Telescopers for differential forms with one parameter
Shaoshi Chen, Ruyong Feng, Ziming Li, Michael F. Singer, Stephen Watt
aa r X i v : . [ c s . S C ] J a n Telescopers for differential forms with oneparameter ∗ Shaoshi Chen , Ruyong Feng , Ziming Li ,Michael F. Singer , Stephen Watt KLMM, AMSS, Chinese Academy of Sciences andSchool of Mathematics, University of Chinese Academy of Sciences,Beijing, 100190, China Department of Mathematics, North Carolina State University,Raleigh, 27695-8205, USA SCG, Faculty of Mathematics, University of Waterloo,Ontario, N2L3G1, Canada { schen, ryfeng } @amss.ac.cn, [email protected]@math.ncsu.edu, [email protected] January 20, 2021
Abstract
Telescopers for a function are linear differential (resp. difference)operators annihilated by the definite integral (resp. definite sum) ofthis function. They play a key role in Wilf-Zeilberger theory and algo-rithms for computing them have been extensively studied in the pastthirty years. In this paper, we introduce the notion of telescopers fordifferential forms with D -finite function coefficients. These telescop-ers appear in several areas of mathematics, for instance parametrizeddifferential Galois theory and mirror symmetry. We give a sufficientand necessary condition for the existence of telescopers for a differentialform and describe a method to compute them if they exist. Algorithmsfor verifying this condition are also given. ∗ S. Chen was partially supported by the NSFC grants 11871067, 11688101, the Fund ofthe Youth Innovation Promotion Association, CAS, and the National Key Research andDevelopment Project 2020YFA0712300, R. Feng was partially supported by the NSFCgrants 11771433, 11688101, Beijing Natural Science Foundation under Grant Z190004,and the National Key Research and Development Project 2020YFA0712300. Introduction
In the Wilf-Zeilberger theory, telescopers usually refer to the operators inthe output of the method of creative telescoping, which are linear differential(resp. difference) operators annihilated by the definite integrals (resp. thedefinite sums) of the input functions. The telescopers have emerged atleast from the work of Euler [17] and have been found many applicationsin the various areas of mathematics such as combinatorics, number theory,knot theory and so on (see Section 7 of [19] for details). In particular,telescopers for a function are often used to prove the identities involvingthis function or even obtain a simpler expression for the definite integral orsum of this function. As a clever and algorithmic process for constructingtelescopers, creative telescoping firstly appeared as a term in the essay ofvan der Poorten on Apr´ey’s proof of the irrationality of ζ (3) [30]. However,it was Zeilberger and his collaborators [3, 28, 35, 36, 38] in the early 1990swho equipped creative telescoping with a concrete meaning and formulatedit as an algorithmic tool. Since then, algorithms for creative telescoping havebeen extensively studied. Based on the techniques used in the algorithms,the existing algorithms are divided into four generations, see [13] for thedetails. Most recent algorithms are called reduction-based algorithms whichwere first introduced by Bostan et al. in [6] and further developed in [7,14, 15, 8] etc. The termination of these algorithms relies on the existenceof telescopers. The question for which input functions the algorithms willterminate has been answered in [37, 1, 2, 16, 10] etc for several classes offunctions such as rational functions and hypergeometric functions and soon. The algorithmic framework for creative telescoping is now called theWilf-Zeilberger theory.Most of algorithms for creative telescoping focus on the case of one bi-variate function as input. There are only a few algorithms which deal withmultivariate case (see [11, 9, 20, 12] etc). It is still a challenge to developthe multivariate analogue of the existing algorithms (see Section 5 of [13]).In the language of differential forms (with m variables and one parameter),the results in [11] and [20] dealt with the cases of differential 1-forms anddifferential m -forms respectively. On the other hand, in the applications toother domains such as mirror symmetry (see [22, 25, 26]), one needs to dealwith the case of differential p -forms with 1 ≤ p ≤ m . Below is an example. Example 1
Consider the following one-parameter family of the quintic poly-nomials W ( t ) = 15 ( x + x + x + x + x ) − tx x x x x here t is a parameter. Set ω = X i =1 ( − i − x i W ( t ) d x ∧ · · · ∧ c d x i ∧ · · · ∧ d x . To obtain the Picard-Fuchs equation for the mirror quintic, the geometriestswant to compute a fourth order linear differential operator L in t and ∂ t such that L ( ω ) = d η for some differential 3-form η . Here one has that L = (1 − t ) ∂ ∂t − t ∂ ∂t − t ∂ ∂t − t ∂∂t − . Set θ t = t∂/∂t . Then ˜ L = − L t = θ t − t (5 θ t + 1)(5 θ t + 2)(5 θ t + 3)(5 θ t + 4) and the equation ˜ L ( y ) = 0 is the required Picard-Fuchs equation. We call the operator L appearing in the above example a telescoper for thedifferential form ω (see Definition 4). In this paper, we study the telescopersfor differential forms with D -finite function coefficients. Instead of the geo-metric method used in [22, 25, 26], we provide an algebraic treatment. Wegive a sufficient and necessary condition guaranteeing the existence of tele-scopers and describe a method to compute them if they exist. Meanwhile,we also present algorithms to verify this condition.The rest of this paper is organized as follows. In Section 2, we recalldifferential forms with D -finite function coefficients and introduce the notionof telescopers for differential forms. In Section 3, we give a sufficient andnecessary condition for the existence of telescopers, which can be consideredas a parametrized version of Poincar´e lemma on differential manifolds. InSection 4, we give two algorithms for verifying the condition presented inSection 3. Notations : The following notations will be frequently used thoughoutthis paper. ∂ t : the usual derivation ∂/∂ t with respect to t , ∂ x i : the usual derivation ∂/∂ x i with respect to x i , x : { x , · · · , x n } ∂ x : { ∂ x , · · · , ∂ x n } , 3he following formulas will also be frequently used: ∂ µx x ν = ( ν ( ν − · · · ( ν − µ + 1) x ν − µ + ∗ ∂ x , ν ≥ µ ∗ ∂ x , ν < µ (1) x µ ∂ νx = ( ( − ν µ ( µ − · · · ( µ − ν + 1) x µ − ν + ∂ x ∗ , µ ≥ ν∂ x ∗ , µ < ν (2)where ∗ ∈ k h x, ∂ x i . D -finite elements and differential forms Throughout this paper, let k be an algebraically closed field of characteristiczero and let K be the differential field k ( t, x , · · · , x n ) with the derivations ∂ t , ∂ x , · · · , ∂ x n . Let D = K h ∂ t , ∂ x i be the ring of linear differential op-erators with coefficients in K . For S ⊂ { t, x , ∂ t , ∂ x } , denote by k h S i thesubalgebra over k of D generated by S . For brevity, we denote k h t, x , ∂ t , ∂ x i by W . Let U be the universal differential extension of K in which everyalgebraic differential equation having a solution in an extension of U has asolution (see page 133 of [18] for more precise description). Definition 2
An element f ∈ U is said to be D -finite over K if for every δ ∈ { ∂ t , ∂ x , · · · , ∂ x n } , there is a nonzero operator L δ ∈ K h δ i such that L δ ( f ) = 0 . Denote by R the ring of D -finite elements over K , and by M a free R -module of rank m with base { a , · · · , a m } . Define a map D × M → M given by L, m X i =1 f i a i ! → L m X i =1 f i a i ! := m X i =1 L ( f i ) a i . This map endows M with a left D -module structure. Let ^ ( M ) = m M i =0 ^ i ( M )be the exterior algebra of M , where V i ( M ) denotes the i -th homogeneouspart of V ( M ) as a graded R -algebra. We call an element in V i ( M ) an i -form. V ( M ) is also a left D -module. Let d : R → M be a map defined asd f = ∂ x ( f ) a + · · · + ∂ x m ( f ) a m f ∈ R . Then d is a derivation over k . Note that for each i = 1 , · · · , m ,d x i = a i . Hence in the rest of this paper we shall use { d x , · · · , d x m } insteadof { a , · · · , a m } . The map d can be extended to a derivation on V ( M ) whichis defined recursively asd( ω ∧ ω ) = d ω ∧ ω + ( − i − ω ∧ d ω for any ω ∈ V i ( M ) and ω ∈ V j ( M ). For detailed definitions on exterioralgebra and differential forms, we refer the readers to Chapter 19 of [21] andChapter 1 of [34] respectively. As the usual differential forms, we introducethe following definition. Definition 3
Let ω ∈ V ( M ) be a form. (1) ω is said to be closed if d ω = 0 , and exact if there is η ∈ V ( M ) suchthat ω = d η . (2) ω is said to be ∂ t -closed ( ∂ t -exact) if there is a nonzero L ∈ k ( t ) h ∂ t i such that L ( ω ) is closed (exact). Definition 4
Assume that ω ∈ V ( M ) . A nonzero L ∈ k ( t ) h ∂ t i is called atelescoper for ω if L ( ω ) is exact. The famous Poincar´e lemma states that if B is an open ball in R n , anysmooth closed i -form ω defined on B is exact, for any integer i with 1 ≤ i ≤ n . In this section, we shall prove the following lemma which can be viewedas a parametrized analogue of Poincar´e lemma for V ( M ). Lemma 5 (Parameterized Poincar´e lemma)
Let ω ∈ V p ( M ) . If ω is ∂ t -closed then it is ∂ t -exact. To the above lemma, we need some lemmas.
Lemma 6 (Lipshitz’s lemma (Lemma 3 of [24]))
Assume that f is a D -finite element over k ( x ) . For each pair ≤ i < j ≤ n , there is a nonzerooperator L ∈ k ( x , x , · · · , x n ) h ∂ x i , ∂ x j i such that L ( f ) = 0 . The following lemma is a generalization of Lipshitz’s lemma.5 emma 7
Assume that f , · · · , f m are D -finite elements over k ( x , t ) and S ⊂ { t, x , · · · , x n , ∂ t , ∂ x , · · · , ∂ x n } with | S | > n + 1 . Then one can compute a nonzero operator T in k h S i suchthat T ( f i ) = 0 for all i = 1 , · · · , m . Proof.
For each δ ∈ { t, ∂ x , · · · , ∂ x n } and i = 1 , · · · , m , let T i be a nonzerooperator in K h δ i such that T i ( f i ) = 0. Set T to be the least common leftmultiple of T , . . . , T m . Then T ( f i ) = 0 for all i = 1 , · · · , m . The lemmathen follows from an argument similar to that in the proof of Lipshitz’slemma. Lemma 8
Assume that f , · · · , f m are D -finite over k ( x , t ) , I, J ⊂ { , · · · , n } and I ∩ J = ∅ . Assume further that V ⊂ { x i , ∂ x i | i ∈ { , · · · , n } \ ( I ∪ J ) } with | V | = n − | I | − | J | . Then one can compute an operator P of the form L + X i ∈ I ∂ x i M i + X j ∈ J N j ∂ x j such that P ( f l ) = 0 for all l = 1 , · · · , m , where L is a nonzero operator in k h{ t, ∂ t } ∪ V }i , M i , N j ∈ W and N j is free of x i for all i ∈ I and j ∈ J . Proof.
Without loss of generality, we assume that I = { , · · · , r } and J = { r + 1 , · · · , r + s } where r = | I | and s = | J | . Let S = { t, ∂ t } ∪ { ∂ x i | i ∈ I } ∪ { x j | j = r + 1 , · · · , r + s } ∪ V. Then | S | = n + 2 > n + 1. By Lemma 7, one can compute a T ∈ k h S i \ { } such that T ( f l ) = 0 for all l = 1 , · · · , m . Write T = X d =( d , ··· ,d r ) ∈ Γ ∂ d x · · · ∂ d r x r T d where T d ∈ k h{ t, ∂ t , x r +1 , · · · , x r + s } ∪ V }i \ { } and Γ is a finite subsetof Z r . Let ¯ d = ( ¯ d , · · · , ¯ d r ) be the minimal element of Γ with respect tothe lex order on Z r . Multiplying T by Q ri =1 x ¯ d i i on the left and using theformula (2) yield that r Y i =1 x ¯ d i i ! T = αT ¯ d + r X i =1 ∂ x i ˜ T i (3)6here α is a nonzero integer and ˜ T i ∈ k h S ∪ { x i | i ∈ I }i . Write T ¯ d = X e =( e , ··· ,e s ) ∈ Γ L e x e r +1 · · · x e s r + s where L e ∈ k h{ t, ∂ t } ∪ V i \ { } and Γ is a finite subset of Z s . Let ¯ e =(¯ e , · · · , ¯ e s ) be the maximal element of Γ with respect to the lex order on Z s . Multiplying T ¯ d by Q si =1 ∂ ¯ e i x r + i on the left and using the formula (1) yieldthat s Y i =1 ∂ ¯ e i x r + i ! T ¯ d = βL ¯ e + X j ∈ J ˜ L j ∂ x j (4)where ˜ L i ∈ k h{ t, ∂ t , x r +1 , · · · , x r + s , ∂ x r +1 , · · · , ∂ x r + s }∪ V i and α is a nonzerointeger. Combining (3) with (4) yields the required operator P . Corollary 9
Assume that f , · · · , f m are D -finite over k ( x , t ) , J is a subsetof { , · · · , n } and V ⊂ { x i , ∂ x i | i ∈ { , · · · , n } \ J } with | V | = n − | J | .Assume further that ∂ x j ( f l ) = 0 for all j ∈ J and l = 1 , · · · , m . Thenone can compute a nonzero L ∈ k h{ t, ∂ t } ∪ V i such that L ( f l ) = 0 for all l = 1 , · · · , m . Proof.
In Lemma 8, set I = ∅ .The main result of this section is the following theorem which can be viewedas a generalization of Corollary 9 to differential forms. To describe and provethis theorem, let us recall some notation from the first chapter of [34]. Forany f ∈ R , we define d ( f ) = 0 andd s ( f ) = ∂ x ( f )d x + · · · + ∂ x s ( f )d x s for s ∈ { , , . . . , n } . We can extend d s to the module V ( M ) in a naturalway. Precisely, let ω = P mi =1 f i m i where m i is a monomial in d x , · · · , d x n .Then d ( ω ) = 0 andd s ( ω ) = m X i =1 s X j =1 ∂ x j ( f i )d x j ∧ m i = s X j =1 d x j ∧ ∂ x j ( ω ) . By definition, one sees thatd s ( u ∧ d x s ) = d s − ( u ) ∧ d x s and d s ( u ) = d s − ( u ) + d x s ∧ ∂ x s ( u ) . heorem 10 Assume that ≤ s ≤ n, V ⊂ { x s +1 , · · · , x n , ∂ x x +1 , · · · , ∂ x n } with | V | = n − s and ω ∈ V p ( M ) . If d s ω = 0 , then one can compute anonzero L ∈ k h{ t, ∂ t } ∪ V i and µ ∈ V p − ( M ) such that L ( ω ) = d s µ. Remark 11
1. If p = 0 , then ω = f ∈ R and d s f = 0 if and only if s = 0 or ∂ x i ( f ) = 0 for all ≤ i ≤ s if s > . Therefore Corollary 9is a special case of Theorem 10.2. Note that the parametrized Poincar´e lemma is just the special case ofTheorem 10 when s = n . Proof.
We proceed by induction on s . Assume that s = 0 and write ω = m X i =1 f i m i where m i a monomial in d x , d x , · · · , d x n and f i ∈ R . By Corollary 9 with I = ∅ , one can compute a nonzero L ∈ k h{ t, ∂ t } ∪ V i such that L ( f i ) = 0for all i = 1 , · · · , m . Then one has that L ( ω ) = m X i =1 L ( f i ) m i = 0 . This proves the base case. Now assume that the theorem holds for s < ℓ and consider the case s = ℓ . Write ω = u ∧ d x ℓ + v where both u and v do not involve d x ℓ . Then the assumption d ℓ ω = 0implies thatd ℓ − u ∧ d x ℓ + d ℓ v = d ℓ − u ∧ d x ℓ + d ℓ − v + d x ℓ ∧ ∂ x l ( v ) = 0 . Since all of d ℓ − u, d ℓ − v, ∂ x ℓ ( v ) do not involve d x ℓ , one has that d ℓ − v = 0and d ℓ − ( u ) − ∂ x ℓ ( v ) = 0. By the induction hypothesis, one can compute anonzero ˜ L ∈ k h{ t, x ℓ , ∂ t } ∪ V i and ˜ µ ∈ V p − ( M ) such that˜ L ( v ) = d ℓ − (˜ µ ) . (5)We claim that ˜ L can be chosen to be free of x ℓ . Write˜ L = d X j =0 N j x dℓ N j ∈ k h{ t, ∂ t } ∪ V i and N d = 0. Multiplying ˜ L by ∂ dx ℓ on the left andusing the formula (2) yield that ∂ dx ℓ ˜ L = d X j =0 N j ∂ dx ℓ x jℓ = αN d + ˜ N ∂ x ℓ (6)where α is a nonzero integer and ˜ N ∈ k h{ t, x ℓ , ∂ t , ∂ x ℓ } ∪ V i . The equalities(5) and (6) together with ∂ x ℓ ( v ) = d ℓ − (˜ u ) yield that N d ( v ) = d ℓ − ( π ) forsome π ∈ V p − ( M ). This proves the claim. Now one has that˜ L ( ω ) = ˜ L ( u ) ∧ d x ℓ + d ℓ − (˜ µ ) = ˜ L ( u ) ∧ d x ℓ + d x ℓ ∧ ∂ x ℓ (˜ µ ) + d ℓ (˜ µ ) . Since ˜ L is free of x , · · · , x ℓ , ˜ L d ℓ = d ℓ ˜ L . This implies that0 = ˜ L (d ℓ ( ω )) = d ℓ ( ˜ L ( ω )) = d ℓ − ( ˜ L ( u )) ∧ d x ℓ + d x ℓ ∧ d ℓ − ( ∂ x ℓ (˜ µ ))= d ℓ − (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17) ∧ d x ℓ . Note that ˜ µ can always be chosen to be free of d x ℓ . Hence one has thatd ℓ − ( ˜ L ( u ) − ∂ x ℓ (˜ µ )) = 0. By the induction hypothesis, one can compute anonzero ¯ L ∈ k h{ t, ∂ x ℓ , ∂ t } ∪ V i and ¯ µ ∈ V p − ( M ) such that¯ L (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17) = d ℓ − (¯ µ ) . (7)Write ¯ L = e X j = e ∂ jx ℓ M j where M j ∈ k h{ t, ∂ t } ∪ V i and M e = 0. Multiplying ¯ L by x e ℓ on the leftand using the formula (2) yield that x e ℓ ¯ L = βM e + ∂ x ℓ ˜ M where β is a nonzero integer and ˜ M ∈ k h{ t, ∂ t , ∂ x ℓ , x ℓ } ∪ V i . Hence applying x e ℓ to the equality (7), one gets that βM e (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17) = d ℓ − ( x e ℓ ¯ µ ) + ∂ x ℓ (cid:16) ˜ M (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17)(cid:17) . Set L = βM e ˜ L . The one has that L ( ω ) = βM e (cid:16) ( ˜ L ( u ) − ∂ x ℓ (˜ µ )) ∧ d x ℓ + d ℓ (˜ µ ) (cid:17) = (cid:16) βM e (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ (cid:17)(cid:17) ∧ d x ℓ + d ℓ ( βM e (˜ µ ))= d ℓ − ( x e ℓ ¯ µ ) ∧ d x ℓ + ∂ x ℓ ˜ M (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17) ∧ d x ℓ + d ℓ ( βM e (˜ µ ))= d ℓ (cid:16) x e ℓ ¯ µ + ˜ M (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17) + βM e (˜ µ ) (cid:17) . ℓ − (cid:16) ˜ M (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17)(cid:17) = ˜ M d ℓ − (cid:16) ˜ L ( u ) − ∂ x ℓ (˜ µ ) (cid:17) = 0 . Remark 12
Lemma 5 can be derived from the finiteness of the de Rhamcohomology groups of D -modules in the Bernstein class. To see this, let ω be a differential s -form with coefficients in R and let M be the D -modulegenerated by all coefficients of ω and all derivatives of these coefficients withrespect to ∂ t . By Proposition 5.2 on page 12 of [5], M is a D -module in theBernstein class. Assume that ω is closed. Then ∂ jt ( ω ) ∈ H sDR ( M ) , the j -thde Rham cohomology group of M , for all nonnegative integer j . By Theorem6.1 on page 16 of [5], H sDR ( M ) is of finite dimension over k ( t ) . This impliesthat there are a , · · · , a m ∈ k ( t ) such that P mj =0 a j ∂ jt ( ω ) = 0 in H sDR ( M ) ,i.e. P mj =0 a j ∂ jt ( ω ) is exact. This proves the existence of telescopers for the ∂ t -closed differential forms. However the proof of Theorem 10 is constructiveand it provides a method to compute a telescoper if it exists. The proof of Theorem 10 can be summarized as the following algorithm.
Algorithm 13
Input: ω ∈ V p ( M ) and V ∈ { x i , ∂ x i | i = s + 1 , · · · , n } satisfying that d s ( ω ) = 0 and | V | = n − s Output: a nonzero L ∈ k h{ t, ∂ t } ∪ V i such that L ( ω ) = d s ( µ ) .1. If ω ∈ R , then by Corollary 9, compute a nonzero L ∈ k h{ t, ∂ t } ∪ V i such that L ( ω ) = 0 . Return L .2. Write ω = u ∧ d x s + v with u, v not involving d x s .3. Call Algorithm 13 with v and V ∪ { x s } as inputs and let ˜ L be theoutput.(a) Write ˜ L = P dj =0 N j x js with N j ∈ k h{ t, ∂ t , x s } ∪ V i and N d = 0 .(b) Compute a ˜ µ ∈ V p − ( M ) such that N d ( v ) = d s − (˜ µ ) .4. Write N d ( ω ) = ( N d ( u ) − ∂ x s (˜ µ )) ∧ d x s + d s (˜ µ ) .5. Call Algorithm 13 with N d ( u ) − ∂ x s (˜ µ ) and V ∪ { ∂ x s } as inputs andlet ¯ L be the output.6. Write ¯ L = P e j = e ∂ jx s M j with M j ∈ k h{ t, ∂ t } ∪ V i and M e = 0 .7. Return M e N d . The existence of telescopers
It is easy to see that if a differential form is ∂ t -exact then it is ∂ t -closed.Therefore Lemma 5 implies that given a ω ∈ V p ( M ), to decide whether ithas a telescoper, it suffices to decide whether there is a nonzero L ∈ k h t, ∂ t i such that L (d ω ) = 0. Suppose thatd ω = X ≤ i < ···
Given an element f ∈ U , decide whether there exists a nonzero L ∈ k h t, ∂ t i such that L ( f ) = 0 . Let P ∈ K h ∂ t i \ { } be the monic operator of minimal order such that P ( f ) = 0. Then f is annihilated by a nonzero L ∈ k ( t ) h ∂ t i if and only if P is a right-hand factor of L , i.e. L = QP for some Q ∈ K h ∂ t i . Such operator P will be called a ( x , t )-separable operator. Problem 14 then is equivalentto the following one. Problem 15
Given a P ∈ K h ∂ t i \ { } , decide whether P is ( x , t ) -separable. The rest of this paper is aimed at developing an algorithm to solve the aboveproblem. Let us first investigate the solutions of ( x , t )-separable operators. Notation 16 C t := { c ∈ U | ∂ t ( c ) = 0 } , C x := { c ∈ U | ∀ x ∈ x , ∂ x ( c ) = 0 } . Assume that L ∈ k ( t ) h ∂ t i \ { } . By Corollary 1.2.12 of [29], the solutionspace of L = 0 in U is a C t -vector space of dimension ord( L ). Moreover wehave the following lemma. Lemma 17 If L ∈ k ( t ) h ∂ t i \ { } , then the solution space of L = 0 in U hasa basis in C x . Proof.
Let d = ord( L ) > { v , · · · , v d } be a basis of the solutionspace of L = 0 in U . For all 1 ≤ i ≤ d and all 1 ≤ l ≤ m , L ( ∂ x l ( v i )) = ∂ x l ( L ( v i )) = 0 . v = ( v , · · · , v d ) t . Then for each l = 1 , · · · , m , ∂ x l ( v ) = A l v , A l ∈ Mat d ( C t ) . Since ∂ x i ∂ x j ( v ) = ∂ x j ∂ x i ( v ) and v , · · · , v d are linearly independent over C x , for all 1 ≤ i < j ≤ m , ∂ x i ( A j ) − ∂ x j ( A i ) = A i A j − A j A i (8)On the other hand, ∂ t ( A i ) = 0 for all 1 ≤ i ≤ d . These together with (8)imply that the system ∂ x ( Y ) = A Y, · · · , ∂ x m ( Y ) = A m Y, ∂ t ( Y ) = 0is integrable. Then there is an invertible matrix G with entries in U satisfyingthis system. Let ¯ v = G − v . As ∂ t ( G − ) = 0, ¯ v is still a basis of the solutionspace of L = 0 in U . Furthermore, for each i = 1 , · · · , m , we have ∂ x i (¯ v ) = ∂ x i ( G − v ) = ∂ x i ( G − ) v + G − A i v = − G − A i v + G − A i v = 0 . Thus ¯ v ∈ C d x .As a consequence, we have the following corollary. Corollary 18
Assume that P ∈ K h ∂ t i \ { } . Then P is ( x , t ) -separable ifand only if the solutions of P ( y ) = 0 in U are of the form s X i =1 g i h i , g i ∈ C t , h i ∈ C x ∩ { f ∈ U | P ( f ) = 0 } . (9) Proof.
The “only if” part is a direct consequence of Lemma 17. For the“if” part, one only need to prove that if h ∈ C x ∩ { f ∈ U | P ( f ) = 0 } then h is annihilated by a nonzero operator in k ( t ) h ∂ t i . Suppose that h ∈ C x ∩ { f ∈U | P ( f ) = 0 } . Let L be the monic operator in K h ∂ t i \ { } which annihilates h and is of minimal order. Write L = ∂ ℓt + ℓ − X i =0 a i ∂ it , a i ∈ K. Then for every j ∈ { , . . . , m } ∂ x j ( L ( h )) = ℓ − X i =0 ∂ x j ( a i ) ∂ it ( h ) + L ( ∂ x j ( h )) = ℓ − X i =0 ∂ x j ( a i ) ∂ it ( h ) . The last equality holds because h ∈ C x . By the miniality of L , one sees that ∂ x j ( a i ) = 0 for all i = 0 , . . . , ℓ − j = 1 , . . . , m . Hence a i ∈ k ( t ) forall i . In other words, L ∈ k ( t ) h ∂ t i . 12or convention, we introduce the following definition. Definition 19 (1)
We say f ∈ U is split if it can be written as the form f = gh where g ∈ C t and h ∈ C x , and say f is semisplit if it is thesum of finitely many split elements. (2) We say a nonzero operator P ∈ K h ∂ t i is semisplit if it is monic andall its coefficients are semisplit. The semisplit operators have the following property.
Lemma 20
Assume that P = Q Q where P, Q , Q are monic operatorsin K h ∂ t i . Assume further that Q ∈ k ( t )[ x , /r ] h ∂ t i where r ∈ k [ x , t ] . Then P ∈ k ( t )[ x , /r ] h ∂ t i if and only if so is Q . Proof.
Comparing the coefficients on both sides of P = Q Q concludesthe lemma.As a direct consequence, we have the following corollary. Corollary 21
Assume that P = Q Q where P, Q , Q are monic operatorsin K h ∂ t i . Assume further that Q is semisplit. Then P is semisplit if andonly if so is Q . In Proposition 10 of [11], we show that given a hyperexponential function h over K , ann( h ) ∩ k ( t ) h ∂ t i 6 = { } if and only if there is a nonzero p ∈ k ( x )[ t ]and r ∈ k ( t ) such that a = ∂ t ( p ) p + r, where a = ∂ t ( h ) /h . Remark that a, p, r with p = 0 satisfy the above equalityif and only if p ( ∂ t − a ) = ( ∂ t − r ) p . Under the notion of ( x , t )-separableand the language of differential operators, Proposition 10 of [11] states that ∂ t − a is ( x , t )-separable if and only if it is similar to a first order operatorin k ( t ) h ∂ t i by some 1 /p with p being nonzero polynomial in t . In thissection, we shall generalize Proposition 10 of [11] to the case of completelyreducible operators. We shall use lclm( Q , Q ) to denote the monic operatorof minimal order which is divisible by both Q and Q on the right. Weshall prove that if P is ( x , t )-separable and completely reducible then thereis a nonzero L ∈ k ( t ) h ∂ t i such that P is the transformation of L by some Q with semisplit coefficients. To this end, we need to introduce some notationsfrom [27]. 13 efinition 22 Assume that
P, Q ∈ K h ∂ t i \ { } .1. We say ˜ P is the transformation of P by Q if ˜ P is the monic operatorsatisfying that ˜ P Q = λ lclm( P, Q ) for some λ ∈ K .2. We say ˜ P is similar to P (by Q ) if there is an operator Q with gcrd( P, Q ) = 1 such that ˜ P is the transformation of P by Q , where gcrd( P, Q ) denotes the greatest common right-hand factor of P and Q . Definition 23
1. We say P ∈ K h ∂ t i is completely reducible if it is thelclm of a family of irreducible operators in K h ∂ t i .2. We say Q ∈ K h ∂ t i is the maximal completely reducible right-handfactor of P ∈ K h ∂ t i if Q is the lclm of all irreducible right-hand factrosof P . Given a P ∈ K h ∂ t i , Theorem 7 of [27] implies that P has the followingunique decomposition called the maximal completely reducible decomposi-tion or the m.c.r. decomposition for short, P = λH r H r − . . . H where λ ∈ K and H i is the maximal completely reducible right-hand factorof H r . . . H i . For an L ∈ k ( t ) h ∂ t i , it has two m.c.r. decompositions viewedit as an operator in k ( t ) h ∂ t i and an operator in K h ∂ t i respectively. Inthe following, we shall prove that these two decompositions coincide. Forconvenience, we shall denote by P x i = c i the operator obtained by replacing x i by c i ∈ k in P . Lemma 24
Assume that
P, L are two monic operators in K h ∂ t i . Assumefurther that P ∈ k ( t )[ x , /r ] h ∂ t i with r ∈ k [ x , t ] , and L ∈ k ( t ) h ∂ t i . Let c ∈ k m be such that r ( c ) = 0 .1. If gcrd( P x = c , L ) = 1 then gcrd( P, L ) = 1 .2. If gcrd(
P, L ) = 1 then there is a ∈ k m such that r ( a ) = 0 and gcrd( P x = a , L ) = 1 . Proof.
1. We shall prove the lemma by induction on m = | x | . Assumethat m = 1, and gcrd( P, L ) = 1. Then there are M, N ∈ k ( t )[ x ] h ∂ t i withord( M ) < ord( L ) such that M P + N L = 0 . Write M = n − X i =0 a i ∂ it , N = s X i =0 b i ∂ it n = ord( L ). If the a i ’s have a common factor c in k ( t )[ x ], then onesees that c is a common factor of the b i ’s. Thus we can cancel this factor c .So without loss of generality, we may assume that the a i ’s have no commonfactor. This implies that M x = c = 0 and M x = c P x = c + N x = c L = 0.Since ord( M x = c ) < ord( L ), gcrd( P x = c , L ) = 1, a contradiction. Forthe general case, set Q = P x = c . Then Q x = c ,...,x m = c m = P x = c . Thisimplies that gcrd( Q x = c ,...,x m = c m , L ) = 1. By the induction hypothesis,gcrd( Q, L ) = 1. Finally, regarding P and L as operators with coefficientsin k ( t, x , . . . , x m )[ x , /r ] and by the induction hypothesis again, we getgcrd( P, L ) = 1.2. Since gcrd(
P, L ) = 1, there are
M, N ∈ K h ∂ t i such that M P + N L = 1. Let a ∈ k m be such that r ( a ) = 0 and both M x = a and N x = a arewell-defined. For such a , one has that M x = a P x = a + N x = a L = 1 and thengcrd( P x = a , L ) = 1. Lemma 25
Let L ∈ k ( t ) h ∂ t i . The m.c.r. decompositions of L viewed as anoperator in k ( t ) h ∂ t i and an operator in K h ∂ t i respectively coincide. Proof.
We first claim that an irreducible operator of k ( t ) h ∂ t i is irreduciblein K h ∂ t i . Let P be a monic irreducible operator in k ( t ) h ∂ t i and assume that Q is a monic right-hand factor of P in K h ∂ t i with 1 ≤ ord( Q ) < ord( P ).Then P = ˜ QQ for some ˜ Q ∈ K h ∂ t i . Suppose that Q ∈ k ( t )[ x , /r ] h ∂ t i . ByLemma 20, ˜ Q belongs to k ( t )[ x , /r ] h ∂ t i . Let c ∈ k m be such that r ( c ) = 0.Then P = ˜ Q x = c Q x = c and 1 ≤ ord( Q x = c ) ≤ ord( P ). These imply that P is reducible, a contradiction. So P is irreducible and thus the claim holds.Let L = λH r H r − . . . H be the m.c.r. decomposition in k ( t ) h ∂ t i . The aboveclaim implies that H viewed as an operator in K h ∂ t i is completely reducible.Assume that H is not the maximal compleltely reducible right-hand factorof L in K h ∂ t i . Let M ∈ K h ∂ t i \ K be a monic irreducible right-hand factorof L satisfying that gcrd( M, H ) = 1. Due to Lemma 24, there is a ∈ k m satisfying that gcrd( M x = a , H ) = 1. Note that M x = a is a right-hand factorof L . Therefore M x = a has some irreducible right-hand factor of L as a right-hand factor. Such irreducible factor must be a right-hand factor of H andthus gcrd( M x = a , H ) = 1, a contradiction. Therefore H is the maximalcompletely reducible right-hand factor of L in K h ∂ t i . Using the inductionon the order, one sees that λH r H r − . . . H is the m.c.r. decomposition of L in K h ∂ t i . Lemma 26
Assume that P is monic, ( x , t ) -separable and completely re-ducible. Assume further that P ∈ k ( t )[ x , /r ] h ∂ t i with r ∈ k [ x , t ] . Let c ∈ k m be such that r ( c ) = 0 . Then P x = c is similar to P . roof. Let ˜ L be a nonzero monic operator in k ( t ) h ∂ t i with P as a right-hand factor. Since P is completely reducible, by Theorem 8 of [27], P is aright-hand factor of the maximal completely reducible right-hand factor of˜ L . By Lemma 25, the maximal completely reducible right-hand factor of ˜ L is in k ( t ) h ∂ t i . Hence we may assume that ˜ L is completely reducible afterreplacing ˜ L by its maximal completely reducible right-hand factor. Assumethat ˜ L = QP for some Q ∈ K h ∂ t i . By Lemma 20, Q ∈ k ( t )[ x , /r ] h ∂ t i .Then ˜ L = Q x = c P x = c , i.e. P x = c is a right-hand factor of ˜ L . We claim thatfor a right-hand factor T of ˜ L , there is a right-hand factor L of ˜ L satisfyingthat gcrd( T, L ) = 1 and lclm(
T, L ) = ˜ L . We prove this claim by inductionon s = ord( ˜ L ) − ord( T ). When s = 0, there is nothing to prove. Assumethat s >
0. Then since ˜ L is completely reducible, there is an irreducibleright-hand factor L of ˜ L such that gcrd( T, L ) = 1. Let N = lclm( T, L ).We have that ord( N ) = ord( T ) + ord( L ). Therefore ord( ˜ L ) − ord( N ) < s .By induction hypothesis, there is a right-hand factor L of ˜ L such thatgcrd( N, L ) = 1 and lclm( N, L ) = ˜ L . Let L = lclm( L , L ). Then˜ L = lclm( N, L ) = lclm( T, L , L ) = lclm( T, L ) . Taking the order of the operators in the above equality yields thatord(lclm(
T, L )) = ord(lclm(
N, L )) = ord( N ) + ord( L )= ord( T ) + ord( L ) + ord( L ) . On the other hand, we haveord(lclm(
T, L )) ≤ ord( T ) + ord( L ) ≤ ord( T ) + ord( L ) + ord( L ) . This implies that ord(lclm(
T, L )) = ord( T ) + ord( L ) . So gcrd(
T, L ) = 1 and then L is a required operator. This proves the claim.Now let L c be a ritht-hand factor of ˜ L satisfying that gcrd( P x = c , L c ) = 1and lclm( P x = c , L c ) = ˜ L . Let M ∈ k ( t ) h ∂ t i be such that ˜ L = M L c . Then P x = c is similar to M . It remains to show that P is also similar to M . Dueto Lemma 24, gcrd( P, L c ) = 1. Thenord(lclm( P, L c )) = ord( P ) + ord( L c ) = ord( P x = c ) + ord( L c ) = ord( ˜ L ) . Note that lclm(
P, L c ) is a right-hand factor of ˜ L . Hence lclm( P, L c ) = ˜ L and thus P is similar to M . 16or the general case, the above lemma is not true anymore as shown inthe following example. Example 27
Let y = x log( t + 1) + x log( t − and P = ∂ t + ( t − x + ( t + 1) x ( t − t − x + ( t + 1) x ) ∂ t . Then P is ( x, t ) -separable since { , y } is a basis of the solution space of P = 0 in U . We claim that P is not similar to P x = c for any c ∈ k \{ (0 , } . Suppose on the contrary that P is similar to P x = c for some c =( c , c ) ∈ k \ { (0 , } , i.e. there are a, b ∈ k ( x , t ) , not all zero, such that gcrd( a∂ t + b, P x = c ) = 1 and P is the transformation of P x = c by a∂ t + b .Denote Q = a∂ t + b . As { , y x = c } is a basis of the solution space of P x = c , { Q (1) , Q ( y x = c ) } is a basis of the solution space of P = 0 . In other words,there is C ∈ GL ( C t ) such that (cid:18) b, a (cid:18) c t + 1 + c t − (cid:19) + by x = c (cid:19) = (1 , y ) C. Note that log( t + 1) , log( t − , are linearly independent over k ( x , x , t ) .We have that b ∈ C t \ { } and bc = ˜ cx , bc = ˜ cx for some ˜ c ∈ C t . Thisimplies that x /x = c /c ∈ k , a contradiction. When the given two operators are of length two, i.e. they are the prod-ucts of two irreducible operators, a criterion for the similarity is presentedin [23]. For the general case, suppose that P is similar to P x = c by Q . Thenthe operator Q is a solution of the following mixed differential equation P z ≡ P x = c . (10)An algorithm for computing all solutions of the above mixed differentialequation is developed in [32]. In the following, we shall show that if P is( x , t )-separable then Q is an operator with semisplit coefficients. Note that Q can be chosen to be of order less than ord( P x = c ) and all solutions of themixed differential equation with order less than ord( P x = c ) form a vectorspace over k ( x ) of finite dimension. Furthermore Q induces an isomorphismfrom the solution space of P x = c ( y ) = 0 to that of P ( y ) = 0. Proposition 28
Assume that P is monic and completely reducible. Assumefurther that P ∈ k ( t )[ x , /r ] h ∂ t i with r ∈ k [ x , t ] . Let c ∈ k m be such that r ( c ) = 0 . Then P is ( x , t ) -separable if and only if P is similar to P x = c byan operator Q with semisplit coefficients. roof. Denote n = ord( P x = c ) = ord( P ). Assume that { α , · · · , α n } is abasis of the solution space of P x = c ( y ) = 0 in C x and P is similar to P x = c by Q . Write Q = P n − i =0 a i ∂ it where a i ∈ K . Then( Q ( α ) , . . . , Q ( α n )) = ( a , . . . , a n − ) α α . . . α n α ′ α ′ . . . α ′ n ... ... ... α ( n − α ( n − . . . α ( n − n and Q ( α ) , . . . , Q ( α n ) form a basis of the solution space of P ( y ) = 0.Now suppose that P is ( x , t )-separable. Due to Lemma 26, P is similar to P x = c by Q . By Corollary 18, the Q ( α i ) are semisplit. The above equalitiesthen imply that the a i are semisplit. Conversely, assume that P is similar to P x = c by Q and the a i are semisplit. It is easy to see the Q ( α i ) are semisplit.By Corollary 18 again, P is ( x , t )-separable.Using the algorithm developed in [32], we can compute a basis of thesolution space over k ( x ) of the equation (10). It is clear that the solutionswith semisplit entries form a subspace. We can compute a basis for this sub-space as follows. Suppose that { Q , . . . , Q ℓ } is a basis of the solution spaceof the equation (10) consisting of solutions with order less than ord( P x = c ).We may identity Q i with a vector g i ∈ K n under the basis 1 , ∂ t , . . . , ∂ n − t .Let q ∈ k ( x )[ t ] be a common denominator of all entries of the g i . Write g i = p i /q for each i = 1 , . . . , ℓ , where p i ∈ k ( x )[ t ] n . Write q = q q where q is split but q is not. Note that a rational function in t with coefficients in k ( x ) is semisplit if and only if its denominator is split. For c , . . . , c ℓ ∈ k ( x ), P ℓi =1 c i g i is semisplit if and only if all entries of P ℓi =1 c i p i are divided by q . For i = 1 , . . . , ℓ , let h i be the vector whose entries are the remaindersof the corresponding entries of p i by q . Then all entries of P ℓi =1 c i p i aredivided by q if and oly if P ℓi =1 c i h i = 0. Let c , . . . , c s be a basis of thesolution space of P ℓi =1 z i h i = 0. Then { ( Q , . . . , Q ℓ ) c i | i = 1 , . . . , s } is therequired basis. Consequently, the required basis can be computed by solvingthe system of linear equations P ℓi =1 z i h i = 0.In the following, for the sake of notations, we assume that { Q , . . . , Q ℓ } isa basis of the solution space of the equation (10) consisting of solutions withsemisplit coefficients. By Proposition 28 and the definition of similarity, P is ( x , t )-separable if and only if there is a nonzero ˜ Q in the space spanned by Q , . . . , Q ℓ such that gcrd( P x = c , ˜ Q ) = 1. Note that ˜ Q induces a homomor-phism from the solutions space of P x = c ( y ) = 0 to that of P ( y ) = 0. More-over, one can easily see that gcrd( P x = c , ˜ Q ) = 1 if and only if ˜ Q is an isomor-phism i.e. ˜ Q ( α ) , . . . , ˜ Q ( α n ) form a basis of the solution space of P ( y ) = 018here { α , . . . , α n } is a basis of the solution space of P x = c ( y ) = 0. Assumethat ˜ Q = P n − i =0 a ,i ∂ it with a ,i ∈ K . Using the relation P x = c ( α j ) = 0 with j = 1 , . . . , n , one has that for all j = 1 , . . . , n ˜ Q ( α j ) ′ = n − X i =0 a ,i α ( i ) j ! ′ = n − X i =0 a ,i α ( i ) j for some a ,i ∈ K . Repeating this process, we can compute a l,i ∈ K suchthat for all j = 1 , . . . , n and l = 1 , . . . , n − Q ( α j ) ( l ) = n − X i =0 a l,i α ( i ) j . Now suppose that ˜ Q = P ℓi =1 z i Q i with z i ∈ k ( x ). One sees that the a l,i arelinear in z , . . . , z ℓ . Set A ( z ) = ( a i,j ) ≤ i,j ≤ n − with z = ( z , . . . , z ℓ ). Thenone has that A ( z ) α . . . α n ... ... α ( n − . . . α ( n − n = ˜ Q ( α ) . . . ˜ Q ( α n )... ...˜ Q ( α ) ( n − . . . ˜ Q ( α n ) ( n − . (11)It is well-known that ˜ Q ( α ) , . . . , ˜ Q ( α n ) form a basis if and only if the right-hand side of the above equality is a nonsingular matrix and thus if and onlyif A ( z ) is nonsingular. In the sequel, one can reduce the problem of theexistence of ˜ Q satisfying gcrd( ˜ Q, P x = c ) = 1 to the problem of the existenceof a ∈ k ( x ) ℓ in k ( x ) such that det( A ( a )) = 0.Suppose now we have already had an operator Q with semisplit coeffi-cients such that P is similar to P x = c by Q . Write Q = P n − i =0 b i ∂ it where b i ∈ K is semisplit. Write further b i = P sj =1 h i,j β j where h i,j ∈ k ( x ) and β j ∈ k ( t ) \ { } . Let L = P x = c and let L i be the transformation of L i − by ∂ t for i = 1 , · · · , n −
1. Then L i annihilates α ( i ) j for all j = 1 , · · · , n and L i β l annihilates β l α ( i ) j for all l = 1 , . . . , s and j = 1 , . . . , n . Set L = lclm (cid:18)(cid:26) L i β l | i = 0 , . . . , n − , l = 1 , . . . , s (cid:27)(cid:19) . Then L annihilates all ˜ Q ( α i ) and thus has P as a right-hand factor. Wesummarize the previous discussion as the following algorithm. Algorithm 29
Input: P ∈ K h ∂ t i that is monic and completely reducible.Output: a nonzero L ∈ k ( t ) h ∂ t i which is divided by P on the right if it exists,otherwise 0. . Write P = ∂ nt + n − X i =0 a i r ∂ it where a i ∈ k ( t )[ x ] , r ∈ k [ x , t ] .2. Pick c ∈ k m such that r ( c ) = 0 . By the algorithm in [32], compute abasis of the solution space V of the equation (10).3. Compute a basis of the subspace of V consisting of operators withsemisplit coefficients, say Q , · · · , Q ℓ .4. Set ˜ Q = P ℓi =1 z i Q i and using ˜ Q , compute the matrix A ( z ) as in (11).5. If det( A ( z )) = 0 then return 0 and the algorithm terminates. Other-wise compute a = ( a , . . . , a ℓ ) ∈ k ℓ such that det( A ( a )) = 0 .6. Set b i to be the coefficient of ∂ it in P ℓj =1 a j Q j and write b i = P sj =1 h i,j β j where h i,j ∈ k ( x ) and β j ∈ k ( t ) . Let L = P x = c and for each i = 1 , · · · , n − compute L i , the transformation of L i − by ∂ t .7. Return lclm (cid:16)n L i β j | i = 0 , . . . , n − , j = 1 , . . . , s o(cid:17) . Assume that P is ( x , t )-separable and P = Q Q where Q , Q ∈ K h ∂ t i .It is clear that Q is also ( x , t )-separable. One may wonder whether Q is also ( x , t )-separable. The following example shows that Q may not be( x , t )-separable. Example 30
Let K = k ( x, t ) and let P = ∂ t . Then P is ( x , t ) -separableand ∂ t = (cid:18) ∂ t + xxt + 1 (cid:19) (cid:18) ∂ t − xxt + 1 (cid:19) . The operator ∂ t + x/ ( xt + 1) is not ( x , t ) -separable, because / ( xt + 1) is oneof its solutions and it is not semisplit. While, the lemma below shows that if Q is semisplit then Q is also ( x , t )-separable. Lemma 31 (1)
Assume that Q , Q ∈ K h ∂ t i \ { } , and Q is semisplit.Then Q Q is ( x , t ) -separable if and only if both Q and Q are ( x , t ) -separable. Assume that P ∈ K h ∂ t i \ { } and L is a nonzero monic operator in k ( t ) h ∂ t i . Then P is ( x , t ) -separable if and only if so is the transfor-mation of P by L . Proof.
Note that the solution space of lclm( P , P ) = 0 is spanned bythose of P = 0 and P = 0. Hence lclm( P , P ) is ( x , t )-separable if andonly if so are both P and P .(1) For the “only if” part, one only need to prove that Q is ( x , t )-separable. Assume that g is a solution of Q = 0 in U . Let f be a solutionof Q ( y ) = g in U . Such f exists because U is the universal differentialextension of K . Then f is a solution of Q Q = 0 in U . By Corollary 18, f is semisplit. Since Q is semisplit, one sees that g = Q ( f ) is semisplit. ByCorollary 18 again, Q is ( x , t )-separable.Now assume that both Q and Q are ( x , t )-separable. Let ˜ Q ∈ K h ∂ t i be such that ˜ QQ = L where L ∈ k ( t ) h ∂ t i is monic. By Corollary 21 andthe “only if” part, ˜ Q is semisplit and ( x , t )-separable. Thus lclm( Q , ˜ Q ) is( x , t )-separable. Assume that lclm( Q , ˜ Q ) = N ˜ Q with N ∈ K h ∂ t i . Since ˜ Q is semisplit, by the “only if” part again, N is ( x , t )-separable. Let M ∈ K h ∂ t i be such that M N is a nonzero operator in k ( t ) h ∂ t i . We have that M lclm( Q , ˜ Q ) Q = M N ˜ QQ = M N L ∈ k ( t ) h ∂ t i . On the other hand, M lclm( Q , ˜ Q ) Q = M ˜ M Q Q for some ˜ M ∈ K h ∂ t i .Hence P = Q Q is ( x , t )-separable.(2) Since L is ( x , t )-separable, we have that P is ( x , t )-separable if andonly if so is lclm( P, L ). Let ˜ P be the transformation of P by L . Then˜ P L = lclm(
P, L ). As L is semisplit, the assertion then follows from (1).Assume that P is a nonzero operator in K h ∂ t i . Let P be an irreducibleright-hand factor of P . By Algorithm 29, we can decide whether P is ( x , t )-separable or not. Now assume that P is ( x , t )-separable. Then we cancompute a nonzero monic operator L ∈ k ( t ) h ∂ t i having P as a right-handfactor. Let P be the transformation of P by L . Lemma 31 implies that P is ( x , t )-separable if and only if so is P . Note thatord( P ) = ord(lclm( P, L )) − ord( L ) ≤ ord( P ) + ord( L ) − ord( P ) − ord( L ) = ord( P ) − ord( P ) . In other words, ord( P ) < ord( P ). Replacing P by P and repeating theabove process yield an algorithm to decide whether P is ( x , t )-separable.21 lgorithm 32 Input: a nonzeor monic P ∈ K h ∂ t i .Output: a nonzero L ∈ k ( t ) h ∂ t i which is divided by P on the right if it exists,otherwise 0.1. If P = 1 then return 1 and the algorithm terminates.2. Compute an irreducible right-hand factor P of P by algorithms devel-oped in [4, 31, 33].3. Apply Algorithm 29 to P and let L be the output.4. If L = 0 then return 0 and the algorithm terminates. Otherwisecompute the transformation of P by L , denoted by P .5. Apply Algorithm 32 to P and let L be the output.6. Return L L . The termination of the algorithm is obvious. Assume that L = 0. Then L = Q P for some Q ∈ K h ∂ t i . We have that P L = lclm( P, L ). There-fore L L = Q P L = Q lclm( P, L ) = Q Q P for some Q ∈ K h ∂ t i . This proves the correctness of the algorithm. References [1] S. A. Abramov. When does Zeilberger’s algorithm succeed? Adv. inAppl. Math., 30(3):424–441, 2003.[2] S. A. Abramov and H. Q. Le. A criterion for the applicability of Zeil-berger’s algorithm to rational functions. Discrete Math., 259(1-3):1–17,2002.[3] G. Almkvist and D. Zeilberger. The method of differentiating underthe integral sign. J. Symbolic Comput., 10(6):571–591, 1990.[4] E. Beke. 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