Separability Problems in Creative Telescoping
aa r X i v : . [ c s . S C ] F e b Separability Problems in Creative Telescoping ∗ Shaoshi Chen a,b , Ruyong Feng a,b ,Pingchuan Ma a,b , and Michael F. Singer ca KLMM, Academy of Mathematics and Systems Science,Chinese Academy of Sciences,Beijing, 100190, China b School of Mathematical Sciences,University of Chinese Academy of Sciences,Beijing 100049, (China) c Department of Mathematics,North Carolina State University,Raleigh, NC 27695, (USA) [email protected], [email protected]@mails.ucas.ac.cn, [email protected]
February 9, 2021
Abstract
For given multivariate functions specified by algebraic, differential ordifference equations, the separability problem is to decide whether theysatisfy linear differential or difference equations in one variable. In thispaper, we will explain how separability problems arise naturally in creativetelescoping and present some criteria for testing the separability for severalclasses of special functions, including rational functions, hyperexponentialfunctions, hypergeometric terms, and algebraic functions. ∗ S. Chen was partially supported by the NSFC grants 11871067, 11688101, the Fundof the Youth Innovation Promotion Association, CAS, and the National Key Research andDevelopment Project 2020YFA0712300. R. Feng was partially supported by the NSFC grants11771433, 11688101, and Beijing Natural Science Foundation under Grant Z190004. P. Mawas partial supported by the NSFC grants 11871067. M.F. Singer was partially supported byby a grant from the Simons Foundation (No. 349357, Michael Singer) Introduction
The method of separation of variables has been used widely in solving differentialequations [21]. In order to solve the one-dimensional heat equation ∂y∂t − c ∂ y∂x = 0 , where c ∈ C , together with the boundary conditions y ( t,
0) = y ( t, L ) = 0. One can try to finda nonzero solution of the form y = u ( t ) v ( x ) , and then substitute this form into the equation to get ∂u ( t ) ∂t u = c ∂ v ( x ) ∂x v . Since both sides only depend on one variable, there exits some constant λ ∈ C such that ∂u∂t − λu = 0 and c ∂ v∂x − λv = 0 . Note that the above two equations are also satisfied by y = u ( t ) v ( x ), whichare linear differential equation in only one variable. After solving these specialequations with the boundary conditions into account, a special solution of theheat equation can be given as y ( t, x ) = ∞ X n =1 d n sin (cid:16) nπxL (cid:17) exp (cid:18) − n π ctL (cid:19) , (1.1)where d n ∈ C are coefficients determined by the initial conditions. Motivatedby this example, one would ask the following natural question. Problem 1.1 (Separability Problem) . Given a multivariate function specifiedby certain equations (e.g. algebraic, differential or difference equations), decidewhether this function satisfies linear differential or difference equations in onevariable.
To make the problem more tractable, we will consider some special classesof functions, such as rational functions, algebraic functions, hyperexponentialfunctions and hypergeometric terms etc.. The main goal of this paper is to showthe close connection between the separability problem and Zeilberger’s methodof creative telescoping [27, 28].The remainder of this paper is organized as follows. We specify the sepa-rability problem and the existence problem of telescopers precisely in Section 2together with the definition of orders and (local) dispersions of rational func-tions. After this, we explain how the separability problems arise naturally increative telescoping for rational functions in Section 3, hyperexponential func-tions and hypergeometric terms in Section 4, and for algebraic functions inSection 5. Separability criteria will be given for these classes of special func-tions. We then conclude our paper with some comments on the separabilityproblem on D-finite functions and P-recursive sequences.2
Preliminaries
Let F be a field of characteristic zero and let E = F ( t, x ) be the field of rationalfunctions in t and x = ( x , . . . , x m ) over F . Let δ t , δ x i be the usual partialderivations ∂/∂ t , ∂/∂ x i with x i ∈ { x , . . . , x m } , respectively. The shift oper-ators σ t and σ x i on E are defined as the F -automorphisms such that for any f ∈ E , σ t ( f ( t, x )) = f ( t + 1 , x ) and σ x i ( f ( t, x )) = f ( t, x , . . . , x i − , x i + 1 , x i +1 , . . . , x m ) . The ring of linear functional operators in t and x over E is denoted by E h ∂ t , ∂ x i ,where ∂ x = ( ∂ x , . . . , ∂ x m ) and ∂ v with v ∈ { t, x } is either the derivation D v such that D v f = f D v + δ v ( f ) or the shift operator S v such that S v f = σ v ( f ) S v for any f ∈ E , and ∂ t and ∂ x i commute. For v ∈ { t, x } , we let ∆ v denote thedifference operator S v − , where stands for the identity map on E . Abusingnotation, we let δ v and σ v denote arbitrary extensions of δ v and σ v to derivationand F -automorphism of E , the algebraic closure of E . The functions we con-sider will be in certain differential or difference extension of E , which is also an E h ∂ t , ∂ x i -module via the action defined by simply interpreting D v , S v by δ v , σ v ,respectively, for v ∈ { t, x } . The ring F ( t ) h ∂ t i is a subring of E h ∂ t , ∂ x i that is alsoa left Euclidean domain. Efficient algorithms for basic operations in F ( t ) h ∂ t i ,such as computing the least common left multiple (LCLM) of operators, havebeen developed in [10, 6]. Definition 2.1 (Separable functions) . Let M be an E h ∂ t , ∂ x i -module and f ∈ M . We say that f ( t, x ) is ∂ t -separable if there exists a nonzero L ∈ F ( t ) h ∂ t i such that L ( f ) = 0 . As an example, the special solution (1.1) of the one-dimensional heat equa-tion is both D t -separable and D x -separable. Note that ∂ t -separable functionsare just the D-finite functions in the differential case and the P-recursive se-quences in the shift case, which are both introduced in [26]. By the closureproperties of D-finite functions and P-recursive sequences, we have the sameclosure properties for ∂ t -separable functions. Proposition 2.2.
Let M be an E h ∂ t , ∂ x i -module. If f, g ∈ M are ∂ t -separable,so are f + g, f · g , and a · f for all a ∈ F ( t ) . We will focus on the separability problem on function in an E h ∂ t , ∂ x i -module. Definition 2.3 (Creative telescoping) . Let M be an E h ∂ t , ∂ x i -module and f ∈ M . A nonzero operator L ∈ F ( t ) h ∂ t i is called a telescoper of type ( ∂ t , ∂ x ) for f if there exist Q , . . . , Q m ∈ E h ∂ t , ∂ x i such that L ( t, ∂ t )( f ) = ∂ x ( Q ( f )) + · · · + ∂ x m ( Q m ( f )) , (2.1) where ∂ t ∈ { D t , S t } and ∂ x i ∈ { D x i , ∆ x i } . V = ( V , . . . , V s ) be any set partition of the variables v = { t, x , . . . , x m } .A rational function f ∈ F ( t, x ) is said to be split with respect to the partition V if f = f · · · f s with f i ∈ F ( V i ) and be semi-split with respect to V if thereare split functions g j ∈ F ( t, x ) such that f = g + · · · + g n . By definition, wehave f = p/q with p, q ∈ F [ t, x ] and gcd( p, q ) = 1 is semi-split with respectto the partition V if and only if the denominator q is a split polynomial withrespect to the partition V . Split rational functions will be used to describe theseparability of given functions.Let K = F ( x ) and p ∈ K [ t ] be an irreducible polynomial in t . For any f ∈ K ( t ), we can write f = p m a/b , where m ∈ Z , a, b ∈ K [ t ] with gcd( a, b ) = 1and p ∤ ab . Conventionally, we set ν p (0) = + ∞ . The integer m is called the order of f at p , denoted by ν p ( f ). We collect some basic properties of valuationsas follows and refer to [9, Chapter 4] for their proofs. Proposition 2.4.
Let f, g ∈ K ( t ) and p ∈ K [ t ] be an irreducible polynomial.Then, ( i ) ν p ( f g ) = ν p ( f ) + ν p ( g ) . ( ii ) ν p ( f + g ) ≥ min { ν p ( f ) , ν p ( g ) } and equality holds if ν p ( f ) = ν p ( g ) . ( iii ) If ν p ( f ) = 0 , then ν p ( D t ( f )) = ν p ( f ) − . In particular, for any i ∈ N , ν p ( D it ( f )) = ν p ( f ) − i if ν p ( f ) < . The dispersion introduced by Abramov in [1] can be viewed as a shift ana-logue of the order. For any polynomial u ∈ K [ t ] with deg t ( u ) ≥
1, the dispersion of u , denoted by dis( u ), is defined as max { k ∈ N | gcd( u, σ kt ( u )) = 1 } , whichis the maximal integer root-distance | α − β | with α, β being roots of u in ¯ K .Define dis( u ) = 0 if u ∈ K \ { } and dis(0) = + ∞ . For a rational function f = a/b ∈ K ( t ) with a, b ∈ K [ t ] and gcd( a, b ) = 1, define dis( f ) = dis( b ). Forlater use, we introduce a local version of Abramov’s dispersion. Let p ∈ K [ t ] bean irreducible polynomial. If σ it ( p ) | u for some i ∈ Z , the local dispersion of u at p , denoted by dis p ( u ), is defined as the maximal integer distance | i − j | with i, j ∈ Z satisfying σ it ( p ) | u and σ jt ( p ) | u ; otherwise we define dis p ( u ) = 0. Con-ventionally, we set dis p (0) = + ∞ . For a rational function f = a/b ∈ K ( t ) with a, b ∈ K [ t ] and gcd( a, b ) = 1, we also define dis p ( f ) = dis p ( b ). By definition, wehave dis( u ) = max { dis p ( u ) | p is an irreducible factor of u } . The set { σ it ( p ) | i ∈ Z } is called the σ t -orbit at p , denoted by [ p ] σ t . Note thatdis p ( u ) = dis q ( u ) if q ∈ [ p ] σ t . So we can define the local dispersion of a rationalfunction f at a σ t -orbit at p , denoted by dis [ p ] σt ( f ).4 xample 2.5. Let u = x ( x + 1)( x − x + 1)( x + 4 x + 5) ∈ Q [ x ] . Then wehave dis x ( u ) = 6 and dis x +1 ( u ) = 2 . Abramov’s dispersion of u is then equalto . We now shows how the local dispersions change under the action of linearrecurrence operators, which was first proved for Abramov’s dispersions in [1, 2]and [23, Section 3.1].
Lemma 2.6.
Let f = a/b ∈ K ( t ) with a, b ∈ K [ t ] and gcd( a, b ) = 1 and let p ∈ K [ t ] be an irreducible factor of b . Let L = P ρi =0 ℓ i S it ∈ K [ t ] h S t i be suchthat ℓ ρ ℓ = 0 and σ it ( p ) does not divide ℓ ρ ℓ for any i ∈ Z . Then dis p ( L ( f )) = dis p ( f ) + ρ . In particular, dis p (∆ t ( f )) = dis p ( f ) + 1 .Proof. Let d = dis p ( b ). Without loss of generality, we may assume that p | b but σ it ( p ) ∤ b for any i <
0. Since gcd( a, b ) = 1 and σ t is a K -automorphism of K [ t ], we have gcd( σ it ( a ) , σ it ( b )) = 1 for any i ∈ Z . Applying L to f yields L ( f ) = ρ X i =0 ℓ i σ it (cid:16) ab (cid:17) = P ρi =0 ℓ i σ it ( a ) u i u , where u = bσ t ( b ) · · · σ ρt ( b ) and u i = u/σ it ( b ). Write L ( f ) = A/B with
A, B ∈ K [ t ] and gcd( A, B ) = 1. Then B | u and dis p ( L ( f )) = dis p ( B ) by definition.Since σ it ( p ) ∤ ℓ and σ it ( p ) ∤ ℓ ρ for any i ∈ Z , we have both p and σ d + ρt ( p ) do notdivide the sum P ρi =0 ℓ i σ it ( a ) u i , but they divide u . So p | B and σ d + ρt ( p ) | B ,which implies that dis p ( B ) ≥ d + ρ . Since B | u , we have dis p ( B ) ≤ dis p ( u ) = d + ρ . Therefore, dis p ( L ( f )) = d + ρ . We first explain how the existence problem of telescopers for rational functionsis naturally connected to the separability problem on this class of functions. Let f ( t, x ) be a bivariate rational function in F ( t, x ). By the Ostrogradsky-Hermitereduction [22, 18], we can decompose f into the form f = D x ( g ) + ab , where g ∈ F ( t, x ) and a, b ∈ F ( t )[ x ] with gcd( a, b ) = 1, deg x ( a ) < deg x ( b ) and b being squarefree in x over F ( t ). Moreover, f = D x ( h ) for some h ∈ F ( t, x ) ifand only if a = 0. Then f has a telescoper of type ( S t , D x ) if and only if a/b does. Applying a nonzero operator L = P ρi =0 ℓ i S it ∈ F ( t ) h S t i to a/b yields L (cid:16) ab (cid:17) = ρ X i =0 ℓ i ( t ) σ it (cid:16) ab (cid:17) = ρ X i =0 ℓ i ( t ) a ( t + i, x ) b ( t + i, x ) = pq , where p, q ∈ F [ t, x ] with gcd( p, q ) = 1. Since the shift operator S t is an F ( x )-automorphism and preserves the degrees in t and x , we have b ( t + i, x ) is square-free in x over F ( t ) for any i ∈ N and deg x ( a ( t + i, x )) < deg x ( b ( t + i, x )). So5eg x ( p ) < deg x ( q ) and q is also squarefree in x over F ( t ). This implies theoperator L is a telescoper of type ( S t , D x ) for a/b , i.e., L ( a/b ) = D x ( g ) for some g ∈ F ( t, x ) if and only if p = 0, i.e., L ( a/b ) = 0. Therefore, we conclude that f has a telescoper of type ( S t , D x ) if and only a/b is S t -separable.We can also consider telescopers of type ( D t , S x ). By Abramov’s reduc-tion [3, 4], we can decompose f ∈ F ( t, x ) into the form f = ∆ x ( g ) + ab , where g ∈ F ( t, x ) and a, b ∈ F ( t )[ x ] with gcd( a, b ) = 1, deg x ( a ) < deg x ( b ) and b being shift-free in x over F ( t ), i.e., gcd( b, σ ix ( b )) = 1 for all nonzero i ∈ Z .Applying a nonzero operator L = P ρi =0 ℓ i ( t ) D it ∈ F ( t ) h D t i to a/b yields L (cid:16) ab (cid:17) = ρ X i =0 ℓ i δ it (cid:16) ab (cid:17) = ρ X i =0 ℓ i ( t ) a i b i +1 = pq , where a i , p, q ∈ F [ t, x ] with deg x ( a i ) < ( i + 1) deg x ( b ) and gcd( p, q ) = 1. Since b is shift-free in x , so is b i for any i ∈ N . Note that any factor of a shift-freepolynomial is still shift-free. So q is shift-free and deg x ( p ) < deg x ( q ). Thisimplies the operator L is a telescoper of type ( D t , S x ) for a/b , i.e., L ( a/b ) =∆ x ( g ) for some g ∈ F ( t, x ) if and only if p = 0, i.e., L ( a/b ) = 0. Then we alsohave that f has a telescoper of type ( D t , S x ) if and only a/b is D t -separable.The next theorem characterizes all possible separable rational functions interms of semi-split rational functions. Theorem 3.1.
A rational function f ∈ F ( t, x ) is ∂ t -separable if and only if f is semi-split in t and x .Proof. Assume that f is semi-split in t and x . Then f = a b + · · · + a n b n , where a i ∈ F ( t ) and b i ∈ F ( x ) for all i with 1 ≤ i ≤ n . Since each a i b i is annihilatedby the operator L i := ∂ t − ∂ t ( a i ) /a i ∈ F ( t ) h ∂ t i , the rational function f isannihilated by LCLM( L , . . . , L n ). So f is ∂ t -separable.For the necessity we assume that f = a/b with a, b ∈ F [ t, x ] and gcd( a, b ) = 1is ∂ t -separable, i.e., there exists a nonzero operator L = P ρi =0 ℓ i ∂ it ∈ F ( t ) h ∂ t i with ℓ ρ = 0 such that L ( f ) = 0. It suffices to show that the denominator b issplit with respect to t and x . Suppose for the sake of contradiction that b isnot split. Then b has at least one irreducible factor p such that p is not split.Now we proceed by a case distinction according to the type of ∂ t . In the casewhen ∂ t = D t , we have ν p ( ℓ i D it ( f )) = ν p ( f ) − i for each i with ℓ i = 0, since ν p ( f ) < ν p ( ℓ i ) = 0, which implies further that ν p ( L ( f )) = ν p ( f ) − ρ byProposition 2.4. But ν p ( L ( f )) = ν p (0) = + ∞ , which leads to an contradiction.In the case when ∂ t = S t , we may always assume that ℓ i ∈ F [ t ] and ℓ = 0 since σ t is an F ( x )-automorphism of F ( t, x ). Since ℓ and ℓ ρ are free of x , we have σ it ( p ) ∤ ℓ ℓ ρ for any i ∈ Z . By Lemma 2.6, we get dis p ( L ( f )) = dis p ( f ) + ρ < ∞ ,which contradicts with dis p ( L ( f )) = dis p (0) = + ∞ . Remark 3.2.
With the above theorem, we can detect easily the ∂ t -separability ofrational functions by the computation of contents and derivatives of multivariatepolynomials in t . The Hyperexponential and HypergeometricCases
The separability problem on hyperexponential functions and hypergeometricterms was first studied in [19], which was later connected to the existence ofparallel telescopers for hyperexponential functions [13]. We motivate this prob-lem by revisiting Zeilberger’s algorithm which computes telescopers for hyper-geometric terms (see [24, Chapter 6]).Let H ( t, x ) be a nonzero hypergeometric term over the rational-functionfield F ( t, x ), i.e., both σ t ( H ) /H and σ x ( H ) /H are in F ( t, x ). If telescopersof type ( S t , S x ) exist for H , Zeilberger’s algorithm starts from an ansatz: forfixed ρ ∈ N , set L = P ρi =0 ℓ i S it ∈ F ( t ) h S t i with the ℓ i ’s being undeterminedcoefficients. Applying L to H yields T := L ( H ) = ρ X i =0 ℓ i σ it ( H ) = ρ X i =0 ℓ i a i H = P ρi =0 ℓ i P i Q H, where a i = σ it ( H ) /H = P i /Q ∈ F ( t, x ) with P i , Q ∈ F [ t, x ]. The second step ofZeilberger’s algorithm is computing the Gosper form of L ( H ) that gives σ x ( L ( H )) L ( H ) = σ x ( P ρi =0 ℓ i P i ) P ρi =0 ℓ i P i σ x ( p ) p qr , where ( p, q, r ) ∈ F ( t )[ x ] is a Gosper form of the rational function Qσ x ( H )( σ x ( Q ) H )satisfying that gcd( q, σ ix ( r )) = 1 for all i ∈ Z . The last step is finding ℓ , . . . , ℓ ρ ∈ F ( t ), not all zero, such that the equation ρ X i =0 ℓ i P i ! p = qσ x ( z ) − σ − x ( r ) z. has a polynomial solution in F ( t )[ x ]. If so, then L = P ρi =0 ℓ i S it is a telescoper for H . It may happen that the final choice of the ℓ i ’s satisfies that P ρi =0 ℓ i P i = 0.This means division by zero may happen in the second step. To avoid this, weshould first detect whether L ( H ) = 0 for some L ∈ F ( t ) h S t i , i.e., the separabilityproblem on hypergeometric terms.The following theorem characterizes all possible separable hyperexponentialfunctions and hypergeometric terms, whose proof was given in [19, Lemma 4]or in [13, Proposition 10]. Theorem 4.1.
Let M be an E h ∂ t , ∂ x i -module and let H ∈ M be such that ∂ t ( H ) = aH and ∂ x i ( H ) = b i H with a, b i ∈ F ( t, x ) . Then we have, i ) Hyperexponential case: H is D t -separable if and only if there exist p ∈ F ( x )[ t ] and r ∈ F ( t ) such that a = δ t ( p ) p + r. ( ii ) Hypergeometric case: H is S t -separable if and only if there exist p ∈ F ( x )[ t ] and r ∈ F ( t ) such that a = σ t ( p ) p · r. Remark 4.2.
The above form for ∂ t ( H ) /H can be detected by algorithms forcomputing the Gosper form and its differential analogue in [17, 7]. In this section, we solve the separability problem on algebraic functions. Wefirst explain the connection between this problem and the following existenceproblem of telescopers for rational functions in three variables.
Problem 5.1.
Given f ∈ F ( t, x, y ) , decide whether there exists a nonzero op-erator L ∈ F ( t ) h D t i such that L ( f ) = ∆ x ( g ) + D y ( h ) for some g, h ∈ F ( t, x, y ) . By applying the Ostrogradsky-Hermite reduction in y and Abramov’s reduc-tion in x to f ∈ F ( t, x, y ), we get f = ∆ x ( u ) + D y ( v ) + r with r = I X i =1 α i y − β i where u, v, r ∈ F ( t, x, y ), α i , β i ∈ F ( t, x ) and β i ’s are in distinct σ x -orbits. Then f has a telescoper of type ( D t , S x , D y ) if and only if r does. By Theorem 4.21in [12] or Theorem 4.43 in [11], we have r has a telescoper of type ( D t , S x , D y )if and only if for each i with 1 ≤ i ≤ I , either α i is D t -separable in F ( t, x ) or β i ∈ F ( t ) and α i ∈ F ( t, x )( β i ) has a telescoper of type ( D t , S x ). The existenceproblem of telescopers of type ( D t , S x ) in F ( t, x )( β ) with β ∈ F ( t ) has beensolved in [15]. To completely solve Problem 5.1, it remains to solve the followingseparability problem. Problem 5.2.
Given an algebraic function f ( t, x ) over F ( t, x ) , decide whether f ( t, x ) is D t -separable. We assume that F is an algebraically closed and computable subfield of C inthe remaining part of this section. 8 .1 A descent theorem We first recall some basic notions and results from the theory of algebraic func-tions of one variable [16]. Let k be a field of characteristic zero and k ( x, y ) bean algebraic function field of one variable over k , i.e., the transcendence degreeof k ( x, y ) over k is one. This means there exists a polynomial f ∈ k [ X, Y ]such that f ( x, y ) = 0. The field of constants of k ( x, y ) is defined as the set ofelements of k ( x, y ) which are algebraic over k . A subring R of k ( x, y ) is calleda valuation ring if k ⊂ R $ k ( x, y ) and for any x ∈ k ( x, y ), either x ∈ R or x − ∈ R . Any valuation ring R of k ( x, y ) is a local ring, whose unique maximalideal p is called a place of k ( x, y ) and the quotient field R/ p is called the residuefield of the place p , denoted by Σ p . Lemma 5.3.
Let k ( x, y ) and f ∈ k [ X, Y ] be as above. Assume that (¯ x, ¯ y ) ∈ k satisfies that f (¯ x, ¯ y ) = 0 and ∂f∂Y (¯ x, ¯ y ) = 0 . Then there is a unique place p of k ( x, y ) containing x − ¯ x and y − ¯ y . Furthermore, the residue field Σ p of p isisomorphic to k and k is the field of constants of k ( x, y ) .Proof. By Corollary 2 of [16, page 8], there is a place of k ( x, y ) containing x − ¯ x and y − ¯ y , say p . Let a be the discrete valuation ring (DVR) with respect to p . It is easy to see that the ring k [ x, y ] is contained in a . Let m be the ideal in k [ x, y ] generated by x − ¯ x and y − ¯ y . Then m is a maximal ideal. Denote by R the localization of k [ x, y ] at m and we still use m to denote the unique maximalideal of R . Rewriting f ( X, Y ) as a polynomial in X − ¯ x, Y − ¯ y yields that (cid:18) ∂f∂Y (¯ x, ¯ y ) + ( Y − ¯ y ) A (cid:19) ( Y − ¯ y ) + ( X − ¯ x ) B for some A, B ∈ k [ X − ¯ x, Y − ¯ y ]. Since ∂f∂Y (¯ x, ¯ y ) = 0, one has that ∂f∂Y (¯ x, ¯ y ) +( y − ¯ y ) A ( x − ¯ x, y − ¯ y ) is invertible in R and so y − ¯ y ∈ ( x − ¯ x ) R . It impliesthat R is a regular local ring, i.e., a DVR. Therefore R = a , since R ⊂ a . Thisconcludes that p is unique.We have that Σ p = R/ m = k [ x, y ] / m ∼ = k . Since the field of constants of k ( x, y ) is a subfield of Σ p under the natural homomorphism, it coincides with k . Remark 5.4.
Let k ( x, y ) and (¯ x, ¯ y ) be as in Lemma 5.3. The above proofimplies that k ( x, y ) can be embedded into the field of formal Laurent series k (( x − ¯ x )) . Theorem 5.5.
Let F ⊆ k ⊆ C be fields with F being algebraically closed. Let f ( t, Y ) be an irreducible polynomial in k [ t, Y ] . Let k ( t, y ) be the quotient fieldof k [ t, Y ] / h f i . Assume that1. the places of k ( t ) that ramify in k ( t, y ) are defined over F , i.e., their uni-formizing parameters can be chosen to be /t or t − c with c ∈ F . . there exists a solution ( a, α ) of the system f ( a, α ) = 0 ,∂f∂Y ( a, α ) = 0 , where a ∈ F and α ∈ k .Then there exists β ∈ F ( t ) such that k ( t, y ) = k ( t, β ) .Proof. Since ( a, α ) is a simple point of f ( t, Y ) = 0 in k , by [25], f ( t, Y ) isabsolutely irreducible over k . This implies that f is irreducible over C , i.e., C [ t, Y ] / h f i is an integral domain. Let C ( t, y ) be the quotient field of C [ t, Y ] / h f i .Then k ( t, y ) can be considered as a subfield of C ( t, y ) under the natural homo-morphism. From Theorem 3 in [16, page 92], none of places of C ( t, y ) is ramifiedwith respect to k ( t, y ). Therefore the condition 1 holds for C ( t, y ). Proposition2.1 in [20, page 10] states that there is β ∈ F ( t ) such that C ( t, y ) = C ( t, β ).Now there are g ( t ) , · · · , g n − ( t ) ∈ C ( t ) such that β = n − X i =0 g i ( t ) y i , (5.1)where n = [ C ( t, y ) : C ( t )]. For each i , let g i = q i /q with q i , q ∈ C [ t ] and let s = max i { deg t q i , deg t q } . Equation (5.1) implies that qβ = P n − i =0 q i y i andtherefore the set (cid:8) t j β, t j y i (cid:9) j =0 ,...s,i =0 ,...n − is linearly dependent over C . This set lies in k ( t, y, β ) and, since it is linearlydependent over D t -constants in a larger differential field, it is linearly dependentover D t -constants in k ( t, y, β ). Denote by ˜ k the set of D t -constants of k ( t, y, β ).If ˜ k = k , then β ∈ k ( t, y ), which will conclude the proposition. Therefore itsuffices to prove that ˜ k = k . It is easy to verify that ˜ k coincides with the field ofconstants of k ( t, y, β ). In the following, we will show that the field of constantsof k ( t, y, β ) is equal to k .From Remark 5.4, k ( t, y ) and C ( t, y ) can be embedded into k (( t − a )) and C (( t − a )) respectively. We will consider them as the subfields of k (( t − a )) and C (( t − a )) respectively. Since β ∈ C ( t, y ) ∩ F ( t ), F is algebraically closed and a ∈ F , β ∈ F (( t − a )). Therefore, k ( t, y, β ) ⊆ k (( t − a )). Since k is algebraicallyclosed in k (( t − a )), the field of constants of k ( t, y, β ) is equal to k . This completesthe proof. Let P = P ni =0 A i Y i ∈ F ( t, x )[ Y ] be the minimal polynomial of y ∈ F ( t, x ). Wecan always pick ( a, α ) ∈ F × F ( x ) such that A n ( x , a ) = 0 , P ( x , a, α ) = 0 and ∂P∂Y ( x , a, α ) = 0 . (5.2)10et K = F ( x , α ) and ℓ = [ K ( t, y ) : K ( t )]. Asume that z ∈ F ( t, x ) also satisfiesthe equation P ( z ) = 0. Then z and y are conjugated over F ( t, x ). By Theorem3.2.4 in [9], any field automorphism of the splitting field of P commutes with thederivation D t . So for any L ∈ F ( t ) h D t i , L ( z ) = 0 if and only if L ( y ) = 0. Thusto detect if there is a nonzero L ∈ F ( t ) h D t i such that L ( y ) = 0, it suffices todetect if there exists such operator for z . In the following, we will characterizeall possible D t -separable algebraic functions.Assume that y is D t -separable, i.e., there exists a nonzero L ∈ F ( t ) h D t i suchthat L ( y ) = 0. Let p be a place of K ( t ) and q a place of K ( t, y ) that is ramifiedwith respect to p . Suppose that p and q are uniformizing parameters of p and q respectively, and e is the corresponding ramification index. Then p = aq e forsome invertible a in the DVR with respect to q . Furthermore assume that p is an irreducible polynomial in K [ t ]. Let ℘ be a place of C ( t, y ) lying above q .Then by Theorem 3 in [16, page 92]), ℘ is not ramified with respect to q andso q is a uniformizing parameter of ℘ . Since p ∈ ℘ , the uniformizing parameterof ℘ ∩ C ( t ) can be selected as a factor of p , say t − c for some c ∈ C . It iseasy to see that p/ ( t − c ) is an invertible element in the DVR with respect to ℘ . It implies that t − c = ¯ aq e for some invertible element ¯ a and thus K ( t, y )can be embedded into C (( t − c ) /e ). Therefore y ∈ C (( t − c ) /e )) and c is asingular point of L . Note that the singular points of L lie in the algebraicallyclosed field F . So c ∈ F and then p = b ( t − c ) for some b ∈ K . In other words, t − c is a uniformizing parameter of p . Hence K ( t, y ) satisfies the condition 1 ofTheorem 5.5. By Theorem 5.5, there is β ∈ F ( t ) such that K ( t, y ) = K ( t, β ).We now characterize separable algebraic functions as follows. Proposition 5.6.
Let P = P ni =0 A i Y i ∈ F [ t, x ][ Y ] with A n = 0 be the minimalpolynomial of y ∈ F ( t, x ) . Let K = F ( x )( α ) with α ∈ F ( x ) be as in (5.2) and β ∈ k ( t ) be such that K ( t, y ) = K ( t, β ) . If y is D t -separable, then (1) A n ( x , t ) is split, i.e., A n ( x , t ) = a ( x ) b ( t ) , where a ( x ) ∈ F [ x ] , b ( t ) ∈ F [ t ] ,and (2) y = 1 b ( t ) q ( t ) ℓ − X i =0 a i ( t ) β i , (5.3) where ℓ = [ K ( t, y ) : K ( t )] , a i ( t ) ∈ K [ t ] and q ( t ) is the discriminant of thebase { , β, · · · , β ℓ − } .Proof. Let r i = A i /A n = p i /q i ∈ F ( t, x ) with 0 ≤ i ≤ n , p i , q i ∈ F [ t, x ] andgcd( p i , q i ) = 1. Since y is D t -separable, so are all of the conjugate roots of P ( Y ) = 0. By Vieta’s formulas, the r i ’s are polynomials of these roots, whichtherefore are also D t -separable by Proposition 2.2. By Theorem 3.1, q i is splitfor all i with 0 ≤ i ≤ n . Since A n is the LCM of the q i ’s, we have A n ( x , t ) isalso split. 11et S be the integral closure of K [ t ] in K ( t, y ). Then β, A n ( x , t ) y ∈ S . Since { , β, · · · , β ℓ − } is a base of K ( t, y ) over K ( t ), one has that A n ( x , t ) y = 1 q ( t ) ℓ − X i =0 g i ( t ) β i , where g i ( t ) ∈ K [ t ]. Setting a i ( t ) = g i ( t ) /a ( x ), we obtain the required expressionfor y .Recall that K = F ( x , α ) and ℓ = [ K ( t, y ) : K ( t )]. Since the i -th derivative of y is also in K ( t, y ) for any i ∈ N , we have that Y = (1 , y, y , · · · , y ℓ − ) t satisfiesa linear differential system of the form Y ′ = AY, where A ∈ Mat ℓ ( K ( t )) . (5.4)We will call (5.4) the associated differential equation of y over K ( t ). The follow-ing proposition will allow us to design an algorithm for testing the separabilityof algebraic functions. Proposition 5.7.
Let y and K be as above. Assume that (5.4) is the associateddifferential equation of y over K ( t ) . Then y is D t -separable if and only if thereis an invertible matrix G with entries in K [ t ] such that G − G ′ − G − AG ∈ Mat ℓ ( F ( t )) . Proof.
Assume that there exists a nonzero L ∈ F ( t ) h D t i such that L ( y ) = 0.Then by Proposition 5.6, y has the form (5.3). Let E be the Galois clo-sure of K ( t, β ) over K ( t ). Let β = β, β , · · · , β ℓ be the conjugates of β and σ i ∈ Gal(
E/K ( t )) such that σ i ( β ) = β i . Then σ ( y ) , · · · , σ ℓ ( y ) are allzeroes of P ( x , t, y ). We will denote the Vandermonde matrix generated by σ ( y ) , · · · , σ ℓ ( y ) by U ( y ) and the one generated by β , · · · , β ℓ by U ( β ). Then U ( y ) is a fundamental matrix of the system (5.4) and U ( β ) is a fundamentalmatrix of a system Y ′ = BY with B ∈ Mat ℓ ( F ( t )). Using the argument similarto that in the proof of Proposition 5.7, we have that for all j with 1 ≤ j ≤ ℓ − y j = 1 b ( t ) j q ( t ) ℓ − X i =0 a i,j ( t ) β i , (5.5)where a i,j ( t ) ∈ K [ t ] and b ( t ) , q ( t ) are as in Proposition 5.6. Applying σ l to bothsides of the equalities (5.5) implies that σ l ( y ) j = 1 b ( t ) j q ( t ) ℓ − X i =0 a i,j ( t ) β il , (5.6)where j = 1 , · · · , ℓ − , l = 1 , · · · , ℓ . Let ˜ a i,j = a i,j b ℓ − − j and G = b ( t ) ℓ − q ( t ) 0 · · · a , ( t ) ˜ a , ( t ) · · · ˜ a ℓ − , ( t )... ... ... ...˜ a ,ℓ − ( t ) ˜ a ,ℓ − ( t ) · · · ˜ a ℓ − ,ℓ − ( t ) ℓ ( K [ t ]). Then the equations (5.6) can be rewritten as U ( y ) = ( GU ( β )) / ( b ( t ) ℓ − q ( t )). Hence G is invertible and an easy calculationyields that U ( β ) ′ = ( b ℓ − qG − U ( y )) ′ = (cid:0) ( b ℓ − q ) ′ − b ℓ − qG − G ′ + b ℓ − qG − AG (cid:1) G − U ( y )= BU ( β ) = b ℓ − qBG − U ( y ) . This implies that G − AG − G − G ′ = B − ( b ℓ − q ) ′ b ℓ − q ∈ Mat ℓ ( F ( t )) . Now we prove the converse. Assume that there is an invertible matrix G ∈ Mat ℓ ( K [ t ]) such that ˜ B = G − AG − G − G ′ ∈ Mat ℓ ( F ( t )) . Then U ( y ) = GF , where F is a fundamental matrix of Y ′ = ˜ BY with entries insome differential extension field of K ( t ). Obviously, the entries of both G and F are annihilated by nonzero operators in F ( t ) h D t i and thus so are the sum ofproducts of entries of G and F , in particular, so is y . Remark 5.8.
Once β is computed, one can obtain the linear differential equa-tions Y ′ = BY satisfied by U ( β ) . We now present an algorithm to decide whether a given algebraic function y ∈ F ( t, x ) is D t -separable or not. For the sake of simplicity, we may take F = ¯ Q ,the field of all algebraic numbers over Q . Let P = P ni =0 A i Y i ∈ F [ t, x ][ Y ] bethe minimal polynomial of y . Furthermore, assume that A n is split. Underthis assumption, y is D t -separable if and only if A n y is D t -separable. Thereforewithout loss of generality, we may assume that P ( x , t, Y ) = Y n + A n − ( x , t ) Y n − + · · · + A ( x , t ) , (5.7)where A i ∈ F [ x , t ]. Let ( a, α ) ∈ F × F ( x ) satisfy P ( x , a, α ) = 0 , ∂P∂Y ( x , a, α ) = 0 , (5.8)and let K = F ( x , α ). Then P ( x , t, Y ) may be factorized into a product ofirreducible polynomials in K [ t, Y ]. There is a unique factor of P ( x , t, Y ) in K [ t, Y ] vanishing at ( a, α ), denoted by ¯ P ( x , α, t, Y ). Let K ( t, y ) be the quotientfield of K [ t, Y ] / h ¯ P ( x , α, t, Y ) i . Furthermore suppose that K ( t, y ) satisfies thecondition 1 of Theorem 5.5. Then Theorem 5.5 implies that there is β ∈ F ( t )such that K ( t, y ) = K ( t, β ). We shall show how to find such β .13et R = F ( t )[ x ] and S the integral closure of R in K ( t, y ). Then α, y ∈ S .Suppose that ¯ P ( x , α, t, Y ) = B ℓ Y ℓ + B ℓ − Y ℓ − + · · · + B , (5.9)where B ℓ ∈ F [ x ] , B i ∈ k [ x , α, t ] with i = 0 , · · · , ℓ −
1. Note that[ K ( t, y ) : F ( x , t )] = [ K ( t, y ) : K ( t )][ K ( t ) : F ( x , t )]= [ K ( t, y ) : K ( t )][ K : F ( x )] = ℓ [ K : F ( x )] . The set (cid:8) α i y j | i = 0 , · · · , [ K : F ( x )] − , j = 0 , · · · , ℓ − (cid:9) is a base of K ( t, y ) over F ( x , t ). Let D ( x , t ) be the discriminant of the above baseand let F ( x , Y ) be an irreducible polynomial in F [ x , Y ] such that F ( x , α ) = 0.Then we have Lemma 5.9.
Let ( c , b ) ∈ F m +1 satisfy F ( c , b ) = 0 and D ( c , t ) B ℓ ( c ) = 0 . Then ¯ P ( c , b, t, Y ) is irreducible in F [ t, Y ] and for any root Y = γ of ¯ P ( c , b, t, Y ) = 0 ,we have that K ( t, y ) is isomorphic to K ( t, γ ) .Proof. Let β ∈ K ( t, y ) be as above. Since β is algebraic over F ( t ) we have that β is integral over R = F ( t )[ x ]. Therefore we may write β = 1 D ( x , t ) X b i,j α i y j , where the b i,j ∈ R . Let ( c , b ) satisfy the hypothesis of the lemma and considerthe ideal p = h x − c , . . . , x m − c m , α − b i ⊳ R [ α ] . Note that p is a maximal ideal. The Going Up Theorem implies that there is amaximal ideal q ⊳ S such that q ∩ R [ α ] = p . In particular, D ( x , t ) / ∈ q . There isa natural map φ : S → S/ q . We will let M denote the field S/ q . The element γ = φ ( y ) is a root of ¯ P ( c , b, t, γ ) = 0. Since the minimal polynomial Q ( t, Y )of β lies in F [ t, Y ], it remains unchanged when we apply φ to its coefficients.Therefore φ ( β ) satisfies Q ( t, φ ( β )) = 0. In particular, the degree of φ ( β ) over F ( t ) is equal to ℓ , the degree of K ( t, β ) over K ( t ). Since φ ( β ) = 1 D ( c , t ) X φ ( b i,j ) φ ( α ) i γ j we have that φ ( β ) ∈ F ( t )( γ ). Note that ¯ P ( c , b, t, Y ) = 0. The element γ satisfies¯ P ( c , b, t, γ ) = 0 and so it has degree at most ℓ over F ( t ). Since φ ( β ) ∈ F ( t, γ ),we have that ℓ ≥ [ F ( t, γ ) : F ( t )] ≥ [ F ( t, φ ( β )) : F ( t )] = [ K ( t, β ) : K ( t )]= [ K ( t, y ) : K ( t )] = ℓ and so [ F ( t, γ ) : F ( t )] = ℓ . Therefore ¯ P ( c , b, t, Y ) is irreducible. Furthermore F ( t, β ) is isomorphic to F ( t, φ ( β )) = F ( t, γ ). This implies that K ( t, y ) is isomor-phic to K ( t, γ ). 14et ¯ P ( x , α, t, Y ) be as above. Lemma 5.9 implies that if y is D t -separablethen one can compute ( c , b ) ∈ F m +1 such that ¯ P ( c , b, t, Y ) is irreducible over F ( t ) and β can be taken to be a zero of ¯ P ( c , b, t, Y ). From ¯ P ( c , b, t, Y ), wecan construct the associated differential equation of β over F ( t ). Denote thisassociated differential equation by Y ′ = BY with B ∈ Mat ℓ ( F ( t )). The proofof Proposition 5.7 implies that if y is D t -separable then there is an invertiblematrix G with entries in K [ t ] such that G ′ = AG − G (cid:18) B − q ′ ( t ) q ( t ) (cid:19) , where q ( t ) is the discriminant of { , β, · · · , β ℓ − } and Y ′ = AY is the associateddifferential equation of y over K ( t ). Here the polynomial b ( t ) in (5.5) disappearsbecause we assume that P is monic in Y . Note that G is a polynomial solutionof the linear differential equation Y ′ = AY − Y ( B − q ( t ) ′ /q ( t )), which can becomputed by algorithms developed in [5, 8].We summarize the above results as the following algorithm. Algorithm 5.10.
Input: An irreducible polynomial P ( t, x , Y ) = A n Y n + A n − Y n − + · · · + A ∈ F [ t, x , Y ] . Output: “Yes” if y is D t -separable, otherwise “No”, where y ∈ k ( x , t ) is a rootof P ( Y ) = 0 . (1) If A n is not split, then y is not D t -separable and return “No”. (2) Transform P ( x , t, Y ) into a monic polynomial by replacing Y by Y /A n and clear the denominators. (3) Compute β : (3 .a ) Find ( a, α ) ∈ F × F ( x ) satisfying the conditions (5.8). (3 .b ) Decompose P into a product of irreducible polynomials over F ( x , α ) .Let ¯ P ( x , α, t, Y ) be the irreducible factor satisfying that ¯ P ( x , α, a, α ) = 0 . (3 .c ) Compute D ( x , t ) , the discriminant of the base { α i ¯ y j } , where ¯ y is azero of ¯ P ( x , α, t, Y ) in F ( t, x ) . (3 .d ) Compute a point ( c , b ) ∈ F m +1 such that D ( c , t ) B ℓ ( c ) = 0 and F ( c , b ) = 0 ,where F is the minimal polynomial of α over F ( x ) and B ℓ ( x ) is theleading coefficient of ¯ P ( x , α, t, Y ) . (3 .e ) Let β be a zero of ¯ P ( c , b, t, Y ) = 0 in F ( t ) . (4) Compute G : .a ) Compute q ( t ) , the discriminant of the base { β j | j = 0 , · · · , ℓ − } andcompute the associated differential equations of y and β , which aredenoted by Y ′ = AY and Y ′ = BY respectively. (4 .b ) By algorithms developed in [5, 8], compute a base of polynomial so-lutions of Z ′ = AZ − Z ( B − q ( t ) ′ /q ( t )) , where Z = ( z ij ) with inde-terminate entries, say { Q , · · · , Q s } . (4 .c ) Compute C = det( z Q + · · · + z s Q s ) with z , · · · , z s being indeter-minates. If C = 0 then return “No”, otherwise return “Yes”. We now show an example to illustrate the main steps of the above algorithm.
Example 5.11.
Let E = ¯ Q ( t, x ) and y be the algebraic function over E definedby P ( x, t, Y ) := Y − xt + 1) Y + ( xt + 1) − t. We are going to decide whether y is D t -separable or not. We will follow theabove algorithm step by step. Since P ( x, t, Y ) is monic in Y . We begin with thethird step, i.e., computing β .(3) Compute β = √ t + 1 :(3.a) Set ( a, α ) = (1 , x ) . One sees that P ( x, , x ) = 0 and ∂P∂Y ( x, , x ) = − = 0 . So ¯ Q ( x, α ) = ¯ Q ( x ) .(3.b) Since P ( x, t, Y ) is irreducible over ¯ Q ( x ) , we take ¯ P ( x, α, t, Y ) to be P ( x, t, Y ) . (3.c) Set D ( x, t ) = 4 t , which is the discriminant of the base { , ¯ y } with P ( x, t, ¯ y ) = 0 .(3.d) One sees that B ( x ) = 1 and F = z − x . So the point (0 , satisfies D (0 , t ) B (0) = 0 and F (0 ,
0) = 0 .(3.e) Set β = √ t + 1 which is a zero of P (0 , t, Y ) = Y − Y + 1 − t .(4) Compute G :(4.a) Set q ( t ) = 4 t , which is the discriminant of the base { , β } , and set A = (cid:18) x − t t (cid:19) , B = (cid:18) − t t (cid:19) . Then Y ′ = AY and Y ′ = BY are the associated differential equationsof y and β respectively.(4.b) Set Z = ( z ij ) ≤ i,j ≤ , and compute a base of the polynomial solutionsof the system Z ′ = AZ − Z ( B − /t ) . One has that (cid:26) Q := (cid:18) t xt + t (cid:19) , Q := (cid:18) − t t (cid:19)(cid:27) is a required base.(4.c) One has that det( z Q + z Q ) = z z t = 0 . So y is D t -separable. Conclusion and future work
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