displaystyle\frac xaD\right)t.
\end{equation*}
%Let $m$ be a positive integer.
There is an \emph{elementary} class consisting of all $(V,\cdot,t)$
such that
\begin{enumerate}
\item 
$(V,\cdot)$ is a Lie ring,
\item
$t\in\End V$,
\item
$(\{Dt\colon D\in V\},\circ)$ is a \textbf{field} $K$,
%(so $K\included(\End V,\circ)$),
\item
for all $f$ and $g$ in $K$ and $D$ in $V$,
\begin{equation*}
  f\circ(Dg)=(f(D))g,
\end{equation*}
\item
$\dim_K(V)\leq m$.
\end{enumerate}
Let $\VL^m$ be the theory of this class.  Then $\VL^m$ has
$\forall\exists$ axioms.


\pagebreak

\begin{theorem}[P]
The theory $\VL^m$ has a model companion, whose models are precisely those
  models $(V,\cdot,t)$ of $\VL^m$ 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

Suppose first $(V,K)$ is just a vector space, in the signature
comprising
\begin{enumerate}
\item 
the signature of abelian groups, for the vectors;
\item
the signature of rings, for the scalars;
\item
a symbol $*$ for the (right) action $(v,x)\mapsto\sca xv$ of $K$ on
$V$. 
\end{enumerate}

Let the theory of such structures 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*}
  \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.  Then $T$ has a model companion if and
  only if the class of its existentially closed models is elementary.
  In this case, the theory of this class is the model companion.
\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:

There is a
chain 
\begin{equation*}
  (V,K)\included(V',K')\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 $(v_s,\dots,v_{s+n-1})$, but
\item
  $v_{s}=\sca{x_s}{v_{s+1}}$ for some $x_s$ in $K^{(s+1)}\setminus
    K^{(s)}$, so 
\item
the union of the chain has dimension $1$.
\end{enumerate}
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*}
%  P^nv_0\dotsb v_{n-1}\iff
  \Exists{x^0}\dotsb\Exists{x^{n-1}}
  \Bigl(\sum_{i<n}\sca{x^i}{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 (even, model \emph{completion}),
  the theory if infinite-dimensional spaces over algebraically closed
  fields. 
  \end{enumerate}
\end{theorem}

\pagebreak

The key is lowering dimension to $n$.

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&-\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 $K$.


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


\pagebreak

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

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}[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} can be defined as any $(V,K)$, where:
\begin{asparaenum}
\item 
$V$ and $K$ are abelian groups, \emph{each} acting on the
other, from the left and right respectively, by
\begin{align*}
  (x,y)&\mapsto \der xy,&
\sca yx&\mapsfrom(x,y).
\end{align*}
\item
The actions are faithful:
\begin{align*}
  \Exists y(\der xy=0&\lto x=0),&
  \Exists x(\sca yx=0&\lto y=0).
\end{align*}
\item
Multiplications are induced, 
\begin{compactenum}[(i)]
  \item
on $V$, by the bracket;
\item
on $K$, by (opposite) composition:
\end{compactenum}
\begin{align*}
  \der{[x,y]}z&=\der x{(\der yz)}-\der y{(\der xz)},&
\sca{(y\cdot z)} x&=\sca z{(\sca yx)}.
\end{align*}
\item
These multiplications are compatible with the actions:
\begin{align*}
  \der{(\sca yx)}z&=(\der xz)\cdot y,&
\sca{(\der yz)}x&=[y,\sca zx]-\sca z{[y,x]}.
\end{align*}
\end{asparaenum}

\pagebreak

Then $V$ does act on $K$ as a Lie ring \textbf{of
  derivations;} that is,
\begin{equation*}
  \der x{(y\cdot z)}=(\der xy)\cdot z+y\cdot(\der xz).
\end{equation*}
Indeed,
  \begin{align*}
&\phantom{{}={}}  \sca{(\der x{(y\cdot z)})}w\\
&=[x,\sca{(y\cdot z)}w]-\sca{(y\cdot z)}{[x,w]}\\
&=  [x,\sca z{(\sca yw)}]-\sca z{(\sca y{[x,w]})}\\
&  \begin{aligned}=
\sca{(\der xz)}{(\sca yw)}  &+\sca z{[x,\sca yw]}\\
                            &-\sca z{[x,\sca yw]}+\sca z{(\sca{(\der
                              xy)}w)}
  \end{aligned}\\
&=\sca{(\der xz)}{(\sca yw)}+\sca z{(\sca{(\der xy)}w)}\\
&=\sca{(y\cdot(\der xz))}w+\sca{((\der xy)\cdot z)}w\\
&=\sca{(y\cdot(\der xz)+(\der xy)\cdot z)}w.
  \end{align*}

We may (asymmetrically!) make $K$ commutative, and make $V$
torsion-free as a $K$-module, so $K$ is an integral domain.

\pagebreak

The multiplications are \textbf{definable.} 
%we don't need symbols for them. 

Indeed, let $V$ and
$K$ act mutually as abelian groups, as before.
% be abelian groups acting on each other as before.  

Then $K$
becomes a sub-ring of $(\End V,\circ)$ and an integral domain when we
require
\begin{gather*}
\Exists w\sca z{(\sca yx)}=\sca wx,\\
  \sca yx=0\lto x=0\lor y=0,\\
\sca z{(\sca yx)}=\sca wx\lto x=0\lor \sca z{(\sca yu)}=\sca wu,\\
\sca z{(\sca yx)}=\sca y{(\sca zx)}
\end{gather*}
Then we can require $V$ to act on $K$ as a \textbf{module} (over $K$) of
\textbf{derivations:}
\begin{multline*}
\sca z{(\sca yx)}=\sca wx\\
\lto
\sca{(\der vw)}x=\sca{(\der vz)}{(\sca yx)}+\sca z{(\sca{(\der vy)}x)}
\end{multline*}
%---even as a \textbf{module} of them:
\begin{gather*}
  \sca{(\der{(\sca zy)}w)}x=\sca z{(\sca{(\der yw)}x)}.
\end{gather*}
%What about the multiplication on $V$?

\pagebreak

However, with no symbol for the bracket on $V$, the theory of
Lie--Rinehart pairs is not inductive.

Indeed, the union of the chain
\begin{equation*}
  (V_0,K_0)\included(V_1,K_1)\included\dotsb
\end{equation*}
of Lie--Rinehart pairs is not a Lie--Rinehart pair when 
\begin{align*}
  K_n&=\Q(t^i\colon i<n),&
 V_n&=\lspan{K_n}{D_i\restriction{K_n}\colon i<n},
\end{align*}
where
\begin{align*}
  D_0&=\sum_{i<\vnn}\partial_i,&
D_1&=\sum_{i<\vnn}(i+1)t^i\partial_{i+1},&
D_n&=\partial_n\text{ if }1<n<\vnn,
\end{align*}
where
%and on each $K_n$, the derivation $\partial_i$ given by
\begin{equation*}
  \partial_it^j=\delta_i^j.
\end{equation*}
For,
\begin{equation*}
  [D_0,D_1]=\sum_{i<\vnn}(i+1)\partial_{i+1}\notin V.
\end{equation*}

\pagebreak

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

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

\begin{theorem}[\"Ozcan Kasal]
The
existentially closed models of $T$ are just those such that
\begin{enumerate}
\item 
$\trdeg{K/\Q}=\infty$;
\item
$(K,v_0,\dots,v_{m-1})\models\DCF^{(m)}_0$ whenever
  $(v_0,\dots,v_{m-1})$ 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 $v$ in $V$,
\begin{equation*}
  \bigwedge_{i<n}\der 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 v\Bigl(\bigwedge_{j\neq i}\der v{x^j}=0\lto \der
v{x^i}=0\Bigr). 
\end{equation*}
Let the new theory be
\begin{equation*}
  T',
\end{equation*}
so
\begin{equation*}
  T'\entails\Forall{\tuple x}\Bigl(\lnot Q_n\tuple
  x\Leftrightarrow\Exists{\tuple 
    v}\bigwedge_{\substack{i<n\\j<n}}\der{v_i}{x^j}=\delta_i^j\Bigr).
\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
$(v_0,\dots,v_{k+\ell-1})$ 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

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

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

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