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\title{Fields with several commuting derivations}
\author{David Pierce}
\revauthor{Pierce, David}

\address{Mathematics Department\\
Middle East Technical University\\
Ankara 06531, Turkey}

\email{dpierce@metu.edu.tr}
\urladdr{http://www.math.metu.edu.tr/\urltilde dpierce}

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\begin{document}

  \begin{abstract}
The existentially closed models of the theory of fields (of arbitrary
    characteristic) with a given finite number of commuting
    derivations can be characterized geometrically, in several ways.
    In each case, the 
    existentially closed models are those models that contain
    points of certain differential varieties, which are
    determined by certain ordinary varieties.
  \end{abstract}

 \maketitle

% \tableofcontents

How can we tell whether a given system of partial differential equations
has a solution?  An answer given in this paper is that, if we
differentiate the equations enough times, and no contradiction arises,
then it never will, and the system is soluble.  Here, the meaning of
`enough times'
can be expressed \emph{uniformly;} this is one way of showing that the
theory, $\mDF$, of fields with a 
finite number $m$ of commuting derivations has a model-companion.
In fact, this theorem is worked out here (as Corollary~\ref{cor:dcf},
of Theorem~\ref{thm:dcf}), not in terms of polynomials, but 
in terms of the varieties that they define, and the function-fields of
these: in a word, the treatment is \emph{geometric.}

The model-companion of $\mDF_0$ (in
characteristic~$0$) has been axiomatized before, explicitly in terms
of differential polynomials: see \S~\ref{sect:several}.  I
attempted in \cite{MR2000487} to characterize its models (namely, the
existentially closed
models of $\mDF_0$) in terms of
differential forms, but I made a mistake.  Here I correct the mistake,
but also work in arbitrary characteristic.  The existence
of a model-companion of $\mDF$ (with no specified characteristic) appears to
be a new result when $m>1$ (despite a remark by Saharon Shelah
\cite[p.~315]{MR0344116}: `I am
quite sure that for characteristic $p$ as well, [making $m$ greater
  than $1$] does not make any essential
difference').

The theory of model-companions and model-completions was worked out
decades ago; perhaps for that very reason, it may be worthwhile to
review the theory here, as I do in \S~\ref{sect:mt}.  I try
to give the original references, when
I have been able to consult them.  
In \S~\ref{sect:one}, I review the various known characterizations of
existentially closed fields with single derivations.  Then
\S\S~\ref{sect:several} and~\ref{sect:resolution} consider how those
characterizations can be 
generalized to allow for several commuting derivations.

\section{Model-theoretic background}\label{sect:mt}

Let $\str M$ be an arbitrary (first-order) structure; its
theory is
\myboxed{\Th{\str M}}.
Let $T$ be an
arbitrary consistent (first-order) theory; its models compose the
class
\myboxed{\Mod T}.
Every class $\bm K$ of structures in some signature has a theory,
\myboxed{\Th{\bm K}}.  Then $\bm K\included\Mod{\Th{\bm K}}$; in case of
equality, $\bm K$ is \defn{elementary}.   Always, $\Th{\Mod T}=T$.

The structure $\str M$ has the universe $M$.  The structure denoted by
$\str M_M$ is the expansion of $\str M$ that has a name for every
element of $M$. 
Then $\str M$ embeds in $\str N$ if and only if $\str M_M$ embeds in
an expansion of $\str N$.  However, although the class of
structures in 
which $\str M$ embeds need not be elementary, the class of
structures in which $\str M_M$ embeds \emph{is} elementary.  The
theory of the latter class is the \defn{diagram}{} of $\str M$, or
\myboxed{\diag{\str M}}: it is
axiomatized by the quantifier-free sentences in $\Th{\str M_M}$
\cite[Thm 2.1.3, p.~24]{MR0153570}. 
The class of structures in which $\str M_M$
embeds \emph{elementarily}
is also elementary, and its theory is just \myboxed{\Th{\str M_M}}. 
The class of
substructures of models of $T$ is elementary, and its theory is
denoted by \myboxed{T_{\forall}}: this is axiomatized by the universal
sentences of $T$ \cite[Thm 3.3.2, p.~71]{MR0153570}.

By a \defn{system over}{} $\str M$, I mean a finite conjunction of
atomic and negated atomic formulas in the signature of $\str M_M$;
likewise, a system \defn{over}{} $T$ is in the signature
of $T$.  A structure $\str M$ \defn{solves}{} a system $\phi(\tuple x)$
if $\str M\models\Exists{\tuple x}\phi(\tuple x)$.  Note well here that
$\tuple x$, in boldface, is a \emph{tuple} of variables, perhaps
$(x^0,\dots,x^{n-1})$.  By an
\defn{extension}{} of a model of $T$, I mean another model of $T$ of
which the first is a substructure.  Two systems over a model $\str M$
of $T$ are \defn{equivalent}{} if they are soluble in the same extensions.

An \defn{existentially closed}{} model of $T$ is a model of $T$ that
solves every system over
itself that is soluble in some extension.  So a model $\str M$ of $T$ is
existentially closed if and only if
$T\cup\diag{\str M}\proves\Th{\str M_M}_{\forall}$, that is, every
extension of $\str M$ is a substructure of an elementary extension
  (\cite[\S~7]{MR0277372} or \cite[\S~2]{MR51:12518}).

A theory is \defn{model-complete}{} if its every
model is existentially closed.  An equivalent formulation explains the
name: $T$ is model-complete if and only if $T\cup\diag{\str M}$ is
complete whenever $\str M\models T$
\cite[Ch.~2]{MR0472504}.  

Suppose every model of $T$ has an existentially closed extension.
Such is the case when $T$ is \defn{inductive}, that is, $\Mod{T}$ is
closed under unions of chains \cite[Thm~7.12]{MR0277372}: equivalently,
$T=T_{\forall\exists}$ \cite{MR0089813,MR0103812}.  Suppose further
that we have a uniform
first-order way to tell when systems over models of $T$ are soluble in
extensions: more precisely, suppose there is a function
\begin{equation}\label{eqn:fn}
\phi(\tuple x,\tuple y)\longmapsto
\robfn{\phi}(\tuple x,\tuple y),
\end{equation}
where 
$\phi(\tuple x,\tuple y)$ ranges over the systems over $T$ (with
variables analyzed as shown),
such that, for every model $\str
M$ of $T$ and every tuple $\tuple a$ of parameters from $M$, the
system $\phi(\tuple x,\tuple a)$ is soluble in some extension of $\str
M$ just in case $\robfn{\phi}(\tuple x,\tuple a)$ is soluble in $\str M$.
Then the existentially closed
models of $T$ compose an elementary class, whose theory $T^*$ is
axiomatized by $T$ together with the sentences
\begin{equation}\label{eqn:sen}
  \Forall{\tuple y}(\Exists{\tuple x}\robfn{\phi}(\tuple x,\tuple
  y)\to
\Exists{\tuple x}\phi(\tuple x,\tuple y)).
\end{equation}
Immediately, $T^*$ is model-complete, so $T^*\cup\diag{\str M}$ is
complete when $\str M\models T^*$.  What is more,
$T^*\cup\diag{\str M}$ is complete whenever $\str M\models
T$ \cite[Thm 5.5.1]{MR0153570}.

In general, $T^*$ is a \defn{model-completion}{} of $T$ if 
$T^*{}_{\forall}\included T\included T^*$
and 
$T^*\cup\diag{\str M}$ is
complete whenever $\str M\models T$.  Model-completions are unique
\cite[(2.8)]{MR0091922}.   We have sketched the proof of part of the
following (the rest is \cite[(3.5)]{MR0091922}):

\begin{lemma}[Robinson's Criterion]
\label{lem:rob-crit} 
  Let $T$ be inductive.  Then $T$ has a model-comple\-tion if and
  only if a function $\phi(\tuple x,\tuple
  y)\mapsto\robfn{\phi}(\tuple x,\tuple y)$ exists as
  in~\eqref{eqn:fn}. 
In this case, the model-completion is axiomatized \emph{modulo} $T$ by
  the sentences in~\eqref{eqn:sen}.
\end{lemma}

If $T_{\forall}=T^*{}_{\forall}$ and $T^*$ is model-complete, then
$T^*$ is a \defn{model-companion}{} of~$T$
(\cite[\S~5]{MR0272613}; \emph{cf.}~\cite[\S~2]{MR0277372}).
Model-completions are model-companions, and model-companions are
unique \cite[Thm 5.3]{MR0272613}.
If $T$ has a
model-companion, then its models are just the existentially closed
models of $T$ \cite[Prop.~7.10]{MR0277372}.
Conversely, if $T$ is inductive,
and the class of existentially closed models 
of $T$ is elementary, then the theory of this class is the
model-companion of $T$ \cite[Cor.~7.13]{MR0277372}.

\section{Fields with one derivation}\label{sect:one}

Let \myboxed{\DF}{} be the theory of fields with a derivation, and let
\myboxed{\DPF}{} be 
the theory of models of $\DF$ that, for each prime $\ell$, satisfy also
\begin{equation*}
  \Forall x\Exists y(\underbrace{1+\dotsb+1}_{\ell}=0\land
  Dx=0\to\underbrace{y\dotsb y}_{\ell}=x).
\end{equation*}
So models of $\DPF$ are \defn{differentially perfect}.  A
subscript on the name of one of these theories will indicate a 
required characteristic for the field.  In particular, we have $\DPF_0$,
which is the same as $\DF_0$. 

Abraham Seidenberg
\cite{MR0082487} shows the existence of the function
in Lemma~\ref{lem:rob-crit} in case
$T$ is $\DPF_p$, where $p$ is prime or $0$.  Consequently:

\begin{theorem}[Robinson]
  $\DF_0$ has a model-completion, called \myboxed{\DCF_0}.
\end{theorem}

\begin{theorem}[Wood \cite{MR48:8227}]
  If $p$ is prime, then $\DF_p$ has a model-companion,
  \myboxed{\DCF_p}, which is the model-completion of $\DPF_p$. 
\end{theorem}

\subsection{Single variables}\label{subsect:one-var}

Since it involves \emph{all} systems over a given theory,
Robinson's criterion yields the crudest possible axiomatization for a
model-completion.  By contrast, though the theory \myboxed{\ACF}{} of
algebraically closed fields is the model-completion of the theory of
fields, its axioms (\emph{modulo} the latter theory) can involve only
systems in one variable (indeed, single equations in one variable).
A generalization of this observation is the following, which can be
extracted from the proof of \cite[Thm~17.2, pp.~89--91]{MR0398817}
(see also \cite{MR0491149}):

\begin{lemma}[Blum's Criterion]\label{lem:blum}
  Say $T^*{}_{\forall}\included T\included T^*$.
  \begin{enumerate}
    \item
The theory $T^*$ is the model-completion of $T$ if and only if the
  commutative diagram
  \begin{equation*}
    \xymatrix{
\str M &\\
\str A \ar[u] \ar[r] & \str B \ar@{.>}[ul]
}
  \end{equation*}
of structures and embeddings
can be completed as indicated when
$\str A$ and $\str B$ are models of $T$ and $\str M$ is a
$\size{B}^+$-saturated model of $T^*$.
\item\label{item:blum2}
If $T=T_{\forall}$, it is enough to assume that $\str B$ is generated
over $\str A$
by a single element.
  \end{enumerate}
\end{lemma}

This allows a refinement of Lemma~\ref{lem:rob-crit} in a special
case:

\begin{lemma}
  Suppose $T=T_{\forall}$.  Then Lemma~\ref{lem:rob-crit} still holds
  when $\phi(\tuple x,\tuple y)$ is replaced with
  $\phi(x,\tuple y)$ (where $x$ is a single variable).
\end{lemma}

From Lemma~\ref{lem:blum},
Lenore Blum obtains Theorem~\ref{thm:dcf-1} below in characteristic
$0$, in which case the first two numbered conditions amount to
$K\models\ACF$ (\cite[pp.~298~ff.]{MR0398817} or~\cite{MR0491149}).
If $p>0$, then $\DF_p$ is not universal, so
part~\eqref{item:blum2} of Blum's criterion does 
\emph{not} apply; Carol Wood instead uses a
primitive-element theorem of Seidenberg \cite{MR0049174} to obtain new
axioms for $\DCF_p$ \cite{MR50:9577}.  These can be combined with
Blum's axioms for $\DCF_0$ to yield the following.  (Here \myboxed{\SCF}{} is
the theory of separably closed fields.)

\begin{theorem}[Blum, Wood]\label{thm:dcf-1}
A model $(K,D)$ of $\DF$ is existentially closed if and only if
\begin{enumerate}
\item
$K\models\SCF$;
  \item
$(K,D)\models\DPF$;
\item
$(K,D)\models\Exists x(f(x,Dx,\dots,D^{n+1}x)=0\land
g(x,Dx,\dots,D^nx)\neq0)$ whenever $f$ and $g$ are ordinary
polynomials over $K$ in tuples $(x^0,\dots, x^{n+1})$ and
$(x^0,\dots,x^n)$ of variables
respectively such that $g\neq0$ and
$\partial f/\partial x^{n+1}\neq0$. 
\end{enumerate}
Hence $\DF$ has a model-companion, \myboxed{\DCF}.
\end{theorem}

There is a similar characterization of the existentially closed
\emph{ordered} differential fields \cite{MR495120}.

\subsection{First derivatives}\label{sect:first-der}

Alternative simplified axioms for $\DCF$ are parallel to
those found for the model-companion \myboxed{\ACFA}{} of the theory of fields
with an automorphism \cite{MR99c:03046,MR2000f:03109}.  Suppose
$(K,D)\models\DPF$ and $K\models\SCF$.
Every system
over $(K,D)$ can be written as
\begin{equation}\label{eqn:sys}
  \bigwedge_ff(\tuple x,D\tuple x,\dots,D^n\tuple x)=0\land g(\tuple
  x,D\tuple x,\dots,D^n\tuple x)\neq0,
\end{equation}
where $g$ and the $f$ are ordinary polynomials over $K$.  This system is
equivalent to one that involves only first derivatives, namely
\begin{multline*}
  \bigwedge_ff(\tuple x_0,\dots,\tuple x_{n-1},\tuple y_{n-1})=0\land 
g(\tuple x_0,\dots,\tuple x_{n-1},\tuple y_{n-1})\neq0\land{}\\
{}\land\bigwedge_{i+1<n}\tuple x_{i+1}=\tuple y_i\land
\bigwedge_{i<n}D\tuple x_i=\tuple y_i.
\end{multline*}
This system, in form, is a special case of the system
\begin{equation}\label{eqn:1st}
  \bigwedge_ff(\tuple x,\tuple y)=0\land g(\tuple x,\tuple
  y)\neq0\land D\tuple x=\tuple y,
\end{equation}
which can also be written as
$\bigwedge_ff(\tuple x,D\tuple x)=0\land g(\tuple x,D\tuple x)\neq0$. 
Suppose~\eqref{eqn:1st} has the solution $(\tuple a,\tuple b)$.
Then $K(\tuple a,\tuple b)/K$ is separable
\cite[Lem.~1.5, p.~1328]{MR2114160}.  Let
$V$ and $W$ be the varieties over $K$ with generic points $\tuple a$
and $(\tuple a,\tuple b)$ respectively, let \myboxed{T_D(V)}{} be the
twisted tangent bundle of $V$, and let $U$ be the open subset of $W$
defined by the inequation $g\neq0$; then the situation can be
depicted thus:
\begin{equation*}
\SelectTips{cm}{}
\newdir{ >}{{}*!/-5pt/@{>}}
  \xymatrix{
U \ar@{ >->}[r] \ar@{-->>}[dr]
 & W \ar@{ >->}[r] \ar@{-->>}[d] & T_D(V) \ar@{-->>}[dl]^{(\tuple
  x,\tuple y)\mapsto\tuple x}\\
& V&
}
\end{equation*}
In characteristic $0$, the model $(K,D)$ of $\DF$ is existentially
closed if and only if, in every such geometric situation, $U$ contains
a $K$-rational point $(\tuple c,D\tuple c)$; this yields the so-called
geometric axioms for $\DCF_0$ found with Anand Pillay
\cite{MR99g:12006}.  In positive characteristic, it is still true
that, if $(\tuple a,\tuple b)$ is a generic point of $V$, then $D$
extends to $K(\tuple a)$ so that $D\tuple a=\tuple b$.  However, an
additional condition is needed to ensure that $D$ extends to all of
$K(\tuple a,\tuple b)$; it is enough to require that the projection of
$T_D(W)$ onto $T_D(V)$ contain a generic point of $W$; this yields
Piotr Kowalski's geometric axioms for $\DCF_p$ \cite{MR2119125}.

In an alternative geometric approach to $\DCF$, instead
of~\eqref{eqn:sys}, it is enough to look
at an arbitrary system of \emph{equations,}
$\bigwedge_ff(\tuple x,D\tuple x,\dots,D^n\tuple x)=0$.
As before, we can take the derivations out of the polynomials,
this time getting the equivalent system
\begin{equation*}
  \bigwedge_ff(\tuple x_0,\dots,\tuple
  x_n)=0\land\bigwedge_{i<n}D\tuple x_i=\tuple x_{i+1}.
\end{equation*}
This is a special case of
\begin{equation}\label{eqn:1-my}
  \bigwedge_ff(x^0,\dots,
  x^{n-1})=0\land\bigwedge_{i<k}Dx^i=g^i(x^0,\dots,x^{n-1}), 
\end{equation}
where $k\leq n$ and the $g^i$ are rational functions over $K$.
Suppose this has solution $\tuple a$, which is a generic point of
a variety $V$.  It is enough to assume that 
$(a^0,\dots,a^{k-1})$ is a separating tran\-scendence-basis of $K(\tuple
a)/K$.  Then we have a dominant, separable rational map $\tuple
x\mapsto(x^0,\dots,x^{k-1})$ or $\phi$ from $V$ onto $\Aff^k$,
and another rational map $\tuple x\mapsto(g^0(\tuple x),\dots,g^{k-1}(\tuple
x))$ or $\psi$ from $V$ to $\Aff^k$.  So $(K,D)$ is existentially
closed if
and only if $V$ always has a $K$-rational point $P$ such that
$D(\phi(P))=\psi(P)$ \cite[Thm~1.6, p.~1328]{MR2114160}. 

\section{Fields with several derivations}\label{sect:several}

Let \myboxed{\mDF}{} be the theory of fields with $m$ commuting derivations.
Tracey McGrail \cite{MR2001h:03066} axiomatizes the model-completion,
\myboxed{\mDCF_0}, of $\mDF_0$.  Alternative axiomatizations arise as
special cases in 
work of Yoav Yaffe \cite{MR1807840} and Marcus Tressl
\cite{MR2159694}.  There is a common theme:  A differential ideal has
a generating set of a special form; in the terminology of Joseph Ritt
\cite[\S~I.5, p.~5]{MR0201431} (when $m=1$) and Ellis Kolchin
\cite[\S~I.10, pp.~81~ff.]{MR58:27929}, this is a \emph{characteristic set.}
There is a first-order way to tell, uniformly in the parameters,
whether a given set of differential 
polynomials is a characteristic set of some differential ideal, and
then to tell, if it \emph{is}
a characteristic set, whether it has a root.  In short, the function
$\phi\mapsto\robfn{\phi}$ in Robinson's criterion
(Lemma~\ref{lem:rob-crit}) is defined for
sufficiently many systems $\phi$.  (Applying Blum's criterion, McGrail
and Yaffe consider only systems in one variable, so they must include
inequations in these systems; Tressl uses only equations, in
arbitrarily many variables.)

I do not give the definition of a characteristic set, as not all
ingredients of the definition are needed for the arguments presented
in \S~\ref{sect:resolution}.  However, some of the ingredients
\emph{are} needed; these are in~\ref{subsect:terms}.

\subsection{Differential forms}\label{subsect:fga}

In \cite{MR2000487} I attempted to apply the geometric approach
described in~\ref{sect:first-der} to $\mDF_0$.  I worked more
generally with $\DF^m_0$, where \myboxed{\DF^m}{} is the theory of structures
$(K,D_0,\dots,D_{m-1})$ such that $(K,D_i)\models\DF$ for each $i$, and
each bracket $[D_j,D_k]$ is a $K$-linear combination of the $D_i$.
(This is roughly what Yaffe did too.)
In
\cite[\S~2]{MR2114160} I made some minor corrections and otherwise adapted
the argument to arbitrary characteristic.  Nonetheless, in May, 2006,
Ehud Hrushovski showed me a counterexample to \cite[Thm~A,
  p.~926]{MR2000487}, a theorem that was an introductory formulation of
 \cite[Thm~5.7, p.~942]{MR2000487}.  Then I found an error at
 the end of the proof of the latter theorem.  Before offering a resolution of
 the problem, let me review the general situation.


Let $(K,D_0,\dots,D_{m-1})\models\DF^m$, and let $\lrv$ be the
$K$-linear span of the $D_i$.  Then $\lrv$ is a Lie-ring, as well as a
vector-space over $K$.  As a vector-space, $\lrv$ has a dual,
\myboxed{\lrv^*}; and there is a derivation \myboxed{\dee}{} from $K$ into
$\lrv^*$ given by
\begin{equation}\label{eqn:dee-def}
  D(\dee x)=Dx.
\end{equation}
Then $\lrv^*$ has a basis $(\dee t^i\colon i<\ell)$ for some $t^i$ in
$K$ and some $\ell$ no greater than $m$ \cite[Lem.~4.4,
  p.~932]{MR2000487}, and this basis is dual to a basis
$(\partial_i\colon i<\ell)$ of $\lrv$, where 
$[\partial_i,\partial_j]=0$ in each case, and $\dee$ is then given by
\begin{equation}\label{eqn:dee}
  \dee x=\sum_{i<\ell}\dee t^i\cdot\partial_ix
\end{equation}
\cite[Lem.~4.7, p.~934]{MR2000487}.
We can use these ideas to find an extension of $(K,D_0,\dots,D_{m-1})$
in which the named derivations are linearly independent over the field.
Indeed, if $\ell<m$, then let $L=K(\alpha^{\ell},\dots,\alpha^{m-1})$,
where
$(\alpha^{\ell},\dots,\alpha^{m-1})$ is algebraically
independent over $K$.
Extend the original $\partial_i$ (where $i<\ell$) to $L$ so that they
are $0$ at the $\alpha^j$;
then, if $\ell\leq k<m$, define $\partial_k$ to be $0$ on $K$ and to be
$\deltaup_k^j$ at $\alpha^j$.  Then $(\partial_i\colon i<m)$ is a
linearly independent $m$-tuple of commuting derivations on $L$; from this, we
obtain linearly independent extensions of the $D_i$ to $L$ such that
the brackets $[D_j,D_k]$ are the same linear combinations of the $D_i$
as before (\cite[Lem.~5.2, p.~937]{MR2000487}---or \cite[Lem.~2.1,
  p.~1930]{MR495120}, by a different method---in characteristic $0$;
generally,  \cite[Lem.~2.4, p.~1334]{MR2114160}).  

Consequently, a model-companion $\DCF^m$ for $\DF^m$ can be obtained from a
model-com\-panion $\mDCF$ for $\mDF$ (\cite[Thm~5.3 and proof]{MR2000487} or
\cite[\S~3]{MR495120}).  This much stands, and differential forms
are convenient for establishing it.

But I tried also to obtain $\DCF^m$
independently as follows.  Suppose now we have
a separably closed field~$K$, along with a Lie-ring and
finite-dimensional space $\lrv$ of derivations 
of $K$; as a space, $\lrv$ has a basis $(\partial_i\colon i<m)$, whose
dual is $(\dee t^i\colon i<m)$, so that the $\partial_i$ commute.  We
may assume  that
$(K,\partial_0,\dots,\partial_{m-1})$ is differentiably perfect
\cite[Lem.~2.4]{MR2114160}.
Every system over $(K,\partial_0,\dots,\partial_{m-1})$
 is equivalent to a system of the
form of~\eqref{eqn:1-my}, generalized to
\begin{equation}\label{eqn:m-my}
  \bigwedge_ff(x^0,\dots,x^{n-1})=0\land
  \bigwedge_{j<k}\bigwedge_{i<m}\partial_ix^j=g_i^j(\tuple x).
\end{equation}
By means of~\eqref{eqn:dee}, we can also write this as
\begin{equation}\label{eqn:many-my}
  \bigwedge_ff(x^0,\dots,x^{n-1})=0\land
  \bigwedge_{j<k}\dee x^j=\sum_{i<m}\dee t^i\cdot g_i^j(\tuple x).
\end{equation}
If $\tuple a$ is a solution (from some extension), it is enough to
assume that $(a^0,\dots,a^\ell)$ is a
separating transcendence-basis of $K(\tuple a)/K$ for some $\ell$ such
that $k\leq\ell<m$.  That we cannot generally assume $k=\ell$ is an important
difference from the case of one derivation; it is what causes the
difficulties in the case of several derivations.  The solution $\tuple a$
to~\eqref{eqn:many-my} can be 
understood as follows.  First
we have the field $K(\tuple a)$, and then~\eqref{eqn:many-my} can be be
written as
\begin{equation}\label{eqn:step-1}
  \bigwedge_{j<k}\dee a^j=\sum_{i<m}\dee t^i\cdot g_i^j(\tuple a).
\end{equation}
A solution of this can be understood as a model
$(L,\tilde{\partial}_0,\dots,\tilde{\partial}_{m-1})$ of $\mDF$
extending $(K,\partial_0,\dots,\partial_{m-1})$ such that $K(\tuple
a)\included L$ and~\eqref{eqn:step-1} holds when $\dee
a^j=\sum_{i<m}\dee t^i\cdot\tilde{\partial}_ia^j$, that is, 
\begin{equation}\label{eqn:jkim}
\bigwedge_{j<k}\bigwedge_{i<m}\tilde{\partial}_ia^j=g_i^j(\tuple a).  
\end{equation}
Since the
$\tilde{\partial}_i$ commute, it is
necessary that 
\begin{equation}\label{eqn:jkhim}
\bigwedge_{j<k}\bigwedge_{h<i<m}\tilde{\partial}_h(g_i^j(\tuple
a))=\tilde{\partial}_i(g_h^j(\tuple a))
\end{equation}
\cite[\S~1, p.~926]{MR2000487}.
Any derivative with respect to $\tilde{\partial}_i$ of an element of
$K(\tuple a)$ is a constant plus a linear combination
of the derivatives  $\tilde{\partial}_ia^j$, where $j<\ell$ (by
\cite[Fact~1.1 (0, 2)]{MR2114160}, for example); we know
what these derivatives $\tilde{\partial}_ia^j$ are when $j<k$,
by~\eqref{eqn:jkim};
so~\eqref{eqn:jkhim} becomes a linear system over $K(\tuple a)$ in the
unknowns $\tilde{\partial}_ia^j$ where $k\leq j<\ell$.   

If $k=\ell$, then this linear system has no variables, so it is true
or false; its truth is a sufficient condition for~\eqref{eqn:step-1}
to have a solution.  If $k<\ell$, then the linear system is soluble or
not.  If it is soluble, then it is possible to extend the $\partial_i$
to derivations $\tilde{\partial}_i$ as required by~\eqref{eqn:jkim}
that commute on
$K(a^0,\dots,a^{k-1})$; but these derivations need not commute on all
of $K(\tuple a)$.  In \cite{MR2000487} I claimed that they could
commute, and that the solubility 
of~\eqref{eqn:jkhim} was sufficient for solubility
of~\eqref{eqn:step-1} in the sense above.  I was wrong.  

\subsection{A counterexample}
In the counterexample supplied by Hrushovski, the constants
$(m,k,\ell)$ of \S~\ref{subsect:fga} are $(2,2,3)$.
Let $(a,b,c)$ be an algebraically independent triple, and
consider the system
\begin{equation}\label{eqn:sys-ex}
  \dee a=\dee t^0\cdot c\cdot c+\dee t^1\cdot c,\qquad
\dee b=\dee t^0\cdot 2a+\dee t^1\cdot c;
\end{equation}
equivalently, by~\eqref{eqn:dee}, the system comprises the equations
  $\partial_0a= c\cdot c$, $\partial_1a=c$,
  $\partial_0b=2a$, and $\partial_1b=c$.
From these, we compute
\begin{equation*}
\partial_1\partial_0a=2c\cdot\partial_1c,\qquad
\partial_0\partial_1a=\partial_0c,\qquad   
\partial_1\partial_0b=2\cdot\partial_1a=2c,\qquad
\partial_0\partial_1b=\partial_0c.
\end{equation*}
Equating $\partial_0\partial_1$ and $\partial_1\partial_0$ yields the
linear system
\begin{equation*}
  \partial_0c-2c\cdot\partial_1c=0,\qquad
  \partial_0c=2c,
\end{equation*}
which has the solution $\partial_0c=2c$, $\partial_1c=1$.  But then we
must have
$\partial_1\partial_0c=2\cdot\partial_1c=2$, while
$\partial_0\partial_1c=\partial_01=0$, which means~\eqref{eqn:sys-ex}
has no solution, contrary to my claim in~\cite{MR2000487}.

For the record, the mistake is at the end of the proof of
\cite[Thm~5.7, p.~942]{MR2000487} and can be seen as follows.
Write the system~\eqref{eqn:sys-ex} as 
$\dee a=\alpha$, $\dee b=\beta$; then
\begin{gather}\label{eqn:dab}
\begin{aligned}
  \dee\alpha
&=\dee(\dee t^0\cdot c\cdot c+\dee t^1\cdot c)\\
&=\dee c\wedge\dee t^0\cdot 2c+\dee c\wedge\dee t^1\\
&=\dee c\wedge(\dee t^0\cdot 2c+\dee t^1),\\
\end{aligned}
\quad
\begin{aligned}
\dee\beta
&=\dee(\dee t^0\cdot 2a+\dee t^1\cdot c)\\
&=\dee a\wedge\dee t^0\cdot 2+\dee c\wedge\dee t^1\\
&=(\dee t^0\cdot
c\cdot c+\dee t^1\cdot c)
\wedge\dee t^0\cdot 2+\dee c\wedge\dee t^1\\
&=-\dee t^0\wedge\dee t^1\cdot 2c+\dee c\wedge\dee t^1\\
&=(\dee c-\dee t^0\cdot 2c)\wedge\dee t^1.
\end{aligned}
\end{gather}
Since also $\dee\beta=\dee^2b=0$,
\eqref{eqn:dab} imposes a condition on $\dee c\wedge\dee
t^1$, hence on $\partial_0c$; in particular, $\partial_0c=2c$, which
is what we found above.  But
there is no apparent
condition on $\partial_1c$, so I try introducing a new transcendental,
$d$, for this.  Then $\dee c=\dee t^0\cdot
2c+\dee t^1\cdot d$.  But this causes a problem, since it allows a
substitution in~\eqref{eqn:dab}, yielding
\begin{align*}
  \dee\alpha
&=(\dee t^0\cdot 2c+\dee t^1\cdot d)\wedge
(\dee t^0\cdot 2c+\dee t^1)\\ 
&=\dee t^0\wedge\dee t^1\cdot 2c(1-d),
\end{align*}
which means $d=1$, contrary to
assumption.  In short, the next to last sentence of the proof of
 \cite[Thm~5.7]{MR2000487} (beginning ``This ideal is linearly disjoint
 from'') is simply wrong.  (I had given no argument that it was
 correct.)

\section{Resolution}\label{sect:resolution}

To resolve the problem, it is better not to introduce differential
forms from the beginning, but to allow equations to involve any number
of applications of the derivations.  Examples (as
in~\ref{subsect:another}) may involve only one variable; but in
contrast to~\ref{subsect:one-var}, there does not seem to be an
advantage in restricting attention to this case.

\subsection{Another example}\label{subsect:another}

Over a
differential field $(K,\partial_0,\partial_1)$, consider the system
\begin{equation}\label{eqn:my-ex}
  \partial_0{}^n\partial_1{}^nx-x=0,\qquad
\partial_0{}^n\partial_1x-\partial_1{}^nx=0,
\end{equation}
where $n\geq2$.
Think of the derivatives $\partial_0{}^i\partial_1{}^jx$ as matrix
entries:
\begin{equation*}
  \begin{matrix}
    x&\partial_1x&\partial_1{}^2x&\dots\\
\partial_0x&\partial_0\partial_1x&\partial_0\partial_1{}^2x&\dots\\
\partial_0{}^2x&\partial_0{}^2\partial_1x&\partial_0{}^2\partial_1{}^2x&\dots\\
\hdotsfor{4}
  \end{matrix}
\end{equation*}
Treat each derivative  $\partial_0{}^i\partial_1{}^jx$ as a new
variable, $x^{(i,j)}$.  In case $n=3$, the system~\eqref{eqn:my-ex}
determines a $4\times4$ matrix $(a^{(i,j)})^{i<4}_{j<4}$ in which
$a^{(3,3)}=a^{(0,0)}$ and $a^{(3,1)}=a^{(0,3)}$; we can depict this as
follows: 
\begin{equation}\label{eqn:matrix}
 \begin{matrix}
    a&*&*&b\\
    *&*&*&*\\
    *&*&*&*\\
    *&b&*&a
  \end{matrix}
\end{equation}
We first ask whether the derivations can be
extended to commuting derivations on the field
$K(a^{(i,j)}\colon(i,j)\leq(3,3))$, where $\leq$ is
the product order on $\vnn^2$, so that
$\partial_0a^{(i,j)}=a^{(i+1,j)}$ when $(i+1,j)\leq(3,3)$ and 
$\partial_1a^{(i,j)}=a^{(i,j+1)}$ when $(i,j+1)\leq(3,3)$.
That is, writing $a^{(4,j)}$ for $\partial_0a^{(3,j)}$, and
$a^{(i,4)}$ for $\partial_1a^{(i,3)}$, we ask whether $\partial_0$ and
$\partial_1$ can map $K(a^{(i,j)}\colon(i,j)\leq(3,3))$ into 
 $K(a^{(i,j)}\colon(i,j)\leq(4,3))$ and
 $K(a^{(i,j)}\colon(i,j)\leq(3,4))$ respectively, in the manner
suggested by the notation; and we ask further whether $\partial_0$ and
$\partial_1$ can still commute, which means there should be
$a^{(4,4)}$ so that 
$\partial_0$ maps
$K(a^{(i,j)}\colon(i,j)\leq(3,4))$, and
$\partial_1$ maps
$K(a^{(i,j)}\colon(i,j)\leq(4,3))$, 
into
$K(a^{(i,j)}\colon(i,j)\leq(4,4))$.
In short, can
the matrix~\eqref{eqn:matrix} be extended by one row and
column?  Since $a^{(3,3)}=a^{(0,0)}$, we must have
$a^{(4,3)}=a^{(1,0)}$, and so forth; none of this causes any problem,
and the new matrix can be depicted:
\begin{equation}\label{eqn:matrix1}
  \begin{array}{cccc|c}
    a     &d     &*     &b     &g     \\
    c     &e     &*     &f     &h     \\
    *     &*     &*     &*     &*     \\
    *     &b     &g     &a     &d     \\\cline{1-4}
    *     &f     &h     &\multicolumn{1}{c}{c}     &\multicolumn{1}{c}{e}
  \end{array}
\end{equation}
To extend the
  derivations further to
$K(a^{(i,j)}\colon(i,j)\leq(4,4))$, we require a new condition on
 the original matrix $(a^{(i,j)})^{i<4}_{j<4}$,
 namely, $a^{(3,0)}=a^{(0,2)}$; this comes out when we try to
  extend~\eqref{eqn:matrix1} by one column:
  \begin{equation}\label{eqn:matrix2}
 \begin{array}{cccc|cc}
    a     &d     &j     &b     &g     &a\\
    c     &e     &*     &f     &h     &c\\
    i     &*     &*     &*     &*     &i\\
    j     &b     &g     &a     &d     &j\\\cline{1-4}
    *     &f     &h
    &\multicolumn{1}{c}{c}&\multicolumn{1}{c}{e}&\multicolumn{1}{c}{*}
   \end{array}
  \end{equation}
Indeed, as $a^{(0,4)}=a^{(3,2)}$, so $a^{(0,5)}=a^{(3,3)}=a^{(0,0)}$,
so column~$5$ must be the same as column~$0$; also
$a^{(0,1)}=a^{(3,4)}$, whence $a^{(0,2)}=a^{(3,5)}=a^{(3,0)}$.
This does not mean that~\eqref{eqn:my-ex} is insoluble; it is.  But
the additional condition found 
in~\eqref{eqn:matrix2} does mean that the associated non-homogeneous system
\begin{equation}\label{eqn:my-ex1}
  \partial_0{}^n\partial_1{}^nx-x=0,\qquad
\partial_0{}^n\partial_1x-\partial_1{}^nx=1
\end{equation}
is insoluble, although this is not clear from~\eqref{eqn:matrix}
or~\eqref{eqn:matrix1}. 
Alternatively, we can work with the differential ideal, looking
for a characteristic set as mentioned at the beginning of this section:
\begin{equation*}
\begin{split}
&\phantom{{}={}}[\partial_0{}^n\partial_1{}^nx-x,\ 
\partial_0{}^n\partial_1x-\partial_1{}^nx-1]\\
&=[\partial_1{}^{2n-1}x-x,\ 
\partial_0{}^n\partial_1x-\partial_1{}^nx-1]\\
&=[\partial_1{}^{2n-1}x-x,\ 
\partial_0{}^n\partial_1x-\partial_1{}^nx-1,\
\partial_1{}^{3n-2}x-\partial_0{}^nx]\\
&=[\partial_1{}^{2n-1}x-x,\ 
\partial_0{}^n\partial_1x-\partial_1{}^nx-1,\
\partial_0{}^nx-\partial_1{}^{n-1}x]\\
&=[\partial_1{}^{2n-1}x-x,\ 
1,\
\partial_0{}^nx-\partial_1{}^{n-1}x]
=[1],
\end{split}
\end{equation*}
so~\eqref{eqn:my-ex1} is insoluble; but higher-order derivatives were
needed to discover this.  

By the same computation, the differential ideal 
$[\partial_0{}^n\partial_1{}^nx-x,\ 
\partial_0{}^n\partial_1x-\partial_1{}^nx]$ has the characteristic set
$\{\partial_1{}^{2n-1}x-x,
\partial_0{}^nx-\partial_1{}^{n-1}x\}$, which, in the manner described above,
determines the following matrix in case $n=2$:
\begin{equation}\label{eqn:matrix-n2}
\begin{matrix}
    a&b&*&a\\
    *&*&*&*\\
    b&*&*&*
  \end{matrix}
\end{equation}
Therefore~\eqref{eqn:my-ex}
has a solution, simply because the ordinary ideal
$(x^{(0,2n-1)}-x,x^{(n,0)}-x^{(0,n-1)})$ 
has a zero.  One way of justifying this conclusion is
\cite[Lem.~3.1.2]{MR2001h:03066}; another way will be
Theorem~\ref{thm:second}, according to
which it is enough to 
observe that the dependencies in~\eqref{eqn:matrix-n2} extend in a `nice'
way to a triangle:
\begin{equation}
\begin{array}{cccc|cc} \label{eqn:triangle}
  a&b&c&a&b&c\\
  d&e&f&d&e& \\
  b&c&a&b& &  \\\cline{1-4}
  e&f&d&\multicolumn{3}{c}{}  \\
  c&a& &\multicolumn{3}{c}{}  \\
  f& & &\multicolumn{3}{c}{}  
\end{array}
\end{equation}

\subsection{Terminology}\label{subsect:terms}

I shall now avoid working
with differential polynomials as such, but shall work instead with the
algebraic dependencies that they determine.

Let $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$.  Higher-order
derivatives with respect to the $\partial_i$ can be indexed by
elements of $\vnn^m$:  so, for
$\partial_0{}^{\sigma(0)}\dotsb\partial_{m-1}{}^{\sigma(m-1)}x$, we
may write \myboxed{\partial^{\sigma}x}.
Let \myboxed{\leq}{} be the product ordering of $\vnn^m$.
Then the derivative $\partial^{\sigma}x$ is
\defn{below}{} $\partial^{\tau}x$ (and the latter is \defn{above}{} the
former) if $\sigma\leq\tau$.  If $n\in\vnn$, then two elements of
$\vnn^n\times n$ will be related by $\leq$ only if they agree in the
last coordinate, so that
\begin{equation*}
  (\sigma,k)\leq(\tau,\ell)\iff\sigma\leq\tau\land k=\ell;
\end{equation*}
we may use the corresponding terminology of above and below, so that
$\partial^{\sigma}x_k$ is below~$\partial^{\tau}x_{\ell}$ if (and only if)
$(\sigma,k)\leq(\tau,\ell)$. 

If
$\sigma\in\vnn^m$, let the sum $\sum_{i<m}\sigma(i)$ be denoted by
\myboxed{\size{\sigma}}: this is the \defn{height}{} of $\sigma$ or of
$\partial^{\sigma}x$.  (Kolchin \cite[\S~I.1, p.~59]{MR58:27929} uses
the word \emph{order}.) 
If $n$ is a positive integer, let $\vnn^m\times n$ be (totally) ordered by
\myboxed{\tleq}, which is taken from 
the left lexicographic ordering 
of $\vnn^{m+1}$ by means of the embedding
\begin{equation*}
  (\xi,k)\longmapsto
(\size{\xi},k,\xi(0),\dots,\xi(m-2))
\end{equation*}  
of $\vnn^m\times n$ in $\vnn^{m+1}$.  
Then $(\vnn^m\times n,\tleq)$ is isomorphic to $(\vnn,\leq)$.  
In case $n=1$, and $m$ is $2$ or $3$, the picture is thus:
\begin{center}
\hfill
  \begin{pspicture}(-1,-2.5)(3,0.5)
    \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}
\hfill
\begin{pspicture}(-2.5,-0.5)(2.5,3)
    \psset{radius=0.1}
    \Cnode(0,2.4){a}
    \uput[u](0,2.4){$(0,0,2)$}
    \Cnode(0.7,1.2){b}
    \uput[ur](0.7,1.2){$(0,1,1)$}
    \ncline{->}{a}{b}
    \Cnode(1.4,0){c}
    \uput[r](1.4,0){$(0,2,0)$}
    \ncline{->}{b}{c}
    \Cnode(-0.7,1.2){d}
    \uput[ul](-0.7,1.2){$(1,0,1)$}
    \ncline{->}{c}{d}
    \Cnode(0,0){e}
    \uput[d](0,0){$(1,1,0)$}
    \ncline{->}{d}{e}
    \Cnode(-1.4,0){f}
    \uput[l](-1.4,0){$(2,0,0)$}
    \ncline{->}{e}{f}  
\end{pspicture}
\hfill
\mbox{}
\end{center}

We may
write $(\sigma,k)\tl\infty$ for all $(\sigma,k)$ in $\vnn^m\times n$.
Suppose $(x_h\colon h<n)$ is a tuple of indeterminates.  By ordering
the formal derivatives $\partial^{\sigma}x_k$ in terms of $(\sigma,k)$ and
$\tleq$, we have Kolchin's example of an \emph{orderly ranking} of
derivatives \cite[\S~I.8, p.~75]{MR58:27929}.  If
$(\sigma,k)\tl(\tau,\ell)$, I shall say that the derivative
$\partial^{\sigma}x_k$ is
\defn{less}{} than $\partial^{\tau}x_{\ell}$ or is a \defn{predecessor}{}
of $\partial^{\tau}x_{\ell}$, and
 $\partial^{\tau}x_{\ell}$ is \defn{greater}{} than 
 $\partial^{\sigma}x_k$; likewise for the expressions $a_k^{\sigma}$
and $a_{\ell}^{\tau}$, introduced in~\eqref{eqn:L} below.  (So, the
terms just defined refer
to the total ordering $\tl$, while `below' and `above' refer to the
partial ordering $\leq$.)

Addition and subtraction on $\vnn$ induce corresponding operations on
$\vnn^m$.  Then 
\begin{gather}\notag
\tau\leq\sigma+\tau,\\\notag
\partial^{\sigma}\partial^{\tau}x_k=
\partial^{\sigma+\tau}x_k,\\\label{eqn:sks}
(\sigma,k)\tleq(\sigma+\tau,k),\\\notag
(\sigma,k)\tleq(\tau,\ell)\Iff 
(\sigma+\rho,k)\tleq(\tau+\rho,\ell).
\end{gather}
If $i<m$, let \myboxed{\ichar}{} denote the
characteristic function of $\{i\}$ in $\vnn^m$, so that
$\partial^{\ichar}=\partial_i$, and more generally
$\partial_i\partial^{\sigma}=\partial^{\sigma+\ichar}$, and
$\partial_i\partial^{\sigma-\ichar}=\partial^{\sigma}$. 

Let $L$ be an extension of $K$ with generators that are indexed by an initial
segment of $(\vnn^m\times n,\tleq)$; that is,
\begin{equation}\label{eqn:L}
  L=K(a_h^{\xi}\colon(\xi,h)\tl(\tau,\ell)),
\end{equation}
where $(\tau,\ell)\in\vnn^m\times n$, or possibly $(\tau,\ell)=\infty$,
in which case $L=K(a_h^{\xi}\colon(\xi,h)\in\vnn^m\times n)$. 
By~\eqref{eqn:sks}, if $a_k^{\sigma+\ichar}$ is one of the generators
of $L/K$, then so is $a_k^{\sigma}$.  Let us say that $L$, with the
given generators, meets the \defn{differential condition}{} if there is
no obstacle to extending each derivation $\partial_i$ to a derivation
$D_i$ on $K(a_h^{\xi}\colon(\xi+\ichar,h)\tl(\tau,\ell))$ such that
\begin{equation}\label{eqn:Dia}
  D_ia_k^{\sigma}=a_k^{\sigma+\ichar}
\end{equation}
whenever $(\sigma+\ichar,k)\tl(\tau,\ell)$.
(So, if the right-hand member of~\eqref{eqn:Dia} is not defined,
then the left need not be defined.)  Formally, the
differential condition is that, if $f$ is a rational function over $K$
in variables $(x_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))$, where
$(\sigma+\ichar,k)\tl(\tau,\ell)$ for some $i$ in $m$, and if
\begin{equation}\label{eqn:fahx}
  f(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))=0,
\end{equation}
then we may apply $D_i$ to this, assuming~\eqref{eqn:Dia}, to get
\begin{equation}\label{eqn:sumeta}
  \sum_{(\eta,g)\tleq(\sigma,k)}
\frac{\partial f}{\partial
  x_g^{\eta}}(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))\cdot
a_g^{\eta+\ichar}+ 
f^{\partial_i}(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))=0.
\end{equation}
(Note well the assumption that
 $(\sigma+\ichar,k)\tl(\tau,\ell)$.  In~\eqref{eqn:sumeta}, each
of the $a_g^{\eta+\ichar}$ must exist, even though the
coefficient $(\partial f/\partial
 x_g^{\eta})(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))$ might be $0$.)
So the differential condition is \emph{necessary} for the
extensibility of the $\partial_i$ as desired (see for example
\cite[Fact~1.1~(0)]{MR2114160}); sufficiency is part of
Lemma~\ref{lem:above} below.

An extension $(M,D_0,\dots,D_{m-1})$ of
$(K,\partial_0,\dots,\partial_{m-1})$ is \defn{compatible}{} with the
extension $L$ of $K$ given in~\eqref{eqn:L} if $L\included M$,
and~\eqref{eqn:Dia} holds whenever
$(\sigma+\ichar,k)\tl(\tau,\ell)$. 

Borrowing some terminology
used for differential polynomials \cite[\S~IX.1, p.~163]{MR0201431},
let us say that a generator $a_k^{\sigma}$ of $L/K$ is a \defn{leader}{} if
it is algebraically dependent over $K$ on its predecessors, that is, 
\begin{equation*}
a_k^{\sigma}\in\alg{K(a_h^{\xi}\colon(\xi,h)\tl(\sigma,k))}.  
\end{equation*}
Then
$a_k^{\sigma}$ is a \defn{separable}{} leader if it is separably
algebraic over $K(a_h^{\xi}\colon(\xi,h)\tl(\sigma,k))$; otherwise, it
is an \defn{inseparable}{} leader.  A separable
leader $a_k^{\sigma}$ is \defn{minimal}{} if
there is no separable leader strictly below it---no separable leader
$a_k^{\rho}$ such that $\rho<\sigma$.

For example, in the field $K(a^{(i,j)}\colon(i,j)\leq(3,3))$ depicted
in~\eqref{eqn:matrix} above, the generator
$a^{(3,3)}$ is a (non-minimal) separable leader, and
$a^{(3,1)}$ is a minimal separable leader.  But here we wanted $a^{(3,3)}$
ultimately to be a derivative of $a^{(3,1)}$, namely
 $\partial_1{}^2a^{(3,1)}$.  Passing to a larger field
in~\eqref{eqn:matrix2}, we found the
condition $a^{(3,0)}=a^{(0,2)}$; then $a^{(3,0)}$ became a new separable
leader, strictly below than the formerly minimal separable leader $a^{(3,1)}$.

\begin{lemma}\label{lem:above}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $L$ is
an extension
$K(a_h^{\xi}\colon(\xi,h)\tl(\tau,\ell))$ of $K$ meeting the
differential condition.  Then the derivations
$\partial_i$ extend to derivations $D_i$ from
$K(a_h^{\xi}\colon(\xi+\ichar,h)\tl(\tau,\ell))$ into $L$
 such that~\eqref{eqn:Dia} holds when $(\sigma+\ichar,k)\tl(\tau,\ell)$.
 If $a_k^{\sigma}$ is a
 separable leader, and $(\sigma+\ichar,k)\tl(\tau,\ell)$, then
 \begin{equation}\label{eqn:aksi}
   a_k^{\sigma+\ichar}\in K(a_h^{\xi}\colon (\xi,h)\tl(\sigma+\ichar,k))
 \end{equation}
(that is, $a_k^{\sigma+\ichar}$ is a rational function over $K$ of its
 predecessors);
in particular, $a_k^{\sigma+\ichar}$ is a separable leader.  Therefore
generators of $L/K$ that are above separable leaders are themselves
separable leaders.
\end{lemma}

\begin{proof}
The claim follows from the basic properties of derivations, such as
are gathered in \cite[\S~1]{MR2114160}.
Let $B$ comprise the generators of $L/K$ that are not leaders.  Then
$B$ is algebraically independent over $K$, so we are free to extend
the $\partial_i$ to $D_i$ on $K(B)$ so that~\eqref{eqn:Dia} holds
whenever it applies.  Then these extensions are uniquely determined,
and there are further unique extensions of the $D_i$
to the separable closure $\sep{K(B)}$.  In particular, suppose
$a_k^{\sigma}$ is a separable leader, and
$(\sigma+\ichar,k)\tl(\tau,\ell)$.   Then $D_ia_k^{\sigma}$ is
obtained by differentiating the minimal polynomial of $a_k^{\sigma}$
over $K(B)$.  That is, $D_ia_k^{\sigma}$ is obtained by
differentiating an equation like~\eqref{eqn:fahx}; by the differential
condition, $D_ia_k^{\sigma}$ must be $a_k^{\sigma+\ichar}$ as given
by~\eqref{eqn:sumeta}; this shows that $a_k^{\sigma+\ichar}$ is a
rational function over $K$ of its predecessors.  

Finally, in a
positive characteristic~$p$, there may be an inseparable leader
$a_k^{\sigma}$.  Then $(a_k^{\sigma})^{p^r}\in \sep{K(a_h^{\xi}\colon
  (\xi,h)\tl(\sigma,k))}$ for some positive $r$.  If
$(\sigma+\ichar,k)\tl(\tau,\ell)$, then we are free to define
$D_ia_k^{\sigma}$ as $a_k^{\sigma+\ichar}$, provided
$D_i((a_k^{\sigma})^{p^r})=0$.  But again this condition is ensured by
the differential condition.  Indeed, we may suppose~\eqref{eqn:fahx}
shows the separable dependence of $(a_k^{\sigma})^{p^r}$ over the
predecessors of $a_k^{\sigma}$.  
That is, we can understand $f(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))$
as $g((a_k^{\sigma})^{p^r})$ for some separable polynomial $g$ over
$K(a_h^{\xi}\colon(\xi,h)\tl(\sigma,k))$. 
Then $D_i((a_k^{\sigma})^{p^r})$ is
obtained from~\eqref{eqn:sumeta}, provided we replace the term
$(\partial f/\partial
x_k^{\sigma})(a_h^{\xi}\colon(\xi,h)\tleq(\sigma,k))\cdot
a_k^{\sigma+\ichar}$ with
$g'((a_k^{\sigma})^{p^r})\cdot
D_i((a_k^{\sigma})^{p^r})$.  But in the present case, the former term
is $0$.  Since,
after the replacement, the resulting 
equation still holds, we must have $D_i((a_k^{\sigma})^{p^r})=0$.
\end{proof}

\subsection{An example}\label{subsect:example}

The differential ideal 
$[\partial^{(1,1)}x-\partial^{(0,2)}x,
\partial^{(1,2)}x-\partial^{(2,0)}x]$,
with two generators,
is equal to $[f,g]$, where
\begin{equation*}
f=\partial^{(1,1)}x-\partial^{(0,2)}x,\qquad
g=\partial^{(0,3)}x-\partial^{(2,0)}x.
\end{equation*}
Then
$\partial_1{}^2f+\partial_1g-\partial_0g=
\partial^{(1,3)}x-\partial^{(0,4)}x+
\partial^{(0,4)}x-\partial^{(2,1)}x-\partial^{(1,3)}x+\partial^{(3,0)}x=h$, 
where
\begin{equation*}
h=\partial^{(3,0)}x-\partial^{(2,1)}x. 
\end{equation*}
Then $[f,g]=[f,g,h]$, and the conditions imposed by
$f$, $g$, and $h$ are as in 
the following triangle, where the minimal separable leaders in the
present sense are
underlined (though we have not yet checked the corresponding field
meets the differential condition):
\begin{equation}\label{eqn:triangle2}
  \begin{matrix}
    *&*&a&\underline b\\
    *&\underline a&*&\\
    b&c&&\\
    \underline c&&&
  \end{matrix}
\end{equation}
These leaders have common derivatives, which however impose no new
conditions. (In the terminology introduced by Azriel Rosenfeld \cite[\S~I.2,
  p.~397]{MR0107642} in characteristic $0$, the set $\{f,g,h\}$ is
\emph{coherent.})  For example,
\begin{multline*}
  \partial^{(3,0)}g-
\partial^{(0,3)}h
=\partial^{(3,0)}(\partial^{(0,3)}x-\partial^{(2,0)}x)-
\partial^{(0,3)}(\partial^{(3,0)}x-\partial^{(2,1)}x)\\
=\partial^{(3,3)}x-\partial^{(5,0)}x-
\partial^{(3,3)}x+\partial^{(2,4)}x
=\partial^{(2,4)}x-\partial^{(5,0)}x,
\end{multline*}
where we used derivatives as great as
$\partial^{(3,3)}x$; but we already had
\begin{multline*}
  \partial^{(2,1)}g-\partial^{(2,0)}h
=\partial^{(2,1)}(\partial^{(0,3)}x-\partial^{(2,0)}x)-
\partial^{(2,0)}(\partial^{(3,0)}x-\partial^{(2,1)}x)\\
=\partial^{(2,4)}x-\partial^{(4,1)}x-
\partial^{(5,0)}x+\partial^{(4,1)}x
=\partial^{(2,4)}-\partial^{(5,0)}x,
\end{multline*}
where no derivative is as great as $\partial^{(3,3)}x$.  This checking
did require some derivatives that were not \emph{below}
$\partial^{(3,3)}x$; but they do all belong to the following larger
triangle, whose corresponding field meets the differential condition
(though note how the $c$ from~\eqref{eqn:triangle2} changes to $b$): 
\begin{equation}\label{eqn:triangle3}
  \begin{array}{cccc|ccc}
    *&*&a&\underline b&b&b&b\\
    *&\underline a&b&b&b&b& \\
    b&b&b&b&b& & \\
    \underline 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}

We shall see that extensibility to triangles as
in~\eqref{eqn:triangle3}, without introduction of new minimal
separable leaders, is sufficient to guarantee 
solutions.

\subsection{A solubility condition}

If $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, then this model
has an extension whose underlying field is the separable closure of
$K$ (as by \cite[Lem.~3.4, p.~930]{MR2000487} and \cite[Lem.~2.4,
  p.~1334]{MR2114160}).  We shall need this in a more general form:

\begin{lemma}\label{lem:comm}
  Suppose a field $M$ has two subfields $L_0$ and $L_1$, which in turn
  have a common subfield $K$.  For each $i$ in $2$, suppose there is a
  derivation $D_i$ mapping $K$ into $L_i$ and $L_{1-i}$ into $M$.
  Then the bracket $[D_0,D_1]$ is a well-defined
  derivation on $K$.  Suppose it is the $0$-derivation, that is, the
  following diagram commutes.
  \begin{equation*}
    \begin{CD}
      K @>{D_1}>> L_0\\
@V{D_0}VV @VV{D_0}V\\
L_1 @>>{D_1}> M
    \end{CD}
  \end{equation*}
Suppose also that $a$ is an element of $M$ that is separably algebraic over
$K$.  Then each $D_i$ extends uniquely to $K(a)$, and $D_ia\in
L_{1-i}(a)$, so $D_{1-i}D_ia$ is also well-defined.  Moreover, $[D_0,D_1]a=0$.
\end{lemma}

\begin{proof}
  The claim follows from standard facts, at least if $L_0=K=L_1$; but
  the proof is the same in the general case.  Indeed, though the
  derivations $D_0$ and $D_1$ are defined on $K$, their bracket
  $[D_0,D_1]$ need not be so, since the compositions $D_0D_1$ and
  $D_1D_0$ need not be so; but if they are, then $[D_0,D_1]$ is a
  \emph{derivation} on $K$.
  A derivation on $K$ extends uniquely to $\sep K$; if the derivation
  is $0$ on $K$, then it is $0$ on $\sep K$
  \cite[Fact~1.1~(2)]{MR2114160}.  
In the present case, as $a\in \sep K$, so $D_ia\in L_{1-i}(a)$, and
  therefore $D_ia\in\sep{L_{1-i}{}}$; hence $D_{1-i}D_ia$ is defined.
Thus $[D_0,D_1]$ is
  defined on $K(a)$, where $a\in \sep K$; and if the bracket is $0$ on
  $K$, then is $0$ at $a$.
\end{proof}

Another example illustrates the following theorem in positive
characteristic.  Over a differential field
$(K,\partial_0,\partial_1)$, where $\Char K=p>0$,
the differential equation
\begin{equation*}
  (\partial_0x)^p+x=\partial_1x
\end{equation*}
determines an extension
$K(a^{\xi}\colon\size{\xi}\leq 2)$ of $K$ that meets the differential
condition: the generators form a triangle thus: 
\begin{equation*}
  \begin{matrix}
    a&b^p+a&b^p+a\\
    b&b    &     \\
    c&     &
  \end{matrix}
\end{equation*}
where $(a,b,c)$ is algebraically independent over $K$.
In particular, $a^{(1,0)}$ (which has the value $b$) is an inseparable leader,
but none of the generators of height $2$ (namely, $a^{(0,2)}$,
$a^{(1,1)}$ and $a^{(2,0)}$) is an inseparable 
leader. 

\begin{theorem}\label{thm:first}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $K$
has an extension $K(a_h^{\xi}\colon\size{\xi}\leq2r\land
h<n)$ meeting the differential condition for some positive integers
$r$ and $n$.  Suppose further that, whenever $a_k^{\sigma}$ is a
minimal separable leader, 
then $\size{\sigma}\leq r$.  Then  $(K,\partial_0,\dots,\partial_{m-1})$
has an extension 
$(M,D_0,\dots,D_{m-1})$
compatible with
$K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$.  
\end{theorem}

\begin{proof}
  The claim can be compared to and perhaps derived from a
  differential-alge\-braic lemma of 
  Rosenfeld \cite[\S~I.2]{MR0107642}, at least in
  characteristic~$0$.  Here I give an independent 
  argument, for arbitrary characteristic.  
We shall obtain $M$ recursively as
  $K(a_h^{\xi}\colon(\xi,h)\in\vnn^m\times n)$, at the same time
  proving inductively that the $\partial_i$ can be extended to $D_i$ so
  that~\eqref{eqn:Dia} holds in all cases.

Let $L=K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$; this is
$K(a_h^{\xi}\colon(\xi,h)\tleq((2r-1,0,\dots,0),n-1))$.  Then
by~\eqref{eqn:sumeta}, the differential
condition requires of the tuple 
$(a_h^{\xi}\colon\size{\xi}=2r\land h<n)$ only that it solve
some linear equations over $L$.  The hypothesis of our claim is that
there \emph{is} a solution, namely
 $(a_h^{\xi}\colon\size{\xi}=2r\land h<n)$.  We may therefore
assume that this tuple is a
\emph{generic} solution of these equations.  In particular, no entry
of this tuple is an inseparable leader.  (If the entries of
$(a_h^{\xi}\colon\size{\xi}=2r\land h<n)$ were chosen from the field $L$,
then this field
would be closed under the desired extensions $D_i$ of $\partial_i$,
and the derivations $D_i$ would commute on the subfield
$K(a_h^{\xi}\colon\size{\xi}+1<2r\land h<n)$; but they might not
commute on all of $L$.) 

Now, as an inductive hypothesis, suppose we have the extension
$K(a_h^{\xi}\colon(\xi,h)\tl(\tau,{\ell}))$ of $K$ meeting the differential
condition, so that there are derivations $D_i$ as given by
Lemma~\ref{lem:above}; suppose also that
\begin{enumerate}
\item
if $a_k^{\sigma}$ is a minimal separable leader, then
$\size{\sigma}\leq r$;
\item
if $a_k^{\sigma}$ is an inseparable leader, then
$\size{\sigma}<2r$.
\end{enumerate}
We need
to choose $a_{\ell}^{\tau}$ in such a way that these conditions still
hold for $K(a_h^{\xi}\colon(\xi,h)\tleq(\tau,{\ell}))$.  
The inductive
hypothesis is correct when $\size{\tau}\leq2r$, and then the desired
conclusion follows; so we may
assume $\size{\tau}>2r$.  
Hence, if $\tau(i)>0$, so that $\tau-\ichar$ is defined, then
$\size{\tau-\ichar}\geq2r$, so
$a_{\ell}^{\tau-\ichar}$ is not an inseparable leader.  

If $a_{\ell}^{\tau-\ichar}$ is not a
leader at all, for any $i$ in $m$, then we may let $a_{\ell}^{\tau}$ be a
new transcendental, and we may define each derivative
$D_ia_{\ell}^{\tau-\ichar}$ as this \cite[Fact~1.1~(1)]{MR2114160}.  

In the other case, $a_{\ell}^{\tau-\ichar}$ is a
separable leader for some $i$.  Then $D_ia_{\ell}^{\tau-\ichar}$ is determined
(Lemma~\ref{lem:above}).  We want to let $a_{\ell}^{\tau}$ be this
derivative.  However, possibly also $a_{\ell}^{\tau-\jchar}$ is a
separable leader, where $i\neq j$.  In this case, we must check that
\begin{equation}\label{eqn:comm}
  D_ja_{\ell}^{\tau-\jchar}=D_ia_{\ell}^{\tau-\ichar},
\end{equation}
that is, $[D_i,D_j]a_{\ell}^{\tau-\ichar-\jchar}=0$. 

There are minimal separable leaders $a_{\ell}^{\pi}$ and $a_{\ell}^{\rho}$
below $a_{\ell}^{\tau-\ichar}$ and
$a_{\ell}^{\tau-\jchar}$ respectively.
Let $\nu$ be $\pi\vee\rho$,
the least upper bound of $\{\pi,\rho\}$ with respect to $\leq$.
Then $\nu\leq\tau$.  But
$\size{\nu}\leq\size{\pi}+\size{\rho}\leq2r<\size{\tau}$; so $\nu<\tau$. 
Hence $\nu\leq\tau-\kchar$ for some $k$ in $m$, which means
$a_{\ell}^{\nu}$ is below $a_{\ell}^{\tau-\kchar}$.
Consequently,
\begin{enumerate}
  \item
$a_{\ell}^{\pi}$ is below both $a_{\ell}^{\tau-\ichar}$ and
    $a_{\ell}^{\tau-\kchar}$;
\item
$a_{\ell}^{\rho}$ is below both $a_{\ell}^{\tau-\jchar}$ and
    $a_{\ell}^{\tau-\kchar}$.
\end{enumerate}
If $k=j$, then $a_{\ell}^{\pi}$ is below
$a_{\ell}^{\tau-\ichar-\jchar}$, so this is a separable leader.  As
$D_i$ and $D_j$ commute on
$K(a_h^{\xi}\colon (\xi,h)\tl(\tau-\ichar-\jchar,{\ell}))$ by the 
differential condition, they must
commute also at $a_{\ell}^{\tau-\ichar-\jchar}$ (Lemma~\ref{lem:comm}),
so~\eqref{eqn:comm} is established.  The argument is the same if
$k=i$.  If $k$ is different from $i$ and $j$, then again the same
argument yields
 $D_ja_{\ell}^{\tau-\jchar}=D_ka_{\ell}^{\tau-\kchar}$ and
 $D_ka_{\ell}^{\tau-\kchar}=D_ia_{\ell}^{\tau-\ichar}$,
so~\eqref{eqn:comm} holds.   

In no case did we introduce a new minimal separable leader or an
  inseparable leader.  This completes the induction and the
  proof.  
\end{proof}

The claim at the end of the last subsection (\ref{subsect:example}) is
now justified.

In terms of differential polynomials and ideals, the theorem can be
understood as follows.  Given the hypothesis of the theorem, let $S$
be the set of differential polynomials
$f(\partial^{\xi}x_h\colon\size{\xi}<2r\land h<n)$, where $f$
ranges over the ordinary polynomials over $K$ such that
$f(a^{\xi}_h\colon\size{\xi}<2r\land h<n)=0$.  Then $S$
includes a characteristic set for the differential ideal that it
generates.  

We can now characterize the existentially closed models of $\mDF$ by
means of the following lemma.  The lemma follows from unproved
statements in \cite[\S~0.17, p.~49]{MR58:27929}; let's just prove it here.  

\begin{lemma}\label{lem:fin}
For every $m$ in $\vnn$ and positive integer $n$, 
  every antichain of $(\vnn^m\times n,\leq)$ is finite.
\end{lemma}

\begin{proof}
The general case
follows from the case when $n=1$, since if $S$ is an antichain of
$(\vnn^m\times n,\leq)$, then 
\begin{equation*}
S=\bigcup_{j<n}\{(\xi,h)\in S\colon h=j\}, 
\end{equation*}
and each component of the union is in bijection with
an antichain of $(\vnn^m,\leq)$.  As an inductive hypothesis,
suppose every antichain of 
$(\vnn^{\ell},\leq)$ is finite; but suppose also, if possible, that
there is an infinite 
antichain $S$ of $(\vnn^{\ell+1},\leq)$.  Then $S$ contains some
$\sigma$.  By inductive hypothesis, the subset
\begin{equation*}
  \bigcup_{j\leq\ell}\bigcup_{i\leq\sigma(j)}\{\xi\in S\colon\xi(j)=i\}
\end{equation*}
of $S$
is a finite union of finite sets, so its complement in
$S$ has infinitely many elements $\tau$; but then $\sigma<\tau$, so
$S$ was not an antichain.
\end{proof}

\begin{theorem}\label{thm:dcf}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$.  Then the
following are equivalent:
\begin{enumerate}
  \item\label{item:ec}
The model
$(K,\partial_0,\dots,\partial_{m-1})$ of $\mDF$ is existentially closed.
\item\label{item:<}
For all positive integers $r$ and $n$, if $K$ has an extension
$K(a_h^{\xi}\colon\size{\xi}\leq2r\land h<n)$ meeting the
differential condition such that $\size{\sigma}\leq r$ whenever
$a_k^{\sigma}$ is a minimal separable leader, then the tuple
$(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$ has a
specialization $(\partial^{\xi}b_h\colon\size{\xi}<2r\land
h<n)$ for some tuple $(b_h\colon h<n)$ of elements of $K$. 
\end{enumerate}
\end{theorem}

\begin{proof}
Assume~\eqref{item:ec} and the hypothesis of~\eqref{item:<}.  Let $S$ be a 
  (finite) generating set of the ideal of
 $(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$ over $K$.  By
  Theorem~\ref{thm:first}, the system 
  \begin{equation*}
    \bigwedge_{f\in S}f(\partial^{\xi}x_h\colon\size{\xi}<2r\land h<n)=0
  \end{equation*}
has a solution in some extension, hence it has a solution
in $K$ itself, which means the conclusion of~\eqref{item:<} holds.
So~\eqref{item:<} is necessary for~\eqref{item:ec}.

Every system over $(K,\partial_0,\dots,\partial_{m-1})$ is equivalent to a
system of equations.  Suppose such a system has a
solution $(a_h\colon h<n)$ in some extension.  
Then the extension $K(\partial^{\xi}a_h\colon(\xi,h)\in(\vnn^m\times
n))$ has a \emph{finite} set of minimal separable leaders, by
Lemma~\ref{lem:fin}, since this
set is indexed by an antichain of $(\vnn^m\times n,\leq)$.
Hence there is $r$ large enough that all of these minimal separable
leaders are also generators of 
$K(a_h^{\xi}\colon\size{\xi}\leq r\land h<n)$.  We may assume also
that $r$ is large enough that $\size{\sigma}\leq r$ for every
derivative $\partial^{\sigma}x_k$ that appears in the original
system.  The hypothesis of~\eqref{item:<} is now satisfied when each
$a^{\sigma}_k$ is taken as $\partial^{\sigma}a_k$.  If the conclusion
of~\eqref{item:<} follows, then $(b_h\colon h<n)$ is a solution of the
original system.
Thus,~\eqref{item:<} is sufficient
for~\eqref{item:ec}. 
\end{proof}

\begin{corollary}\label{cor:dcf}
  The theory $\mDF$ has a model-companion, $\mDCF$.
\end{corollary}

\begin{proof}
We consider all possible situations in which
the hypothesis of~\eqref{item:<} in the theorem is satisfied.
In each case, there is a finite tuple $\tuple c$ of parameters from
$K$, there is a universal formula $\phi(\tuple y)$, and there is a system
$\psi(\tuple x,\tuple y)$ such that:
\begin{enumerate}
  \item
$(K,\partial_0,\dots,\partial_{m-1})\models\phi(\tuple c)$;
\item
$(K,\partial_0,\dots,\partial_{m-1})\models\psi(\tuple b,\tuple c)$ if
  and only if $\tuple b$ is a tuple $(b_h\colon h<n)$ as in the
  conclusion of~\eqref{item:<}; 
\item
for all models $\str M$ of $\mDF$ and all $\tuple d$ from $M$, if
$\str M\models\phi(\tuple d)$, then the system $\psi(\tuple x,\tuple
d)$ is soluble in some extension of $\str M$.
\end{enumerate}
Then $\mDCF$ is axiomatized \emph{modulo} $\mDF$ by all of the sentences
\begin{equation*}
  \Forall{\tuple x}(\phi(\tuple x)\to\Exists{\tuple y}\psi(\tuple
  y,\tuple x))
\end{equation*}
that can arise in this way.  To check this in detail, we note that
$\phi(\tuple c)$ and $\psi(\tuple x,\tuple c)$ can be such that:
\begin{enumerate}
  \item
the system $\psi(\tuple x,\tuple c)$ includes equations
\begin{equation}\label{eqn:ideal}
f(\partial^{\xi}x_h\colon\size{\xi}< 2r\land h<n)=0, 
\end{equation}
where
    the (ordinary) polynomials $f$ generate the ideal of
    $(a_h^{\xi}\colon\size{\xi}< 2r\land h<n)$ over $K$;
\item
for each leader of $K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$, there
is an equation~\eqref{eqn:ideal} in $\psi(\tuple x,\tuple c)$ such
that the variables appearing in $f$ correspond only to that leader and
its predecessors that are \emph{not} leaders, so that (since these
non-leading predecessors are algebraically independent) $f$ is in
effect a constant multiple of the minimal polynomial of the leader
over (the extension of $K$ generated by) those predecessors;
\item
then $\phi(\tuple c)$ says that those $f$ are irreducible (this is why
$\phi$ is universal: quantification is over the coefficients in
potential factors of $f$);
\item
also, $\phi(\tuple c)$ shows explicitly how nothing new results when
the equations~\eqref{eqn:ideal} are differentiated: if this
differentiation does not introduce a new derivative (namely, some
$\partial^{\sigma}x_k$ where $\size{\sigma}=2r$), then the new
equation is an explicit linear combination of the original equations;
in the cases where new derivatives are introduced, the resulting
linear system has an explicit solution.
\end{enumerate}
This can all have been done so that $\tuple c$ can be extracted to yield
$\phi(\tuple x)$ and $\psi(\tuple y,\tuple x)$ as desired.
\end{proof}

\subsection{Differential forms again}

The condition in Theorem~\ref{thm:first} can be adjusted to yield the
following: 

\begin{theorem}\label{thm:second}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $K$
has an extension $K(a_h^{\xi}\colon\size{\xi}\leq\size{\mu}\land
h<n)$ meeting the differential condition for some $\mu$ in $\vnn^m$
and some positive integer $n$.  Suppose further that, if $a_k^{\sigma}$ is a
minimal separable leader, 
then $\sigma\leq\mu$.  Then  $(K,\partial_0,\dots,\partial_{m-1})$
has an extension 
compatible with
$K(a_h^{\xi}\colon\size{\xi}<\size{\mu}\land h<n)$.  
\end{theorem}

\begin{proof}
  The proof is as for Theorem~\ref{thm:first}, \emph{mutatis
  mutandis.}  What needs adjusting is the choosing of
  $a_{\ell}^{\tau}$ in case both $a_{\ell}^{\tau-\ichar}$ and
  $a_{\ell}^{\tau-\jchar}$ are separable leaders.  Again we have
  minimal separable leaders
 $a_{\ell}^{\pi}$ and $a_{\ell}^{\rho}$
below $a_{\ell}^{\tau-\ichar}$ and
$a_{\ell}^{\tau-\jchar}$ respectively.  Since we may assume
  $\size{\mu}<\size{\tau}$, there is some $k$ in $m$
  such that $\mu(k)<\tau(k)$.  If $k=j$, then
  $\pi(j)\leq\mu(j)<\tau(j)$, so
  $\pi(j)\leq(\tau-\jchar)(j)=(\tau-\ichar-\jchar)(j)$.  Then
  $\pi\leq\tau-\ichar-\jchar$, so $a_{\ell}^{\pi}$ is below
  $a_{\ell}^{\tau-\ichar-\jchar}$.  Now we can proceed as before. 
\end{proof}

As Theorem~\ref{thm:first} yields Theorem~\ref{thm:dcf}, so
Theorem~\ref{thm:second} yields a characterization of the existentially
closed models of $\mDF$.  Moreover,
Theorems~\ref{thm:first} and~\ref{thm:second} can be combined in the
following way:

\begin{theorem}\label{thm:third}
Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $K$
has an extension $K(a_h^{\xi}\colon\size{\xi}\leq2r\land
h<n)$ meeting the differential condition for some positive integers $n$
and $r$.  Suppose further that, for each $k$ in $m$, either
 $\sigma\leq r$ whenever $a_k^{\sigma}$ is a 
minimal separable leader, or else there is some $\tau$ in $\vnn^m$
such that $\size{\tau}=2r$, and
 $\size{\sigma}\leq\size{\tau}$ whenever $a_k^{\sigma}$ is a 
minimal separable leader.
Then  $(K,\partial_0,\dots,\partial_{m-1})$
has an extension 
compatible with
$K(a_h^{\xi}\colon\size{\xi}<2r\land h<n)$.    
\end{theorem}

\begin{proof}
  Combine the proofs of
Theorems~\ref{thm:first} and~\ref{thm:second}.
\end{proof}

Again there is a corresponding characterization of the models of
$\mDCF$. 

Suppose $K(a_h^{\xi}\colon\size{\xi}\leq\size{\mu}\land h<n)$ is as in
the hypothesis of Theorem~\ref{thm:second}, and $\size{\mu}>2$.  If
$\size{\tau}\leq2$ and $\size{\sigma+\tau}\leq\size{\mu}$, let
$a_{(\sigma,k)}^{\tau}$ denote $a_k^{\sigma+\tau}$.  Then the
extension of $K$ can be written as
\begin{equation}\label{eqn:ext}
  K(a_{(\xi,h)}^{\eta}\colon
  \size{\eta}\leq2\land(\xi,h)\tl((0,\dots,0,\size{\mu}-1),0)),
\end{equation}
but this no longer need satisfy the conditions in
Theorem~\ref{thm:third}.  For example, in case $m=2$ and $n=1$ and
$\mu=(3,0)$, the original extension might determine the following
triangle, with the single leader underlined:
\begin{equation}\label{eqn:a-k}
  \begin{matrix}
    a&b&d&g\\
    c&e&h& \\
    f&k& & \\
    \underline a& & &
  \end{matrix}
\end{equation}
Then the derived extension in~\eqref{eqn:ext} determines three
triangles thus, with minimal separable leaders underlined:
\begin{equation*}
  \begin{matrix}
    a&\underline b&d\\
    \underline c&e& \\
    f           & &
  \end{matrix}
\qquad\qquad
\begin{matrix}
  b&d&g\\
  e&h& \\
  k& &
\end{matrix}
\qquad\qquad
\begin{matrix}
  c&\underline e&h\\
  f&k&            \\
  \underline a& &
\end{matrix}
\end{equation*}
We know that the two minimal separable leaders in the last triangle
will not cause a problem; but we don't know this directly from the
tree small triangles.  In the present context, the
system~\eqref{eqn:step-1} is
\begin{align*}
  \dee a&=\dee t^0\cdot c+\dee t^1\cdot b,&
  \dee b&=\dee t^0\cdot e+\dee t^1\cdot d,&
  \dee c&=\dee t^0\cdot f+\dee t^1\cdot e,\\
  \dee d&=\dee t^0\cdot h+\dee t^1\cdot g,&
  \dee e&=\dee t^0\cdot k+\dee t^1\cdot h,&
  \dee f&=\dee t^0\cdot a+\dee t^1\cdot k,
\end{align*}
and then~\eqref{eqn:jkim} consists of $\partial_0g=\partial_1h$,
$\partial_0h=\partial_1k$, and
$\partial_0k=b$.  As linear equations, these are soluble, but they
don't carry the information in~\eqref{eqn:a-k} that lets us know that
the corresponding differential system is soluble.

\subsection{Another sufficient condition}

If $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and
$K(a_h^{\xi}\colon\size{\xi}\leq\size{\pi}\land h<n)$ is an extension $L$
of $K$ meeting the differential condition, this by itself is not enough
to ensure that $(K,\partial_0,\dots,\partial_{m-1})$
has an extension compatible with
$L$.  However, if such
an extension does exist, then its existence can be shown by means of the
Theorem~\ref{thm:first}, provided $\size{\pi}$ can be made large
enough:  this is
Theorem~\ref{thm:s} below, which relies on the existence of
bounds as in the following.

\begin{lemma}
  For all positive integers $m$ and $n$, for all sequences $(a_i\colon
  i\in\vnn)$ of positive integers, there is a 
  bound on the length of strictly increasing chains
  \begin{equation}\label{eqn:chain}
    S_0\pincluded S_1\pincluded S_2\pincluded\dotsb
  \end{equation}
of antichains $S_k$ of $(\vnn^m\times n,\leq)$, where also
$S_k\included\{(\xi,h)\colon\size{\xi}\leq a_k\}$.
\end{lemma}

\begin{proof}
Divide and conquer.
  First reduce to the case when $n=1$.  Indeed, suppose the claim
  does hold in this case.  Suppose also, as an inductive hypothesis,
  that the claim holds when $n=\ell$.  Now fix $m$ and the sequence
  $(a_i\colon i\in\vnn)$ or rather  $(a(i)\colon i\in\vnn)$, and
  consider arbitrary chains as 
  in~\eqref{eqn:chain}, where $n=\ell+1$.  Analyze each $S_k$ as
  $S_k'\cup S_k''$, where
  \begin{align*}
    S_k' &=\{(\xi,h)\in S_k\colon h<\ell\},\\
    S_k''&=\{(\xi,h)\in S_k\colon h=\ell\}.
  \end{align*}
For each $k$ such that $S_{k+1}$ exists, at least one of the
inclusions $S_k'\included S_{k+1}'$ and $S_k''\included S_{k+1}''$
is strict; also, by our assumption, there is an upper bound
$f(k)$ on those $r$ such that
\begin{equation}\label{eqn:sk''}
  S_k''\pincluded S_{k+1}''\pincluded\dotsb\pincluded S_{r-1}''.
\end{equation}
The function $f$ depends only on
$m$, $n$, and $(a_i\colon i\in\vnn)$), not on the choice
of chain in~\eqref{eqn:chain}.

Let $k(0)=0$, and if $k(i)$ has been chosen, let $k(i+1)$ be the least
$r$, if it exists, such that $S_{k(i)}'\pincluded S_r'$.  
Here $k$ does depend on the chain.
But if $r$ is maximal in~\eqref{eqn:sk''}, and $S_r'$ exists,
then $S_k'\pincluded S_r'$.
Hence
$k(i+1)\leq f(k(i))$.  Since the function $f$ is not necessarily
increasing, we derive from it the increasing function $g$, where
$g(k)=\max_{i\leq k}f(i)$.  Then $x\leq y\implies g(x)\leq g(y)$, so
\begin{equation}\label{eqn:krg}
  k(r)\leq f(k(r-1))\leq g(k(r-1))\leq g\circ g(k(r-2))\leq\dotsb
\leq
\overbrace{g\circ\dotsb\circ g}^r(0)=g^r(0).
\end{equation}
In particular,
$S_{k(r)}\included\{(\xi,h)\colon\size{\xi}\leq
a(g^r(0))\}$.  The sequence $(a(g^i(0))\colon i\in\vnn)$ does not
depend on the original chain.
Hence the inductive hypothesis applies to the chain
\begin{equation}\label{eqn:chain2}
  S_{k(0)}'\pincluded S_{k(1)}'\pincluded\dotsb,
\end{equation}
showing
that there is $s$ (independent of the original chain) such that
$k(s)$ is defined, and $r\leq s$ for all entries $S_{k(r)}'$
in~\eqref{eqn:chain2}.  
Hence also, by~\eqref{eqn:krg}, if
$S_r$ is an entry in~\eqref{eqn:chain2}, then $r\leq k(s)\leq
g^s(0)$. 

Now suppose $S_r'$ is the final entry in~\eqref{eqn:chain2}.  Then
$S_r''\pincluded S_{r+1}''\pincluded\dotsb$; but if $S_t''$ is an
entry of this chain, then $t<f(r)\leq g(r)\leq g(g^s(0))=g^{s+1}(0)$. 

Therefore the
original chain in~\eqref{eqn:chain} has a final entry $S_t$, where
$t<g^{s+1}(0)$.  Thus the claim holds when $n=\ell+1$.  By induction, the
claim holds for all positive $n$, provided it holds when $n=1$.

It remains to show that, for all positive $m$, for all sequences
$(a_i\colon i\in\vnn)$, there is a bound on the length of chains
\begin{equation}\label{eqn:chain3}
  S_0\pincluded S_1\pincluded S_2\pincluded\dotsb
\end{equation}
of antichains $S_k$ of $(\vnn^m,\leq)$,
where $S_k\included\{\xi\colon\size{\xi}\leq a_k\}$.  The claim is
trivially true when $m=1$.  Suppose it is true when $m=\ell$.  Now let
$m=\ell+1$, and suppose we have a chain as in~\eqref{eqn:chain3}.  We
may assume that $S_0$ contains some $\sigma$.  If $i<m$ and
$j\in\vnn$, let
\begin{equation*}
  S_k^{i,\,j}=\{\xi\in S_k\colon\xi(i)=j\}.
\end{equation*}
Then the inductive hypothesis applies to chains of the form
\begin{equation*}
  S_{k(0)}^{i,\,j}\pincluded
  S_{k(1)}^{i,\,j}\pincluded
  S_{k(2)}^{i,\,j}\pincluded\dotsb.
\end{equation*}
Moreover, if $\tau\in S_k$, then $\tau(i)\leq\sigma(i)$ for some $i$
in $m$ (since
$\sigma$ is also in $S_k$, and this is an antichain).  Hence
\begin{equation*}
  S_k=\bigcup_{i<m}\bigcup_{j\leq\sigma(i)}S_k^{i,\,j},
\end{equation*}
a union of no more than $\size{\sigma}+m$-many sets, hence no more than
$a_0+m$-many 
sets.  So the proof can proceed as in the reduction to $n=1$: for each
$k$ such that $S_{k+1}$ exists, one of the inclusions
$S_k^{i,\,j}\included S_{k+1}^{i,\,j}$ is strict, and so forth.
\end{proof}

\begin{theorem}\label{thm:s}
  Suppose $(K,\partial_0,\dots,\partial_{m-1})\models\mDF$, and $r$
  and $n$ are positive integers.  Then there is a positive integer $s$
  such that $r\leq s$, and if $K(a_h^{\xi}\colon\size{\xi}\leq s\land
  h<n)$ meets the differential condition, then
  $(K,\partial_0,\dots,\partial_{m-1})$ has an extension that is
  compatible with $K(a_h^{\xi}\colon\size{\xi}\leq
  r\land h<n)$.
\end{theorem}

\begin{proof}
Suppose $K(a_h^{\xi}\colon\size{\xi}\leq 2^tr\land h<n)$ 
meets the differential condition for some~$t$.  When $u\leq t$, let
  $K_u=K(a_h^{\xi}\colon\size{\xi}\leq 2^ur\land h<n)$, and let $S_u$
  be the set of minimal separable leaders of $K_u$.  Then we have an
  increasing chain $S_0\included S_1\included\dots\included S_t$.
  By the preceding lemma, there is a value of $t$, depending only on
  $m$, $r$, and $n$, large
  enough that this chain cannot be strictly increasing.  Then
  $S_u=S_{u+1}$ for some $u$ less than this~$t$.  
  Then $K_{u+1}$ satisfies the hypothesis of Theorem~\ref{thm:first}.  
So $(K,\partial_0,\dots,\partial_{m-1})$ has an extension compatible
  with $K(a_h^{\xi}\colon\size{\xi}<2^{u+1}r\land h<n)$, and \emph{a
  fortiori} with 
 $K(a_h^{\xi}\colon\size{\xi}\leq r\land h<n)$.  In short, the desired
  $s$ is $2^tr$.
\end{proof}

This theorem yields another yet characterization of the models of
$\mDCF$.

%\bibliographystyle{asl}

%\bibliography{revisited}

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\end{document}

