Darboux-Jouanolou Integrability of Polynomial Differential Forms
aa r X i v : . [ n li n . S I] F e b DARBOUX-JOUANOLOU INTEGRABILITY OF POLYNOMIALDIFFERENTIAL FORMS
EDILENO DE ALMEIDA SANTOS AND SERGIO RODRIGUES
Abstract.
We prove a Darboux-Jouanolou type theorem on the algebraicintegrability of polynomial differential r -forms over arbitrary fields ( r ≥ ). Introduction
In his seminar memoir [3], G. Darboux (1878) showed the fascinating relation-ships between integrability and the existence of algebraic solutions for a planarpolynomial differential system. The classical approach of Darboux shows that, fora planar polynomial vector field of degree d , on R or C , from (cid:0) d +12 (cid:1) + 1 invariantalgebraic curves we can deduce an analytic (possibly multivaluated) first integral.This method was studied by H. Poincaré (1891) in [17], where he observes thedifficulty to obtain algorithmically that invariant curves.A general result of J.-P. Jouanolou (1979) in [8] shows that if K is a field ofcharacteristic and ω is a polynomial -form of degree d on K n admitting at least (cid:0) d − nn (cid:1) · (cid:0) n (cid:1) + 2 invariant irreducible algebraic hypersurfaces, then ω has a rationalfirst integral, computed in terms of the invariant hypersurfaces.The Darboux Integration Method have been successfully used in Physics (see [7],[12], [13], [22], [24]). For example, C. G. Hewitt (1991) in [7] study some newsolutions to the Einstein field equations.For holomorphic foliations on compact complex manifolds, E. Ghys (2000) in [4]gives an extended version of Jouanolou’s Theorem. And B. Scárdua (2011) showedin [19] a local Darboux-Jouanolou type theorem for germs of integrable -formson ( C n , . Over fields of characteristic zero, a Darbou-Jouanolou type theorem isproved in [2] for polynomial differential r -forms, r ≥ .In positive characteristic, M. Brunella and M. Nicollau (1999) proved in [1] thatif ω is a rational -form on a smooth projective variety over a field K of positivecharacteristic p > with infinitely many invariant hypersurfaces, then ω admits arational first integral.In sharp contrast with characteristic case, where a theorem of Jouanolou saysthat a generic vector field on the complex plane does not admit any invariantalgebraic curve, J. V. Pereira (2001) shows in [15] that a generic vector field onan 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 [2] ( r -forms over characteristic , r > ) and [18] ( -forms over arbitrary fields) giving a general account of Darboux-Jouanolou integrability for polynomial r -forms on K n ( r ≥ ), where K is an Mathematics Subject Classification.
Key words and phrases.
Positive Characteristic, Vector Fields, Codes. arbitrary field. We obtain a general version of
Darboux-Jouanolou Criterion ([8],Théorème 3.3, p. 102): for each polynomial r -form ω of degree d on K n , with K of characteristic p ≥ , 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 A (Theorem 5.3) . Let ω ∈ Ω ( K n ) be a polynomial r -form of degree d over an arbitrary field K . If ω has N K ( n, d − , r +1)+ r +1 invariant hypersurfaces,then ω admits a rational first integral. The number N K ( n, d − ,
2) + r + 1 has the property that in characteristic holds N K ( n, d − ,
2) + r + 1 = (cid:0) d − nn (cid:1) · (cid:0) nr +1 (cid:1) + r + 1 but in characteristic p > we have N K ( n, d − , r + 1) + r + 1 < (cid:18) d − nn (cid:19) · (cid:18) nr + 1 (cid:19) + r + 1 Another approach is to look at integrating factors , that is, a function f for the r -form ω such that d ( f ω ) = 0 . In this direction we obtain Theorem B (Theorem 6.4) . Let ω ∈ Ω rd ( K n ) be a rational r -form of degree d . If ω has N K ( n, d − , r + 1) invariant irreducible polynomials, then there is a logarithmic -form η = 0 such that ω ∧ η = 0 or ω ∧ η = dω . If k = Q or K = Z p , p > , then ω has a rational first integral or a rational integration factor. Finally, we give an alternative definition of "closedness" for r -forms (" p -closedness", see Definition 7.1), in such a way that was possible to obtain thealgebraic general version of Poincaré’s Lemma bellow Theorem C (Theorem 7.2) . Let K be an arbitrary field and consider a rational(respectively polynomial) r -form ω ∈ Ω rK ( z ) /K (respectively ω ∈ Ω rK [ z ] /K ). Then ω is exact if and only if ω is p -closed. Differential Forms
We remember first some definitions and results from reference [18], where thereader can obtain more information about vector fields and differential forms overarbitrary fields.
Definition 2.1.
Let K be a field of arbitrary characteristic p ( p = 0 or p is a primeinteger). The field of differential constants is the sub-field of K ( z ) = K ( z , ..., z n ) given by K ( z p ) = { f : f ∈ K ( z ) , df = 0 } Note that in characteristic we have K ( z p ) = K ( z ) = K . Otherwise, in primecharacteristic p > we obtain K ( z p ) as the K -vector subspace of K ( z ) generatedby { g p : g ∈ K ( z ) } .We call the elements of K ( z p ) of ∂ -constants . Remark 2.2.
Note that K ( z p ) is the kernel of the d operator, and K ( z ) is infinitedimensional as a K ( z p ) -vector space.We can look at K [ z ] as a K [ z p ] -algebra and we note that K [ z p ] − K ∗ is a K -submodule of the module K [ z ] . To make a K ( z p ) -algebra from K [ z ] we define M := K [ z ] ⊗ K [ z p ] K ( z p ) ∼ = K [ z ]( K [ z p ] − K ∗ ) ⊗ K K ( z p ) ∼ = M ≤ i ,...,i n ≤ p − K ( z p ) · z i ...z i n n ARBOUX-JOUANOLOU INTEGRABILITY OF POLYNOMIAL DIFFERENTIAL FORMS 3
Then M is also a finite dimensional K ( z p ) -vector space.In general, for an integer s > we can also define K [ z s ] as the K -algebragenerated by z s ,..., z sn and M s := K [ z ]( K [ z s ] − K ∗ ) ⊗ K K ( z p ) ∼ = M ≤ i ,...,i n ≤ s − K ( z p ) · z i ...z i n n Then M s is also a finite dimensional K ( z p ) -vector space. In particular,we have M = K ( z p ) and M p = M . Proposition 2.3.
Let K be an arbitrary field. Then K ( z ) is isomorphic, as K ( z p ) -vector space, to M p , and hence it is finite dimensional over K ( z p ) .Proof. A base for K ( z ) over K ( z p ) is given by the monomials z i ...z i n n , where ≤ i , ..., i n ≤ p − . In fact, if f = PQ ∈ K ( z ) , we can write f = PQ = ( 1 Q ) p · Q p − P and the polynomial Q p − P can be expressed as a K ( z p ) -linear combination of thatmonomials z i ...z i n n . (cid:3) Let A be a ring and let B be an A -algebra We denote by Ω B/A the moduleof differential forms of B over A (see [11], page 210). We can easily see that Ω K [ z ] /K ≃ Ω K [ z ] /K [ z p ] as K [ z p ] -modules and Ω K ( z ) /K ≃ Ω K ( z ) /K ( z p ) as K ( z p ) -modules. Definition 2.4.
We define Ω rK [ z ] /K := Ω K [ z ] /K ∧ ... ∧ Ω K [ z ] /K and Ω rK ( z ) /K :=Ω K ( z ) /K ∧ ... ∧ Ω K ( z ) /K ( r times) and we indicate by Ω rd ( K n ) ⊂ Ω rK [ z ] /K the K -vector subspace of polynomial differentials r -forms of degree ≤ d . Finally b Ω rd ( K n ) := Ω r ( K n ) ⊗ K M min { d +1 ,p } and N K ( n, d, r ) := dim K ( z p ) ( b Ω rd ( K n )) ≤ dim K (Ω rd ( K n )) These two modules of r -forms have natural decomposition by degree: Ω r ( K n ) ⊂ Ω r ( K n ) ⊂ ... ⊂ Ω rd ( K n ) ⊂ ... Ω rK [ z ] /K = ∞ [ d =0 Ω rd ( K n ) and b Ω r ( K n ) ⊂ b Ω r ( K n ) ⊂ ... ⊂ b Ω rp − ( K n )Ω rK ( z ) /K = p − [ e =0 b Ω re ( K n ) We note that for each d ≥ there is a natural e such that ≤ e ≤ p − and b Ω rd ′ ( K n ) = b Ω rd ( K n ) ⇒ e ≤ d ′ . This number e is, by definition, the degree of b Ω rd ( K n ) = b Ω re ( K n ) .For every d ≥ we have a natural inclusion Ω rd ( K n ) → b Ω rd ( K n ) , and we canwrite the commutative diagram bellow. E. A. SANTOS AND S. RODRIGUES / / K d / / K [ z ] d / / (cid:15) (cid:15) Ω d ( K n ) d / / (cid:15) (cid:15) ... d / / Ω r ( K n ) / / (cid:15) (cid:15) / / K d / / K ( z ) d / / b Ω d ( K n ) d / / ... d / / b Ω r ( K n ) / / Hence for every r > we have an inclusion homomorphism of K -vector spaces (infact also of K [ z p ] -modules) Ω rK [ z p ] /K = S ∞ d =0 Ω rd ( K n ) → Ω rK ( z p ) /K = S p − d =0 b Ω rd ( K n ) compatible with the d operator. Finally, we obtain an inclusion homomorphismbetween the anticommutative graded algebras Ω ∗ K [ z p ] /K = ∞ M r =0 Ω rK [ z p ] /K → Ω ∗ K ( z p ) /K = ∞ M r =0 b Ω rK ( z p ) /K Logarithmic 1-forms and Residues
Definition 3.1.
Let F ,..., F m be a collection of irreducible polynomials and λ ,..., λ m ∈ K ( z p ) be ∂ -constants. The linear combination η = λ dF F + ... + λ m dF m F m will be called a ∂ -logarithmic -form (or simply a logarithmic -form).Note that every ∂ -logarithmic -form η is closed , that is, dη = 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 ) = { z ∈ ( K a ) n : f ( z ) = 0 , ∀ f ∈ I } be the algebraic variety associated to I ,where K a is the algebraic closure of K . Let P be a polynomial in K [ z ] such that P ( c ) = 0 for every zero ( c ) = ( c , ..., c n ) ∈ V ( I ) . Then there is an integer m > such that P m ∈ I . Using the above theorem, Jouanolou proved his important lemma ([8], Lemme3.3.2, p. 102)
Lemma 3.3. If S is a finite representative system of primes in K [ z ] = K [ z , ..., z n ] (that is, S is a finite collection of distinct irreducible polynomials), then the K -linearmap K ( S ) −→ Ω K ( z ) /K ( λ j ) F j ∈S X F j ∈S λ j dF j F j is injective. The following two lemmas can be find in [18].
Lemma 3.4.
Let g ∈ K [ x ] , g (0) = 0 , be a polynomial function in one variable,where K is a field of positive characteristic p > . If α ∈ K ( x p ) , then Res( α · dgg ,
0) = 0
ARBOUX-JOUANOLOU INTEGRABILITY OF POLYNOMIAL DIFFERENTIAL FORMS 5
Lemma 3.5 (Jouanolou’s Lemma for Arbitrary Fields) . If S is a finite represen-tative system of primes in K [ z ] = K [ z , ..., z n ] (that is, S is a finite collection ofdistinct irreducible polynomials), then the K ( z p ) -linear map K ( z p ) ( S ) −→ Ω K ( z ) /K ( λ j ) F j ∈S X F j ∈S λ j dF j F j is injective. Definition 3.6 (Following [5]) . Let ω ∈ Ω rK ( z ) /K be a rational r -form. We definethe space of tangent -forms of ω by E ∗ ( ω ) = { η ∈ Ω K ( z ) /K ; ω ∧ η = 0 } Lemma 3.7 ([2], Lemma 2.1) . Let ω ∈ Ω r ( K n ) . If there are r elements η ,..., η r ∈ E ∗ ( ω ) linearly independent over K ( z ) , then there is R ∈ K ( z ) such that ω = R · η ∧ ... ∧ η r Proof.
We can complete { η , ..., η r } to make a K ( z ) -basis { η , ..., η r , η r +1 ..., η M } for Ω r ( K n ) , where M = dim K ( z ) Ω r ( K n ) . So we can write ω = X ≤ i <...
Definition 4.1.
Let F ∈ K [ z ] and V = { F = 0 } be an algebraic hypersurface in ( K a ) n and ω ∈ Ω rd ( K n ) be a polynomial r -form. We say that F is invariant by ω if F divides ω ∧ dF , that is, ω ∧ dF = F Θ F , Θ F ∈ Ω r − d +1 ( K n ) . By Hilbert’sNullstellensatz this is equivalent to say that ( ω ∧ dF ) | V = 0 . Definition 4.2.
We say that the rational function f is a rational first integral for ω if df = 0 and ω ∧ df = 0 Just like in [18] (Proposition 6.1), we can easily prove:
Proposition 4.3.
Let ω ∈ Ω rK [ z ] /K be a r -form, where K is an arbitrary field.If there are invariant irreducible polynomials F ,..., F m in K [ z ] and ∂ -constants λ ,..., λ m in K ( z p ) such that λ Θ F + ... + λ m Θ F m = 0 then there is a logarithmic -form η = 0 such that ω ∧ η = 0 .Proof. We can consider the logarithmic -form η = m X i =1 λ i · dF i F i Therefore ω ∧ η = m X i =1 λ i ω ∧ dF i F i = m X i =1 λ i Θ F i = 0 (cid:3) E. A. SANTOS AND S. RODRIGUES
Corollary 4.4.
Let ω ∈ Ω rd ( K n ) be a polynomial r -form of degree d . If ω has N K ( n, d − , r + 1) + 1 invariant irreducible polynomials, then there is a logarithmic -form η = 0 such that ω ∧ η = 0 .Proof. Let F ,..., F m be the m := N K ( n, d − , r + 1) + 1 invariant polynomials.The m associated cofactors Θ F ,..., Θ F m can be seen as elements of b Ω r +1 d − ( K n ) =Ω r +10 ( K n ) ⊗ K M min { d,p } and since m = dim K ( z p ) ( b Ω r +1 d − ( K n )) + 1 we can obtain λ ,..., λ m such that λ Θ F + ... + λ m Θ F m = 0 Then the result follow by Proposition 4.3, that is, the logarithmic -form η := λ dF F + ... + λ m dF m F m satisfies ω ∧ η = λ Θ F + ... + λ m Θ F m = 0 (cid:3) Definition 4.5.
Let ω ∈ Ω K ( z ) /z be a rational r -form. The p -degree of ω is thesmaller ǫ such that ω ∈ b Ω rǫ ( K n ) .We can hope to improve the above corollary by interchange the degree by the p -degree, that is, replacing b Ω rd ( K n ) by b Ω rǫ ( K n ) , but, if e is the degree of b Ω rd ( K n ) ,in general ǫ < e and the proof just presented will fall. Example 4.6.
Consider the -form ω = αxdy ∧ dz + βydz ∧ dx + γzdx ∧ dy , where α , β , γ are ∂ -constants in K ( z p ) ∗ . We can easily see that F = x , G = y and H = z are invariant. The associated cofactors are Θ x = ω ∧ dxx = αdx ∧ dy ∧ dz , Θ y = ω ∧ dyy = βdx ∧ dy ∧ dz and Θ z = ω ∧ dzz = γdx ∧ dy ∧ dz . If λ , λ , λ are ∂ -constants not all zero satisfying α · λ + β · λ + γ · λ = 0 then λ Θ x + λ Θ y + λ Θ z = 0 and we obtain a logarithmic -form η = λ dxx + λ dyy + λ dzz tangent to ω . Proposition 4.7.
Let ω ∈ Ω rK [ z ] /K be a polynomial r -form. If there are invariantirreducible polynomials F ,..., F m in K [ z ] and constants δ ,..., δ m in Z (or in theprime sub-field Z p of K if char ( K ) = p > ) such that δ Θ F + ... + δ m Θ F m = 0 then ω has a rational (resp. polynomial) first integral.Proof. If we take the -form η = δ dF F + ... + δ m dF m F m , then ω ∧ η = m X i =1 δ i ω ∧ dF i F i = m X i =1 δ i Θ F i = 0 Hence, since dF δ i = δ i F δ i − i dF i F i , defining the rational function (or polynomial) f = F δ ...F δ m m , we have df = f η and so ω ∧ df = f ω ∧ η = 0 that is, f is a first integral for ω . (cid:3) ARBOUX-JOUANOLOU INTEGRABILITY OF POLYNOMIAL DIFFERENTIAL FORMS 7 Darboux-Jouanolou Criterion
Lemma 5.1.
Let ω ∈ Ω rK ( z ) /K be a rational -form and let η ,..., η r ∈ E ∗ ( ω ) be K ( z p ) -linearly independent -forms. If η ∧ ... ∧ η r = 0 , then ω has a rational firstintegral.Proof. Since η ∧ ... ∧ η r = 0 , we have that η ,..., η r are linearly dependent over K ( z ) .Let m be the largest positive integer such that η ,..., η m are linearly independentover K ( z ) . Then η m +1 = m X i =1 f i · η i with f ,..., f m ∈ K ( z ) . Since η i is closed for i = 1 , ..., m + 1 , we obtain m X i =1 df i ∧ η i For each j , multiplying the above expression by η ∧ ... ∧ b η j ∧ ... ∧ η m , we obtain m X i =1 df i ∧ η i ∧ η ∧ ... ∧ b η j ∧ ... ∧ η m = ( − j +1 df j ∧ η ∧ ... ∧ η m Since η ,..., η m are K ( z ) -linearly independent, then, for each j = 1 ,..., m , thereexist g j ,..., g jm ∈ K ( z ) such that df j = P mi =1 g ji · η i . The -forms η ,..., η m , η m +1 are K ( z p ) -linearly independent, hence there exists j ∈ { , ..., m } such that df j = 0 ,that is, R j ∈ K ( z ) − K ( z p ) . Therefore ω ∧ df j = m X i =1 g j i · ω ∧ η i = 0 That is, f j is a rational first integral for ω . (cid:3) Lemma 5.2.
Let ω ∈ Ω rK ( z ) /K be a rational -form and let η ,..., η r , η r +1 , ∈ E ∗ ( ω ) be K ( z p ) -linearly independent -forms. If η ∧ ... ∧ η r = 0 and η ∧ ... ∧ η r +1 = 0 ,then ω has a rational first integral.Proof. By hypotheses { η , ..., η r } and { η , ..., η r +1 } are two sets of K ( z ) -linearlyindependent logarithmic -forms in E ∗ ( ω ) . Writing τ = η ∧ ... ∧ η r and τ = η ∧ ... ∧ η r +1 , we have distinct polar sets | τ | ∞ = | τ | ∞ and τ − τ = ( η + ( − r η r +1 ) ∧ η ∧ ... ∧ η r By Lemma 3.7 there exist rational functions h and h such that τ i = h i · ω , i = 1 , . Hence, writing f = h h , we have τ = f · τ , that is, τ − f · τ = ( η + ( − r f · η r +1 ) ∧ η ∧ ... ∧ η r and therefore η + ( − r f · η r +1 = r X i =2 f i · η i for rational functions f ,..., f r ∈ K ( z ) , that is, if f r +1 = ( − r +1 f , we obtain η = r +1 X i =2 f i · η i E. A. SANTOS AND S. RODRIGUES
Therefore, just like in Lemma 5.1, for each j = 1 ,..., r + 1 , we have df j ∧ ω = 0 , df j ∧ η ∧ ... ∧ η r +1 = 0 and for some j f j ∈ K ( z ) − K ( z p ) , that is, f j is a rational first integral for ω .(Alternatively, we can conclude using directly the Lemma 5.1. If f ∈ K ( z ) − K ( z p ) is not ∂ -constant, then f is a rational first integral for ω . On the other hand, if f ∈ K ( z p ) , then the -forms µ := η + ( − r f · η r +1 , η ,..., η r ∈ E ∗ ( ω ) are K ( z p ) -linearly independent and then we conclude the existence of rational first integralfor ω by Lemma 5.1 again.) (cid:3) Theorem 5.3.
Let ω ∈ Ω rK [ z ] /K be a polynomial r -form of degree d and N = dim K ( z p ) ( b Ω r +1 d − ( K n )) . If ω admits N + r + 1 invariant polynomials, then ω has a rational first integral.Proof. Suppose that the polynomial r -form ω admits N + r + 1 invariant irreduciblepolynomials F ,..., F N + r +1 . Then, for each i ∈ { , , ..., N + r + 1 } , there exists Θ F i ∈ Ω r +1 d − ( K n ) such that ω ∧ dF i = F i · Θ F i Since dim K ( z p ) ( b Ω r +1 d − ( K n )) = N , for each s = 1 ,..., r + 1 , we can choose λ ss ,..., λ ss + N ∈ K [ z p ] such that λ ss = 0 for s < r + 1 , λ r +1 N + r +1 = 0 and s + N X j = s λ sj · Θ F j = 0 For each s = 1 ,..., r + 1 , we can define the logarithmic -form η s = s + N X j = s λ sj · dF j F j Therefore ω ∧ η s = 0 s = 1 ,..., r + 1 .Define τ = η ∧ ... ∧ η r and τ = η ∧ ... ∧ η r +1 = ( − r η r +1 ∧ η ∧ ... ∧ η r If τ = 0 or τ = 0 , we obtain a rational first integral for ω by Lemma 5.1.If τ = 0 and τ = 0 , we obtain a rational first integral for ω by Lemma 5.2. (cid:3) Corollary 5.4.
Let X be a polynomial vector field of degree d on K n and N thedimension of the K ( z p ) -vector space b Ω nd − ( K n ) . If X admits N + n invariant irreducible polynomials, then X has a rational first integral. ARBOUX-JOUANOLOU INTEGRABILITY OF POLYNOMIAL DIFFERENTIAL FORMS 9
Proof.
Let X = P ni =1 P i ∂∂ i be a polynomial vector field of degree d . We can asso-ciate to it the ( n − -form ω X = n X i =1 ( − i +1 P i dz ∧ ... ∧ c dz i ∧ ... ∧ dz n that is, ω X is the rotational of X .Then F is X -invariant if and only if F is ω -invariant.Now we can apply the above Theorem. (cid:3) Corollary 5.5.
Let ω ∈ Ω rK [ z ] /K be a polynomial r -form over an infinity field K .Then ω has rational first integral if and only if it has an infinite number of invariantpolynomials.Proof. Suppose that f = PQ is a rational first integral for ω . Consider f λ = P − λQ ,where λ ∈ K . Then ω ∧ df λ = ω ∧ ( dP − λdQ ) = ω ∧ dP − λω ∧ dQ We have ω ∧ d ( PQ ) = 0 and in particular ω ∧ ( QdP − P dQ ) = 0
Hence ω ∧ dPP = ω ∧ dQQ , that is, ω ∧ dP = P ω ∧ dQQ and ω ∧ dQ = Qω ∧ dPP In this way ω ∧ df λ = ( P − λQ ) ω ∧ dPP = ( P − λQ ) ω ∧ dQQ Therefore ω ∧ df λ f λ is a polynomial ( r + 1) -form, that is, the irreducible factors of f λ are invariant.Reciprocally we can apply Theorem 5.3. (cid:3) Integrating Factors and Darboux’s Method
Definition 6.1.
A rational function f is said to be an integration factor of ω if f ω is closed, that is, d ( f ω ) = 0 . Proposition 6.2.
Let ω ∈ Ω rK [ z ] /K be a polynomial r -form. If there are invariantirreducible polynomials F ,..., F m in K [ z ] and ∂ -constants λ ,..., λ m in K ( z p ) −{ } such that λ Θ F + λ Θ F + ... + λ m Θ F m = ( − r +1 dω then there is a logarithmic -form η = 0 such that dω = ω ∧ η . Proof.
Consider the rational -form η = λ dF F + λ dF F + ... + λ m dF m F m = 0 Obviously η is closed ( dη = 0 ). Also η satisfies ω ∧ η = m X i =1 λ i ω ∧ dF i F i = m X i =1 λ i Θ F i = dω (cid:3) Proposition 6.3.
Let ω ∈ Ω rK [ z ] /K be a polynomial r -form. If there are invariantirreducible polynomials F ,..., F m in K [ z ] and constants δ ,..., δ m in Q (or in theprime sub-field Z p of K if char ( K ) = p > ) such that δ Θ F + δ Θ F + ... + δ m Θ F m = ( − r +1 dω then ω has a rational (respectively polynomial) factor of integration.Proof. Up to multiplying ω by a integer, we can suppose that δ ,..., δ m ∈ Z if char ( K ) = 0 . Just as above, the rational -form η = δ dF F + δ dF F + ... + δ m dF m F m = 0 is closed and η satisfies ω ∧ η = m X i =1 δ i ω ∧ dF i F i = m X i =1 δ i Θ F i = ( − r +1 dω Define the rational (respectively polynomial) function G = F δ F δ ...F δ m m We see that dG = Gη , hence ω ∧ dG = Gω ∧ η = ( − r +1 dω and d ( Gω ) = dG ∧ ω + Gdω = 0 (cid:3)
Theorem 6.4.
Let ω ∈ Ω rd ( K n ) be a rational r -form of degree d . If ω has N K ( n, d − , r + 1) invariant irreducible polynomials, then there is a logarithmic -form η = 0 such that ω ∧ η = 0 or ω ∧ η = dω . If k = Q or K = Z p , p > , then ω has arational first integral or a rational integration factor.Proof. Suppose that there are m = N K ( n, d − , r + 1) invariant irreducible poly-nomials F , F ,..., F m .The cofactors associated to the invariant polynomials are differential ( r + 1) -forms of degree lass than or equal to d − . Since the K ( z p ) -vector subspace (in Ω K ( z ) /K ) generated by differential ( r + 1) -forms of degree lass than or equal to d − has K ( z p ) -dimension equal to N K ( n, d − , r + 1) , we have that the cofactorsassociated to the polynomials F i , i = 1 ,..., m , are linearly dependent or form a baseto the K ( z p ) -subspace generated by polynomial differential ( r + 1) -forms of degreelass than or equal to d − . We have two possibilities, as follow. ARBOUX-JOUANOLOU INTEGRABILITY OF POLYNOMIAL DIFFERENTIAL FORMS 11
Case . The cofactors associated to the polynomials F i , i = 1 ,..., m , are K ( z p ) -linearly dependent . Then there are ∂ -constants α ,..., α m in K such that α Θ F + α Θ F + ... + α m Θ F m = 0 hence for η = m X i =1 δ dF i F i we have ω ∧ η = 0 . Case . The cofactors associated to the polynomials F i , i = 1 ,..., m , form a baseto the K ( z p ) -subspace generated by polynomial differential ( r + 1) -forms of degreelass than or equal to d − . Then the ( r +1) -form dω can be writen as a K ( z p ) -linearcombination of the cofactors Θ F i , and we also obtain the logarithmic -form η suchthat ω ∧ η = dω . (cid:3) Example 6.5 (Lotka-Volterra Equation) . Consider the -form ω = y ( γ − δx ) dx + x ( α − βy ) dy We can easily see that the polynomials F = x and G = y are invariant by ω , andtheir cofactors are Θ x = ω ∧ dxx = ( βy − α ) dx ∧ dy and Θ y = ω ∧ dyy = ( γy − δ ) dx ∧ dy Taking the differential of ω , we have dω = [( α − βy ) − ( γ − δx )] dx ∧ dy = − Θ x − Θ y and therefore dω = ω ∧ ( − dxx − dyy ) We see that ( xy ) − is a rational factor of integration for ω . Proposition 6.6.
Let ω ∈ Ω K [ z ] /K be a -form. If there are two rational closed -forms η and η such that dω = ω ∧ η i i ∈ { , } , then ω admits a rational integration factor.Proof. Consider the -form η = η − η . It follows that ω ∧ η = 0 . Thereforethere is a rational function h such that hω = η . As η is closed, we have that h isa rational factor of integration for ω . (cid:3) In characteristic (for example, K = R or K = C ) it is well know that a r -form is closed if and only if it is exact: dω = 0 ⇔ ω = dη (Poincaré Lemma).But it doesn’t work in positive characteristic, therefore from an integration factorobtained here we cannot conclude the exactness. The next section will develop asimilar result that works in arbitrary characteristic. Poincaré Lemma
The d operator on r -forms over R , where R = K [ z ] or R = K ( z ) , gives us achain sequence: −→ K d −→ R d −→ Ω R/K d −→ ... d −→ Ω nR/K −→ Over characteristic , this sequence is proved to be exact (see [6], Proposition(7.1) (Poincaré lemma) , page 53), that is, a r -form is exact if and only if it is closed.In order to give the positive characteristic version of this important theorem, weneed more then simply the closedness of forms. The following definition will providethe necessary and sufficient condition to exactness. Definition 7.1.
Let K be a field of positive characteristic p > . A p -closed r -form ω ∈ Ω r ( K n ) is defined inductively by the following conditions.(1) If r = 0 , that is, ω is a rational function, then we must have ω = 0 .(2) If r > , then, for every i = 1 , ..., n , we require that ω = dz i ∧ ( ω i + η i ) + τ i where ω i is p -closed, ∂ p − η i ∂z p − i = ( ∂∂z i ) p − η i = 0 and τ i is free of dz i .A sketch of proof of the following theorem in the case r = 1 can be found in [20].In order to obtain a general statement, we introduce the definition above, whichgeneralizes inductively that one of [20]. Theorem 7.2 (Positive Characteristic Poincaré Lemma) . Let K be a field of pos-itive characteristic p > . A r -form ω ∈ Ω rR/K is exact if and only if ω is p -closed.Proof. Since for every ω ∈ K ( z ) , there is a ∂ -constant λ ∈ K [ z p ] such that λ · ω ∈ K [ z ] , we can suppose R = K [ z ] .The "exactness imply p -closedness" part is more easy to prove by induction on r . Consider ω = dη exact. For r = 0 we must have ω = 0 ∈ K and then ω istrivially p -closed. Suppose the property true for r ≥ . Let ω = dη be an exact ( r + 1) -form. We want to show that ω is p -closed. We can write η = dz i ∧ α i + β i ,where β i has no term with dz i , and ω = dη = dz i ∧ dα i + dβ i Also dβ i = dz i ∧ η i + τ i with τ i free of dz i . Then ω = dz i ∧ ( dα i + η i ) + τ i where τ i is free of dz i and ω i := dα i is, by the induction hypothesis, p -closed. Hence ω is p -closed.Now the " p -closedness imply exactness". Suppose that ω is p -closed. We proceedby induction on n and r . For r = 0 we have that ω = 0 ∈ R and there is nothing todo. Suppose the property true to r -forms, where r ≥ . We now prove by inductionon n that the every ( r + 1) -form is exact if and only if is p -closed. The case n = 0 is trivial. Let ω be a ( r + 1) -form and n ≥ . Since ω is p -closed, we can write ω = dz ∧ ( ω + η ) + τ = dz ∧ ω + dz ∧ η + τ where ω is p -closed, ( ∂∂z ) p − η = 0 and τ is free of dz . ARBOUX-JOUANOLOU INTEGRABILITY OF POLYNOMIAL DIFFERENTIAL FORMS 13
By induction hypothesis, ω is exact. Then dz ∧ ω is also exact and hence also p -closed. In this way, we have that ω − dz ∧ ω is p -closed and we can replace ω by it (that is, we can suppose ω = 0 ). We define θ = Z η dz (in the obvious algebraic "integration" sense, which is possible because ( ∂∂z ) p − η = 0 ). Then θ ∈ Ω r − R/K and dθ = dz ∧ η + θ where θ does not involve dz .Replacing ω by ω − dθ = τ − θ , we reduce to the case where ω is free of dz .Since dω = 0 , we conclude that in fact we can see ω as an element of Ω r +1 S/K , where S = K [ z , ..., z n ] . By our induction hypothesis (over n ) the result is true for ω andtherefore the proof is complete. (cid:3) In the following corollary we use the standard multi-index notation.
Corollary 7.3.
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