# Positive Characteristic Darboux-Jouanolou Integrability of Differential
Forms

Edileno de Almeida SANTOS  Instituto de Ciência, Engenharia e Tecnologia
(ICET) - Universidade Federal dos Vales do Jequitinhonha e Mucuri (UFVJM),
Campus do Mucuri, Teófilo Otoni - MG, Brazil<EMAIL_ADDRESS>and
Sergio RODRIGUES  Faculdade de Ciências Exatas e Tecnologia (FACET) -
Universidade Federal da Grande Dourados (UFGD), Rodovia Dourados - Itahum, Km
12 - Cidade Universitátia, Dourados - MS, Brazil<EMAIL_ADDRESS>
###### Abstract.

We prove a Darboux-Jouanolou type theorem on the algebraic integrability of
polynomial differential $r$-forms over arbitrary fields ($r\geq 1$). We also
investigate the Darboux’s method for producing integrating factors.

###### Key words and phrases:

Positive Characteristic, Vector Fields, Codes

###### 1991 Mathematics Subject Classification:

37F75

##  1\. Introduction

In his seminar memoir [5], G. Darboux (1878) showed the fascinating
relationships between integrability and the existence of algebraic solutions
for a planar polynomial differential system. The classical approach of Darboux
shows that, for a planar polynomial vector field of degree $d$, on
$\mathbb{R}^{2}$ or $\mathbb{C}^{2}$, from $\binom{d+1}{2}+1$ invariant
algebraic curves we can deduce an analytic (possibly multivaluated) first
integral. This method was studied by H. Poincaré (1891) in [16], where he
observes the difficulty to obtain algorithmically invariant curves.

A general result of J.-P. Jouanolou (1979)in [9] shows that if $K$ is a field
of characteristic $0$ and $\omega$ is a polynomial $1$-form of degree $d$ on
$K^{n}$ admitting at least $\binom{d-1+n}{n}\cdot\binom{n}{2}+2$ irreducible
invariant algebraic hypersurfaces, then $\omega$ has a rational first
integral, computed in terms of the invariant hypersurfaces.

The Darboux Integration Method have been successfully used in Physics (see
[8], [12], [13], [20], [21]). For example, C. G. Hewitt (1991) in [8] study
some new solutions to the Einstein field equations.

For holomorphic foliations on compact complex manifolds, E. Ghys (2000) in [6]
gives an extended version of Jouanolou’s Theorem. Over fields of
characteristic zero, a Darbou-Jouanolou type theorem is proved in [4] for
polynomial differential $r$-forms, $r\geq 1$.

In positive characteristic, M. Brunella and M. Nicollau (1999) proved in [3]
that if $\omega$ is a rational $1$-form on a smooth projective variety over an
algebraically closed field $K$ of positive characteristic $p>0$ with
infinitely many invariant hypersurfaces, then $\omega$ admits a rational first
integral.

In sharp contrast with the characteristic $0$ case, where a theorem of
Jouanolou says that a generic vector field on the complex plane does not admit
any invariant algebraic curve, J. V. Pereira (2001) shows in [14] that a
generic vector field on an affine space of positive characteristic admits an
invariant algebraic hypersurface (the generic condition is that the divergent
of the vector field must be zero).

Our goal in this paper is to extend results of [4] ($r$-forms over
characteristic $0$, $r>0$) and [17] ($1$-forms over arbitrary fields) giving a
general account of Darboux-Jouanolou integrability for polynomial $r$-forms on
$K^{n}$ ($r\geq 1$), where $K$ is an arbitrary field. We obtain a general
version of Darboux-Jouanolou Criterion ([9], Théorème 3.3, p. 102): for each
polynomial $r$-form $\omega$ of degree $d$ on $K^{n}$, with $K$ of
characteristic $p\geq 0$, we define a natural number $N_{K}(n,d,r)$ (see
Definition 2.4) that depends only on $n$, $r$, $d$ and $p$, and we prove the
following theorem.

######  Theorem A (Theorem 5.3).

Let $\omega\in\Omega^{r}_{d}(K^{n})$ be a polynomial $r$-form of degree $d$
over an arbitrary field $K$ and $N=N_{K}(n,d-1,r+1)$. If $\omega$ has $N+r+1$
invariant hypersurfaces, then $\omega$ admits a rational first integral.

The number $N_{K}(n,d,r)$ has the property that in characteristic $0$ holds
$N_{K}(n,d,r)=\binom{d+n}{n}\cdot\binom{n}{r}$ but in characteristic $p>0$ we
have only $N_{K}(n,d,r)\leq\binom{d+n}{n}\cdot\binom{n}{r}$.

Another approach is to look at integrating factors, that is, a function $f$
for a given $r$-form $\omega$ such that $d(f\omega)=0$. In this direction we
obtain the result below.

######  Theorem B (Theorem 6.6).

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form of degree $d$ and
$N=N_{K}(n,d-1,r+1)$. If $\omega$ admits $N+r$ invariant hypersurfaces, then
$\omega$ has a rational first integral or a rational integration factor.

Our exposition was heavily influenced by the excellent monograph [15] of J. V.
Pereira, where the Darboux-Jouanolou theory is appreciated on
$\mathbb{C}^{2}$.

##  2\. Differential Forms

We remember first some definitions and results from reference [17], where the
reader can obtain more information about vector fields and differential forms
over arbitrary fields.

######  Definition 2.1.

Let $K$ be a field of arbitrary characteristic $p>0$, where $p$ is a prime
integer. The ring of differential constants is the sub-ring of
$K[z]:=K[z_{1},...,z_{n}]$ given by

$K[z^{p}]:=K[z_{1}^{p},...,z_{n}^{p}]$

and the field of differential constants is the sub-field of
$K(z):=K(z_{1},...,z_{n})$ given by

$K(z^{p}):=K(z_{1}^{p},...,z_{n}^{p})$

We have natural inclusions $K[z]\subset K(z)$ and $K[z^{p}]\subset K(z^{p})$.
In fact, we have more: $K[z^{p}]$ is a $K$-sub-module of $K[z]$ generated by
$\\{G^{p}:G\in K[z]\\}$ and $K(z^{p})$ is a $K$-vector subspace of $K(z)$
generated by $\\{g^{p}:g\in K(z)\\}$. The elements of $K(z^{p})$ are called
$\partial$-constants.

The ring of polynomials $K[z]$ is a $K[z^{p}]$-algebra. If $n\geq 0$ then, for
each $j=1$,…, $n$,
$z_{j}^{n}=z_{j}^{pq_{j}+i_{j}}=(z_{j}^{q_{j}})^{p}z_{j}^{i_{j}}$ with $0\leq
i_{j}<p$, hence $K[z]$ is finitely generated as a $K[z^{p}]$-algebra. In the
next proposition we will prove that $K(z)$ is also finitely generated as a
vector space over $K(z^{p})$.

Note that $K[z^{p}]-K^{*}$ is a $K$-sub-module of the module $K[z]$, and we
have the following isomorphisms

$\mathcal{M}:=K[z]\otimes_{K[z^{p}]}K[z^{p}]\cong\frac{K[z]}{(K[z^{p}]-K^{*})}\otimes_{K}K[z^{p}]\cong\bigoplus_{0\leq
i_{1},...,i_{n}\leq p-1}K[z^{p}]\cdot z_{1}^{i_{1}}...z_{n}^{i_{n}}$

then $K[z]\simeq\mathcal{M}$. To make a $K(z^{p})$-algebra from $K[z]$ we
define

$\mathcal{L}:=K[z]\otimes_{K[z^{p}]}K(z^{p})\cong\frac{K[z]}{(K[z^{p}]-K^{*})}\otimes_{K}K(z^{p})\cong\bigoplus_{0\leq
i_{1},...,i_{n}\leq p-1}K(z^{p})\cdot z_{1}^{i_{1}}...z_{n}^{i_{n}}$

######  Proposition 2.2 (see also [2], 1.1.1 Lemma, p. 3).

Let $K$ be a field of characteristic $p>0$. Then, as $K(z^{p})$-vector spaces,
$K(z)$ is isomorphic to $\mathcal{L}$, and hence $K(z)$ is a finite
dimensional vector space over $K(z^{p})$, and $\dim_{K(z^{p})}(K(z))=p^{n}$.

###### Proof.

A base for $K(z)$ over $K(z^{p})$ is given by the monomials
$z_{1}^{i_{1}}...z_{n}^{i_{n}}$, where $0\leq i_{1},...,i_{n}\leq p-1$. In
fact, if $f=\frac{P}{Q}\in K(z)$, we can write

$f=\frac{P}{Q}=\left(\frac{1}{Q}\right)^{p}\cdot Q^{p-1}P$

and the polynomial $Q^{p-1}P$ can be expressed as a $K(z^{p})$-linear
combination of the monomials $z_{1}^{i_{1}}...z_{n}^{i_{n}}$.

∎

In what follows we consider multi-indexes $I=(i_{1},...,i_{n})$. If
$J=(j_{1},...,j_{n})$ is another multi-index of length $n$, we write $J<I$
(resp. $J\leq I$) if $j_{1}<i_{1}$,…, $j_{n}<i_{n}$ (resp. $j_{1}\leq
i_{1}$,…, $j_{n}\leq i_{n}$). There are two special multi-indexes:
$\overrightarrow{0}=(0,...,0)$ and $\overrightarrow{p}=(p,...,p)$.

######  Proposition 2.3.

Let $K$ be a field of characteristic $p>0$. Then

$K(z^{p})=\\{f\in K(z):df=0\\}$

and

$K[z^{p}]=\\{F\in K[z]:dF=0\\}$

###### Proof.

A base for $K[z]$ over $K[z^{p}]$ is given by the monomials
$z_{1}^{i_{1}}...z_{n}^{i_{n}}$, where $0\leq i_{1},...,i_{n}\leq p-1$, and
this is also a base for $K(z)$ over $K(z^{p})$, by the above proposition. It
is clear that $K(z^{p})\subset\\{f\in K(z):df=0\\}$. Let $f\in K(z)$ be a
rational function then, by the above proposition,

$f(z)=\sum_{\overrightarrow{0}\leq I<\overrightarrow{p}}g_{I}(z^{p})z^{I}$

Hence, writing $I=(I(1),...,I(n))$,

$\frac{\partial f}{\partial
z_{j}}=\sum_{\begin{array}[]{c}\overrightarrow{0}\leq
I\leq\overrightarrow{p}\\\ I(j)>0\end{array}}g_{I}(z^{p})I(j)z^{I}z_{j}^{-1}$

and, fixing $j$, the monomials $z^{I}z_{j}^{-1}$ are $K(z^{p})$-linearly
independent. Therefore, if $df=0$, then all $g_{I}$ with
$\overrightarrow{0}<I$ must be $0$, so we must have
$f=g_{\overrightarrow{0}}(z^{p})\in K(z^{p})$. In particular,
$K[z^{p}]=\\{F\in K(z):dF=0\\}$.

∎

Define the $K[z^{p}]$-sub-module of $K[z^{p}]$

$\mathcal{M}_{d}:=\bigoplus_{\begin{array}[]{c}0\leq i_{1},...,i_{n}\leq
p-1\\\ 0\leq i_{1}+...+i_{n}\leq d\end{array}}K[z^{p}]\cdot
z_{1}^{i_{1}}...z_{n}^{i_{n}}$

and the $K(z^{p})$-vector subspace of $K(z)$

$\mathcal{L}_{d}:=\bigoplus_{\begin{array}[]{c}0\leq i_{1},...,i_{n}\leq
p-1\\\ 0\leq i_{1}+...+i_{n}\leq d\end{array}}K(z^{p})\cdot
z_{1}^{i_{1}}...z_{n}^{i_{n}}$

where the number $d$ is called the $p$-degree of the module (or vector space)
if $d\leq n(p-1)$.

We can write a decomposition by $p$-degree

$K[z^{p}]=\mathcal{M}_{0}\subset\mathcal{M}_{1}\subset...\subset\mathcal{M}_{n(p-1)}=K[z]$

and

$K(z^{p})=\mathcal{L}_{0}\subset\mathcal{L}_{1}\subset...\subset\mathcal{L}_{n(p-1)}=K(z)$

Let $A$ be a ring and let $B$ be an $A$-algebra. We denote by
$\Omega^{1}_{B/A}$ the module of relative differential $1$-forms of $B$ over
$A$, that is, if $E$ is the free $A$-module generated by $db$ with $b\in B$
and $F$ is the submodule generated by $da$ with $a\in A$, then
$\Omega^{1}_{B/A}:=E/F$ with an $A$-derivation
$d:B\rightarrow\Omega^{1}_{B/A}$ (see [11], page 210 ). By using the exterior
product, we obtain
$\Omega^{r}_{B/A}:=\Omega^{1}_{B/A}\wedge...\wedge\Omega^{1}_{B/A}$ ($r$
times). If $M\subset B$ is an $A$-sub-module of $B$, then we define

$\Omega^{1}_{M/A}:=d(M)\subset\Omega^{1}_{B/A}$

We can easily deduce the isomorphisms
$\Omega^{1}_{K[z]/K}\simeq\Omega^{1}_{K[z]/K[z^{p}]}$ as $K[z^{p}]$-modules
and $\Omega^{1}_{K(z)/K}\simeq\Omega^{1}_{K(z)/K(z^{p})}$ as $K(z^{p})$-vector
spaces.

######  Definition 2.4.

If $d\geq 0$ is an integer, then we denote by
$\Omega^{r}_{d}(K^{n})\subset\Omega^{r}_{K[z]/K}$ the $K$-vector subspace of
polynomial differential $r$-forms of degree $\leq d$. Then

$\Omega_{\mathcal{M}_{d}/K[z^{p}]}^{r}:=\Omega^{1}_{\mathcal{M}_{d}/K[z^{p}]}\wedge...\wedge\Omega^{1}_{\mathcal{M}_{d}/K[z^{p}]}\simeq\Omega_{0}^{r}(K^{n})\otimes_{K}\mathcal{M}_{d}=\Omega_{0}^{r}(K^{n})\otimes_{K}\mathcal{M}_{s(d)}$

and

$\Omega_{\mathcal{L}_{d}/K(z^{p})}^{r}:=\Omega_{\mathcal{L}_{d}/K(z^{p})}^{1}\wedge...\wedge\Omega_{\mathcal{L}_{d}/K(z^{p})}^{1}\simeq\Omega_{0}^{r}(K^{n})\otimes_{K}\mathcal{L}_{d}=\Omega_{0}^{r}(K^{n})\otimes_{K}\mathcal{L}_{s(d)}$

where $s(d)=\min\\{d,n(p-1)\\}$, and we define

$N_{K}(n,d,r)=N(p,n,d,r):=\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d}/K(z^{p})}^{r})\leq\min\\{\binom{n}{r}\cdot
p^{n},\dim_{K}(\Omega_{d}^{r}(K^{n}))\\}$

These two modules of $r$-forms have natural decomposition by $p$-degree:

$\Omega^{r}_{\mathcal{M}_{0}/K[z^{p}]}\subset\Omega^{r}_{\mathcal{M}_{1}/K[z^{p}]}\subset...\subset\Omega^{r}_{\mathcal{M}_{n(p-1)}/K[z^{p}]}=\Omega^{r}_{K[z]/K}$

and

$\Omega^{r}_{\mathcal{L}_{0}/K(z^{p})}\subset\Omega^{r}_{\mathcal{L}_{1}/K(z^{p})}\subset...\subset\Omega^{r}_{\mathcal{L}_{n(p-1)}/K(z^{p})}=\Omega^{r}_{K(z)/K}$

For every $d\geq 0$ we have a natural inclusion
${\Omega}_{d}^{r}(K^{n})\rightarrow\Omega^{r}_{\mathcal{L}_{s(d)}/K(z^{p})}$.

##  3\. Logarithmic 1-forms and Residues

######  Definition 3.1.

Let $F_{1}$,…, $F_{m}$ be a collection of irreducible polynomials and
$\lambda_{1}$,…, $\lambda_{m}\in K(z^{p})$ be $\partial$-constants. The linear
combination

$\eta=\lambda_{1}\frac{dF_{1}}{F_{1}}+...+\lambda_{m}\frac{dF_{m}}{F_{m}}$

will be called a $\partial$-logarithmic $1$-form (or simply a logarithmic
$1$-form). Note that every $\partial$-logarithmic $1$-form $\eta$ is closed,
that is, $d\eta=0$.

Remember also the Hilbert’s Nullstellensatz:

######  Theorem 3.2 ([10], Theorem 1.5, p. 380).

Let $I$ be an ideal of $K[z]$ and let
$V(I)=\\{c\in(\overline{K})^{n}:f(c)=0,\forall f\in I\\}$ be the algebraic
variety associated to $I$, where $\overline{K}$ is the algebraic closure of
$K$. Let $P$ be a polynomial in $K[z]$ such that $P(c)=0$ for every zero
$(c)=(c_{1},...,c_{n})\in V(I)$. Then there is an integer $m>0$ such that
$P^{m}\in I$.

Using the above theorem, Jouanolou proved the following important lemma ([9],
Lemme 3.3.2, p. 102) in the case that the characteristic of the field $K$ is
0.

######  Lemma 3.3.

If $\mathcal{S}$ is a finite representative system of primes in
$K[z]=K[z_{1},...,z_{n}]$ (that is, $\mathcal{S}$ is a finite collection of
distinct irreducible polynomials), then the $K$-linear map

$K^{(\mathcal{S})}\longrightarrow\Omega^{1}_{K(z)/K}$
$(\lambda_{j})_{F_{j}\in\mathcal{S}}\longmapsto\sum_{F_{j}\in\mathcal{S}}\lambda_{j}\frac{dF_{j}}{F_{j}}$

is injective.

######  Remark 3.4.

Two irreducible polynomials $P$, $Q\in K[z]$ are distinct if there is no
constant $\lambda\in K^{*}$ such that $P=\lambda Q$, that is, the rational
function $f=\frac{P}{Q}\in K(z)$ is not a constant ($f\notin K$).

The following lemma is a residue formula over positive characteristic $p>0$.

######  Lemma 3.5 ([17], Lemma 5.2).

Let $G\in K[x]$, $G(0)\neq 0$, be a polynomial function in one variable, where
$K$ is a field of positive characteristic $p>0$. If $\alpha\in K(x^{p})$, then

$Res(\alpha\cdot\frac{dG}{G},0)=0$

The following lemma is the Positive Characteristic Jouanolou’s Lemma.

######  Lemma 3.6 ([17], Lemma 5.3).

If $\mathcal{S}$ is a finite representative system of non-$\partial$-constant
primes in $K[z]-K[z^{p}]$ (that is, $\mathcal{S}$ is a finite collection of
distinct non-$\partial$-constant irreducible polynomials), then the
$K(z^{p})$-linear map

$K(z^{p})^{(\mathcal{S})}\longrightarrow\Omega^{1}_{K(z)/K}$
$(\lambda_{j})_{F_{j}\in\mathcal{S}}\longmapsto\sum_{F_{j}\in\mathcal{S}}\lambda_{j}\frac{dF_{j}}{F_{j}}$

is injective.

######  Remark 3.7.

If $K$ is a perfect field, then every irreducible polynomial in $K[z]$ is
non-$\partial$-constant. Now let $K$ be a non perfect field, that is, there is
$\lambda\in K$ such that $\lambda$ is not a $p$-power, where $p>0$ is the
characteristic of $K$. Then $P(x):=x^{p}-\lambda$ is an irreducible polynomial
in $K[x]$, but $dP=0$.

######  Definition 3.8 ([7], 2.4 Définition, p. 116).

Let $\omega\in\Omega^{r}_{K(z)/K}$ be a rational $r$-form. We define the space
of tangent $1$-forms of $\omega$ by

$\mathcal{E}^{*}(\omega)=\\{\eta\in\Omega^{1}_{K(z)/K}:\omega\wedge\eta=0\\}$

######  Lemma 3.9 ([4], Lemma 2.1).

Let $\omega\in\Omega^{r}_{K(z)/K}$. If there are $r$ elements $\eta_{1}$,…,
$\eta_{r}\in\mathcal{E}^{*}(\omega)$ linearly independent over $K(z)$, then
there is $h\in K(z)$ such that

$\omega=h\cdot\eta_{1}\wedge...\wedge\eta_{r}$

###### Proof.

We can complete $\\{\eta_{1},...,\eta_{r}\\}$ to make a $K(z)$-basis
$\\{\eta_{1},...,\eta_{r},\eta_{r+1}...,\eta_{M}\\}$ for
$\Omega^{r}_{K(z)/K}$, where $M=\dim_{K(z)}\Omega^{r}_{K(z)/K}$. So we can
write

$\omega=\sum_{1\leq i_{1}<...<i_{r}\leq
M}h_{i_{1}...i_{r}}\cdot\eta_{i_{1}}\wedge...\wedge\eta_{i_{r}}$

Since $\omega\wedge\eta_{i}=0$ for $i=1$,…, $r$, we have
$\omega=h_{1...r}\cdot\eta_{1}\wedge...\wedge\eta_{r}$.

∎

##  4\. Darboux Method

######  Definition 4.1.

Let $F\in K[z]$ and $\omega\in\Omega_{d}^{r}(K^{n})$ be a polynomial $r$-form
of degree $d$. We say that $F$ is invariant by $\omega$ if $dF\neq 0$ and $F$
divides $\omega\wedge dF$, that is,

$\omega\wedge dF=F\cdot\Theta_{F}$

where $\Theta_{F}\in\Omega_{d-1}^{r+1}(K^{n})$ is called the associated
cofactor of $F$. If $\overline{K}$ is the algebraic closure of $K$, by
Hilbert’s Nullstellensatz this is equivalent to saying that

$(\omega\wedge dF)\mid_{V}=0$

where $V=\\{F=0\\}$ is the associated algebraic hypersurface in
$(\overline{K})^{n}$.

A rational function $f$ is said to be a rational first integral for $\omega$
if $df\neq 0$ and $\omega\wedge df=0$, that is, $df$ is tangent do $\omega$.

The following proposition is similar to Proposition 6.1 of [17].

######  Proposition 4.2.

Let $\omega\in\Omega^{r}_{K[z]/K}$ be an $r$-form, where $K$ is an arbitrary
field. If there are invariant irreducible polynomials $F_{1}$,…, $F_{m}$ in
$K[z]$ and $\partial$-constants $\lambda_{1}$,…, $\lambda_{m}$ in
$K(z^{p})^{*}$ such that

$\lambda_{1}\Theta_{F_{1}}+...+\lambda_{m}\Theta_{F_{m}}=0$

then there is a logarithmic $1$-form $\eta\neq 0$ such that
$\omega\wedge\eta=0$.

###### Proof.

We can consider the logarithmic $1$-form

$\eta=\sum_{i=1}^{m}\lambda_{i}\cdot\frac{dF_{i}}{F_{i}}$

Therefore

$\omega\wedge\eta=\sum_{i=1}^{m}\lambda_{i}\omega\wedge\frac{dF_{i}}{F_{i}}=\sum_{i=1}^{m}\lambda_{i}\Theta_{F_{i}}=0$

∎

######  Corollary 4.3.

Let $\omega\in\Omega_{d}^{r}(K^{n})$ be a polynomial $r$-form of degree $d$
and
$N=N_{K}(n,d-1,r+1)=\dim_{K(z^{p})}(\Omega^{r+1}_{\mathcal{L}_{d-1}/K(z^{p})})$.
If $\omega$ has $N+1$ invariant irreducible polynomials, then there is a
logarithmic $1$-form $\eta\neq 0$ such that $\omega\wedge\eta=0$.

###### Proof.

By hypothesis, $\omega$ has $F_{1}$,…, $F_{m}$ as invariant irreducible
polynomials, where $m=N+1$. The $m$ associated cofactors $\Theta_{F_{1}}$,…,
$\Theta_{F_{m}}$ are elements of
$\Omega^{r+1}_{\mathcal{L}_{d-1}/K(z^{p})}=\Omega_{0}^{r+1}(K^{n})\otimes_{K}\mathcal{M}_{s(d)}$
and since $m=\dim_{K(z^{p})}(\Omega^{r+1}_{\mathcal{L}_{d-1}/K(z^{p})})+1$
then there exist $\lambda_{1}$,…, $\lambda_{m}\in K(z^{p})$, not all zero,
such that

$\lambda_{1}\Theta_{F_{1}}+...+\lambda_{m}\Theta_{F_{m}}=0$

By Proposition 4.2, the logarithmic $1$-form
$\eta:=\lambda_{1}\frac{dF_{1}}{F_{1}}+...+\lambda_{m}\frac{dF_{m}}{F_{m}}$
satisfies

$\omega\wedge\eta=\lambda_{1}\Theta_{F_{1}}+...+\lambda_{m}\Theta_{F_{m}}=0$

∎

######  Example 4.4.

Consider the $2$-form $\omega=\alpha xdy\wedge dz+\beta ydz\wedge dx+\gamma
zdx\wedge dy$, where $\alpha$, $\beta$, $\gamma$ are $\partial$-constants in
$K(z^{p})^{*}$. We can easily see that $F=x$, $G=y$ and $H=z$ are invariant.
The associated cofactors are $\Theta_{x}=\omega\wedge\frac{dx}{x}=\alpha
dx\wedge dy\wedge dz$, $\Theta_{y}=\omega\wedge\frac{dy}{y}=\beta dx\wedge
dy\wedge dz$ and $\Theta_{z}=\omega\wedge\frac{dz}{z}=\gamma dx\wedge dy\wedge
dz$. If $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$ are $\partial$-constants
not all zero satisfying
$\alpha\cdot\lambda_{1}+\beta\cdot\lambda_{2}+\gamma\cdot\lambda_{3}=0$ then

$\lambda_{1}\Theta_{x}+\lambda_{2}\Theta_{y}+\lambda_{3}\Theta_{z}=0$

and the logarithmic $1$-form
$\eta=\lambda_{1}\frac{dx}{x}+\lambda_{2}\frac{dy}{y}+\lambda_{3}\frac{dz}{z}$
is tangent to $\omega$.

######  Definition 4.5.

Let $\omega\in\Omega^{r}_{K(z)/K}$ be a rational $r$-form. The $p$-degree of
$\omega$ is the smaller $\epsilon$ such that
$\omega\in\Omega_{\mathcal{L}_{\epsilon}/K(z^{p})}^{r}$.

######  Remark 4.6.

In Corollary 4.3 it is not possible to exchange
$\Omega_{\mathcal{L}_{d}/K(z^{p})}^{r}=\Omega_{\mathcal{L}_{s(d)}/K(z^{p})}^{r}$
by $\Omega_{\mathcal{L}_{\epsilon}/K(z^{p})}^{r}$, where $\epsilon$ is the
$p$-degree of $\omega$, because in general we have
$\epsilon<s(d)=\min\\{d,n(p-1)\\}$ and the proof of the corollary will fail.

######  Proposition 4.7.

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form. If there are
invariant irreducible polynomials $F_{1}$,…, $F_{m}$ in $K[z]$ and constants
$\delta_{1}$,…, $\delta_{m}$ in $\mathbb{Z}$ (or in the prime sub-field
$\mathbb{Z}_{p}$ of $K$ if char$(K)=p>0$) not all zero, such that

$\delta_{1}\Theta_{F_{1}}+...+\delta_{m}\Theta_{F_{m}}=0$

then $\omega$ has a rational (resp. polynomial) first integral.

###### Proof.

If we take the $1$-form
$\eta=\delta_{1}\frac{dF_{1}}{F_{1}}+...+\delta_{m}\frac{dF_{m}}{F_{m}}$, then

$\omega\wedge\eta=\sum_{i=1}^{m}\delta_{i}\omega\wedge\frac{dF_{i}}{F_{i}}=\sum_{i=1}^{m}\delta_{i}\Theta_{F_{i}}=0$

Since $dF_{i}^{\delta_{i}}=\delta_{i}F_{i}^{\delta_{i}}\frac{dF_{i}}{F_{i}}$,
then the rational function (or polynomial)
$f=F_{1}^{\delta_{1}}...F_{m}^{\delta_{m}}$ is such that $df=f\eta$ and then

$\omega\wedge df=f\omega\wedge\eta=0$

that is, $f$ is a first integral for $\omega$. ∎

##  5\. Darboux-Jouanolou Criterion

######  Lemma 5.1.

Let $\omega\in\Omega^{r}_{K(z)/K}$ be a rational $r$-form and let
$\eta_{1}$,…, $\eta_{r}\in\mathcal{E}^{*}(\omega)$ be $K(z^{p})$-linearly
independent closed $1$-forms. If $\eta_{1}\wedge...\wedge\eta_{r}=0$, then
$\omega$ has a rational first integral.

###### Proof.

Since $\eta_{1}\wedge...\wedge\eta_{r}=0$, then $\eta_{1}$,…, $\eta_{r}$ are
linearly dependent over $K(z)$. Let $m<r$ be the largest positive integer such
that $\eta_{1}$,…, $\eta_{m}$ are linearly independent over $K(z)$. Then

$\eta_{m+1}=\sum_{i=1}^{m}f_{i}\cdot\eta_{i}$

with $f_{1}$,…, $f_{m}\in K(z)$.

The $1$-forms $\eta_{1}$,…, $\eta_{m}$, $\eta_{m+1}$ are $K(z^{p})$-linearly
independent, hence there exists $j_{0}\in\\{1,...,m\\}$ such that
$df_{j_{0}}\neq 0$, that is, $f_{j_{0}}\in K(z)-K(z^{p})$

By hypothesis, $\eta_{i}$ is closed for $i=1,...,m+1$, then

$0=\sum_{i=1}^{m}df_{i}\wedge\eta_{i}$

Multiplying the above expression by
$\eta_{1}\wedge...\wedge\widehat{\eta_{j_{0}}}\wedge...\wedge\eta_{m}$, we
obtain

$0=\sum_{i=1}^{m}df_{i}\wedge\eta_{i}\wedge\eta_{1}\wedge...\wedge\widehat{\eta_{j_{0}}}\wedge...\wedge\eta_{m}=(-1)^{{j_{0}}+1}df_{j_{0}}\wedge\eta_{1}\wedge...\wedge\eta_{m}$

Since $\eta_{1}$,…, $\eta_{m}$ are $K(z)$-linearly independent, then there
exist $g_{1}$,…, $g_{m}\in K(z)$ such that
$df_{j_{0}}=\sum_{i=1}^{m}g_{i}\cdot\eta_{i}$ and, by hypothesis, for each
$j=1$,…, $m$, $\omega\wedge\eta_{j}=0$. Therefore

$\omega\wedge df_{j_{0}}=\sum_{i=1}^{m}g_{i}\cdot\omega\wedge\eta_{i}=0,$

that is, $f_{j_{0}}$ is a rational first integral for $\omega$.

∎

######  Lemma 5.2.

Let $\omega\in\Omega^{r}_{K(z)/K}$ be a rational $r$-form, with $1\leq r<n$,
and let $\eta_{1}$,…, $\eta_{r}$, $\eta_{r+1}$, $\in\mathcal{E}^{*}(\omega)$
be $K(z^{p})$-linearly independent closed $1$-forms. If
$\eta_{1}\wedge...\wedge\eta_{r}\neq 0$ and
$\eta_{2}\wedge...\wedge\eta_{r+1}\neq 0$, then $\omega$ has a rational first
integral.

###### Proof.

By hypothesis $\\{\eta_{1},...,\eta_{r}\\}$ and
$\\{\eta_{2},...,\eta_{r+1}\\}$ are two sets of $K(z)$-linearly independent
closed $1$-forms in $\mathcal{E}^{*}(\omega)$. Write
$\tau_{1}=\eta_{1}\wedge...\wedge\eta_{r}$ and
$\tau_{2}=\eta_{2}\wedge...\wedge\eta_{r+1}$.

By Lemma 3.9 there exist rational functions $h_{1}$ and $h_{2}$ such that
$\tau_{i}=h_{i}\cdot\omega$, $i=1,2$. Writing $f=\frac{h_{1}}{h_{2}}$, we have
$\tau_{1}={f}\cdot\tau_{2}$. Hence

$0=\tau_{1}-\tau_{1}=\tau_{1}-{f}\cdot\tau_{2}=(\eta_{1}+(-1)^{r}f\cdot\eta_{r+1})\wedge\eta_{2}\wedge...\wedge\eta_{r}$

Moreover, using that $d\tau_{1}=d\tau_{2}=0$, we have

$0=d\tau_{1}=df\wedge\tau_{2}=h_{2}df\wedge\omega$

and then $\omega\wedge df=0$.

Now we have two possibilities for this rational function: $f\in K(z)-K(z^{p})$
or $f\in K(z^{p})$. If $f\in K(z)-K(z^{p})$, then $f$ is a rational first
integral for $\omega$. On the other hand, suppose that $f\in K(z^{p})$. By
hypothesis, the closed $1$-forms $\eta_{1}$, $\eta_{2}$,…,
$\eta_{r}\in\mathcal{E}^{*}(\omega)$ are $K(z^{p})$-linearly independent,
which imply that $\mu:=\eta_{1}+(-1)^{r}f\cdot\eta_{r+1}$, $\eta_{2}$,…,
$\eta_{r}\in\mathcal{E}^{*}(\omega)$ are also $K(z^{p})$-linearly independent
closed $1$-forms. Then we conclude the existence of rational first integral
for $\omega$ by Lemma 5.1.

∎

######  Theorem 5.3 (Theorem A).

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form of degree $d$ and
$N=\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1})$. If $\omega$
admits $N+r+1$ invariant polynomials, then $\omega$ has a rational first
integral.

###### Proof.

Suppose that the polynomial $r$-form $\omega$ admits $N+r+1$ invariant
irreducible polynomials $F_{1}$,…, $F_{N+r+1}$. Then, for each
$i\in\\{1,2,...,N+r+1\\}$, there exists
$\Theta_{F_{i}}\in\Omega_{d-1}^{r+1}(K^{n})$ such that

$\omega\wedge dF_{i}=F_{i}\cdot\Theta_{F_{i}}$

Since $\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1})=N$, for each
$s=1$,…, $r+1$, we can choose $\lambda_{s}^{s}$,…, $\lambda_{s+N}^{s}\in
K(z^{p})$ such that $\lambda_{s}^{s}\neq 0$ and

$\sum_{j=s}^{s+N}\lambda_{j}^{s}\cdot\Theta_{F_{j}}=0$

For each $s=1$,…, $r+1$, we can define the logarithmic $1$-form

$\eta_{s}=\sum_{j=s}^{s+N}\lambda_{j}^{s}\cdot\frac{dF_{j}}{F_{j}}$

and then $\omega\wedge\eta_{s}=0$.

Define

$\tau_{1}=\eta_{1}\wedge...\wedge\eta_{r}$

and

$\tau_{2}=\eta_{2}\wedge...\wedge\eta_{r+1}$

By Lemma 3.6, $\\{\eta_{1},..,\eta_{r}\\}$ and $\\{\eta_{2},..,\eta_{r+1}\\}$
are two sets of logarithmic $1$-forms linearly independent over $K(z^{p})$.
Then:

  * (i)

if $\tau_{1}=0$ or $\tau_{2}=0$, we obtain a rational first integral for
$\omega$ by Lemma 5.1;

  * (ii)

if $\tau_{1}\neq 0$ and $\tau_{2}\neq 0$, we obtain a rational first integral
for $\omega$ by Lemma 5.2.

∎

We say that a polynomial $F$ is invariant by the vector field $X$ if $dF\neq
0$ and $X(F)$ is divisible by $F$. If $f$ is a rational function, we say that
$f$ is a rational first integral for $X$ if $df\neq 0$ and $X(f)=0$.

We can associate to the vector field
$X=\sum_{i=1}^{n}P_{i}\frac{\partial}{\partial z_{i}}$ the $(n-1)$-form

$\omega_{X}=\sum_{i=1}^{n}(-1)^{i+1}P_{i}dz_{1}\wedge...\wedge\widehat{dz_{i}}\wedge...\wedge
dz_{n}=i_{X}(dz_{1}\wedge...\wedge dz_{n}),$

that is, $\omega_{X}$ is the rotational of $X$. Since

$\omega_{X}\wedge df=X(f)dz_{1}\wedge...\wedge dz_{n}$

we have that $f$ (or $F$) is a rational first integral (or an invariant
polynomial) for $X$ if and only if it is the same for $\omega_{X}$.

######  Corollary 5.4.

Let $X$ be a polynomial vector field of degree $d$ on $K^{n}$ and
$N=\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{n})$. If $X$ admits
$N+n$ invariant irreducible polynomials, then $X$ has a rational first
integral.

###### Proof.

Let $X=\sum_{i=1}^{n}P_{i}\frac{\partial}{\partial z_{i}}$ be a polynomial
vector field of degree $d$. Then $F$ is $X$-invariant if and only if $F$ is
$\omega_{X}$-invariant. By Theorem 5.3, $\omega_{X}$ has a rational first
integral $f\in K(z)$, which will be also a rational first integral for $X$.

∎

Now we will prove that if $\omega$ has a rational first integral, then
$\omega$ has infinitely many invariant polynomials. We will denote the
greatest common factor between two polynomials $P$, $Q\in K[z]$ by $gcd(P,Q)$.
In this way, $P$ and $Q$ are relatively primes if and only if $gcd(P,Q)=1$.

######  Lemma 5.5.

Let $f=\frac{P}{Q}\in K(z)$ be a rational function such that $df\neq 0$. Let
$H\in K[z]$ be a polynomial. If $\lambda=(HP)^{p}\in K[z^{p}]$ then $P+\lambda
Q$ has a non-$\partial$-constant factor $G$ with $gcd(H,G)=1$.

###### Proof.

If $\lambda=(HP)^{p}$, then

$P+\lambda Q=P+H^{p}P^{p}Q=P(1+H^{p}P^{p-1}Q)$

Hence, if $G=1+H^{p}P^{p-1}Q$, then $gcd(H,G)=1$ and

$dG=d(1+H^{p}P^{p-1}Q)=H^{p}d(P^{p-1}Q)=H^{p}(P^{p-1}dQ+(p-1)P^{p-2}QdP)=$
$=H^{p}(P^{p-1}dQ-P^{p-2}QdP)=H^{p}P^{p-2}(PdQ-QdP)=-H^{p}P^{p-2}Q^{2}df\neq
0$

∎

######  Lemma 5.6.

Let $f=\frac{P}{Q}\in K(z)$ be a rational function such that $df\neq 0$. Then
there are infinitely many non-$\partial$-constant irreducible factors of
polynomials $P+\lambda Q\in K[z]$, $\lambda\in K[z^{p}]$.

###### Proof.

Let $H_{1}$,…, $H_{m}$ be non-$\partial$-constant irreducible factors of
$P+\lambda_{1}Q$,…, $P+\lambda_{m}Q$, respectively. Define
$H=H_{1}\cdot...\cdot H_{m}$. By the above lemma, there is $\lambda\in
K[z^{p}]$ such that $P+\lambda Q$ has a non-$\partial$-constant factor $G$
such that $gcd(H,G)=1$. Therefore there is a non-$\partial$-constant
irreducible factor $H_{m+1}$ of $G$. This proves the lemma.

∎

######  Remark 5.7.

For any field $K$, if $P$, $Q\in K[z]$ are two coprime polynomials,

$\lambda_{1}\neq\lambda_{2}\in K\Rightarrow
gcd(P+\lambda_{1}Q,P+\lambda_{2}Q)=1$

This imply the existence of infinitely many factors as claimed by Lemma 5.6 if
$K$ is an infinite field (in particular if $K$ has characteristic $0$). But
this property does not work for $\lambda_{1}\neq\lambda_{2}\in K[z^{p}]$. For
example, suppose that $K$ has characteristic $p=2$. If
$P=z_{1}+z_{2}^{3}=z_{1}+z_{2}^{2}\cdot z_{2}$ and $Q=z_{2}$, then $P$ and $Q$
are irreducible polynomials with $gcd(P,Q)=1$. If $\lambda_{1}=z_{2}^{2}$,
$\lambda_{2}=(z_{1}+z_{2})^{2}\in K(z^{2})$,

$F_{1}=P+\lambda_{1}Q=z_{1}$

and

$F_{2}=P+\lambda_{2}Q=z_{1}(1+z_{1}z_{2})$

then $gcd(F_{1},F_{2})=z_{1}$.

######  Theorem 5.8.

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form over a field $K$.
Then $\omega$ has a rational first integral if and only if it has an infinite
number of invariant irreducible polynomials.

###### Proof.

Suppose that $f=\frac{P}{Q}$ is a rational first integral for $\omega$ and
$gcd(P,Q)=1$. We have $\omega\wedge d(\frac{P}{Q})=0$ and in particular

$\omega\wedge(QdP-PdQ)=0$

which imply $Q\omega\wedge\frac{dP}{P}=\omega\wedge dQ$ and
$P\omega\wedge\frac{dQ}{Q}=\omega\wedge dP$. Since $P$ and $Q$ are coprimes,
then $\omega\wedge\frac{dP}{P}$ and $\omega\wedge\frac{dQ}{Q}$ are polynomial
$(r+1)$-forms.

Consider $F_{\lambda}:=P+\lambda Q$, where $\lambda\in K[z^{p}]$. Then

$\omega\wedge dF_{\lambda}=\omega\wedge(dP+\lambda dQ)=\omega\wedge
dP+\lambda\omega\wedge dQ=\omega\wedge dP+\lambda Q\omega\wedge\frac{dP}{P}=$
$=\omega\wedge(P\frac{dP}{P}+\lambda Q\frac{dP}{P})=(P+\lambda
Q)\cdot\omega\wedge\frac{dP}{P}=F_{\lambda}\cdot\omega\wedge\frac{dP}{P}=F_{\lambda}\cdot\omega\wedge\frac{dQ}{Q}$

Therefore $\omega\wedge\frac{dF_{\lambda}}{F_{\lambda}}$ is a polynomial
$(r+1)$-form, that is, for every $\lambda\in K[z^{p}]$ the
non-$\partial$-constants irreducible factors of $F_{\lambda}$ are invariant.
By the above lemma, we conclude the existence of infinitely many invariant
polynomials for $\omega$.

Reciprocally, if there are infinitely many invariant polynomials we can apply
Theorem 5.3 to conclude the existence of rational first integral for $\omega$.

∎

##  6\. Integrating Factors and Darboux Method

######  Definition 6.1.

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form. A rational
function $f$ is said to be an integration factor of $\omega$ if $f\omega$ is
closed, that is, $d(f\omega)=0$.

The above definition of integration factor is different from those we find in
the books of elementary differential equations in the plane (for example [1],
Chapter 2, p. 99), where it is required $f\omega$ to be exact, that is,
$f\omega=d\eta$. In characteristic $0$ the existence of a rational such $\eta$
is equivalent to the condition $d(f\omega)=0$ (Poincaré’s Lemma), but that
does not work in positive characteristic. The relationship between closed and
exact rational differential forms over fields of characteristic $p>0$ is
explored in our pre-print [18].

######  Proposition 6.2.

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form. If there are
invariant irreducible polynomials $F_{1}$,…, $F_{m}$ in $K[z]$ and
$\partial$-constants $\lambda_{1}$,…, $\lambda_{m}$ in $K(z^{p})^{*}$ such
that

$\lambda_{1}\Theta_{F_{1}}+\lambda_{2}\Theta_{F_{2}}+...+\lambda_{m}\Theta_{F_{m}}=d\omega$

then there is a logarithmic $1$-form $\eta\neq 0$ such that
$\omega\wedge\eta=d\omega$.

###### Proof.

Consider the logarithmic $1$-form

$\eta=\lambda_{1}\frac{dF_{1}}{F_{1}}+\lambda_{2}\frac{dF_{2}}{F_{2}}+...+\lambda_{m}\frac{dF_{m}}{F_{m}}\neq
0$

Then

$\omega\wedge\eta=\sum_{i=1}^{m}\lambda_{i}\omega\wedge\frac{dF_{i}}{F_{i}}=\sum_{i=1}^{m}\lambda_{i}\Theta_{F_{i}}=d\omega$

∎

######  Proposition 6.3.

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form. If there are
invariant irreducible polynomials $F_{1}$,…, $F_{m}$ in $K[z]$ and constants
$\delta_{1}$,…, $\delta_{m}$ in $\mathbb{Q}$ (or in the prime sub-field
$\mathbb{Z}_{p}$ of $K$ if $char(K)=p>0$), not all zero, such that

$\delta_{1}\Theta_{F_{1}}+\delta_{2}\Theta_{F_{2}}+...+\delta_{m}\Theta_{F_{m}}=d\omega$

then $\omega$ has a rational (respectively polynomial) integration factor.

###### Proof.

As in the previous proposition, define the logarithmic $1$-form

$\eta=\delta_{1}\frac{dF_{1}}{F_{1}}+\delta_{2}\frac{dF_{2}}{F_{2}}+...+\delta_{m}\frac{dF_{m}}{F_{m}}\neq
0$

Up to multiplying each $\delta_{1}$,…, $\delta_{m}$ by $-1$ if necessary, we
can suppose that

$\omega\wedge\eta=\sum_{i=1}^{m}\delta_{i}\omega\wedge\frac{dF_{i}}{F_{i}}=\sum_{i=1}^{m}\delta_{i}\Theta_{F_{i}}=\sigma
d\omega$

where $\sigma=(-1)^{r+1}$, and hence $(-1)^{r}\omega\wedge\eta=-d\omega$.

Define the rational (respectively polynomial) function

$f=F_{1}^{\delta_{1}}F_{2}^{\delta_{2}}...F_{m}^{\delta_{m}}$

By differentiation we have $df=f\eta$, and then

$d(f\omega)=df\wedge\omega+fd\omega=(-1)^{r}\omega\wedge
df+fd\omega=(-1)^{r}f\omega\wedge\eta+fd\omega=-fd\omega+fd\omega=0$

∎

######  Theorem 6.4.

Let $\omega\in\Omega_{d}^{r}(K^{n})$ be a rational $r$-form of degree $d$. If
$\omega$ has
$N=N_{K}(n,d-1,r+1)=\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1})$
invariant irreducible polynomials $F_{1}$, $F_{2}$,…, $F_{N}$, then there is a
logarithmic $1$-form $\eta\neq 0$ such that $\omega\wedge\eta=0$ or
$\omega\wedge\eta=d\omega$. If $K=\mathbb{Q}$ or $K=\mathbb{Z}_{p}$, $p>0$,
then $\omega$ has a rational first integral or a rational integration factor.

###### Proof.

The cofactors associated to the invariant polynomials are differential
$(r+1)$-forms of degree lass than or equal to $d-1$. By the natural inclusion
${\Omega}_{d-1}^{r+1}(K^{n})\rightarrow\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1}$
and since we have
$N=\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1})$ cofactors
associated to the polynomials $F_{i}$, $i=1$,…, $N$, then that cofactors are
linearly dependent or form a base to the $K(z^{p})$-vector space
$\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1}$. Hence there are two
possibilities:

Case $1$. The cofactors associated to the polynomials $F_{i}$, $i=1$,…, $N$,
are $K(z^{p})$-linearly dependent. Then there are $\partial$-constants
$\lambda_{1}$,…, $\lambda_{N}$ in $K$, not all zero, such that

$\lambda_{1}\Theta_{F_{1}}+\lambda_{2}\Theta_{F_{2}}+...+\lambda_{N}\Theta_{F_{N}}=0$

hence for the logarithmic $1$-form

$\eta=\sum_{i=1}^{N}\lambda_{i}\frac{dF_{i}}{F_{i}}$

as in Proposition 4.7, we have $\omega\wedge\eta=0$, and if $K=\mathbb{Q}$ or
$K=\mathbb{Z}$, then $f=F_{1}^{\lambda_{1}}...F_{N}^{\lambda_{N}}$ defines a
first integral for $\omega$.

Case $2$. The cofactors associated to the polynomials $F_{i}$, $i=1$,…, $N$,
form a base to the $K(z^{p})$-vector space
$\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1}$. Then, as in Proposition 6.3, the
$(r+1)$-form $d\omega$ can be written as a $K(z^{p})$-linear combination of
the cofactors $\Theta_{F_{i}}$ and we obtain a logarithmic $1$-form $\eta$
such that $\omega\wedge\eta=d\omega$, and if $K=\mathbb{Q}$ or $K=\mathbb{Z}$,
then $f=F_{1}^{\lambda_{1}}...F_{N}^{\lambda_{N}}$ defines an integrating
factor for $\omega$.

∎

######  Example 6.5 (Lotka-Volterra Equation).

Consider the $1$-form

$\omega=y(\gamma-\delta x)dx+x(\alpha-\beta y)dy$

We can easily see that the polynomials $F=x$ and $G=y$ are invariant by
$\omega$, and their cofactors are

$\Theta_{x}=\omega\wedge\frac{dx}{x}=(\beta y-\alpha)dx\wedge dy$

and

$\Theta_{y}=\omega\wedge\frac{dy}{y}=(\gamma y-\delta)dx\wedge dy$

Taking the differential of $\omega$, we have

$d\omega=[(\alpha-\beta y)-(\gamma-\delta x)]dx\wedge
dy=-\Theta_{x}-\Theta_{y}$

and therefore

$d\omega=\omega\wedge(-\frac{dx}{x}-\frac{dy}{y})$

and then $(xy)^{-1}$ is a rational integration factor for $\omega$.

######  Theorem 6.6 (Theorem B).

Let $\omega\in\Omega^{r}_{K[z]/K}$ be a polynomial $r$-form of degree $d$ and
$N=N_{K}(n,d-1,r+1)=\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1})$.
If $\omega$ admits $N+r$ invariant irreducible polynomials, then $\omega$ has
a rational first integral or a rational integration factor.

###### Proof.

Suppose that the polynomial $r$-form $\omega$ admits $N+r$ invariant
irreducible polynomials $F_{1}$,…, $F_{N+r}$. Then, for each
$i\in\\{1,2,...,N+r\\}$, there exists
$\Theta_{F_{i}}\in\Omega_{d-1}^{r+1}(K^{n})$ such that

$\omega\wedge dF_{i}=F_{i}\cdot\Theta_{F_{i}}$

Since $\dim_{K(z^{p})}(\Omega_{\mathcal{L}_{d-1}/K(z^{p})}^{r+1})=N$, for each
$s=1$,…, $r$, we can choose $\lambda_{s}^{s}$,…, $\lambda_{s+N}^{s}\in
K[z^{p}]$, not all zero, such that $\lambda_{s}^{s}\neq 0$ and

$\sum_{j=s}^{s+N}\lambda_{j}^{s}\cdot\Theta_{F_{j}}=0$

For each $s=1$,…, $r$, we can define the logarithmic $1$-form

$\eta_{s}=\sum_{j=s}^{s+N}\lambda_{j}^{s}\cdot\frac{dF_{j}}{F_{j}}$

and then $\omega\wedge\eta_{s}=0$. Define

$\tau=\eta_{1}\wedge...\wedge\eta_{r}$

By Lemma 3.6, $\eta_{1},...,\eta_{r}$ are linearly independent over $K(z^{p})$
and:

  * (i)

if $\tau=0$ then, by Lemma 5.1, we obtain a rational first integral for
$\omega$;

  * (ii)

if $\tau\neq 0$ then, by Lemma 3.9, there is a rational function $h$ such that
$\omega=h\cdot\tau$; therefore, since $\tau$ is closed, with $f=\dfrac{1}{h}$,
we have

$d(f\cdot\omega)=d\tau=0,$

that is, $f$ is an integration factor for $\omega$.

∎

######  Remark 6.7.

If $\omega\in\Omega^{r}_{d}(K^{n})$ has a rational first integral, then
$\omega$ admits infinitely many invariant polynomials, so $\omega$ also has a
rational integration factor.

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