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\begin{document}
\title{Differential fields}
\author{David Pierce}
\date{Tabriz, August, 2012}
\publishers{Mathematics Department\\
Mimar Sinan Fine Arts University\\
Istanbul\\
\url{http://mat.msgsu.edu.tr/~dpierce/}\\
\url{dpierce@msgsu.edu.tr}}

{\Large
\maketitle}
\relscale{2}

\section{Model-theory}

\textbf{Model-theory:}
\begin{itemize}
\item
is `foundational', like category theory or set theory;
\item
 is a kind of mathematics done self-consciously;
\item
pays attention to the language of mathematics;
\item
is a study of \textbf{structures} as \textbf{models} of \textbf{theories};
\item
is a study of the relation of \textbf{truth} between structures and sentences.
\end{itemize}
The truth-relation is usually symbolized by
\begin{equation}
\models
\end{equation}
(which is \url{\models} in \LaTeX).

\pagebreak

\textbf{Examples of the truth-relation.}
\begin{equation}
\C\models\Exists xx^2+1=0,
\end{equation}
but
\begin{subequations}
\begin{equation}
\R\nmodels\Exists xx^2+1=0,
\end{equation}
that is,
\begin{equation}
\R\models\lnot\Exists xx^2+1=0,
\end{equation}
or equivalently
\begin{equation}
\R\models\Forall xx^2+1\neq0.
\end{equation}
\end{subequations}
Also,
\begin{equation}
\Q\models\Forall x\Forall y(x<y\lto\Exists z(x<z\land z<y)),
\end{equation}
but
\begin{equation}
\Z\models\Exists x\Exists y(x<y\land\Forall z(x\not<z\lor z\not<y)).
\end{equation}

\pagebreak

Let $\sig$ be a \textbf{signature,} such as the signature $\{+,-,\cdot,0,1,\leq\}$ of ordered fields:
it is a set of
\begin{itemize}
\item
 \textbf{operation-symbols} (or \emph{function-symbols,} here $+$, $-$, and $\cdot$),
 \item
\textbf{constants} (or \emph{constant-symbols,} here $0$ and $1$),
\item
\textbf{predicates} (or \emph{relation-symbols,} here $\leq$).
\end{itemize}
There is a class
\begin{equation}
\Mod
\end{equation}
of all \textbf{structures} $\str A$ with signature $\sig$.  Then $\str A$ consists of:
\begin{itemize}
\item
a set $A$, called the \textbf{universe} of the structure (or there may be several `universes', as in a vector-space; these are \textbf{sorts});
\item
an \textbf{interpretation} of $\sig$: a function $s\mapsto s^{\str A}$ on $\sig$ assigning to each operation-symbol in $\sig$ an operation on $A$, and so on.
\end{itemize}
This is just a formalization of the idea that, for example, the same symbol $+$ denotes a different operation in different abelian groups.

\pagebreak

Again $\sig$ is a signature.  There is a set
\begin{equation}
\Sn
\end{equation}
of \textbf{sentences} of first-order logic in the signature $\sig$.  A sentence is a \textbf{formula} with no free variables.  Symbols used in formulas of $\sig$ are:
\begin{itemize}
\item
the symbols in $\sig$;
\item
\textbf{variables,} as $x$ and $y$, standing for \emph{individuals} (not sets as such);
\item
\textbf{Boolean connectives,} as $\lnot$, $\land$, $\lor$, $\lto$, and $\liff$;
\item
\textbf{quantification symbols,} as $\exists$ and $\forall$;
\item
\textbf{brackets} (parentheses), as needed.
\end{itemize}
Then $\models$ is a relation between $\Mod$ and $\Sn$, namely
\begin{equation}
\{(\str A,\sigma)\in\Mod\times\Sn\colon\sigma\text{ is true in }\str A\}.
\end{equation}

\pagebreak

\textbf{Non-examples.}
The following are \emph{not} sentences of first-order logic:
\begin{itemize}
\item
the formula
\begin{equation}
\Exists yx\cdot y=1
\end{equation}
defining the group of units of a commutative ring (there is a free variable, $x$);
\item
the induction axiom for $\N$,
\begin{equation}
\Forall X(1\in X\land\Forall y(y\in X\lto y+1\in X)\lto\Forall yy\in X)
\end{equation}
(here $X$ ranges over sets, not individuals);
\item
the axiom distinguishing \emph{torsion} groups among abelian groups, 
\begin{equation}
%\Forall x\bigvee_{n\in\N}nx=0,
\Forall x(x=0\lor2x=0\lor3x=0\lor\dotsb)
\end{equation}
(sentences must be finite).
\end{itemize}

\pagebreak

We restrict to first-order logic for \textbf{compactness:}

The relation $\models$ gives us a Galois correspondence, as do:
\begin{itemize}
\item
$\{(x,\sigma)\in K\times\Aut K\colon x^{\sigma}=x\}$ in field-theory,
\item
$\{(f,\vec a)\in K[\vec X]\times K^n\colon f(\vec a)=0\}$ in algebraic geometry.
\end{itemize}
If $\mathcal K\included\Mod$, we define the \textbf{theory} of $\mathcal K$:
\begin{equation}
\Th{\mathcal K}=\bigcap_{\str A\in\mathcal K}\{\sigma\in\Sn\colon\str A\models\sigma\},
\end{equation}
the set of sentences true in every structure in $\mathcal K$.

If $\Gamma\included\Sn$, we define the class of \textbf{models} of $\Gamma$:
\begin{equation}
\Mod[\Gamma]=\bigcap_{\sigma\in\Gamma}\{\str A\in\Mod\colon\str A\models\sigma\}.
\end{equation}
Such a class is called \textbf{elementary.}  So there is a one-to-one correspondence between theories and elementary classes.

\pagebreak

The \textbf{Compactness Theorem} is that for all subsets $\Gamma$ of $\Sn$, if every finite subset of $\Gamma$ has a model, then so does $\Gamma$ itself.

If $\sigma\in\Th{\Mod[\Gamma]}$, that is, $\Gamma$ \textbf{entails} $\sigma$, we may write
\begin{equation}
\Gamma\entails\sigma.
\end{equation}
The \textbf{filter} $F$ of $\Sn$ generated by $\Gamma$ is the smallest subset of $\Sn$ that
\begin{itemize}
\item
includes $\Gamma$,
\item
 is closed under entailment,
 \item
 is closed under conjunction ($\land$).
 \end{itemize}
 Then
 \begin{equation}
F\included\Th{\Mod[\Gamma]}.
\end{equation}
The Compactness Theorem is
 \begin{equation}
F=\Th{\Mod[\Gamma]}.
\end{equation}

\pagebreak

The Compactness Theorem fails in second-order logic:  The sentences
\begin{subequations}
\begin{gather}
\Forall X(1\in X\land\Forall y(y\in X\lto y+1\in X)\lto\Forall yy\in X),\\
\Forall xx+1\neq 1,\\
\Forall x\Forall y(x+1=y+1\lto x=y),\\
c\neq1,\quad c\neq1+1,\quad c\neq1+1+1,\quad\cdots	
\end{gather}
\end{subequations}
say there is an infinite natural number (denoted by $c$).  They have no model, but every finite subset of them does.

Also by Compactness, the class of torsion groups is not elementary, since for every $k$ in $\N$, the sentence
\begin{equation}
\Exists x(x\neq0\land2x\neq0\land\dots\land kx\neq0)
\end{equation}
is true in the torsion group $\Z/(k+1)\Z$.  
%Therefore $\Mod[\Th{\text{torsion groups}}]$ has a member with an element $c$ such that
%\begin{equation}
%c\neq0,\quad2c\neq0,\quad3c\neq0,\quad\dots
%\end{equation}

\pagebreak

\section{Model-completeness}

In the following table, $\mathcal K$ and $\mathcal K^*$ are subclasses of $\Mod$:
\begin{center}
\renewcommand{\arraystretch}{1.5}
\makebox[0pt][c]{
\begin{tabular}{c|c|c}
$\sig$&$\mathcal K$&$\mathcal K^*$\\\hline
$\emptyset$&sets&infinite sets\\
$\{<\}$&linear orders&dense linear orders w/o endpoints\\
$\{+,-,0\}\cup\F_p$&vector-spaces/$\F_p$&infinite vector-spaces/$\F_p$\\
$\{+,-,\cdot,0,1\}$&fields&algebraically closed fields
\end{tabular}
}
\end{center}
In each case, $\mathcal K^*$ is the class of \textbf{existentially closed} members of $\mathcal K$.  That is, if $\str A\in\mathcal K^*$, and $\str B\in\mathcal K$, and
\begin{equation}
\str A\included\str B,
\end{equation}
then every \emph{existential} sentence true in $\str B$ is already true in $\str A$.

\pagebreak

Two more examples:
\begin{enumerate}
\item
$\sig=\{+,-,\cdot,0,1,F\}$,\\
 $\mathcal K$ is the class of fields with algebraically closed subfield $F$, and\\
  $\mathcal K^*$ is the class of algebraically closed extensions of transcendence-degree $1$ of an algebraically closed field $F$.
\item
$\sig=\{\bm+,\bm-,\bm0\}\cup\{+,-,\cdot,0,1\}\cup\{*\}$,\\
$\mathcal K$ is the class of vector-spaces, and\\
$\mathcal K^*$ is the class of vector-spaces of dimension $1$ over an algebraically closed field.
\end{enumerate}
As the ground field grows larger, dimension can only go down.

\pagebreak

If $T=\Th{\Mod[\Gamma]}$, then $T$ is the theory \textbf{axiomatized} by $\Gamma$.

The classes of groups, rings, and fields are \textbf{closed under unions of chains.}
Their theories are axiomatized by sets of $\forall\exists$ sentences (such as the field axiom $\Forall x\Exists y(x=0\lor xy=1)$).

\begin{theorem}[Chang, \L o\'s--Suszko, 1950s]
Let $T$ be a theory.  TFAE:
\begin{enumerate}
\item
$\Mod[T]$ is closed under unions of chains.
\item
$T$ is axiomatized by a set of $\forall\exists$ sentences.
\end{enumerate}
\end{theorem} 
In general,
\begin{equation}
T_{\forall\exists}=\Th{\Mod[\{\sigma\in T\colon\sigma\text{ is }\forall\exists\}]}.
\end{equation}
The equivalent conditions of the theorem can be written as
\begin{equation}
T=T_{\forall\exists}.
\end{equation}
In this case, the class of existentially closed models of $T$ is nonempty.  
If it is elementary, then its theory is the \textbf{model-companion} of $T$.

\pagebreak

The theory of fields with an algebraically closed subfield has no model-companion: having transcendence-degree $1$ is not an `elementary' or first-order property.

A stronger notion than model-companion is \textbf{model-completion.}

\begin{theorem}[P.\footnote{D. Pierce, \emph{Model-theory of vector-spaces over unspecified fields} (Arch.\ Math.\ Logic [2009] 48:421--436).}]
The theory of vector-spaces of dimension $1$ over an algebraically closed field is:
\begin{enumerate}[1)]
\item
the model-companion of the theory of vector-spaces,
\item
not the model-completion of the theory of vector-spaces,
\item
the model-completion of the theory of vector-spaces of dimension at most $1$.
\end{enumerate}
\end{theorem}

\pagebreak

Technical details:  A subset $\Gamma$ of $\Sn$ is \textbf{complete} if
\begin{equation}
\Gamma\nentails\sigma\iff\Gamma\entails\lnot\sigma
\end{equation}
for all $\sigma$ in $\Sn$, that is,
\begin{equation}
\Th{\Mod[\Gamma]}=\Th{\{\str A\}}=\Th{\str A}
\end{equation}
 for some $\str A$ in $\Mod$.  By
\begin{equation}
\str A_A
\end{equation}
we mean $\str A$, considered as an element of $\sig\cup A$.  
Then we define
\begin{equation}
\diag=\{\sigma\in\Th{\str A_A}\colon\sigma\text{ is quantifier-free}\}.
\end{equation}

\begin{theorem}
$\Mod[\diag]$ is the class of structures of $\sig\cup A$ in which $\str A_A$ embeds.
\end{theorem}

\pagebreak

If $T$ is a theory of $\sig$, then
\begin{equation}
T_{\forall}=\Th{\Mod[\{\sigma\in T\colon\sigma\text{ is universal}\}]}.
\end{equation}

\begin{theorem}
The class of substructures of models of $T$ is elementary, and its theory is $T_{\forall}$. \end{theorem}
For example, if $T$ is field-theory, then $T_{\forall}$ is the theory of integral domains.

Let $T$ and $T^*$ be theories of $\sig$ such that
$T_{\forall}=T^*{}_{\forall}$.  Then $T^*$:
\begin{enumerate}[1)]
\item
is the \textbf{model-companion} of $T$ if $T\cup\diag$ is complete for all models $\str A$ of $T^*$;
\item
is the \textbf{model-completion} of $T$ if $T\cup\diag$ is complete for all models $\str A$ of $T$;
\item
admits \textbf{quantifier-elimination} if $T\cup\diag$ is complete for all substructures $\str A$ of models of $T^*$ (or $T$).
\end{enumerate}

\newpage

\begin{theorem}[Robinson, $\leq$1963]
Let $T$ be a theory.  TFAE:
  \begin{itemize}
    \item
$T$ has a model-completion.
\item
$T=T_{\forall\exists}$, and there is $\phi\mapsto\hat{\phi}$ on
  existential (or just primitive) formulas such that, if $\str
  M\models T$ and $\vec a\in M^n$.
  \begin{equation}
    \str M\models\hat{\phi}(\vec a)\iff\str M\included\str N\models
    T\cup\{\phi(\vec a)\} \text{ for some $\str N$},
  \end{equation}
\item
the model-completion of $T$ is axiomatized by
\begin{equation}
  T\cup\{\forall\vec x\;(\hat{\phi}(\vec x)\lto\phi(\vec
  x))\colon\phi\text{ existential}\}.
\end{equation}
  \end{itemize}
\end{theorem}

The immediate example is the theory $\DF$ of \textbf{differential fields:}

\pagebreak

A \textbf{differential field} is a pair $(K,D)$, where
\begin{itemize}
\item
$K$ is a field (for simplicity, of characteristic $0$ here),
\item
$D\colon K\to K$, with
\begin{align}
D(x+y)&=Dx+Dy,&
D(xy)&=Dx\cdot y+x\cdot Dy.
\end{align}
\end{itemize}

\begin{example}
$(\C(X),f\mapsto f')$.
\end{example}

\begin{theorem}[Seidenberg]
  $\phi\mapsto\hat{\phi}$ as in Robinson's Theorem exists when $T$
  is $\DF$.
\end{theorem}

\begin{corollary}[Robinson, $\leq$1963]
  $\DF$ has a model-completion.
\end{corollary}

The  model-completion of $\DF$ is called
\begin{equation}
\DCF.
\end{equation}

\pagebreak

For more comprehensible axioms for $\DCF$, one can use:

\begin{theorem}[Blum, $\leq$1977]
  TFAE:
  \begin{itemize}
    \item
$T^*$ is the model-completion of $T$,
\item
If $\str A,\str B\models T$; $\str M\models T^*$; $\str M$ is $\lvert
B\rvert^+$-saturated:
\begin{equation}
  \xymatrix{
&\str M\\
\str A\ar[ur]\ar[r]&\str B\ar@{.>}[u]_{\exists}
}
\end{equation}
  \end{itemize}
If, further, $T=T_{\forall}$, so substructures of models are models,
then the embedding of $\str A$ in $\str B$ can be analyzed as
\begin{equation}
  \str A\to\str A(a_1)\to\str A(a_1,a_2)\to\dots\to\str B
\end{equation}
where each structure is a model of $T$; so $\str B=\str A(a)$
suffices \emph{(Blum's Criterion)}.
\end{theorem}

Since $\DF_{\forall}$ is `close enough' to $\DF$ is universal, Blum gets nice axioms for $\DCF$:

\pagebreak

\begin{theorem}[Blum, $\leq$1977]
  $(K,D)\in\Mod[\DCF]$ if and only if:
  \begin{itemize}
    \item
$(K,D)\in\Mod[\DF]$,
\item
$K=K^{\operatorname{alg}}$,
\item
for all ordinary polynomials $f$ and $g$ over $K$, if $g\neq0$ and $\partial_{n+1}f\neq0$, then
\begin{multline}
(K,D)\models\exists x\;(f(x,Dx,\dots,D^{n+1}x)=0\land{}\\
{}\land  g(x,Dx,\dots,D^nx)\neq0).
\end{multline}
  \end{itemize}
\end{theorem}

There is an alternative approach to simplifying the axioms of $\DCF$:  


\pagebreak

We want to understand when systems of differential polynomial equations and inequations have solutions in some extension.

We can eliminate inequalities by the usual trick:
\begin{equation}
x\neq0\iff\exists y\;xy=1.
\end{equation}

Over a model $(K,D)$ of $\DF$, TFAE:
\begin{gather}
  \exists\vec x\;\bigwedge_ff(\vec x,D\vec x,\dots,D^n\vec x)=0,\\
\exists(\vec x_0,\dots,\vec x_n)\;(\bigwedge_ff(\vec x_0,\dots,\vec
x_n)=0\land\bigwedge_{i<n}D\vec x_i=\vec x_{i+1}).
\end{gather}
The latter is an instance of
\begin{equation}
  \exists{\vec x}\;(\bigwedge_ff(\vec
  x)=0\land\bigwedge_{i<k}Dx_i=g_i(\vec x)),
\end{equation}
where $\vec x=(x_0,\dots,x_{n-1})$ and $k\leq n$.

\pagebreak

\begin{theorem}[P.--Pillay 1998, P. 2004]
$(K,D)\in\Mod[\DCF]$ if and only if:
\begin{itemize}
\item
$(K,D)\in\Mod[\DF]$,
\item
$K=K^{\operatorname{alg}}$,
\item
for all $f$ in $K[X_0,\dots,X_{n-1}]$ and $g_i$ in $K(\vec X)$,
\begin{equation}
  \exists{\vec x}\;(\bigwedge_ff(\vec
  x)=0\land\bigwedge_{i<k}Dx_i=g_i(\vec x)),
\end{equation}
provided the $f$ impose no algebraic condition on $(x_0,\dots,x_{n-1})$.
\end{itemize}
In the last condition, it is enough that the $f$ define an irreducible variety with generic point $\vec a$ such that $(a_0,\dots,a_{k-1})$ is a transcendence-basis of $K(\vec a)/K$.
\end{theorem}


How do these ideas work in case of several derivations?

\pagebreak

Let $\DF^m$ be the theory of fields (of characteristic zero) with $m$ commuting derivations.

\begin{example}
$(\C(X_0,\dots,X_{m-1}),\partial_0,\dots,\partial_{m-1})$.
\end{example}

So $\DF^m$ is the theory of $(m+1)$-tuples $(K,\partial_0,\dots,\partial_{m-1})$, where
\begin{itemize}
\item
$\partial_i\in\Der K$, that is, $(D,\partial_i)\in\Mod[\DF]$;
\item
$[\partial_i,\partial_j]=0$.
\end{itemize}

\begin{theorem}[McGrail, 2000]
  $\DF^m$ has a model-completion, called $\DCF^m$.
\end{theorem}

To prove this, McGrail uses Blum's Criterion (and so considers only
systems in one variable).

Let
\begin{equation}
  \upomega=\{0,1,2,\dots\};
\end{equation}
if $\sigma\in\upomega^m$, this means
\begin{equation}
  \sigma=(\sigma(0),\dots,\sigma(m-1)).
\end{equation}

\pagebreak

For simplicity, we assume $m=2$.  If $\sigma\in\upomega^2$, we write
\begin{equation}
\partial^{\sigma}x=
  \partial_0{}^{\sigma(0)}\partial_{1}{}^{\sigma(1)}x.
\end{equation}
Suppose
\begin{equation} (K,\partial_0,\partial_1)\included(L,\tilde{\partial}_0,\tilde{\partial}_1),
\end{equation}
both being models of $\DF$, and $a\in L$.  We define
\begin{equation}
K\{a\}=K[\partial^{\sigma}a\colon\sigma\in\upomega^2],
\end{equation}
the differential ring generated by $a$ over $K$.  In particular, there is a differential polynomial ring $K\{X\}$.  Then we can define
\begin{equation}
I(a)=\{f\in K\{X\}\colon f(a)=0\};
\end{equation}
this is a prime differential ideal.  Kolchin identifies a kind of finite subset $\Lambda$ of this ideal, called a \textbf{characteristic set} of the ideal, that determines the ideal in the following sense.

\pagebreak

Again if $\sigma\in\upomega^2$, let
\begin{equation}
  \abs{\sigma}=\sigma(0)+\sigma(1).
\end{equation}
Since the characteristic set $\Lambda$ of $I(a)$ is finite, for some $n$ in $\upomega$,
\begin{equation}
\Lambda\included K[\partial^{\sigma}X\colon\abs{\sigma}\leq n].
\end{equation}
For each $f$ in this ring there is $\tilde f$ in $K[X_{\sigma}\colon\abs{\sigma}\leq n]$ such that
\begin{equation}
f=\tilde f(\partial^{\sigma}X\colon\abs{\sigma}\leq n).
\end{equation}
If $(b_{\sigma}\colon\abs{\sigma}\leq n)$ is a generic zero of $\{\tilde f\colon f\in\Lambda\}$ from some extension of $K$, then there are $b_{\sigma}$ for all additional $\sigma$ in $\upomega^2$ such that we can extend the $\partial_i$ to derivations $\tilde{\partial}_i$ on $K(b_{\sigma}\colon\sigma\in\upomega^2)$ such that
\begin{align}
\tilde{\partial}_0b_{\sigma}&=b_{\sigma+(1,0)},&
\tilde{\partial}_1b_{\sigma}&=b_{\sigma+(0,1)},&
I(b_{(0,0)})&=I(a).
\end{align}
This is the sense in which $\Lambda$ determines $I(a)$; but Kolchin gives an algebraic description.

\pagebreak

Again $\Lambda$ is a characteristic set of $I(a)$.
Suppose $g\in K\{X\}\setminus I(a)$.  Then $\Lambda$ and $g$ involve parameters from $K$.  We can put these in a finite list $\vec c$ and write $\Lambda$ as $\Lambda_{\vec c}$ and $g$ as $g_{\vec c}$.  

McGrail shows there is a formula $\phi(\vec y)$ such that
\begin{equation}
(K,\partial_0,\partial_1)\models\phi(\vec c),
\end{equation}
and whenever $(L,\partial_0,\partial_1)\in\Mod[\DF]$ and $(L,\partial_0,\partial_1)\models\phi(\vec d)$, then the differential system
\begin{equation}
\bigwedge_{f\in\Lambda_{\vec d}}f=0\land g_{\vec d}\neq0
\end{equation}
is soluble in some extension of $(L,\partial_0,\partial_1)$.  Then one of the axioms of $\DCF$ is
\begin{equation}
\Forall{\vec y}\Bigl(\phi(\vec y)\lto\Exists x\bigl(\bigwedge_{f\in\Lambda_{\vec y}}f(x)=0\land g_{\vec y}(x)\neq0\bigr)\Bigr).
\end{equation}


\pagebreak

We can see what is going on as follows.  We ask whether the set of differential polynomials
\begin{align}
	\partial^{(2,2)}X&-X,&\partial^{(1,1)}X&-\partial^{(0,2)}X
\end{align}
has a zero.  These polynomials belong to $K[\partial^{\sigma}X\colon\abs{\sigma}\leq4]$.  We depict a potential zero as
\begin{equation}
\begin{matrix}
\partial^{(0,0)}x&\partial^{(1,0)}x&\partial^{(2,0)}x&\partial^{(3,0)}x&\partial^{(4,0)}x\\
\partial^{(0,1)}x&\partial^{(1,1)}x&\partial^{(2,1)}x&\partial^{(3,1)}x&                 \\
\partial^{(0,2)}x&\partial^{(1,2)}x&\partial^{(2,2)}x&                &                 \\
\partial^{(0,3)}x&\partial^{(1,3)}x&                 &                &                 \\
\partial^{(0,4)}x&                 &                 &                &                 
\end{matrix}
\end{equation}
A generic zero of the polynomials
\begin{align}
X_{(2,2)}&-X_{(0,0)},&X_{(1,1)}&-X_{(0,2)}
\end{align}
in $K[X_{\sigma}\colon\abs{\sigma}\leq4]$ can be depicted as
\begin{equation}
\begin{matrix}
a&*&b&*&*\\
*&b&*&*& \\
*&*&a& & \\
*&*& & & \\
*& & & &
\end{matrix}
\end{equation}
Closing under differentiation imposes further conditions:
\begin{equation}
\begin{matrix}
a&*&b&c&\underline a\\
*&\underline b&c&a& \\
*&c&a& & \\
*&a& & & \\
*& & & &
\end{matrix}
\end{equation}
But the underlined entries should have a common derivative---say $d$---outside the triangle:
\begin{equation}
\begin{matrix}
a&*&b&c&\underline a\\
*&\underline b&c&a&d\\
*&c&a& & \\
*&a& & & \\
*& & & &
\end{matrix}
\end{equation}
This imposes further conditions \emph{inside} the triangle:
\begin{equation}
\begin{matrix}
a&d&b&c&a\\
d&b&c&a&d\\
b&c&a& & \\
c&a& & & \\
a& & & &
\end{matrix}
\end{equation}
That's fine, we can extend this diagram indefinitely; so the original differential polynomials have a zero.

But suppose we started instead with
\begin{align}
	\partial^{(2,2)}X&-X,&\partial^{(1,1)}X&-\partial^{(0,2)}X-1.
\end{align}
This gives us (writing $b'$ for $b+1$)
\begin{align}
&\begin{matrix}
a&*&b&*&*\\
*&b'&*&*& \\
*&*&a& & \\
*&*& & & \\
*& & & &
\end{matrix}&
&\begin{matrix}
a&*&b&c&a\\
*&b'&c&a& \\
*&c&a& & \\
*&a& & & \\
*& & & &
\end{matrix}&
&\begin{matrix}
a&d&b&c&a\\
d&b'&c&a&d\\
*&c&a& & \\
*&a& & & \\
*& & & &
\end{matrix}
\end{align}
so $\partial_1d$ must be both $b$ and $b'$, which is absurd.

This test works generally as follows.



\pagebreak




We define a strict partial ordering $\lessdot$ on $\upomega^2$ by
\begin{equation}
\sigma\lessdot\tau\iff(\abs{\sigma},\sigma(0))<_{\ell}(\abs{\tau},\tau(0))
\end{equation}
where $<_{\ell}$ is the left lexicographic ordering on $\upomega^2$.
So $\lessdot$ is thus:
\begin{center}
\psset{unit=15mm}
  \begin{pspicture}(-1,-2.5)(3,0.5)
  %\psgrid
    \psset{radius=0.1}
    \Cnode(0,0){a}
    \uput[ul](0,0){$(0,0)$}
    \Cnode(1,0){b}
    \uput[u](1,0){$(0,1)$}
    \ncline{->}{a}{b}
    \Cnode(0,-1){c}
    \uput[l](0,-1){$(1,0)$}
    \ncline{->}{b}{c}
    \Cnode(2,0){d}
    \uput[ur](2,0){$(0,2)$}
    \ncline{->}{c}{d}
    \Cnode(1,-1){e}
    \uput[dr](1,-1){$(1,1)$}
    \ncline{->}{d}{e}
    \Cnode(0,-2){f}
    \uput[d](0,-2){$(2,0)$}
    \ncline{->}{e}{f}
  \end{pspicture}
\end{center}
Let $\leq$ be the product ordering of $\upomega^2$, so
\begin{equation}
  \sigma\leq\tau\iff\sigma(0)\leq\tau(0)\land\sigma(1)\leq\tau(1).
  \end{equation}
Then we always have
\begin{equation}
\sigma\leq\sigma+\tau.
\end{equation}

\pagebreak

Suppose again
\begin{equation} (K,\partial_0,\dots,\partial_{m-1})\included(L,\tilde{\partial}_0,\dots,\tilde{\partial}_{m-1}),
\end{equation}
both being models of $\DF$, and $a\in L$.  Then
\begin{equation}
\tilde{\partial}^{\sigma}a\in K(\tilde{\partial}^{\rho}a\colon\rho\lessdot\sigma)\alg
\implies 
\tilde{\partial}^{\sigma+\tau}a\in K(\tilde{\partial}^{\rho}a\colon\rho\lessdot\sigma+\tau)\alg.
\end{equation}
A tuple $(a_{\sigma}\colon\abs{\sigma}\leq n)$ of elements of some field-extension of $K$ will be called \textbf{soluble} if the $a_{\sigma}$ belong to some $L$ as above so that
\begin{align}
\tilde{\partial}_0a_{\sigma}&=a_{\sigma+(1,0)},&
\tilde{\partial}_1a_{\sigma}&=a_{\sigma+(0,1)}
\end{align}
when these make sense.  For solubility, it is \emph{necessary} that, for all $f$ in $K[\partial^{\sigma}\colon\abs{\sigma}<n]$,
\begin{equation}
\tilde f(a_{\sigma}\colon\abs{\sigma}<n)=0
\implies
\widetilde{\partial_if}(a_{\sigma}\colon\abs{\sigma}\leq n)=0.
\end{equation}
Call this the \textbf{differential condition.}  The last example shows it is not \emph{sufficient} for solubility.

\pagebreak

Again we have the tuple 
$(a_{\sigma}\colon\abs{\sigma}\leq n)$ is a tuple
of elements of some field-extension of $K$.
We ask what is sufficient for solubility.

Call $\sigma$ a \textbf{leader} if it is $\leq$-minimal among those $\tau$ such that
\begin{equation}
a_{\tau}\in K(a_{\rho}\colon\rho\lessdot\tau)\alg.
\end{equation}


\begin{theorem}[P.]
For $(a_{\sigma}\colon\abs{\sigma}\leq n)$ to be soluble, it is sufficient that 
  \begin{enumerate}[1)]
  \item 
  it meet the differential condition, and
\item
for every leader $a_{\sigma}$, we have $\abs{\sigma}\leq n/2$.
  \end{enumerate}
Moreover, if $(a_{\sigma}\colon\abs{\sigma}\leq n)$ is soluble, then $n$ can be chosen large
enough so that the foregoing conditions are met; and the new $n$
depends only on the original $n$.
\end{theorem}

We can make adjustments to allow each $a_{\sigma}$ to be a \emph{tuple} $(a_{\sigma}^j\colon j<k)$ and to allow $K$ to have arbitrary characteristic.

This leads to a model-completion of the theory of fields (of unspecified characteristic) with $m$ commuting derivations.

\end{document}

\pagebreak

TENTATIVE END

\pagebreak

For example, the system
\begin{align}
  \partial^{(1,1)}x&=\partial^{(0,2)}x,&\partial^{(1,2)}x&=\partial^{(2,0)}x
\end{align}
determines a tuple $(a_{\sigma}\colon\abs{\sigma}\leq3)$ that can be
depicted thus, with leader underlined:
\begin{equation}
  \begin{array}{cccc}
    *&*&a&*\\
    *&\underline a&b& \\
    b&*& & \\
    *& & &
  \end{array}
\end{equation}
But then differentiating $\partial^{(1,1)}x=\partial^{(0,2)}x$ by
$\partial_1$ puts another condition on $\partial^{(1,2)}x$: we get
\begin{equation}
  \partial^{(1,2)}x=\partial^{(0,3)}x
\end{equation}
and therefore
\begin{equation}
  \partial^{(2,0)}x=\partial^{(0,3)}x.
\end{equation}
Now the conditions on $(a_{\sigma}\colon\abs{\sigma}\leq3)$ are thus:
\begin{equation}
  \begin{array}{cccc}
    *&*&a&\underline b\\
    *&\underline a&*& \\
    b&*& & \\
    *& & &
  \end{array}
\end{equation}
The two leaders should have a common derivative; this imposes a new
condition:
\begin{gather}
\partial^{(1,3)}x=\partial^{(0,4)}x,\\
\partial^{(1,3)}x=\partial^{(3,0)}x,  
\end{gather}
and therefore
\begin{equation}
  \partial^{(3,0)}x=\partial^{(0,4)}x.
\end{equation}
\begin{equation}
  \begin{array}{ccccc}
    *&           *&a&\underline b&c\\
    *&\underline a&*&           *& \\
    b&           *&*&            & \\
    c&           *& &            & \\
    *&            & &            &
  \end{array}
\end{equation}
but then also
\begin{equation}
  \partial^{(2,1)}x=\partial^{(0,4)}x=\partial^{(3,0)}x.
\end{equation}
\begin{equation}
  \begin{array}{ccccc}
    *&           *&a&\underline b&c\\
    *&\underline a&*&           *& \\
    b&           c&*&            & \\
    \underline c&           *& &            & \\
    *&            & &            &
  \end{array}
\end{equation}
Further differentiation yields the picture
\begin{equation}
  \begin{array}{cccc|ccc}
    *&*&a&\underline b&b&b&b\\
    *&\underline a&b&b&b&b& \\
    b&b&b&b&b& & \\
    b&b&b&b& & & \\\cline{1-4}
    b&b&b&\multicolumn{4}{c}{} \\
    b&b& &\multicolumn{4}{c}{} \\
    b& & &\multicolumn{4}{c}{} 
  \end{array}
\end{equation}
By the theorem (and obviously!) the picture can be continued
indefinitely; so the original differential system is soluble.

The theorem can be translated into axioms for $\DCF$.

The argument can be adjusted to allow fields of positive characteristic.


\pagebreak

(Where do these fit?)


Say $K$ is the universe of a model of $\DF^2$, and $a$ is from some extension.  We may write
\begin{equation}
K\{a\}=K[\partial^{\sigma}a\colon\sigma\in\upomega^2]
\end{equation}
(the least ring over $K$ containing the $\partial^{\sigma}a$) and
\begin{equation}
K\langle a\rangle=K(\partial^{\sigma}a\colon\sigma\in\upomega^2)
\end{equation}
(the least field over $K$ containing the $\partial^{\sigma}a$).
So $a$ is a zero of the set 
\begin{equation}
\{f\in K\{X\}\colon f(a)=0\}
\end{equation}
of differential polynomials; but this is an infinite set.

*************

For each element $f$ of $K\{X\}$, there is $n$ in $\upomega$ such that
\begin{equation}
f=g(\partial^{\sigma}X\colon\abs{\sigma}\leq n)
\end{equation}
for some $g$ such that  
\begin{equation}
  g\in K[X_{\sigma}\colon\abs{\sigma}\leq n].
\end{equation}
An \textbf{algebraic zero} of $f$ is just a zero of $g$.

Write
\begin{equation}
  V=\{f\in K\{X\}\colon f(a)=0\}.
\end{equation}
We want a finite subset $V'$ of this such that every \emph{generic}
algebraic zero of $V'$ is a zero of $V$.  Moreover, this should be a
first-order condition on the parameters from $K$ appearing in $V'$.

********************


Given some tuple $(a_{\sigma}\colon\abs{\sigma}\leq n)$ from a
field-extension of $K$,

\end{document} THE REAL END IS EARLIER!