A recombination algorithm for the decomposition of multivariate rational functions
aa r X i v : . [ c s . S C ] N ov A RECOMBINATION ALGORITHM FOR THEDECOMPOSITION OF MULTIVARIATE RATIONAL FUNCTIONS
GUILLAUME CH`EZE
Abstract.
In this paper we show how we can compute in a deterministic waythe decomposition of a multivariate rational function with a recombinationstrategy. The key point of our recombination strategy is the used of Darbouxpolynomials. We study the complexity of this strategy and we show that thismethod improves the previous ones. In appendix, we explain how the strategyproposed recently by J. Berthomieu and G. Lecerf for the sparse factorizationcan be used in the decomposition setting. Then we deduce a decompositionalgorithm in the sparse bivariate case and we give its complexity.
Introduction
The decomposition of univariate polynomials has been widely studied since 1922,see [Rit22], and efficient algorithms are known, see [AT85, BZ85, KL89, Gat90a,Gat90b, Gie88, Kl¨u99]. There also exist results and algorithms in the multivariatecase [Dic87, Gat90a, Gie88, GGR03].The decomposition of rational functions has also been studied, [Gie88, Zip91,AGR95, GW95]. In the multivariate case the situation is the following:Let f ( X , . . . , X n ) = f ( X , . . . , X n ) /f ( X , . . . , X n ) ∈ K ( X , . . . , X n ) be a ratio-nal function, where K is a field and n ≥
2. It is commonly said to be composite if itcan be written f = u ( h ) where h ( X , . . . , X n ) ∈ K ( X , . . . , X n ) and u ∈ K ( T ) suchthat deg( u ) ≥ f is saidto be non-composite. In this paper, we give an algorithm which computes a non-composite rational function h ∈ K ( X , . . . , X n ) and a rational function u ∈ K ( T )such that f = u ( h ).In [Ch`e10], the author shows that we can reduce the decomposition problem to afactorization problem and gives a probabilistic and a deterministic algorithm. Theprobabilistic algorithm is nearly optimal: it performs ˜ O ( d n ) arithmetic operations.The deterministic one computes O ( d ) absolute factorizations and then performs˜ O ( d n + ω +2 ) arithmetic operations, where d is the degree of f and ω is the feasiblematrix multiplication exponent as defined in [GG03, Chapter 12]. We recall that2 ≤ ω ≤ . d tends to infinity and n is fixed. We use the classical O and ˜ O (“soft O ”) notation in the neighborhoodof infinity as defined in [GG03, Chapter 25.7]. Informally speaking, “soft O ”s areused for readability in order to hide logarithmic factors in complexity estimates.In this paper we improve the complexity of the deterministic algorithm. Withthis algorithm we just compute two factorizations in K [ X , . . . , X n ] and then we Date : June 19, 2018. use a recombination strategy. Under some hypotheses this new method performs˜ O ( d n + ω − ) arithmetic operations.The decomposition of multivariate rational functions appears when we study thekernel of a derivation, see [MO04]. In [MO04] the author uses Darboux polynomialsand gives an algorithm which works with ˜ O ( d ωn ) arithmetic operations.In this paper, we are also going to use Darboux polynomials (see Section 1 for adefinition) and we add a recombination strategy. Roughly speaking, we are goingto factorize the numerator and the denominator and then thanks to a property ofDarboux polynomials we are going to show that we can recombine the factors anddeduce the decomposition.The decomposition of multivariate rational functions also appears when we studyintermediate fields of an unirational field and the extended L¨uroth’s Theorem, see[GRS01, Ch`e10], and when we study the spectrum of a rational function, see [Ch`e10]and the references therein.The study of decomposition is an active area of research: for a study on mul-tivariate polynomial systems see e.g. [FP09, FGP10], for a study on symbolicpolynomials see e.g. [Wat09], for a study on Laurent polynomials see e.g. [Wat08],for effective results on the reduction modulo a prime number of a non-compositepolynomial or a rational function see e.g. [CN10, BDN09, BCN], for combinatorialresults see e.g. [Gat08].In this paper, we improve the strategy proposed in [MO04]. As in [MO04], weconsider fields with characteristic zero. Furthermore, as we want to give precisecomplexity estimate we are going to suppose that:Hypothesis (C): K is a number field: K = Q [ α ], α is an algebraic number of degree r .As in [Ch`e10], we are going to suppose that the following hypothesis is satisfied:Hypothesis (H): ( ( i ) deg( f + Λ f ) = deg X n ( f + Λ f ) , where Λ is a new variable , ( ii ) R (Λ) = Res X n (cid:16) f (0 , X n ) + Λ f (0 , X n ) , ∂ X n f (0 , X n ) + Λ ∂ X n f (0 , X n ) (cid:17) = 0 in K [Λ] . where deg X n f represents the partial degree of f in the variable X n , deg f is thetotal degree of f and Res X n denotes the resultant relatively to the variable X n .This hypothesis is necessary, because we will use the factorization algorithmsproposed in [Lec07], where this kind of hypothesis is needed. Actually, in [Lec07]the author studies the factorization of a polynomial F and uses hypothesis (L),where (L) is the following: Hypothesis (L): ( ( i ) deg X n F = deg F, and F is monic in X n , ( ii ) Res X n (cid:0) F (0 , X n ) , ∂F∂X n (0 , X n ) (cid:17) = 0 . If F is squarefree, then hypothesis (L) is not restrictive since it can be assured bymeans of a generic linear change of variables, but we will not discuss this questionhere (for a complete treatment in the bivariate case, see [CL07, Proposition 1]). ECOMBINATION FOR DECOMPOSITION 3
Roughly speaking, our hypothesis (H) is the hypothesis (L) applied to the poly-nomial f + Λ f . In (H, i ) we do not assume that f + Λ f is monic in X n . Indeed,the leading coefficient relatively to X n can be written: a + Λ b , with a, b ∈ K . Inour algorithm, we evaluate Λ to λ such that a + λb = 0. Then we can considerthe monic part of f + λf and we get a polynomial satisfying (L, i ). Then (H, i )is sufficient in our situation. Furthermore, in this paper, we assume f /f to bereduced, i.e. f and f are coprime. We recall in Lemma 9 that in this situation f + Λ f is squarefree. Thus hypothesis (H) is not restrictive.Furthermore, hypothesis (H) will also be useful in a preprocessing step, see Sec-tion 2. In this preprocessing step we reduce the decomposition to two factorizationsof squarefree polynomials. Complexity model.
In this paper the complexity estimates charge a constantcost for each arithmetic operation (+, − , × , ÷ ) and the equality test. All theconstants in the base fields are thought to be freely at our disposal.In this paper we suppose that the number of variables n is fixed and that thedegree d tends to infinity.Polynomials are represented by dense vectors of their coefficients in the usualmonomial basis. For each integer d , we assume that we are given a computationtree that computes the product of two univariate polynomials of degree at most d with at most ˜ O ( d ) operations, independently of the base ring, see [GG03, The-orem 8.23]. Then with a Kronecker substitution we can compute the product oftwo multivariate polynomials with degree d with n variables with ˜ O ( d n ) arithmeticoperations. We also recall, see [GG03, Corollary 11.8], that if K is an algebraicextension of Q of degree r then each field operation in K takes ˜ O ( r ) arithmeticoperations in Q .We use the constant ω to denote a feasible matrix multiplication exponent as definedin [GG03, Chapter 12]: two n × n matrices over K can be multiplied with O ( n ω )field operations. As in [BP94] we require that 2 ≤ ω ≤ . m equations and d ≤ m un-knowns over K takes O ( md ω − ) operations in K [BP94, Chapter 2] (see also [Sto00,Theorem 2.10]).In [Lec06, Lec07] the author gives a deterministic algorithm for the multivariate ra-tional factorization. The rational factorization of a polynomial f is the factorizationin K [ X ], where K is the coefficient field of f . This algorithm uses one factorizationof a univariate polynomial of degree d and ˜ O ( d n + ω − ) arithmetic operations, where d is the total degree of the polynomial and n ≥ Main Theorem.
The following theorem gives the complexity result of our algo-rithm.
Theorem 1.
Let f be a multivariate rational function in Q [ α ]( X , . . . , X n ) ofdegree d , where α is an algebraic number of degree r . Under hypotheses (C) and (H),we can compute in a deterministic way the decomposition of f with ˜ O ( rd n + ω − ) arithmetic operations over Q plus two factorizations of univariate polynomials ofdegree d with coefficients in Q [ α ] . Comparison with other algorithms.
There already exist several algorithms forthe decomposition of rational functions. They all use the same global strategy: firstcompute h , and then deduce u . The first step is the difficult part of the problem. G. CH`EZE
In [Ch`e10], we explain how we can perform the second step, i.e. compute u from h and f , with ˜ O ( d n ) arithmetics operations.In [GRS01], the authors provide two algorithms to decompose a multivariate ratio-nal function. These algorithms run in exponential time in the worst case. In the firstone we have to factorize polynomials with 2 n variables f ( X ) f ( Y ) − f ( Y ) f ( X )and to look for factors of the following kind h ( X ) h ( Y ) − h ( Y ) h ( X ). The au-thors say that in the worst case the number of candidates to be tested is exponentialin d = deg( f /f ). Indeed, the authors test all the possible factors.In the second algorithm, for each pair of factors ( h , h ) of f and f (i.e. h di-vides f and h divides f ), we have to test if there exists u ∈ K ( T ) such that f /f = u ( h /h ). Thus in the worst case we also have an exponential number ofcandidates to be tested.To the author’s knowledge, the first polynomial time algorithm is due to J. Moulin-Ollagnier, see [MO04] . This algorithm relies on the study of the kernel of thefollowing derivation: δ ω ( F ) = ω ∧ dF , where F ∈ K [ X ] and ω = f df − f df . In[MO04] the author shows that we can reduce the decomposition of a rational func-tion to linear algebra. The bottleneck of this algorithm is the computation of thekernel of a matrix. The size of this matrix is O ( d n ) × O ( d n ), then the complexityof this deterministic algorithm belongs to O ( d nω ).The reduction of the decomposition problem to a factorization problem is clas-sical, see e.g. [Kl¨u99, Gie88, GW95, GRS01]. In [Ch`e10] the author shows thatif we choose a probabilistic approach then two factorizations in K [ X , . . . , X n ] aresufficient to get h and furthermore we do not have a recombination problem. Thisgives a nearly optimal algorithm. For the deterministic approach the author uses aproperty on the pencil f − λf and shows that with O ( d ) absolute factorization(i.e. factorization in the algebraic closure of K ) we can get h . This deterministicstrategy works with ˜ O ( d n + ω +2 ) arithmetic operations.In this paper, we are going to show that we can obtain a deterministic algorithmwith just two factorizations in K [ X , . . . , X n ] and a recombination strategy. Ouralgorithm uses at most ˜ O ( d n + ω − ) arithmetic operations. This cost corresponds tothe cost of the factorization and the recombination step.Our recombination problem comes from this factorization:If f /f = u /u ( h /h ) then f − λf = e ( h − t h ) · · · ( h − t k h )where λ, e ∈ K , k = deg( u /u ) and t i are the roots of the univariate polynomial u ( T ) − λu ( T ), see Lemma 10.Thus with the factors h − t h and h − t h we can deduce h . Unfortunately thesefactors are not necessarily in K [ X , . . . , X n ] and are not necessarily irreducible. Inthis paper we show how we can reduce the problem to a factorization problem in K [ X , . . . , X n ] and how we can recombine the irreducible factors of f − λf to get h .We can see our recombination scheme as a logarithmic derivative method . Roughlyspeaking, the logarithmic derivative method works as follow:If F ( X, Y ) = Q tj =1 F j ( X, Y ), where F j ( X, Y ) ∈ A and A ⊃ K [ X, Y ] (for example A = K [[ X ]][ Y ]), then we can write the irreducible factors F i ( X, Y ) ∈ K [ X, Y ] of F ECOMBINATION FOR DECOMPOSITION 5 in the following way: F i = Q tj =1 F e i,j j , where e i,j ∈ { , } . Thus we just have tocompute the exponents e i,j to deduce F i . We compute these exponents thanks tothis relation: ∂ X F i F i = t X j =1 e i,j ∂ X F j F j . With this relation the exponents e i,j are now coefficients, and we can compute themwith linear algebra.This strategy has already been used by several authors in order to factorize poly-nomials see e.g. [BHKS09, BLS +
04, Lec06, CL07, Wei10]. Here, we use this kindof technique for the decomposition problem. With this strategy the recombinationpart of our algorithm corresponds to the computation of the kernel of a O ( d n ) ×O ( d )matrix.In our context, we do not use exactly a logarithmic derivative. We use a moregeneral derivation, but we use the same idea: if a mathematical object transformsa product into a sum then the recombination problem becomes a linear algebraproblem. In this paper this mathematical object is the cofactor, see Proposition 8. Structure of this paper.
In Section 1, we recall some results about the Jacobianderivative and Darboux polynomials. In Section 2, we describe a reduction stepwhich eases the recombination strategy. In other words we explain how we canreduce the decomposition problem to a factorization problem. In Section 3, weexplain how we can get h with a recombination strategy. In Section 4, we describeour algorithm with two examples. In Section 5 we conclude this paper with aremark on Darboux method and the logarithmic derivative method. In appendix,we explain how the strategy proposed recently by J. Berthomieu and G. Lecerf in[BL10] for the sparse factorization can be used in the decomposition setting. Thenwe deduce a decomposition algorithm in the sparse bivariate case and we give itscomplexity. Notations.
All the rational functions are supposed to be reduced.Given a polynomial f , deg( f ) denotes its total degree.Given a rational function f = f /f , deg( f ) denotes max (cid:0) deg( f ) , deg( f ) (cid:1) .For the sake of simplicity, sometimes we write K [ X ] instead of K [ X , . . . , X n ], for n ≥ u ◦ h means u ( h ). Res ( A, B ) denotes the resultant of two univariate polynomials A and B . | S | is the cardinal of the set S .1. Derivation and Darboux polynomials
We introduce the main tool of our algorithm.
Definition 2. A K -derivation D of the polynomial ring K [ X , . . . , X n ] is a K -linearmap from K [ X , . . . , X n ] to itself that satisfies the Leibniz rule for the product D ( f.g ) = D ( f ) .g + f.D ( g ) . A K -derivation has a unique extension to K ( X , . . . , X n ) and then we will alsodenote by D the extended derivation. G. CH`EZE
Definition 3.
Given a rational function f /f , the Jacobian derivative associatedto f /f is the following vector derivation, i.e. an ( n − D f /f : K [ X , . . . , X n ] −→ (cid:16) K [ X , . . . , X n ] (cid:17) n − F f . ∂ X (cid:0) f /f ) ∂ X F − ∂ X ( f /f ) ∂ X F ... ∂ X (cid:0) f /f ) ∂ X n F − ∂ X n ( f /f ) ∂ X F . The Jacobian derivative has the following property:
Proposition 4.
Given f = f /f and g ∈ K ( X , . . . , X n ) \ K the following propo-sitions are equivalent: (1) The rank of the Jacobian matrix
Jac ( f, g ) = ∂f∂X · · · ∂f∂X n ∂g∂X · · · ∂g∂X n is equal to one; (2) D f /f ( g ) = 0 ; (3) there exists h in K ( X , . . . , X n ) such that f = u ( h ) and g = v ( h ) for u, v ∈ K ( T ) .Proof. See [PI07] for a proof. In [PI07], K is supposed to be algebraically closed.However, we can remove this hypothesis because we have the equivalence: f iscomposite over K if and only if f is composite over K , see e.g. [BCN, Theorem13]. (cid:3) Definition 5.
Given D a vector derivation i.e. an m -tuple of derivations, a polyno-mial F ∈ K [ X ] is said to be a Darboux polynomial of D if there exists G ∈ (cid:0) K [ X ] (cid:1) m such that D ( F ) = F. G . G is called the cofactor of F for the derivation D .We deduce easily the following classical propositions. Proposition 6. f and f are Darboux polynomials of D f /f . Proposition 7. D f /f ( h /h ) = 0 if and only if h and h are Darboux polyno-mials with the same cofactor. The following proposition is the main tool of our algorithm. Indeed, this propo-sition shows that cofactors transform a product into a sum. Then thanks to thecofactors it will be possible to apply a kind of logarithmic derivative recombinationscheme.
Proposition 8.
Let F ∈ K [ X , . . . , X n ] be a polynomial and let F = F e · · · F e r r be its irreducible factorization in K [ X , . . . , X n ] . Then: F is a Darboux polynomial with cofactor G F if and only if all the F i are Darbouxpolynomials with cofactor G F i . Furthermore, G F = e G F + · · · + e r G F r .Proof. See for example Lemma 8.3 page 216 in [DLA06]. (cid:3)
ECOMBINATION FOR DECOMPOSITION 7 Reduction to a rational factorization problem
In this section, we recall how the decomposition problem can be reduced to afactorization problem. Furthermore, we show that we can reduce our problem to asituation where f and f are squarefree. First, we recall some useful lemmas. Lemma 9. If f /f is reduced in K ( X , . . . , X n ) , where n ≥ and Λ is a variable,then f + Λ f is squarefree. Lemma 10.
Let h = h /h be a rational function in K ( X ) , u = u /u a rationalfunction in K ( T ) and set f = u ◦ h with f = f /f ∈ K ( X ) . For all λ ∈ K suchthat deg( u − λu ) = deg u , we have f − λf = e ( h − t h ) · · · ( h − t k h ) where e ∈ K , k = deg u and t i are the roots of the univariate polynomial u ( T ) − λu ( T ) .Proof. See [Ch`e10, Lemma 8, Lemma 39]. (cid:3)
Remark . If λ = f ( a ) /f ( a ), where a = ( a , . . . , a n ) ∈ K n , then we can supposethat t ∈ K . Indeed, t = h ( a ) /h ( a ) ∈ K .The following lemma says that we can always suppose that deg u = deg u =deg u . Lemma 12.
Let h = h /h be a rational function in K ( X ) , u = u /u a rationalfunction in K ( T ) and set f = u ◦ h with f = f /f ∈ K ( X ) . There exists anhomography H ( T ) = ( aT + b ) / ( αT + β ) ∈ K ( T ) such that: u ◦ H = ˜ u / ˜ u , deg ˜ u = deg ˜ u , and f = ˜ u ˜ u ◦ ˜ h , where ˜ h = H − ◦ h and H − isthe inverse of H for the composition.Proof. If deg u = deg u then we set H ( T ) = T .If deg u > deg u then we have: u u (cid:0) H ( T ) (cid:1) = Q deg u i =1 (cid:0) aT + b − λ i ( αT + β ) (cid:1)Q deg u i =1 (cid:0) aT + b − µ i ( αT + β ) (cid:1) . ( αT + β ) deg u − deg u , where u ( λ i ) = 0 and u ( µ i ) = 0.We set: ˜ u ( T ) = ( αT + β ) deg u − deg u . deg u Y i =1 (cid:0) aT + b − λ i ( αT + β ) (cid:1) = u (cid:0) H ( T ) (cid:1) . ( αT + β ) deg u ∈ K [ T ]˜ u ( T ) = deg u Y i =1 (cid:0) aT + b − µ i ( αT + β ) (cid:1) = u (cid:0) H ( T ) (cid:1) . ( αT + β ) deg u ∈ K [ T ] . If a − λ i α = 0, α = 0, and a − µ i α = 0 then we get deg ˜ u = deg u = deg ˜ u .To conclude the proof we just have to remark that deg H = 1, thus H is invertiblefor the composition. (cid:3) G. CH`EZE
In order to ease the recombination scheme we reduce our problem to a situationwhere the rational function is squarefree, i.e. the numerator and the denominatorare squarefree . The following algorithm shows that if f or f are not squarefreethen we can compute an homography U ( T ) ∈ K ( T ) such that U ( f /f ) is square-free. Furthermore, if we know a decomposition U ( f /f ) = u ( h ) then we can easilydeduce a decomposition f /f = U − (cid:0) u ( h ) (cid:1) . We recall that U is invertible for thecomposition because deg U = 1. Now, we describe an algorithm which computes agood homography. Good homography
Input: f = f /f ∈ K ( X , . . . , X n ) of degree d , such that (C) and (H) are satisfiedand a finite subset S of K n such that | S | = 2 d + 2 d . Output: U ( T ) = ( T − λ a ) / ( T − λ b ) such that U ( f ) is squarefree, λ a = f /f ( a ) , λ b = f /f ( b ) where a, b ∈ K n , λ a = λ b , and deg X n ( f − λ a f ) = deg X n ( f − λ b f ) = d .(1) Compute f ( X n ) := f (0 , X n ), and f ( X n ) := f (0 , X n ).(2) Construct an empty list L .(3) For i from 1 to 2 d + 2 d do:(a) Compute f := f ( i ) /f ( i ),.(b) If f L then L := concatenate( L, [ f ]).(4) Construct an empty list L .(5) For k from 1 to 2 d + 2 do:(a) Compute R := Res X n (cid:16) f ( X n ) − L [ k ] f ( X n ) , ∂ X n f ( X n ) − L [ k ] ∂ X n f ( X n ) (cid:17) .(b) If R = 0 and deg X n ( f − L [ k ] f ) = d , then L := concatenate( L , [ L [ k ]]).(6) λ a := L [1], λ b := L [2].(7) Return U ( T ) = ( T − λ a ) / ( T − λ b ). Proposition 13.
The algorithm
Good homography is correct.Proof.
In Step 3 we construct a list with at least 2 d + 2 distinct elements becausedeg( f ) = d .By hypothesis (H), R (Λ) = 0 and by [GG03, Theorem 6.22], deg( R ) ≤ d −
1. Thus L contains at least two distincts elements.As R ( λ a ) and R ( λ b ) are not equal to zero, and thanks to Step 5b the condition onthe degree is satisfied, we deduce that f − λ a f and f − λ b f are squarefree. (cid:3) Proposition 14.
The algorithm
Good homography can be performed with at most ˜ O ( d n ) arithmetic operations over K .Proof. Step 1 can be done with ˜ O ( d n ) arithmetic operations with Horner’s method.In Step 3 we use a fast multipoint evaluation strategy, then we can perform thisstep with at most ˜ O ( d ) arithmetic operations, see [GG03, Corollary 10.8].In Step 5, the computation of the resultant can be done with ˜ O ( d ) arithmeticoperations, see [GG03, Corollary 11.16]. Thus Step 5 can be done with ˜ O ( d )arithmetic operations.In conclusion the algorithm can be performed with the desired complexity. (cid:3) Remark . Suppose f /f = v /v ( h ). With the algorithm Good homography wecan write U ( f /f ) = u /u ( h ) with u /u ∈ K ( T ), h ∈ K ( X , . . . , X n ), and u (resp. u ) has a root α (resp. α ) in K . Indeed, we have u = v − λ a v (resp. ECOMBINATION FOR DECOMPOSITION 9 u = v − λ b v ) and λ a = f /f ( a ) (resp. λ b = f /f ( b )) then we deduce that α = h /h ( a ) (resp. α = h /h ( b )).3. The recombination method
In this section we describe our recombination method. First, we introduce somenotations. By Proposition 6, F and F are Darboux polynomials of D F /F .Wedenote by G F k = ( G (2) F k , . . . , G ( n ) F k )the cofactor of F k , where k = 1 ,
2, and G ( l ) F k ∈ K [ X , . . . , X n ]. We set: F k = s k Y j =1 F k,j for k = 1 ,
2, and G F k,j = ( G (2) F k,j , . . . , G ( n ) F k,j ) . In Q [ α ][ X , . . . , X n ] polynomials are denoted in the following way: P = X | τ |≤ d r − X ǫ =0 a ǫ,τ α ǫ X τ · · · X τ n n ∈ Q [ α ][ X , . . . , X n ] , where α is an algebraic number of degree r , τ = ( τ , . . . , τ n ), | τ | = τ + · · · + τ n ,and a ǫ,τ ∈ Q . We set coef (cid:0) P , α ǫ X τ (cid:1) = a ǫ,τ . Now we define the linear system S : S := s X j =1 x ,j coef (cid:0) G ( l ) F ,j , α ǫ X τ (cid:1) − s X j =1 x ,j coef (cid:0) G ( l ) F ,j , α ǫ X τ (cid:1) = 0 , where | τ | ≤ d , 0 ≤ ǫ ≤ r −
1, and 2 ≤ l ≤ n .We denote by ker S the kernel of this linear system, and we remark that x = ( x , , . . . , x ,s ) ∈ ker S ⇐⇒ s X j =1 x ,j G F ,j − s X j =1 x ,j G F ,j = 0 . We define the following maps: π : K s + s −→ K s ( x , , . . . , x ,s ) ( x , , . . . , x ,s ) π : K s + s −→ K s ( x , , . . . , x ,s ) ( x , , . . . , x ,s )The following proposition will be the key of our algorithm: Proposition 16.
Suppose that F /F ∈ K ( X , . . . , X n ) comes from the algorithm Good Homography and F /F = u ( h ) where h = h /h ∈ K ( X , . . . , X n ) is a non-composite reduced rational function and u = u /u ∈ K ( T ) is a reduced rationalfunction, with deg u = deg u .We denote by u k = Q t k i =1 u k,i the factorization of u k in K [ T ] , where k = 1 , 2.We denote by F k = Q s k j =1 F k,j the factorization of F k in K [ X , . . . , X n ] , where k = 1 , 2.Then: (1) u k,i (cid:16) h h (cid:17) .h deg u k,i = s k Y j =1 F e k,i,j k,j ∈ K [ X , . . . , X n ] and e k,i,j ∈ { , } .Furthermore, if we set e k,i := ( e k,i, , . . . , e k,i,s k ) , thenthe vectors e k,i , i = 1 , . . . , t k , are orthogonal for the usal scalar product. (2) We have e k,i ∈ π k (ker S ) . (3) { e k, , . . . , e k,t k } is a basis of π k (ker S ) .Proof. (1) By Lemma 10 applied to F /F (resp. F /F ) with λ = 0, we get F k = u k ( h /h ) .h deg u k = t k Y i =1 u k,i ( h /h ) .h deg u k,i . Then we deduce u k,i ( h /h ) .h deg u k,i = s k Y j =1 F e k,i,j k,j in K [ X , . . . , X n ]with e k,i,j ∈ { , } because F k are squarefree. Furthermore, the vectors e k,i are orthogonal for the usual scalar product because F k are squarefree.(2) We show this item for k = 1, the case k = 2 can be proved in a similar way.As F /F comes from the algorithm Good Homography and as explained inRemark 15 we can suppose that: u k, ( T ) = ( T − α k ) , with α k ∈ K . The previous item allows us to write: u ,i u deg u ,i , h h ! = (cid:16) Q s j =1 F e ,i,j ,j (cid:17) . (cid:16) h (cid:17) deg u ,i (cid:16) Q s j =1 F e , ,j ,j (cid:17) deg u ,i . (cid:16) h (cid:17) deg u ,i = Q s j =1 F e ,i,j ,j (cid:16) Q s j =1 F e , ,j ,j (cid:17) deg u ,i . By Proposition 4 applied to u ,i u deg u ,i , (cid:16) h h (cid:17) , we get then: D F /F Q s j =1 F e ,i,j ,j Q s j =1 F e , ,j . deg u ,i ,j ! = 0 . Now, we recall that F k,j are Darboux polynomials, see Proposition 6 andProposition 8. Then by Proposition 8, we deduce s X j =1 e ,i,j G F ,j − deg( u ,i ) s X j =1 e , ,j G F ,j = 0 . It follows ( e ,i, , . . . , e ,i,s , deg( u ,i ) .e , , , . . . , deg( u ,i ) .e , ,s ) ∈ ker S . Thus, e ,i ∈ π (ker S ).(3) The vectors e k, , . . . , e k,t k are linearly independant because they are or-thogonal. We just have to prove that these vectors generate π k (ker S ).Suppose that ρ = ( ρ , . . . , ρ s + s ) ∈ ker S . First, we clear the denominatorsand we suppose that ρ ∈ Z s + s instead of Q s + s .In a first time we explain the strategy of the proof for this item, and in a ECOMBINATION FOR DECOMPOSITION 11 second time we will detail the proof.We set F F = Q s i =1 F ρ j ,j Q s j =1 F ρ s j ,j , where F , F ∈ K [ X ] and F / F is a reduced rational function.Our goal is to get this kind of equality:( E ) , F F = Q s j =1 F ρ j ,j Q s j =1 F ρ s j ,j = Q k =1 Q ( i,k ) ∈ I num (cid:16) Q kj =1 F e k,i,j k,j (cid:17) m uk,i Q k =1 Q ( i,k ) ∈ I den (cid:16) Q kj =1 F e k,i,j k,j (cid:17) m uk,i , where m u k,i ∈ N , I = { (1 , , . . . , ( t , , (1 , , . . . , ( t , } , I num ⊂ I , I den ⊂ I and I num ∩ I den = ∅ .By the unicity of the factorization in irreducible factors we deduce: π ( ρ ) = X ( i, ∈ I num m u ,i e ,i − X ( i, ∈ I den m u ,i e ,i ,π ( ρ ) = X ( i, ∈ I num m u ,i e ,i − X ( i, ∈ I den m u ,i e ,i . We get: { e k, , . . . , e k,t k } generates π k (ker S ), and this is the desired re-sult.Now we detail the proof with four steps:(a) We remark: F F = u u ( h ) = Q deg ui =1 ( h − µ ,i h ) Q deg ui =1 ( h − µ ,i h ) , where µ k,i are roots of u k .(b) We have: F F = Q d j =1 ( h − λ j h ) m j Q d + d j = d +1 ( h − λ j h ) m j .h κ , with κ ∈ Z , m j ∈ N . Indeed, as ρ ∈ ker S , we have s X j =1 ρ j G F ,j − s X j =1 ρ s + j G F ,j = 0 . Thus Q s j =1 F ρ j ,j and Q s j =1 F ρ s j ,j are Darboux polynomials with thesame cofactor. By Proposition 7, we deduce: D F /F Q s j =1 F ρ j ,j Q s j =1 F ρ s j ,j ! = 0 . Then D F /F ( F / F ) = 0 and thus F / F = v /v ( h ) by Proposition4. We denote by λ j the roots of v and v and we get the desiredresult. (c) We claim: F F = Q k =1 Q ( i,k ) ∈ I num (cid:16) u k,i ( h /h ) h deg u k,i (cid:17) m uk,i Q k =1 Q ( i,k ) ∈ I den (cid:16) u k,i ( h /h ) h deg u k,i (cid:17) m uk,i , where m u k ,i ∈ N . Indeed, we have: for all j there exists δ ( j ) such that λ j = µ δ ( j ) .(To prove this remark we suppose the converse: There exists j suchthat λ j = µ k,i , for k = 1 , i = 1 , . . . , deg u .By definition of F k and by step 3b, there exists ( k , j ) such that F k ,j and h − λ j h have a common factor in C [ X ]. We call P this commonfactor.By step 3a, there exists ( k , i ) such that P is a factor of h − µ k ,i h .Thus h − λ j h and h − µ k ,i h have a common factor. As λ j = µ k ,i we deduce that P divides h and h . This is absurd because h /h is reduced.)Thus κ = 0, and for all j there exists k ( j ) ∈ { , } and such that u k ( j ) ( λ j ) = 0.As v , v ∈ K [ T ], by conjugation, we deduce that if λ j and λ j ′ areroots of the same irreducible polynomial u k,i ∈ K [ T ] then m j = m j ′ .We denote by m u k,i this common value.This gives the claimed equality with I num ∩ I den = ∅ , because F / F is reduced.(d) Now we can prove equality ( E ). F F = Q k =1 Q ( i,k ) ∈ I num (cid:16) u k,i ( h /h ) h deg u k,i (cid:17) m uk,i Q k =1 Q ( i,k ) ∈ I den (cid:16) u k,i ( h /h ) h deg u k,i (cid:17) m uk,i , by step 3c , = Q k =1 Q ( i,k ) ∈ I num (cid:16) Q kj =1 F e k,i,j k,j (cid:17) m uk,i Q k =1 Q ( i,k ) ∈ I den (cid:16) Q kj =1 F e k,i,j k,j (cid:17) m uk,i , by the first item . This gives the desired equality ( E ). (cid:3) Now we describe our recombination algorithm:
Recombination for Decomposition
Input: f = f /f ∈ K ( X , . . . , X n ), such that (C) and (H) are satisfied. Output:
A decomposition of f if it exists, with f = u ◦ h , u = u /u with deg u ≥ h = h /h non-composite.(1) Compute F = F /F := U ( f ) with the algorithm Good homography .(2) For k=1, 2, factorize F k = Q s k i =1 F k,i in K [ X ] with F k,i irreducible.(3) For each F k,i compute the corresponding cofactor G F k,i := D F /F ( F k,i ) /F k,i .(4) Build the system S and compute the basis in reduced row echelon form B of π (ker S ) and B of π (ker S ).(5) For k=1, 2, find v k = ( v k, , . . . , v k,s k ) ∈ B k such that: P s k i =1 v k,i deg F k,i = min w ∈B k P s k i =1 w i deg F k,i , where w := ( w , . . . , w s k ).(6) For k=1, 2, compute H k := Q s k i =1 F v k,i k,i . ECOMBINATION FOR DECOMPOSITION 13 (7) Set H := H /H .(8) Compute u such that u ( H ) = f .(9) Return H , and u . Proposition 17.
The algorithm
Recombination for Decomposition is correct.Proof.
Consider F /F := U ( f ). As we want to decompose f /f , we just have todecompose F /F , because deg U = 1 and then U is invertible.As F /F comes from the algorithm Good Homography we can suppose, see Remark15, that u k, ( T ) = ( T − α k ) with α k ∈ K , and k = 1 ,
2. Furthermore, by Lemma12 we can also suppose that deg u = deg u .Then by Proposition 16, the basis B k of π k (ker S ) are { e k, , . . . , e k,t k } .The vector e k,i gives the polynomial H k,i = Q s k j =1 F e k,i,j k,j = u k,i ( h ) h deg u k,i .Furthermore deg H k,i = P s k j =1 e k,i,j deg F k,j = deg u k,i deg h . Thus in Step 5min w ∈B k s k X i =1 w i deg F k,i = deg h, because this minimum is reached with e k, ∈ B k . Hence v k in Step 6 gives H k = u k,i ( k ) ( h ) h deg u k,i ( k ) with deg u k,i ( k ) = 1.It follows H = ( h − αh ) / ( h − βh ) with α, β ∈ K . Thus H = v ( h ) with deg v = 1,then the algorithm is correct. (cid:3) Proposition 18.
The algorithm
Recombination for Decomposition can be performedwith ˜ O ( rd n + ω − ) arithmetic operations over Q and two factorizations of univariatepolynomials of degree d with coefficients in K . We recall that in our complexity analysis the number of variables is fixed andthe degree d tends to infinity. Proof.
Step 1 uses ˜ O ( d n ) arithmetic operations over K by Proposition 14, thus ituses ˜ O ( rd n ) arithmetic operations over Q .Step 2 uses ˜ O ( d n + ω − ) arithmetic operations over K because we can use Lecerf’salgorithm, see [Lec07]. Thus we use ˜ O ( rd n + ω − ) arithmetic operations over Q andtwo factorizations of univariate polynomials of degree d with coefficients in K .In Step 3, we compute D F /F ( F k,i ), thus we perform 2( n −
1) multiplications ofmultivariate polynomials. We can do this with a fast multiplication technique,and then this computation costs ˜ O ( nrd n ) arithmetic operations over Q . Then wedivide D F /F ( F k,i ) by F k,i . We have to perform n − n − O ( rd n ). As s and s are smaller than d ,Step 3 costs ˜ O ( nrd n +1 ) arithmetic operations over Q .Step 4 needs ˜ O ( nrd n d ω − ) arithmetic operations over Q with Storjohann’s method,see [Sto00, Theorem 2.10]. Indeed, S has O (( n − rd n ) equations and s + s unknowns, thus at most 2 d unknowns.Step 5 has a negligeable cost because dim Q π k (ker S ) = t k is smaller than d and s k is also smaller than d .In Step 6, we use a fast multiplication technique and we compute H k with ˜ O ( rd n )arithmetic operations over Q .Step 8 can be done with ˜ O ( rd n ) arithmetic operations over Q , see [Ch`e10]. Thus the global cost of the algorithm belongs to ˜ O ( rd n + ω − ) arithmetic operationsover Q . (cid:3) Examples
In this section we show the behavior of the algorithm
Recombination for Decom-position with two examples. We consider bivariate rational functions with rationalcoefficients. Thus hypothesis (C) is satisfied.4.1. f is non-composite. We set: f = (cid:0) X + Y (cid:1) ( X + Y ) = X + X + XY + Y + Y X + Y ,f = f − (cid:0) Y − X − (cid:1) ( Y − X + 1) = − X + 3 XY + 2 Y + 2 Y X − Y + 1We have deg( f + Λ f ) = deg Y ( f + Λ f ) = 3, and Res Y (cid:16) f (0 , Y )+Λ f (0 , Y ) , ∂ Y f (0 , Y )+Λ ∂ Y f (0 , Y ) (cid:17) = − −
24 Λ −
92 Λ −
64 Λ +8 Λ . Thus hypothesis (H) is satisfied.The algorithm
Good homography gives: λ a = f (0 , /f (0 ,
0) = 0 and λ b = f (0 , /f (0 ,
1) = 1.Then F = (cid:0) X + Y (cid:1) ( X + Y ) ,F , = 1 + X + Y ,F , = X + YF = (cid:0) Y − X − (cid:1) ( Y − X + 1) ,F , = Y − X − F , = Y − X + 1The cofactors are: G F , = 3 X + 8 Y X + 2 X − Y X + 7 XY − Y − Y − Y + 2 Y G F , = 3 X + 8 Y X + 4 Y X + 6 X − Y − Y + 3 − Y − Y G F , = 3 X + 8 Y X + XY − Y X + 2 X − − Y − Y − Y − Y G F , = 3 X + 8 Y X + 6 XY + 8 Y X + 6 X − Y + 8 Y − Y + 3 ECOMBINATION FOR DECOMPOSITION 15
The linear system S is the following: − − − − − − − −
11 87 0 1 60 0 0 0 − − − −
20 0 0 0 − − − − A basis of ker( S ) is given by: { ( − , − , , } .Then it follows that f /f is non-composite.4.2. f is composite. Here we set: h = (cid:0) X + Y (cid:1) ( X + Y ) h = h − (cid:0) Y − X − (cid:1) ( Y − X + 1) u = T. ( T − u = T + 1 f /f = u /u ( h /h ) . We have constructed a composite rational function f /f and now we illustratehow our algorithm computes a decomposition. We can already remark that in theprevious example we have shown that h /h is non-composite.In this situation the hypothesis (H) is satisfied and the algorithm Good Homog-raphy gives: λ a = f (0 , /f (0 ,
0) = 0 and λ b = f (0 , /f (0 ,
2) = 90 / F , = 1 + X + Y F , = 2 X − Y − F , = Y − X − F , = X + YF , = 2 X + 11 X + 9 + 29 XY + 29 Y + 38 XY − Y + 11 Y F , = 11 X + X − − Y X − Y − XY + 10 Y + Y
36 G. CH`EZE
The basis in reduced row echelon form of π (ker S ) (resp. π (ker S )) is { (1 , , , , , , } (resp. { (1 , , } ).Step 5 in the algorithm Recombination for Decomposition gives: v = (1 , , ,
1) and v = (1 , H := F , .F , and H := F , .We remark that H = h and that H = 11 h + 9 h . Then H /H = w ( h /h ),where w ( T ) = T / (11 T + 9). As h /h is non-composite and deg w = 1, we get acorrect output. 5. Conclusion
In conclusion, we summarize our algorithm with a “derivation point of view”.In order to decompose f /f , we have computed with Darboux method a rationalfirst integral of D f /f with minimum degree. That is to say we have computed h /h ∈ K ( X , . . . , X n ) such that D f /f ( h /h ) = 0 and deg( h /h ) is minimum.In a general setting, Darboux method works as follows: If we want to compute a ra-tional first integral of a derivation D , first we compute all the Darboux polynomials F i and their associated cofactors G F i , second we solve the linear system X i e i G F i = 0 . Then thanks to Proposition 8, we deduce that Q i F e i i is a first integral, i.e. D ( Q i F e i i ) = 0.When we consider the derivation D f /f the computation of Darboux polynomi-als is reduced to the factorization of f + λf . Thus this step can be done efficiently.In the general setting, we can also reduce the computation of Darboux polynomialsto a factorization problem, see [Ch`e11].During the second step, we compute the kernel of P i e i G F i = 0. It is actually arecombination step. Indeed, this system explains how we have to recombine F i inorder to get a rational first integral. Furthermore, the cofactor G F i = D ( F i ) /F i canbe viewed as a logarithmic derivative.In conclusion, the recombination scheme used in this paper is called nowadays thelogarithmic derivative method, but this method is Darboux original method. Appendix A. Convex-dense bivariate decomposition
In this appendix we give complexity results for the decomposition of sparse bivari-ate rational functions. These results rely on a strategy proposed by J. Berthomieuand G. Lecerf in [BL10].Given a polynomial f ( X, Y ) ∈ K [ X, Y ], its support is the set S f of integer points( i ; j ) such that the monomial X i Y j appears in f with a non zero coefficient. Theconvex hull, in the real space R of S f is denoted by N ( f ) and called the Newton’spolygon of f . We denote by | N ( f ) | the number of integral points of N ( f ). Wecalled | N ( f ) | the convex-size of f .Roughly speaking, the transformation proposed in [BL10] consists in a monomialtransformation that preserves the convex-size but decreases the dense size. Theconsidered transformation T can be described in the following way: T = B ◦ L , where ECOMBINATION FOR DECOMPOSITION 17 B ( X i Y j ) = X i + b Y j + b , b , b ∈ Z , L ( X i Y j ) = X a i + a j Y a i + a j , a a − a a = ± . T can be defined on K [ X, Y, X − , Y − ], and we define: T ( P i,j f i,j X i Y j ) = P i,j f i,j T ( X i Y j ).The transformation L corresponds to the linear map: ( i, j )
7→ A t ( i, j ), where A = (cid:18) a a a a (cid:19) . We denote by L − the transformation corresponding to A − .If f ( X, Y ) ∈ K [ X, Y ], then L ( f ) ∈ K [ X, Y, X − , Y − ] and L ( f ) can be written L ( f ) = c L ( f ) . L ( f ), where L ( f ) ∈ K [ X, Y ] and c L ( f ) = X i Y j ∈ K [ X, Y, X − , Y − ].Furthermore, we also have L ( F .F ) = L ( F ) . L ( F ).Let S be a finite subset of Z . Set S is said to be normalized if it belongs to N and if it contains at least one point in { } × N , and also at least one point in N × { } . For such a normalized set, we write d x (resp. d y ) for the largest abscissa(resp. ordinate) involved in S , so that the bounding rectangle is R = [0 , d x ] × [0 , d y ].The following result is proved in [BL10, Theorem 2]: For any normalized finite subset S of Z , of cardinality σ , convex-size π , and bound-ing rectangle [0 , d x ] × [0 , d y ] , and dense size δ = ( d x +1)( d y +1) , one can compute anaffine map T = B ◦ L , with O ( σ log δ ) bit-operations, such that T ( S ) is normalizedof dense size at most π . We are going to use this transformation in order to prove:
Theorem 19.
Let f /f ( X, Y ) ∈ K ( X, Y ) such that deg( f /f ) = d , N ( f ) ⊂ N , N ( f ) ⊂ N and N is normalized. Then (1) If K is field with characteristic or at least d ( d −
1) + 1 and (H) is satisfied,then there exists a probabilistic algorithm which computes the decompositionof f /f with at most ˜ O ( | N | , ) operations in K and two factorizations ofa univariate polynomial of degree at most | N | over K . (2) If (C) and (H) are satisfied, then there exists a deterministic algorithmwhich computes the decomposition of f /f with at most ˜ O ( r. | N | ( ω +1) / ) operations over Q and two factorizations of an univariate polynomials ofdegree at most | N | over Q [ α ] . Now, we explain how we use the transformation T in the decomposition setting. Proposition 20. If f /f = u ( h /h ) then T ( f ) / T ( f ) = u (cid:0) L ( h ) / L ( h ) (cid:1) .If T ( f ) / T ( f ) = u ( H /H ) then f /f = u (cid:0) L − ( H ) / L − ( H ) (cid:1) .Proof. We prove the first item, the second can be proved in a similar way.We have: f f = Q i ( h − µ ,i h ) Q j ( h − µ ,j h ) , where µ k,i are roots of u k . Then, T ( f ) T ( f ) = B ◦ L ( f ) B ◦ L ( f ) = X b Y b L ( f ) X b Y b L ( f )= L ( f ) L ( f ) = Q i (cid:0) L ( h ) − µ ,i L ( h ) (cid:1)Q j (cid:0) L ( h ) − µ ,j L ( h ) (cid:1) = u (cid:0) L ( h ) / L ( h ) (cid:1) u (cid:0) L ( h ) / L ( h ) (cid:1) = u (cid:0) L ( h ) / L ( h ) (cid:1) (cid:3) This gives the following algorithm:
Convex bivariate decomposition
Input: f = f /f ∈ K ( X, Y ), where N ( f ) ⊂ N , N ( f ) ⊂ N and N is normalized. Output:
A decomposition of f if it exists, with f = u ◦ h , u = u /u with deg u ≥ h = h /h non-composite.(1) Compute F = T ( f ) / T ( f ).(2) Decompose F = u ( H ).(3) Return f = u ( h ), where h = L − ( H ) L − ( H ) = c L − ( H ) . L − ( H ) c L − ( H ) . L − ( H ) ∈ K ( X, Y ). Proposition 21.
The algorithm
Convex bivariate decomposition is correct.Proof.
This follows from Proposition 20. (cid:3)
Proposition 22.
The algorithm
Convex bivariate decomposition uses one decompo-sition of a rational function of degree at most | N | and O ( σ δ ) bit operations.Proof. We apply [BL10, Theorem 2] to N . (cid:3) The proof of Theorem 19 comes from Proposition 21 and Proposition 22 andcomplexity results given in Theorem 1 and [Ch`e10, Theorem 2].
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Institut de Math´ematiques de Toulouse, Universit´e Paul Sabatier Toulouse 3, MIPBˆat 1R3, 31 062 TOULOUSE cedex 9, France
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