\documentclass[12pt]{amsart} \headheight=8pt \topmargin=0pt \textheight=624pt \textwidth=432pt \oddsidemargin=18pt \evensidemargin=18pt \usepackage{graphicx,calc,amssymb} \newtheorem{itheorem}{Theorem} \newtheorem{theorem}{Theorem}[subsection] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem*{remarks}{Remarks} \newtheorem{remark}[theorem]{Remark} \newtheorem{rmk}[theorem]{} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{defi}[theorem]{Definition} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % SPECIAL MATH WORDS AND SYMBOLS \newcommand{\X}{{}{\ensuremath{X_{ij}}}} \newcommand{\Xk}{{}{\ensuremath{X_{kj}}}} \newcommand{\Y}{{}{\ensuremath{Y_{ij}}}} \newcommand{\C}{\operatorname{\mathcal{C}}} \newcommand{\f}{{}{\ensuremath{f_{ij}}}} \newcommand{\fk}{{}{\ensuremath{f_{ik}}}} \newcommand{\Yk}{{}{\ensuremath{Y_{ik}}}} \newcommand{\GLn}{\operatorname{GL_n}} \newcommand{\SL}{\operatorname{SL_n}} \newcommand{\GL}{{}{\ensuremath{\GLn(C)}}} \newcommand{\M}{\operatorname{M_n}} \newcommand{\bx}{{}{\ensuremath{\mathbf{X}}}} \newcommand{\ba}{{}{\ensuremath{\mathbf{\alpha}}}} \newcommand{\bb}{{}{\ensuremath{\mathbf{\beta}}}} \newcommand{\Lie}{\operatorname{Lie}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \title{On a generic inverse differential Galois problem for GL$_\mathbf n$} \author{Lourdes Juan} \address{Mathematical Sciences Research Institute\\ 1000 Centennial Dr.\\ Berkeley, California 94720\\ USA} \email{ljuan@msri.org} \urladdr{www.msri.org/people/members/ljuan/} \subjclass{Primary 12H05; Secondary 12F12, 12Y05, 20G15} \date{\today} \thanks{ This paper was written while the author held an NSF-funded postoctoral fellowship at the MSRI. The author wishes to thank the MSRI for its hospitality.} \keywords{Constants, differential field, differential Galois group, differential specialization, inverse differential Galois problem, generic, multivariable division algorithm, new constant, Picard-Vessiot extension, wronskian. } \begin{abstract} In this paper we construct a generic Picard-Vessiot extension for the general linear groups. In the case when the differential base field has finite transcendence degree over its field of constants we provide necessary and sufficient conditions for solving the inverse differential Galois problem for this groups via specialization from our generic extension. \end{abstract} \maketitle \markboth{LOURDES JUAN}{ON A GENERIC INVERSE DIFFERENTIAL GALOIS PROBLEM FOR GL$_\mathbf n$} \section*{Introduction} \label{sec:intro} Le $F$ be a differential field of characteristic zero with algebraically closed field of constants $C$. In this paper we give an affirmative answer, for the group $\GL$, to the following {\bf Generic Inverse Differential Galois Problem}: \emph{For a connected algebraic group $G$ over $C$ find a generic Picard-Vessiot extension of $F$ with differential Galois group $G$.} By \emph{generic extension} we mean a Picard-Vessiot extension of a \emph{generic field} that contains $F$ and such that every Picard-Vessiot extension of $F$ for $G$ in the usual sense can be obtained from the generic one by specialization. We point out that any such specialization will provide a solution to the inverse differential Galois problem in the usual sense, namely, to determine, given $F$ and $C$ as above, and a linear algebraic group $G$ over $C$, what differential field extensions $E\supset F$ are Picard-Vessiot extensions with differential Galois group $G$ and, in particular, whether there are any. We use the terminology of A. Magid's book \cite{M}. In \cite{M} the reader may also find definitions and proofs of some results from differential Galois theory that will be recalled. Our construction of a generic Picard-Vessiot extension of $F$ with group $\GL$ may be summarized as follows: Let $F\{\Y\}$ be the ring of differential polynomials over $F$ in the differential indeterminates $\Y$, $1\le i, j\le n$. Our \begin{emph}{generic base field}\end{emph} will be the differential quotient field $F\langle \Y\rangle$ of $F\{\Y\}$. Let $\X$, $1\leq i,j\leq n$, be algebraically independent over $F\langle \Y\rangle$. Extend the derivation on $F\langle \Y\rangle$ to the polynomial ring $F\langle \Y\rangle[\X]$ by letting $D(\X)=\sum_{\ell=1}^n Y_{i\ell}X_{\ell j}$. This derivation extends in a natural way to the quotient field $F\langle \Y\rangle(\X)$ (note that $F\langle \Y\rangle(\X)$ is also the function field of the group obtained from $\GL$ by extending scalars from $C$ to $F\langle\Y\rangle$). Then \begin{itheorem}\label{ige} The differential field extension $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ is a generic Picard-Vessiot extension of $F$ with differential Galois group $\GL$.\end{itheorem} The main step in proving Theorem ~\ref{ige} is to show that $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ is a no-new-constant extension. A new constant in $F\langle \Y\rangle(\X)$ must be the quotient of two relatively prime Darboux polynomials, that is, two polynomials $p_1,p_2\in F\{\Y\}[\X]$ that satisfy $D(p_i)=qp_i$ for some $q\in F\{\Y\}[\X]$. We use Gr\"obner basis techniques to show that the only Darboux polynomials in $F\{\Y\}[\X]$ are those of the form $\ell\det[\X]^a$, where $\ell\in F$ and $a\in \mathbb N$. This implies that there are no two such relatively prime Darboux polynomials and, therefore, no new constants. To show that the extension is generic we let $E\supseteq F$ be any Picard-Vessiot extension of $F$ with group $\GL$. Then $E$ is isomorphic to $F(\X)$, the function field of the group obtained from $\GL$ by extending scalars from $C$ to $F$, as a $\GL$-module and as an $F$-module, and there are $\f\in F$ such that the derivation on $E$ is given by $D_E=\sum_{\ell=1}^n f_{i\ell}X_{\ell j}$. In this situation, the Picard-Vessiot extension $F(\X)\supseteq F$ is obtained from the Picard-Vessiot extension $C\langle \f\rangle(\X)\supseteq C\langle \f\rangle$ by extension of scalars from $C$ to $F$. Therefore, any Picard-Vessiot extension $E\supset F$ can be obtained from $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ via the specialization $\Y\mapsto \f$. That is, $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ is a generic Picard-Vessiot extension of $F$ for $\GL$. Now, suppose that $F$ has finite transcendence degree over $C$ say, $$F=C(t_1,\dots,t_m)[z_1,\dots,z_k],$$ where the $t_i$ are algebraically independent over $C$ and the $z_i$ are algebraic over $C(t_1,\dots,t_m)$. Consider the differential field $F(\X)$ with derivation $D_E$ as above. Let $\C$ denote its field of constants. Let $F\{\Y\}[\X]$ be the differential ring with the derivation defined above. For $k\geq 1$ let $\mathbb T_k$ denote the set of monomials in the both the $t_i$ and $\X$ of total degree less than or equal to $k$. Fix a term order on the set $\mathbb T$ of monomials in the $t_i$ and the $\X$ and let $W_k(\Y)$ denote the wronskian of $\mathbb T_k$ relative to that order (the order will only affect the wronskian by a sign). The following theorem summarizes our specialization results: \begin{itheorem}\label{isp} $F(\X)\supset F$ is a Picard-Vessiot extension for $\GL$ if and only if all the wronskians $W_k(\Y)$ map to nonzero elements in $F(\X)$ via the specialization $\Y\mapsto\f\in F$. \end{itheorem} The above condition on the wronskians means that all the sets $\mathbb T_k$, for $k\geq 1$, are linearly independent over $\C$. This is in turn equivalent to the fact that the set of all the $t_i$ and all the $\X$ are algebraically independent over $\C$. Unfortunately, Theorem ~\ref{isp} gives infinitely many conditions. We do not know at present how to use these conditions to effectively construct solutions to the inverse problem, and this constitutes an interesting open problem. A specialization as in Theorem~\ref{isp}, however, is known to exist by a result of C. Mitschi and M. Singer \cite{M-S2}. They give a constructive algebraic solution to the inverse problem for all connected linear algebraic groups (and, in particular, for $\GL$) when $F$ has finite transcendence degree over $C$. An interesting direction of research in connection with the previous open problem is to try to fully describe all possible solutions (isomorphic and non-isomorphic) that may arise in this situation. The work of Mitschi and Singer in \cite{M-S2} makes use of the logarithmic derivative and an inductive technique developed by Kovacic \cite{K1}, \cite{K2}, to lift a solution to the inverse problem from $G/R_u$, where $R_u$ is the unipotent radical of $G$, to the full group $G$. Using this machinery Kovacic proved that it is enough to find a solution to the inverse problem for reductive groups (observe that $G/R_u$ is reductive). In \cite{P}, van der Put explains and partly proves the results in \cite{M-S2}. In the introduction of \cite{M-S2} the authors briefly review previous work on the inverse problem such as results of Bialynicki-Birula in \cite{BB}, Kovacic \cite{K1}, \cite{K2}, Ramis \cite{ramis1}, \cite{ramis2}, Singer \cite{S2}, Tretkoff and Tretkoff \cite{TT}, Beukers and Heckman \cite{B-H}, Katz \cite{K}, Duval and Mitschi \cite{dm}, Mitschi \cite{M1}, \cite{M2}, Duval and Loday-Richaud \cite{dlr}, Ulmer and Weil \cite{uw} and Singer and Ulmer \cite{su}. A more extensive survey on the inverse problem can be found in M. Singer's \cite{S}. The constructive algebraic solutions to the inverse differential Galois problem for connected algebraic groups that are currently available are based on Kolchin's Main Structure Theorem for Picard-Vessiot extensions (see Theorem ~\ref{sthm} below). In particular, a corollary to this theorem (see Theorem ~\ref{stgln}) establishes that if $E\supset F$ is Picard-Vessiot and $G$ is, for example, unipotent or solvable or $G=\GLn$ or $G=\SL$, then E is isomorphic as an $F$-module and as a $G$-module to the function field of the group $G_F$ obtained from $G$ by extension of scalars from $C$ to $F$. Therefore, to get a Picard-Vessiot extension $E\supset F$ with group $G$ (if it exists) one can begin by taking $E$ to be the function field of $G_F$ and then the problem reduces to extending the derivation from $F$ to $E$ in such a way that $E\supset F$ is Picard-Vessiot for that derivation. In this paper we use this approach for our construction. The idea of tackling the inverse problem by constructing generic extensions is inpired by the works of E. Noether \cite{EN} for the Galois theory of algebraic equations. Following her approach, L. Goldman in \cite{goldman} introduced the notion of a \begin{emph}{generic differential equation with group $G$}\end{emph}. Goldman explicitly constructed a generic equation with group $G$ for some groups. However, after specializing Goldman's equation the group of the new equation obtained is a subgroup of the original group. In order to solve the inverse problem by this means, we need to keep the original group as the group of the equation after specialization. Goldman's generic equation for $\GLn$ is equivalent to Magid's \begin{emph}{general equation of order $n$}\end{emph} (Example 5.26 in \cite{M}). More work in the spirit of Goldman's generic equation came some years later in J. Miller's dissertation \cite{miller}. He defined the notion of hilbertian differential field and gave a sufficient condition for the generic equation with group $G$ to specialize to an equation over such a field with group $G$ as well. However, as pointed out by Mitschi and Singer in \cite{M-S2}, his condition was stronger than the analogous one for algebraic equations and this made the theory especially difficult to apply for those groups that were not already known to be Galois groups. This paper contains the results of the author's Ph.D. dissertation \cite{J}. I wish to thank my Ph.D. advisor Andy Magid for the many valuable research meetings that we had. I am also grateful to Michael Singer for many enlightening conversations on the inverse problem. \section{Preliminaries} \subsection{Notation and some basic results from Differential Galois Theory} We fix a differential field $F$ with algebraically closed field of constants $C$. If $E\supseteq F$ is a differential field extension then the group of differential automorphisms of $E$ over $F$ is denoted by $G(E/F)$. If $G$ is a linear algebraic group over $C$ and $K$ is an overfield of $C$ we denote by $G_K$ the group obtained from $G$ by extending scalars from $C$ to $K$. We will show that the differential field $F\langle\Y\rangle(\X)$ to be defined in ~\ref{diffields} is a Picard-Vessiot extension of $F\langle\Y\rangle$ with differential Galois group $\GL$. Note that \linebreak $F\langle\Y\rangle(\X)$ is the function field of $G_K$ with $G=\GL$ and $K=F \langle\Y\rangle$. The following two results provide the rationale for choosing such function field for the generic Picard-Vessiot extension. The proofs can be found in \cite{M} (Theorem 5.12 and Corollary 5.29 respectively). \begin{theorem}[Kolchin Structure Theorem]\label{sthm} Let $E \supseteq F$ be a Picard-Vessiot extension, let $G\leq G(E/F)$ be a Zariski closed subgroup and let $T$ be the set of all $f$ in $E$ that satisfy a linear homogeneous differential equation over $K=E^G$. Then $T$ is a finitely generated $G$-stable differential $K$-algebra with function field $E$, and if $\overline K$ denotes the algebraic closure of $K$, then there is a $G$-algebra isomorphism $$\overline K\otimes_K T\rightarrow \overline K\otimes_C C[G].$$ Note that $C[G]$ denotes the affine coordinate ring of $G$ and that the target of the above isomorphism is the affine coordinate ring of the group $G_{\overline K}$ obtained from $G$ by extension of scalars from $C$ to $\overline K$. \end{theorem} \begin{theorem}\label{stgln} Let $E \supseteq F$ be a Picard-Vessiot extension, let $G\leq G(E/F)$ be a Zariski closed subgroup with $E^G=F$. Let $\overline F$ be an algebraic closure of $F$, and suppose the Galois cohomology $H^1(\overline F/F,G(\overline F))$ is a singleton. Let $T(E/F)$ be the set of all $f$ in $E$ that satisfy a linear homogeneous differential equation over $F$. Then there are $F$- and $G$-isomorphisms $T(E/F)\rightarrow F[G_F]$ and $E\rightarrow F(G_F)$. In particular, this holds if $G$ is unipotent or solvable, or if $G=\GL$ or if $G=\SL$.\end{theorem} We will use the following characterization of Picard-Vessiot extension given in \cite{M} (Proposition 3.9): \begin{theorem}\label{cpve} Let $E \supseteq F$ be a differential field extension. Then $E$ is a Picard-Vessiot extension if and only if: \begin{enumerate} \item $E=F\langle V\rangle$, where $V\subset E$ is a finite-dimensional vector space over $C$; \item There is a group $G$ of differential automorphisms of $E$ with $G(V)\supseteq V$ and $E^G=F$; \item $E\supset F$ has no new constants. \end{enumerate} In particular, if the above conditions hold and if $\{y_1,\dots,y_n\}$ is a $C$-basis of $V$, then $E$ is a Picard-Vessiot extension of $F$ for the linear homogeneous differential operator $$L(Y)=\frac{w(Y,y_1,\dots,y_n)}{w(y_1,\dots,y_n)}$$ where $w(-)$ denotes the wronskian determinant and $L^{-1}(0)=V$.\end{theorem} In our case the base field is $F\langle \Y\rangle$ and $G=\GL$. We want to show that $F\langle \Y\rangle (\X)\supset F\langle \Y\rangle$ is Picard-Vessiot. We first show (Corollary ~\ref{constfy}) that the field of constants of $F\langle\Y\rangle$ is $C$. Then, conditions 1. and 2. in Theorem ~\ref{cpve} are easily verified with $V$ the $C$-span of the $\X$ and $G=\GL$. Therefore, our main goal henceforth will be to show that $F\langle \Y\rangle (\X)\supset F\langle \Y\rangle$ is a no-new-constant extesion. \subsection{The differential fields $\mathbf {F\langle \Y\rangle (\X)}$ and $\mathbf {F(\X)}$}\label{diffields} Let $\Y$, $1\leq i,j\leq n$, be differential indeterminates over $F$. For convenience, denote the $k$-th derivative $D^{(k)}(\Y)$ by $Y_{ij,k}$, for $k\geq 0$, so that $D(Y_{ij,k})=Y_{ij,k+1}$, $k\geq 0$. As usual, $D^{(0)}(\Y)=Y_{ij,0}$ represents the original element $\Y$. In this situation we will omit the $k$-subindex and write $\Y$ instead of $Y_{ij,0}$. Let $F\{\Y\}$ be the ring of differential polynomials in the $\Y$ and $F\langle\Y\rangle$ its differential quotient field. By that we mean the usual quotient field endowed with the natural derivation: $$\mathcal{D}\Big( \frac pq\Big)= \frac{D(p)q-pD(q)}{q^2}.$$ for $p, q\in F\{\Y\}$, where $D$ is the derivation on $F\{\Y\}$. Next let $\X$, $1\leq i,j\leq n$, be algebraically independent over $F\langle \Y\rangle$. We consider the differential ring $R=F\{\Y\}[\X]$ with derivation extending the derivation on $F\{\Y\}$ by a formula $$D(\X)=\sum_{\ell=1}^nY_{i\ell}X_{{\ell}j}.$$ As above, this derivation extends to the quotient field $$Q=F\langle \Y\rangle(\X)$$ in a natural way. Henceforth we fix the differential field $Q$ as defined above. Likewise we will regard the polynomial ring $F[\X]$ as a differential ring with derivation extending the derivation on $F$ by a formula $$D(\X)=\sum_{\ell=1}^nf_{i\ell}X_{{\ell}j}$$ with $\f\in F.$ Then, as before, we extend this derivation to the quotient field $F(\X)$ in a natural way. The multinomial notation $a_{\ba}\bx^{\ba}$ will be used to denote a term of the form\linebreak $a_{\alpha_{11}\cdots \alpha_{k\ell}}X^{\alpha_{11}}_{11}\cdots X_{k\ell}^{\alpha_{k\ell}}.$ The ring $F[\X]$ is assumed to be ordered with the degree reverse lexicographical order (\emph{degrevlex}). That is, the set $$\mathbb T^{n^2}=\lbrace \bx^{\bb}\,| \, \bx=(\X),\,\bb=(\beta_{ij})\in \mathbb N^{n^2}\rbrace$$ of the power products in the $\X$ is ordered by $X_{11}>\dots>X_{1n}>\dots> X_{n1}>\dots >X_{nn},$ and $$\bx^{\ba}< \bx^{\bb}\Longleftrightarrow\cases \sum_{j=1}^n\sum_{i=1}^n \alpha_{ij}<\sum_{j=1}^n\sum_{i=1}^n \beta_{ij}\\ \text{\it or} \\ \sum_{j=1}^n\sum_{i=1}^n \alpha_{ij}=\sum_{j=1}^n\sum_{i=1}^n \beta_{ij}, \text{\it and the first coordinates}\\ \text{\it $\alpha_{ij}$, $\beta_{ij}$ from the right which are different satisfy $\alpha_{ij}>\beta_{ij}$.}\endcases$$ Henceforth the leading power product of a polynomial in $F[\X]$ is assumed to be with respect to this order. The above definitions as well as the \emph{Multivariable Division Algorithm} that will be used in Remark ~\ref{rmk4}, can be found in \cite{A-L}. \subsection{ Darboux polynomials and the constants of $\mathbf {F\langle \Y\rangle (\X)}$} We need to show that the field of constants $\C$ of $Q=F\langle \Y\rangle (\X)$ coincides with the field of constants $C$ of $F$. In this section we will show (Corollary ~\ref{maincor}) that this can be reduced to proving that the only \begin{emph}{ Darboux polynomials}\end{emph} in $R$ are, up to a scalar multiple in $F$, powers of $\det[\X]$. \begin{defi}\label{dp} Let $D$ be a derivation on the polynomial ring $A=k[Y_1,$ $\dots,Y_s]$. A polyomial $p\in A$ is called a Darboux polynomial if there is a polynomial $q\in A$ such that $D(p)=qp$. That is, $p$ divides $D(p)$.\end{defi} Darboux polynomials correspond to generators of principal differential ideals in $A$. Chapter I of J.A. Weil's Thesis \cite{W} is devoted to constants and Darboux polynomials in Differential Algebra. The following basic proposition (proven in \cite{W} for $A$ as in Definition ~\ref{dp}) characterizes new constants for the extension $Q\supset F$ in terms of Darboux polynomials: \begin{proposition}\label{weil2} Let $p_1,p_2\in R=F\{\Y\}[\X]$, $p_1,p_2\ne 0$, be relatively prime. Then $D(\frac {p_1}{p_2})=0$, if and only if $p_1$ and $p_2$ are Darboux polynomials. Moreover, if $q_1,q_2\in R$ are such that $D(p_1)=q_1p_1$ and $D(p_2)=q_2p_2$, then $q_1=q_2$. \end{proposition} \begin{proof} For the necessity of the condition we have $$D\left( \frac {p_1}{p_2}\right)= \frac{D(p_1)p_2-p_1D(p_2)}{p_2^2} =0,$$ thus $D(p_1)p_2-p_1D(p_2)=0$, that is \begin{equation}D(p_1)p_2=p_1D(p_2). \tag{1}\end{equation} Since $p_1$ and $p_2$ are relatively prime, the last equation implies that $p_1$ divides $D(p_1)$ and $p_2$ divides $D(p_2)$. Now, let $q_1,q_2\in R$ be such that $D(p_1)=q_1p_1$ and $D(p_2)=q_2p_2$, respectively. Then it follows from (1) that $$ q_1p_1p_2=q_2p_1p_2.$$ Hence, $q_1=q_2$. The proof of the converse is obvious. \end{proof} \begin{proposition}\label{detdp} Let $p=\ell\det[\X]^a$, with $\ell\in F$. Then $p$ is a Darboux polynomial in $R$ with $q=\frac {\ell'}{\ell}+a\,\sum_{i=1}^n Y_{ii}.$ \end{proposition} \begin{proof} We have $$D(p)=\ell'\det[\X]^a+\ell( a\det[\X]^{a-1}D(\det[\X])).$$ And, $$D(\det[\X])=\Big(\sum_{i=1}^n Y_{ii}\Big) \det[\X].$$ Thus, \begin{eqnarray*} D(p)&=&\ell'\det[\X]^a+ \ell( a\det[\X]^{a-1}\Big(\sum_{i=1}^n Y_{ii}\Big)\det[\X]\\ &=&\Big(\frac {\ell'}{\ell}+a\,\sum_{i=1}^n Y_{ii}\Big)\,\ell\,\det[\X]^a. \end{eqnarray*} That is, $$p=\ell\det[\X]^a$$ is a Darboux polynomial in $R$ with $$q=\frac {\ell'}{\ell}+a\,\sum_{i=1}^n Y_{ii}.$$\end{proof} \begin{corollary}\label{maincor}Suppose that if $p\in R$ is a Darboux polynomial then $p=\ell\det[\X]^a$, with $\ell\in F$, $a\in\mathbb N$ and $q=\frac {\ell'}{\ell}+a\,\sum_{i=1}^n Y_{ii}.$ Then $F\langle\Y\rangle(\X)\supset F$ is a no-new-constant extension. \end{corollary} \begin{proof} $F\langle\Y\rangle(\X)$ is the fraction field of $F\{\Y\}[\X]$. Thus, if an element $f\in F\langle\Y\rangle(\X)$ satisfies $D(f)=0$ and $f\notin F$ then, by proposition ~\ref{weil2}, $f= {p_1}/{p_2}$ with $p_1,p_2\in R$ relatively prime Darboux polynomials. If the hypothesis is true, this contradicts the fact that $p_1$ and $p_2$ are relatively prime.\end{proof} Next we show that the constants of $F\langle\Y\rangle$ coincide with the constants of $F$. For simplicity, if $h(Y)\in F\{\Y\}$, we will use the notation $h'(Y)$ for $D(h(Y))$. Notice that this is not the usual meaning $h'(Y)=\sum h'_\alpha\mathbf Y^\alpha$. \begin{proposition}\label{nodpy} If $h(Y)\in F\{\Y\}$ satisfies $h'(Y)=g(Y)h(Y)$ for some $g(Y)\in F\{\Y\}$ then $h(Y)\in F$. That is, there are no non-trivial Darboux polynomials in $F\{\Y\}$.\end{proposition} \begin{proof} Suppose that the hypothesis of the proposition is true. According to our notation $D^{(k)}(\Y)=Y_{ij,k}$. Consider the set of subindices $\{ij,k\}$, $i,j,k\in\mathbb N$, ordered with the lexicographical order. That is, $\{i_1 j_1,k_1\}>\{i_2j_2,k_2\}$ if and only if the first coordinates $s_1$ and $s_2$ from the left, for $s=i,j,k$ above, which are different satisfy $s_1>s_2$. Let $\{mn,t\}$ be the largest subindex such that $Y_{mn,t}$ occurs in $h(Y)$. Let $h(Y)= \sum_\alpha a_{\alpha}Y^{\alpha_{11}}_{11}\cdots Y_{mn,t}^{\alpha_{mn,t}}.$ Then \begin{eqnarray*} h'(Y)&=& \sum_\alpha a'_{\alpha}Y^{\alpha_{11}}_{11}\cdots Y_{mn,t}^{\alpha_{mn,t}} + \sum_\alpha a_{\alpha}\alpha_{11}Y^{\alpha_{11}- 1}_{11}Y^{\alpha_{11,1}+1}_{11,1}\cdots Y_{mn,t}^{\alpha_{mn,t}}\\ & &\qquad\qquad\qquad\qquad+\dots + \sum_\alpha a_{\alpha}\alpha_{mn,t}Y^{\alpha_{11}}_{11}\cdots Y_{mn,t}^{\alpha_{mn,t}-1}Y_{mn,t+1}\\ &=&h_1(Y_{11},\cdots ,Y_{mn,t})+\Big( \sum_\alpha a_{\alpha}\alpha_{mn,t}Y^{\alpha_{11}}_{11}\cdots Y_{mn,t}^{\alpha_{mn,t}- 1}\Big)Y_{mn,t+1}\\ &=&g(Y)h(Y).\end{eqnarray*} For $Y_{mn,t+1}=D(Y_{mn,t})$ we have $\{mn,t+1\}>\{mn,t\}$, thus it does not occur in $h(Y)$ by the choice of $\{mn,t\}$. Also, it does not occur in $h_1(Y_{11},\cdots,$ $Y_{mn,t})$. Thus the above equation implies that $Y_{mn,t+1}$ must occur in $g(Y)$. Let $g_{t+1}(Y)$ be its coefficient in $g(Y)$ and let $$h_2(Y)= \sum_\alpha a_{\alpha}\alpha_{mn,t}Y^{\alpha_{11}}_{11}\cdots Y_{mn,t}^{\alpha_{mn,t}-1}.$$ Then we have $$h(Y)g_{t+1}(Y)Y_{mn,t+1}=h_2(Y)Y_{mn,t+1}$$ or $$h(Y)g_{t+1}(Y)=h_2(Y).$$ But the total degree of $h_2(Y)$ is strictly less than the total degree of $h(Y)$. This forces $h(Y)\in F.$\end{proof} \begin{corollary}\label{constfy}The field of constants of $F\langle\Y\rangle$ coincides with $C$, the field of constants of $F$.\end{corollary} \begin{proof} This is a consequence of Propositions ~\ref{weil2} and ~\ref{nodpy}.\end{proof} \subsection{Darboux polynomials in $\mathbf{R=F\{\Y\}[\X]}$} Proposition ~\ref{detdp} shows that scalar multiples of $\det[\X]$ and its powers are Darboux polynomials in $R$. Corollary ~\ref{maincor} implies that if these are the only Darboux polynomials in $R$ then we are done since consequently there will be no new constants in $Q$. In this section we will show that that is the case, namely, the only Darboux polynomials in $R$ are those of the form $p=\ell\det[\X]^a $, with $\ell\in F$, $a\in\mathbb N$ and $q=\frac {\ell'}{\ell}+a\,\sum_{i=1}^n Y_{ii}.$ \begin{remarks} \begin{rmk}[Derivative of a power product in the $\X$]\label{rmk1} Let $$\bx^\alpha= X_{11}^{\alpha_{11}}\cdots X_{1n}^{\alpha_{1n}}\cdots X_{n1}^{\alpha_{n1}}\cdots X_{nn}^{\alpha_{nn}},$$ then \begin{eqnarray*} \lefteqn{D(\bx^\alpha)= \Big(\sum_{i=1}^n\sum_{j=1}^n \alpha_{ij}Y_{ii}\Big)\, \bx^\alpha}\\ && + \sum_{i=1}^n\sum_{j=1}^n\Big(\sum_{\ell>i}\alpha_{ij}Y_{i\ell}X_{11}^{a_{11}}\cdots \X^{\alpha_{ij}-1}\cdots X_{\ell j}^{\alpha_{\ell j}+1}\cdots X_{nn}^{\alpha_{nn}}\\ & &\qquad\qquad+\sum_{\ellt$}\endcases$$ for $1\leq r,s\leq n$, $t\ne r$, and $\bx^{\alpha}$ itself.\end{rmk} \begin{rmk}\label{rmk3} Let $p\in R.$ Since $D(\X)=\sum_{\ell=1}^nY_{i\ell}X_{\ell j}$, then the total degree of $p$ with respect to the $\X$ does not change after differentiation. Therefore, if $D(p)=qp$ then $q\in F\{\Y\}.$\end{rmk}\end{remarks} \begin{proposition}\label{coeffmon} Let $p\in R$. Write it as $p=\sum_\alpha p_\alpha(Y)\bx^\alpha$, with $p_\alpha(Y)\in F\{\Y\}$. Then for any $\alpha$ with $p_\alpha(Y)\ne 0$, the coefficient of $\bx^\alpha$ in $D(p)$ is $$p'_{\alpha}(Y)+ p_{\alpha}(Y)\sum_{i=1}^n\sum_{j=1}^n \alpha_{ij}Y_{ii} +\sum_{i=1}^n\sum_{j=1}^n(\alpha_{ij}+1)\sum _{\ell\ne i}p_{\alpha_{ij,\ell}}(Y)Y_{i\ell},$$ where $\alpha_{ij,\ell}$ is the exponent vector of the power product $$\bx^{\alpha_{ij,\ell}}=\cases X_{11}^{\alpha_{11}}\cdots \X^{\alpha_{ij}+1}\cdots X_{\ell j}^{\alpha_{\ell j}-1}\cdots X_{nn}^{\alpha_{nn}} \,\,\,&\text{if $\,\,\,i<\ell$}\\ X_{11}^{\alpha_{11}}\cdots X_{\ell j}^{\alpha_{\ell j}-1}\cdots \X^{\alpha_{ij}+1}\cdots X_{nn}^{\alpha_{nn}}\,\,\,&\text{if $\,\,\,\ell>i$}\endcases$$ as in Remark ~\ref{rmk2}.\end{proposition} \begin{proof} This is a direct consequence of Remarks ~\ref{rmk1} and ~\ref{rmk2}.\end{proof} \begin{proposition}\label{pinx} Let $p\in R$ and suppose that $D(p)=qp$, for some $q\in F\{\Y\}$. Then $p\in F[\X].$\end{proposition} \begin{proof} Let $p=\sum_\alpha p_{\alpha}(Y)\bx^{\alpha}.$ Then \begin{eqnarray*} D(p)&=&\sum_\alpha p'_{\alpha}(Y) \bx^{\alpha} +p_{\alpha}(Y) D(\bx^{\alpha})\\ &=&qp\\ &=&\sum_\alpha q(Y)p_{\alpha}(Y) \bx^{\alpha}.\end{eqnarray*} By Proposition ~\ref{coeffmon}, for each $\alpha$ with $p_\alpha(Y)\neq 0$ the corresponding coefficient of $\bx^\alpha$ in $D(p)$ is \begin{eqnarray*} D(p)_\alpha=p'_{\alpha}(Y)&+& p_{\alpha}(Y) \sum_{i=1}^n\sum_{j=1}^n \alpha_{ij}Y_{ii}\\ &+& \sum_{i=1}^n\sum_{j=1}^n(\alpha_{ij}+1) \sum _{\ell\ne i}p_{\alpha_{ij,\ell}}(Y)Y_{i\ell}.\end{eqnarray*} Since $D(p)=qp$, it must be $D(p)_\alpha=q(Y)p_\alpha(Y)$ or, equivalently, \begin{eqnarray*} q(Y)p_{\alpha}(Y)=p'_{\alpha}(Y)&+&p_{\alpha} (Y) \sum_{i=1}^n\sum_{j=1}^n \alpha_{ij}Y_{ii}\\ &+& \sum_{i=1}^n\sum_{j=1}^n(\alpha_{ij}+1)\sum _{\ell\ne i}p_{\alpha_{ij,\ell}}(Y)Y_{i\ell}.\end{eqnarray*} This means that for each $\alpha$, the coefficient $p_{\alpha}(Y)$ of $\bx^\alpha$ in $p$ divides the expression $$ p'_{\alpha}(Y)+\sum_{i=1}^n\sum_{j=1}^n(\alpha_{ij}+1) \sum _{\ell\ne i}p_{\alpha_{ij,\ell}}(Y)Y_{i\ell}.$$ Thus, for each $\alpha$, there is $u_{\alpha}(Y)$ such that $$p_{\alpha}(Y)u_{\alpha}(Y)= p'_{\alpha}(Y)+ \sum_{i=1}^n\sum_{j=1}^n(\alpha_{ij}+1) \sum _{\ell\ne i}p_{\alpha_{ij,\ell}}(Y)Y_{i\ell}.$$ As in the proof of Proposition ~\ref{nodpy}, order the triples $\{ij,k\}$, $i,j,k\in\mathbb N$, with the lexicographical order. Let $\{mn,t\}$ be the largest subindex such that $Y_{mn,t}$ occurs in $p$. We have $D(Y_{mn,t})=Y_{mn,t+1}$ and $\{mn,t+1\}>\{mn,t\}$. Now, for each $\alpha$ such that $Y_{mn,t}$ occurs in $p_{\alpha}(Y)$ we have that $Y_{mn,t+1}$ will occur in $p'_{\alpha}(Y)$ but not in $p_\alpha(Y)$ or in $$ \sum_{i=1}^n\sum_{j=1}^n(\alpha_{ij}+1)\sum _{\ell\ne i}p_{\alpha_{ij,\ell}}(Y)Y_{i\ell}$$ by the choice of $\{mn,t\}$. Therefore, it must occur in $p_\alpha(Y)u_\alpha(Y)$. Let $$p_{\alpha}(Y)=\sum a_{\beta}Y^{\beta_{11}}_{11} Y^{\beta_{12}}_{12}\cdots Y^{\beta_{mn,t}}_{mn,t}$$ then \begin{eqnarray*}\lefteqn{p'_{\alpha}(Y) =\sum\,\, a'_{\beta}Y^{\beta_{11}}_{11}\cdots Y^{\beta_{mn,t}}_{mn,t}}\\ &&\qquad\quad+\sum a_{\beta}\,\,{\beta_{11}}\,\,Y^{\beta_{11}- 1}_{11}Y^{\beta_{11,1}+1}_{11,1}\cdots Y^{\beta_{mn,t}}_{mn,t}+\dots\\ & &\qquad\quad\quad+\sum a_{\beta}\,\,{\beta_{mn,t}}\,\,Y^{\beta_{11}}_{11}\cdots Y^{\beta_{mn,t}- 1}_{mn,t}Y_{mn,t+1}.\end{eqnarray*} So $Y_{mn,t+1}$ occurs in $p'_{\alpha}(Y)$ only in \begin{eqnarray*}\lefteqn{ \sum \,\,a_{\beta}\,\,{\beta_{mn,t}}\,\,Y^{\beta_{11}}_{11}\cdots Y^{\beta_{mn,t}- 1}_{mn,t}Y_{mn,t+1}}\\ &&=\Big(\sum a_{\beta}\,\,{\beta_{mn,t}}\,\,Y^{\beta_{11}}_{11}\cdots Y^{\beta_{mn,t}- 1}_{mn,t}\Big)Y_{mn,t+1}\\ &&=v(Y)Y_{mn,t+1}.\end{eqnarray*} Since $Y_{mn,t+1}$ occurs in $p_\alpha(Y)u_\alpha(Y)$ and not in $p_\alpha(Y)$ it must occur in $u_\alpha(Y)$. Let $u_{\alpha,t+1}(Y)$ be the coefficient of $Y_{mn,t+1}$ in $u_\alpha(Y)$. Then it has to be $$p_\alpha(Y)u_{\alpha,t+1}(Y)Y_{mn,t+1}=v(Y)Y_{mn,t+1}.$$ The above equation implies that $p_\alpha(Y)$ divides $v(Y)$. But this is impossible since the total degree of $v(Y)$ is strictly less than the total degree of $p_{\alpha}(Y)$. This contradiction yields the result.\end{proof} \begin{lemma}\label{coeffy} Let $p\in F[\X]$ and suppose that there is $q \in F\{\Y\}$ such that $D(p)=qp.$ Then $q$ is a linear polynomial in the $\Y$. If $\beta=(\beta_{ij})$ is sucht that $\bx^{\beta}$ occurs in $p$, then for $1\leq i\leq n$ the coefficient of $Y_{ii}$ in $q$ is $ \sum_{j=1}^n \beta_{ij}$. In particular, the sums $\sum_{j=1}^n \beta_{ij}$, for $1\leq i\leq n$, are independent of the choice of $\bx^\beta$.\end{lemma} \begin{proof} We have $p=\sum a_{\beta}\bx^{\beta},$ with $a_{\beta}\in F.$ Thus, \begin{eqnarray*} D(p)&=&\sum a'_{\beta} \bx^{\beta} +a_{\beta}D(\bx^{\beta})\\ &=&qp\\ &=&\sum q(Y)a_{\beta} \bx^{\beta}. \end{eqnarray*} By Proposition ~\ref{coeffmon}, the coefficient of $\bx^\beta$ in $D(p)$ is $$a'_{\beta}+ a_{\beta} \sum_{i=1}^n\sum_{j=1}^n \beta_{ij}Y_{ii}+ \sum_{i=1}^n\sum_{j=1}^n(\beta_{ij}+1)\sum _{\ell\ne i}a_{\beta_{ij,\ell}}Y_{i\ell}.$$ Hence, it must be $$ q(Y)a_{\beta}=a'_{\beta}+ a_{\beta}\Big( \sum_{i=1}^n\sum_{j=1}^n \beta_{ij}Y_{ii} + \sum_{i=1}^n\sum_{j=1}^n(\beta_{ij}+1)\sum _{\ell\ne i}a_{\beta_{ij,\ell}}Y_{i\ell}\Big). $$ From this, $$ q(Y)={\frac{a'_{\beta}}{a_{\beta}}}+ \sum_{i=1}^n\sum_{j=1}^n \beta_{ij}Y_{ii}+ \sum_{i=1}^n\sum_{j=1}^n(\beta_{ij}+1)\sum _{\ell\ne i}{\frac{a_{\beta_{ij,\ell}}}{a_{\beta}}}Y_{i\ell}. $$ The coefficient of $Y_{ii}$ in the above expression is $ \sum_{j=1}^n\beta_{ij}$, for $1\leq i\leq n$. Since this expression for $q$ is valid for any index $\beta$, the ``in particular" part follows immediately.\end{proof} \begin{corollary}\label{corcoeffy} Let $p$ be as in Lemma ~\ref{coeffy}. Let $\bx^\alpha$ be the leading power product of $p$. Let $\bx^{\beta}$ be any power product with non-zero coefficient in $p$. Then $ \sum_{j=1}^n \beta_{ij}=\sum_{j=1}^n \alpha_{ij}$, for $1\leq i\leq n$. Thus $p$ is homogeneous of degree $ \sum_{j=1}^n\sum_{i=1}^n \alpha_{ij}.$ \end{corollary} \begin{proof} This is an immediate consequence of the ``in particular" part in Lemma ~\ref{coeffy}.\end{proof} \begin{corollary}\label{corcoeffy2} Let $p\in F[\X]$ and suppose that $D(p)=qp$, for some $q\in F\{\Y\}$. Let $\bx^\alpha$ be the leading power product of $p$, and let $\ell\in F$ be its coefficient. Then $$q={\frac{\ell'}{\ell}}+ \sum_{i=1}^n\sum_{j=1}^n\alpha_{ij}Y_{ii}. $$ \end{corollary} \begin{proof} By Proposition ~\ref{coeffmon} and since $D(p)=qp$, the coefficient of $\bx^\alpha$ in $D(p)$ is \begin{equation} \ell q=\ell'+\ell\,\Big( \sum_{i=1}^n\sum_{j=1}^n\alpha_{ij}Y_{ii}+ \sum_{i=1}^n\sum_{j=1}^n(\alpha_{ij}+1)\sum _{\ell\ne i}p_{\alpha_{ij,\ell}}Y_{i\ell}\Big). \tag{1}\end{equation} The $p_{\alpha_{ij,k}}$ are the coefficients of the power products $\bx^{\alpha_{ij,k}}$ in$p$, with ${\alpha_{ij,k}}\neq\alpha$, such that $D(\bx^{\alpha_{ij,k}})$ contains an expression of the form $Y_{st}\bx^\alpha$. By Remark ~\ref{rmk2}, these power products are $$\bx^{\alpha_{rs,t}}=\cases X_{11}^{\alpha_{11}}\cdots X_{rs}^{\alpha_{rs}+1}\cdots X_{ts}^{\alpha_{ts}-1}\cdots X_{nn}^{\alpha_{nn}}\,\,\,&\text{if $\,\,\,rt$}\endcases,$$ all of which violate Corollary ~\ref{corcoeffy} for $i=r$ and $i=t$. Therefore it must be $p_{\alpha_{ij,k}}=0$, for all $1\leq i,j\leq n$; $k\ne i$. But now, substituting back in (1), we see that $$ \ell q= \ell'+\ell \sum_{i=1}^n\sum_{j=1}^n\alpha_{ij}Y_{ii}.$$ Hence, $$q=\frac{\ell'}{\ell}+ \sum_{i=1}^n\sum_{j=1}^n\alpha_{ij} Y_{ii}.$$\end{proof} Our next step in order to show that the Darboux polynomials $p\in R$ have the desired form will be to show that such a $p$ \begin{emph}{is not reduced with respect to}\end{emph} $\det[\X]$. For that we will show that the leading power product of $p$ is a power of the leading power product of $\det[\X]$. First, we have \begin{lemma}\label{alphalemma} Let $p\in F[\X]$ be such that $D(p)=qp, q\in F\{\Y\}$. Let $\bx^\alpha$ be its leading power product. Then $\alpha_{ij}=0$ for $j\ne n-i+1$ and $\alpha_{i,n-i+1}>0$, $1\leq i\leq n$. That is, $\bx^\alpha=X_{1n}^{\alpha_{1n}}X_{2,n-1}^{\alpha_{2,n-1}} \cdots X_{n1}^{\alpha_{n1}}$.\end{lemma} \begin{proof} To prove that $\alpha_{ij}=0$ for $j\ne n-i+1$ we first show that $\alpha_{ij}=0$ for $j>n-k+1,$ $ i\geq k,$ $2\leq k\leq n$. Indeed, for $k=2$ we have $j>n-1$, so $j=n$ and $$D(\bx^\alpha)=\alpha_{nn} \sum_{k=1}^{n-1} Y_{nk}X_{11}^{\alpha_{11}}\cdots X_{kn}^{\alpha_{kn}+1}\cdots X_{nn}^{\alpha_{nn}-1}\,+\dots$$ Since $q$ has no $\Y$ with $i\ne j$, each term in $D(\bx^\alpha)$ containing such a $\Y$ must be cancelled. In particular we need to cancel the terms containing $$Y_{nj}X_{11}^{\alpha_{11}}\cdots X_{jn}^{\alpha_{jn}+1}\cdots X_{nn}^{\alpha_{nn}-1}$$ for $1\leq j\leq n-1$ above. For that we can only use the derivatives of power products of the form \begin{eqnarray*}\lefteqn{\bx^{\alpha_{nl,j}}=}\\ &&X_{11}^{\alpha_{11}}\cdots X_{j\ell}^{\alpha_{j\ell}-1}\cdots X_{jn}^{\alpha_{jn}+1}\cdots X_{n1}^{\alpha_{n1}}\cdots X_{n\ell}^{\alpha_{n\ell}+1}\cdots X_{nn}^{\alpha_{nn}-1}, \ \ \ \ \ \ \ \ell2$ be such that $\alpha_{in}=0$ for $i\geq k$. Then \begin{eqnarray*}\lefteqn{\bx^\alpha=}\\ &&X_{11}^{\alpha_{11}}\cdots X_{k-1,n}^{\alpha_{k-1,n}} \cdots X_{k,n-1}^{\alpha_{k,n-1}}X_{k+1,1}^{\alpha_{k+1,1}} \cdots X_{k+1,n-1}^{\alpha_{k+1,n-1}}\cdots X_{n,n-1}^{\alpha_{n,n-1}}\end{eqnarray*} and \begin{eqnarray*}\lefteqn{D(\bx^\alpha)= \alpha_{k-1,n}\Big( \sum_{ik-1}Y_{k-1,i}X_{11}^{\alpha_{11}}\cdots X_{k-1,n}^{\alpha_{k-1,n}-1}\cdots X_{in}^{\alpha_{in}+1}\cdots X_{n,n-1}^{\alpha_{n,n-1}}\Big)+\dots \end{eqnarray*} Likewise, we need to cancel all the terms in $D(\bx^\alpha)$ that contain $Y_{k-1,i}$, for $i\ne k-1$. In particular, we need to cancel $$Y_{k-1,i}X_{11}^{\alpha_{11}}\cdots X_{in}^{\alpha_{in}+1} \cdots X_{k-1,n}^{\alpha_{k-1,n}-1}\cdots X_{n,n-1}^{\alpha_{n,n-1}},$$ for $i< k-1$. For that we can only use the power products of the form \begin{eqnarray*}\lefteqn{\bx^{\alpha_{k-1,\ell,i}}=}\\ &&X_{11}^{\alpha_{11}}\cdots X_{i\ell}^{\alpha_{i\ell}-1} \cdots X_{in}^{\alpha_{in}+1}\cdots X_{k-1,\ell}^{\alpha_{k-1,\ell}+1} \cdots X_{k-1,n}^{\alpha_{k-1,n}-1}\cdots X_{n,n-1}^{\alpha_{n,n-1}},\end{eqnarray*} for $i2$, it follows that $\alpha_{kn}=0$, for $2\leq k\leq n$. This makes the statement that $\alpha_{ij}=0$ for $j>n-k+1$, $i\geq k$, true for $k=2$. Now assume that $k$ is such that $\alpha_{ij}=0$ for $j>n-k+1, i\geq k$. So \begin{eqnarray*}\lefteqn{\bx^\alpha =}\\ && X_{11}^{\alpha_{11}}\cdots X_{1n}^{\alpha_{1n}}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}} X_{k+1,1}^{\alpha_{k+1,1}}\cdots X_{k+1,n-k+1}^{\alpha_{k+1,n-k+1}}\cdots X_{n,n-k+1}^{\alpha_{n,n-k+1}}\end{eqnarray*} and for $i> k$ $$\alpha_{i,n-k+1}\Y X_{11}^{\alpha_{11}}\cdots X_{1n}^{\alpha_{1n}}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}+1}\cdots X_{i,n-k+1}^{\alpha_{i,n-k+1}-1}\cdots X_{n,n-k+1}^{\alpha_{n,n-k+1}}$$ occurs in $D(\bx^\alpha)$. Thus we need to cancel it. For that we can only use the derivatives of power products of the form \begin{eqnarray*}\lefteqn{\bx^{\alpha_{ij,k}}=}\\ &&X_{11}^{\alpha_{11}}\cdots X_{kj}^{\alpha_{kj}-1}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}+1}\cdots \X^{\alpha_{ij}+1} \cdots X_{i,n-k+1}^{\alpha_{i,n-k+1}-1}\cdots X_{n,n-k+1}^{\alpha_{n,n-k+1}}\end{eqnarray*} with $jn-k+1$ by hypothesis. But all such power products are strictly greater than $\bx^\alpha$ and therefore they cannot occur in $p$. This forces $\alpha_{i,n-k+1}=0$ for $i>k$. We can repeat this process until $k=n$ and get $\alpha_{ij}=0$ for all $j>n-k+1$, $i\geq k$, $2\leq k\leq n$, that is, \begin{eqnarray*}\lefteqn{\bx^\alpha=}\\ && X_{11}^{\alpha_{11}}\cdots X_{1n}^{\alpha_{1n}} X_{21}^{\alpha_{21}}\cdots X_{2,n-1}^{\alpha_{2,n-1}}X_{31}^{\alpha_{31}}\cdots X_{n-1,2}^{\alpha_{n-1,2}} X_{n1}^{\alpha_{n1}}.\end{eqnarray*} Now we show that $\alpha_{ij}=0$ for $ji$ that $$\alpha_{i1}Y_{i\ell} X_{11}^{\alpha_{11}}\cdots X_{i1}^{\alpha_{i1}-1}\cdots X_{\ell 1}^{\alpha_{\ell 1}+1}\cdots X_{n1}^{\alpha_{n1}} $$ occurs in $D(\bx^\alpha)$. So, in order to cancel it, we need to use the derivatives of power products of the form \begin{eqnarray*}\lefteqn{\bx^{\alpha_{ij,\ell}}=}\\ &&X_{11}^{\alpha_{11}}\cdots X_{i1}^{\alpha_{i1}-1}\cdots \X^{\alpha_{ij}+1}\cdots X_{\ell 1}^{\alpha_{\ell 1}+1}\cdots X_{\ell j}^{\alpha_{\ell j}-1}\cdots X_{n1}^{\alpha_{n1}}\end{eqnarray*} with $j>1$, all of which are strictly greater than $\bx^\alpha$ if $\elli$, we have that $$\alpha_{i,n-k+1}Y_{i\ell}X_{1,n-k+1}^{\alpha_{1,n-k+1}}\cdots X_{i,n-k+1}^{\alpha_{i,n-k+1}-1} \cdots X_{\ell, n- k+1}^{\alpha_{\ell, n-k+1}+1}\cdots X_{n1}^{\alpha_{n1}}$$ occurs in $D(\bx^\alpha)$ and in order to cancel it we only have the derivatives of power products of the form \begin{eqnarray*}\lefteqn{\bx^{\alpha_{ij,\ell}}=} \\ & &X_{1,n-k+1}^{\alpha_{1,n-k+1}}\cdots X_{i,n-k+1}^{\alpha_{i,n-k+1}-1} \cdots \X^{\alpha_{ij}+1} \cdots X_{\ell,n-k+1}^{\alpha_{\ell, n-k+1}+1}\cdots X_{\ell j}^{\alpha_{\ell j}-1} \cdots X_{n1}^{\alpha_{n1}}\end{eqnarray*} with $j>n-k+1$ since $\alpha_{ij}=0$ for $i\leq k$, $jn-k+1$. Thus it has to be $\alpha_{i,n-k+1}=0$ for $i\leq k-1$. We can repeat this process until $k=n-1$ and get $\alpha_{ij}=0$, $jj$}$$ or $$\alpha_{j,n-j+1}Y_{ji}X_{1n}^{\alpha_{1n}}\cdots X_{i,n-j+1}\cdots X_{j,n-j+1}^{\alpha_{j,n-j+1}-1}\cdots X_{n1}^{\alpha_{n1}} +\dots \quad\quad\quad\quad\text{if $i1$, that is, if $a=\alpha_{1n}$, then $$\bx^\alpha=(X_{1n}X_{2,n-1}\cdots X_{n1})^a.$$ \end{lemma} \begin{proof} Let $\ell$ be the coefficient of $\bx^\alpha$ in $p$. We have \begin{equation*}\begin{split}&\quad D(\ell X_{1n}^{\alpha_{1n}}X_{2,n-1}^{\alpha_{2,n-1}}\cdots X_{n1}^{\alpha_{n1}}) =\\ &\qquad\Big( \sum_{i=1}^n \alpha_{i,n-i+1}\ell Y_{ii}\Big)X_{1n}^{\alpha_{1n}} X_{2,n-1}^{\alpha_{2,n-1}}\cdots X_{n1}^{\alpha_{n1}}\\ &\quad\qquad +\alpha_{1n}\ell \sum_{k\ne 1}Y_{1k} X_{1n}^{\alpha_{1n}-1}\cdots X_{k,n- k+1}^{\alpha_{k,n-k+1}}\cdots X_{kn}\cdots X_{n1}^{\alpha_{n1}}\\ &+\ell \sum_{1i}\Y X_{1n}^{\alpha_{1n}}\cdots X_{i,n-i+1}^{\alpha_{i,n-i+1}-1}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}} \cdots X_{k,n-i+1}\cdots X_{n1}^{\alpha_{n1}}\\ &+\ell \sum_{1i} \Y X_{1n}^{\alpha_{1n}} \cdots X_{k,n-i+1}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}}\cdots X_{i,n-i+1}^{\alpha_{i,n-i+1}-1}\cdots X_{n1}^{\alpha_{n1}}\\ &+\ell'X_{1n}^{\alpha_{1n}}X_{2,n-1}^{\alpha_{2,n-1}}\cdots X_{n1}^{\alpha_{n1}}.\end{split}\end{equation*} In order to cancel $$\alpha_{1n}\ell Y_{1k} X_{1n}^{\alpha_{1n}-1}\cdots X_{k,n- k+1}^{\alpha_{k,n-k+1}}\cdots X_{kn}\cdots X_{n1}^{\alpha_{n1}}, \ \ \ \ \ k\ne 1,$$ above, we can only use the derivatives of the power product \begin{eqnarray*}\lefteqn{\bx^{\alpha_{1,n-k+1,k}}=}\\ &&X_{1,n-k+1}\cdots X_{1n}^{\alpha_{1n}-1}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}-1} \cdots X_{kn}\cdots X_{n1}^{\alpha_{n1}},\end{eqnarray*} since for $j\ne n-k+1$ we have $\alpha_{kj}=0$. Let $a_{\alpha_{1,n-k+1,k}}$ be the coefficient of $X^{\alpha_{1,n-k+1,k}}$ in $p$. Then \begin{equation}a_{\alpha_{1,n-k+1,k}}=-\ell \alpha_{1n} \tag{2}\end{equation} On the other hand, in order to cancel $$\alpha_{k,n-k+1}\ell Y_{k1}X_{1,n-k+1}\cdots X_{1n}^{\alpha_{1n}}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}- 1}\cdots X_{n1}^{\alpha_{n1}},\ \ \ \ \ k\ne 1$$ above, the only power product that we can use is, again, \begin{eqnarray*} \bx^{\alpha_{kn,1}} &=&X_{1,n-k+1}\cdots X_{1n}^{\alpha_{1n}-1}\cdots X_{k,n-k+1}^{\alpha_{k,n-k+1}-1}\cdots X_{kn}\cdots X_{n1}^{\alpha_{n1}}\\ &=& X^{\alpha_{1,n-k+1,k}},\end{eqnarray*} since $\alpha_{1j}=0$ for $j\ne n$. Thus it must be \begin{equation}a_{\alpha_{1,n-k+1,k}}=-\ell \alpha_{k,n-k+1} \tag{3}\end{equation} as well. From (2) and (3) it follows that, for $k\ne 1$, $\alpha_{1n}=\alpha_{k,n-k+1}.$ \end{proof} As a consequence of the above results we obtain the following expression for $q$: \begin{corollary}\label{corq} Let $p\in F[\X]$ and suppose that $D(p)=qp$, $q\in F\{\Y\}$. Let $\bx^\alpha$ be the leading power product of $p$. Let $a\in\mathbb N$ be such that $$\bx^\alpha=(X_{1n}X_{2,n-1}\cdots X_{n1})^a$$ and let $\ell\in F$ be the coefficient of $\bx^\alpha$ in $p$. Then $$q=\frac{\ell'}{ \ell}+ a\, \sum_{i=1}^nY_{ii}. $$ \end{corollary} \begin{proof} This is a consequence of Corollary ~\ref{corcoeffy2} and Lemma ~\ref{lp(p)}.\end{proof} \begin{corollary}\label{phomogen} Let $p$ be as in Corollary ~\ref{corq}. Then $p$ is homogeneous of degree $na.$\end{corollary} \begin{proof} This is a consequence of Corollary ~\ref{coeffy} and Lemma ~\ref{lp(p)}\end{proof} Lemma ~\ref{lp(p)} implies that $p$ is not reduced with respect to $\det[\X]$. Since this is a key point in the proof of our main result we restate it as the following \begin{theorem}\label{bigthm} Let $p\in F[\X]$ be such that $D(p)=qp$, $q\in F\{\Y\}$. Let $\bx^\alpha$ be its leading power product. Then $$\bx^\alpha=(X_{1n}X_{2,n-1}\cdots X_{n1})^a= \text{\rm lp}(\det[\X])^a.$$ Thus $p$ is not reduced with respect to $\det[\X]$. \end{theorem} {\bf Note.} If $f$ is a polynomial, $\text{\rm lp}(f)$ denotes its leading power product with respect to a given order. \begin{proof} This is just a restatement of Lemma ~\ref{lp(p)}.\end{proof} \begin{remark}\label{rmk4} Let $p_1,p_2\in F[\X]$ be two polynomials such that $\text{lp}(p_1)=X^\alpha=\text{lp}(p_2)$. Then we can write $p_1=f\,p_2+r$ where $f\in F$ and $r$ is reduced with respect to $p_2$. Indeed, since $\text{lp}(p_1)=\text{lp}(p_2)$, we have that $\text{lp}(p_2)$ divides $\text{lp}(p_1)$. So $p_1$ is not reduced with respect to $p_2$ . We may apply the Multivariable Division Algorithm to $p_1$ and $p_2$, to get $f,r\in F[\X]$, such that $p_1=f\,p_2+r$, with $r$ reduced with respect to $p_2$ and $\text{lp}(p_1)=\text{lp}(f)\text{lp}(p_2)$. The last equation implies that $\text{lp}(f)=1$. Hence, $f\in F$.\end{remark} We are now ready to prove our main result on the form of the Darboux polynomials in $R$: \begin{theorem}\label{bgstthm} Let $p\in F[\X]$ and $q\in F\{\Y\}$ be polynomials in $R$ that satisfy the Darboux condition $D(p)=qp$. Then there is $a\in \mathbb N$ and $\ell\in F$ such that $$p=\ell\,\det[\X]^a$$ and $$q=\frac{\ell'}{ \ell}+a\, \sum_{i=1}^nY_{ii}. $$ \end{theorem} \begin{proof} Let $q_1=\sum_{i=1}^n a\,Y_{ii},$ so that, $$ D(\det[\X]^a)=q_1\det[\X]^a =(q-\frac{\ell'}{ \ell})\det[\X]^a.$$ By Remark ~\ref{rmk4} we can write $p=\ell\det[\X]^a+r$, with $r$ reduced with respect to $\det[\X]^a$. Now, \begin{eqnarray*} D(p)&=&D(\ell\det[\X]^a) +D(r)\\ &=&\ell'\det[\X]^a+\ell(q-\frac{\ell'}{\ell})\det[\X]^a+D(r)\\ &=&\ell'\det[\X]^a+q\ell\det[\X]^a-\ell'\det[\X]^a+D(r)\\ &=&q\ell\det[\X]^a+D(r).\end{eqnarray*} On the other hand, we have \begin{eqnarray*} D(p)&=&qp\\ &=&q\ell\det[\X]^a+qr.\end{eqnarray*} Therefore, it has to be $D(r)=qr$. But $r$ is reduced with respect to $\det[\X]^a$. It follows, by Theorem ~\ref{bigthm}, that $r=0$. The statement about the form of $q$ is just the content of Corollary ~\ref{corq}.\end{proof} Now, for $1\leq i,j\leq n$, let $D_{E(ij)}\in \Lie(\GL)$ be the derivation given by multiplication by the matrix $E(ij)$, with 1 in position (i,j) and zero elsewhere. The set $\{D_{E(ij)}|\,1\leq i,j\leq n\}$ is a basis for $\Lie(\GL)$ with respect to which the derivation in Theorem ~\ref{bgstthm} is expressed. We show that the result does not depend on the basis chosen on $\Lie(\GL)$: \begin{theorem}\label{thmingen} Let ${\mathcal D}_{st}$, $1\leq s,t\leq n$, be any basis of $\Lie(\GL)$. Define a derivation in the ring $R=F\{\Y\}[\X]$ by ${\mathcal D}= \sum Y_{st}{\mathcal D}_{st}$. Let $p$ and $q$ be polynomials in $R$ that satisfy the Darboux condition ${\mathcal D}(p)=qp$. Then there is $a\in\mathbb N$ and $\ell\in F$ such that $p=\ell\det[\X]^a$ and $q=\frac{\ell'}{\ell}+a\, \sum_{i=1}^nY_{ii}$.\end{theorem} \begin{proof} Since $\{D_{E(ij)}|\,1\leq i,j\leq n\}$ is a basis of $\Lie(\GL)$ we have $${\mathcal D}_{st}= \sum c_{st,ij}D_{E(ij)}, $$ with $c_{st,ij}\in C$. Thus, \begin{eqnarray*} {\mathcal D}&=& \sum_{s,t} Y_{st}{\mathcal D}_{st}\\ &=& \sum_{s,t} Y_{st} \sum_{i,j} c_{st,ij}D_{E(ij)}\\ &=& \sum_{i,j}\sum_{s,t}c_{st,ij}Y_{st}D_{E(ij)}\\ &=& \sum_{i,j}Z_{ij}D_{E(ij)},\end{eqnarray*} where $Z_{ij}= \sum_{s,t}c_{st,ij}Y_{st}$. Now, $[c_{st,ij}]$ is a matrix of change of basis so it is invertible. Also the $c_{st,ij}$ are contants for $D$, thus the map $Z_{ij,k}\rightarrow Y_{ij,k}$ is a differential bijection. In other words, the differential rings $$R=F\{\Y\}[\X], D$$ and $$R'=F\{Z_{ij}\}[\X], {\mathcal D}$$ are isomorphic and therefore we can apply Theorem ~\ref{bgstthm} to $R'.$\end{proof} %%New Stuff \section{A Generic Picard-Vessiot Extension for $\GL$} In this section we prove the statement about the generic Picard-Vessiot extension for $F$ with differential Galois group $\GL$. We provide some specialization properties in the case when $F$ has finite transcendence degree over $C$. \subsection{The generic extension} \begin{theorem}\label{thmgpve} $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ is a generic Picard-Vessiot extension with differential Galois group $\GL$. \end{theorem} \begin{proof} First we need to show that $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ is a Picard-Vessiot extension with differential Galois group $\GL$. We will use the characterization of Theorem ~\ref{cpve}. We have \begin{enumerate} \item $F\langle \Y\rangle(\X)=F\langle \Y\rangle\langle V\rangle$, where $V\subset F\langle \Y\rangle(\X)$ is the finite dimensional vector space over $C$ spanned by the $\X$. \item The group $G=\GL$ acts as a group of differential automorphisms of\linebreak $F\langle \Y\rangle(\X)$ with $G(V)\subseteq V$ and $F\langle \Y\rangle(\X)^G=F\langle \Y\rangle$. This follows from the fact that $F\langle \Y\rangle(\X)$ is the function field of $\GL_{F\langle \Y\rangle}$. \item $F\langle \Y\rangle(\X)\supseteq F\langle \Y\rangle$ has no new constants. This is a consequence of Proposition ~\ref{weil2}, Corollary ~\ref{maincor} and Theorem ~\ref{bgstthm}.\end{enumerate} Now, suppose that $E\supseteq F$ is a Picard-Vessiot extension of $F$ with differential Galois group $\GL$. By Theorems ~\ref{sthm} and ~\ref{stgln}, we have that in this situation $E$ is isomorphic to $F(\X)$ (the function field of $\GL_F$) as a $\GL$-module and as an $F$-module. Any $\GL$ equivariant derivation $D_E$ on $F(\X)$ extends the derivation on $F$ in such a way that $$D_E(\X)=\sum_{\ell=1}^n f_{i\ell}X_{\ell j}$$ with $\f\in F$. Since $E\supset F$ is a Picard-Vessiot extension for $\GL$, then so is $C\langle \f\rangle(\X)\supset C\langle \f\rangle$, the derivation on $C\langle \f\rangle(\X)$ being the corresponding restriction of $D_E$. From this Picard-Vessiot extension one can retrieve $F(\X)\supset F$ by extension of scalars from $C$ to $F$. In this way, any Picard-Vessiot extension $E\supset F$ with differential Galois group $\GL$ can be obtained from $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ via the specialization $\Y\mapsto \f$. This means that $F\langle \Y\rangle(\X)\supset F\langle \Y\rangle$ is a generic Picard-Vessiot extension of $F$ for $\GL$.\end{proof} \subsection{Specializing to a Picard-Vessiot extension of $\mathbf F$}\label{sectsp} In this section we give necessary and sufficient conditions for a specialization $\Y\mapsto \f$, $\f\in F$, with $C\langle \f\rangle(\X)\supset C\langle \f\rangle$ a Picard-Vessiot extension, to exist. We restrict ourselves to the case when $F$ has finite transcendence degree over $C$. Our goal is to find $f_{ij}\in F$ such that the specialization (homomorphism) from $C\{\Y\}$ to $F$ given by $\Y\mapsto \f$ is such that $C\langle \f\rangle(\X)\supset C\langle \f\rangle$, with derivation given by $D(\X)= \sum_{\ell=1}^nf_{i\ell}X_{\ell j}$, has no new constants. We have: \begin{theorem}\label{specializ2} Let $F=C(t_1,\dots,t_m)[z_1,\dots,z_k]$ where the $t_i$ are algebraically independent over $C$ and the $z_i$ are algebraic over $C(t_1,\dots,t_m)$. Assume that the derivation on $F$ has field of constants $C$ and that it extends to $F(\X)$ so that $D(f\otimes \X)= D(f)\otimes \X+f\otimes \sum_{\ell=1}^n f_{i\ell}X_{\ell j}$ on $F\otimes C[\X]$. Let $\C$ be the field of constants of $F(\X)$. Then $\C=C$ if and only if the set of all the $t_i$ and all the $\X$ are algebraically independent over $\C$.\end{theorem} \begin{proof} (Sufficiency) Suppose that $\C$ properly contains $C$. Let $r$ be the transcendence degree of $\C$ over $C$. Since $C$ is algebraically closed, $r$ has to be at least one. We have the tower of fields $$C\subset\C\subset\C(\X)\subset F(\X)$$ where the transcendence degree of $\C\subset\C(\X)$ is $n^2$ and the transcendence degree of $C\subset F(\X)$ is $n^2+m$. Since $r\geq 1$ the transcendence degree $\ell$ of $\C(\X)\subset F(\X)$ has to be $\ell