A Subresultant Theory for Ore Polynomials with Applications 1)

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1 MM Research Preprints, No. 17, Dec Beijing 91 A Subresultant Theory for Ore Polynomials with Applications 1) Ziming Li Institute of Systems Science, Academia Sinica currently in GMD-SCAI D Sankt Augustin, Germany. ziming.li@gmd.de Abstract. The subresultant theory for univariate commutative polynomials is generalized to Ore polynomials. The generalization includes: the subresultant theorem, gap structure, and subresultant algorithm. Using this generalization, we define Sylvester s resultant of two Ore polynomials, derive the respective determinantal formulas for the greatest common right divisor and least common left multiple of two Ore polynomials, and present a fraction-free version of the non-commutative extended Euclidean algorithm. 1. Introduction Greatest common right divisors (abbreviated as: gcrd) and least common left multiples (abbreviated as: lclm) are basic objects in the theory and computation of Ore polynomials [?,?,?]. For example, the gcrd of linear ordinary differential (shift) operators represents the intersection of their solution spaces, and the lclm of these operators represents the sum of their solution spaces. The non-commutative Euclidean and extended Euclidean algorithms are used to compute gcrds and lclms, respectively. Naive applications of these two algorithms lead to inefficient implementations because of the coefficient growth of intermediate polynomials, as seen in the commutative case. Motivated by the improvements made on the commutative Euclidean algorithm, we generalize the subresultant theory for univariate commutative polynomials to univariate Ore polynomials. This generalization provides a way to control the coefficient growth in the non-commutative Euclidean and extended Euclidean algorithms. The commutative subresultant theory has undergone a quite intensive study since the work of Collins, Brown and Traub [?,?,?,?]. Chardin presented a subresultant theory for linear ordinary differential operators [?]. In this paper we present a subresultant theory for Ore polynomials over a commutative domain, which includes not only linear ordinary differential operators but also linear shift operators, q-difference operators, etc. The subresultant theory can be further extended to linear inhomogeneous differential and difference equations [?], but we will not present this extension because most of the applications of subresultants are found in an Ore polynomial ring. To extend the commutative subresultant theory, we will overcome two difficulties. First, we need to find new techniques to prove essentially the same statements as those in the commutative case without assuming the commutativity of multiplication. This problem is 1) This report has been published in the Proc. of ISSAC 98, Rostock, Germany, O. Gloor (ed.), ACM Press.

2 92 Z. M. Li solved by Lemmas?? and??. Second, we need to simplify σ-factorial expressions in order to remove extraneous factors. We also remark that it looks rather complicated to adapt the approach in [?,?] to extend the commutative subresultant theory, because it is not trivial to build up a generic coefficient domain for Ore polynomials. Applications of the subresultant theory include: computing gcrds by modular techniques [?]; extending Sylvester s resultant to Ore polynomials (Definition??), expressing the gcrd and lclm of two Ore polynomials by determinants (Proposition??), and computing lclms (Proposition??). The correspondence between subresultants and intermediate polynomials occurring in the Euclidean algorithm is also useful to estimate coefficient and degree bounds, and to analyze complexities. The organization of this paper is as follows. Section?? introduces Ore polynomial rings and specify the notation that will be used later. The definition of subresultants is given in Section??. Section?? presents the row-reduction formula for subresultants, which makes it possible to extend the techniques for establishing commutative subresultant theory in [?,?]. The subresultant theorem for Ore polynomials is proved in Section??. The subresultant algorithm is described in Section??. Applications are presented in Section??. 2. Preliminaries Let R be a commutative domain and σ an injective endomorphism from R to itself, which is called a conjugate operator. An endomorphism δ of the additive group (R, +, 0) is called a pseudo-derivation with respect to σ if δ(ab) = σ(a)δ(b) + δ(a)b, for all a, b R. The (non-commutative) multiplication in R[X] is defined by the commutation rule Xa = σ(a)x + δ(a), for all a R. (1) The triple (R[X], σ, δ) is called an Ore polynomial ring. For A, B R[X], the product of A and B is denoted by AB and the degree of AB is equal to the sum of the degrees of A and B. Example 1 If D is a derivation operator on R, then D is a pseudo-derivation with respect to the identity mapping 1 because D(rs) = rd(s) + D(r)s, for r, s R. Hence, (R[X], 1, D) is the ring with the multiplication given by Xr = rx + D(r), for r R. This ring is isomorphic to the ring of linear homogeneous differential polynomials in one differential indeterminate over R. Example 2 Let E be an injective endomorphism of the domain R and δ be the null mapping 0. Then (R[X], E, 0) is the ring with the multiplication given by Xr = E(r)X, for r R. This ring is isomorphic to the ring of linear homogeneous difference polynomials in one difference indeterminate (with respect to E) over R. Notation We denote (R[X], σ, δ), σ(r) and δ(r) by R[X], σr and δr, respectively. The next lemma extends differential Leibniz s rule and will play an important role in the sequel.

3 Ore Polynomials 93 Lemma 2.1 For all r R, A R[X], and n Z +, the polynomial X n (ra) (σ n r)x n A is an R-linear combination of X n 1 A,, XA, A. Proof If n = 1, then the lemma follows from (??). The lemma is then proved by an easy induction on n. For A R[X], the leading coefficient of A is denoted by lc(a). Lemma?? implies that lc(x n A) = σ n lc(a), for all n N. (2) Definition 2.1 Let r R. The nth σ-factorial of r is the product n 1 i=0 σ i r, for all n Z +, which is denoted by r [n]. In addition, r [0] is set to be 1. Lemma 2.2 If r, s R, and m, n N, then 1. (rs) [m] = r [m] s [m], 2. r [m+n] = r [m] (σ m r) [n], 3. (r [m] ) [n] = (r [n] ) [m], 4. r [m+1][n+1] = r [m+n+1] (σr) [m][n]. Proof All the assertions are immediate from the fact that σ is a ring homomorphism. Definition 2.2 For A, B R[X] with deg A = m and deg B = n 0, a pseudo-remainder of A and B is either A, if m < n; or C R[X] such that deg C < deg B and ( m n i=0 lc(x i B) ) A = QB + C, where Q is in R[X] and is called the (left) pseudo-quotient of A and B. The pseudo-remainder of A and B is unique and denoted by prem(a, B). Using the σ-factorial expressions and (??), we get the pseudo-remainder formula lc(b) [m n+1] A = QB + prem(a, B). (3) Let A 1, A 2,..., A k be a sequence of non-zero elements of R[X] such that A i is R-linearly dependent on prem(a i 2, A i 1 ), for i = 3,... k, and either A k R\{0} or prem(a k 1, A k ) = 0. Such a sequence is called a polynomial remainder sequence (abbreviated as: p.r.s.) of A 1 and A 2. We will show that any member of a p.r.s. of A 1 and A 2 is R-linearly dependent on a subresultant of A 1 and A 2 (see Corollary??).

4 94 Z. M. Li 3. Subresultants In this section we extend the notion of subresultants of commutative polynomials to Ore polynomials. First let us recall the definition of determinant polynomials [?,?]. Definition 3.1 Let M be an r c matrix over R. Assume that r c. The determinant polynomial of M is c r M = det(m i )X i, i=0 where M i is the r r matrix whose first (r 1) columns are the first (r 1) columns of M and whose last column is the (c i)th column of M, for i = 0,..., c r. In the paper we will encounter determinants whose last columns contain polynomials in R[X] \ R. When expanding such a determinant, we place the product of entries in R on the left-hand side of the entry in R[X] \ R. With this convention we can express determinant polynomials by determinants. Let the matrix M in Definition?? be and, for i = 1,..., r, let Then m 11 m 1,r 1 m 1r m 1c m 21 m 2,r 1 m 2r m 2c m r1 m r,r 1 m r,r+1 m rc H i = m i1 X c m ir X c r + + m ic. M = det m 11 m 1,r 1 H 1 m 21 m 2,r 1 H 2 m r1 m r,r 1 H r. (4) Thus a determinant polynomial is alternative and multi-linear for its rows. Let A : A 1, A 2,..., A r be a sequence of polynomials in R[X] and let the maximum of the degrees of the A i s be d. The matrix associated with A, denoted by mat(a), is the r (d + 1) matrix whose entry in the ith row and jth column is the coefficient of X d+1 j in A i, where 1 i r and 1 j d + 1. If r d + 1, then the determinant polynomial of A is defined to be mat(a), which is denoted by A 1, A 2,..., A r or A. Lemma 3.1 Let r R with r 0, and A R[X]. If H m =..., X m (ra), X m 1 (ra),..., X(rA), ra,... is a determinant polynomial of a polynomial sequence in R[X], where m N, then H m = r [m+1]..., X m A, X m 1 A,..., XA, A,.... Proof We proceed by induction on m. The assertion is trivial when m = 0. Assume that m > 0 and the lemma is true for m 1. Thus, H m = r [m]..., X m (ra), X m 1 A,..., XA, A,.... (5) It follows from Lemma?? that X m (ra) in (??) can be replaced by (σ m r)(x m A).

5 Ore Polynomials 95 Definition 3.2 Let A, B R[X] with deg A = m and deg B = n, where m n. The nth subresultant of A and B is B. For j = n 1, n 2,..., 0, the jth subresultant of A and B, sres j (A, B), is X n j 1 A,..., XA, A, }{{} n j X m j 1 B,..., XB, B, }{{} m j The sequence S(A, B): A, B, sres n 1 (A, B),..., sres 0 (A, B), is called the subresultant sequence of A and B. Example 3 Let A = a 2 X 2 + a 1 X + a 0 and B = b 2 X 2 + b 1 X + b 0. Then a sres 1 (A, B) = A, B = 2 A B, b 2 and sres 0 (A, B) = XA, A, XB, B σa 2 δa 2 + σa 1 δa 1 + σa 0 XA 0 a = 2 a 1 A. σb 2 δb 2 + σb 1 δb 1 + σb 0 XB 0 b 2 b 1 B Note that the jth subresultant has degree no greater than j, and that all the subresultants of A and B are contained in the left idea generated by A and B because of (??). The next lemma links sres n 1 (A, B) and prem(a, B). Lemma 3.2 Let A and B be the same as in Definition??. Then sres n 1 (A, B) = ( 1) m n+1 prem(a, B). Proof The following calculation proves the lemma lc(b) [m n+1] sres n 1 (A, B) = lc(b) [m n+1] A, X m n B,..., B = prem(a, B), X m n B,..., B (by (??)) = ( 1) m n+1 X m n B,..., B, prem(a, B) = ( 1) m n+1 lc(b) [m n+1] prem(a, B). 4. Row-Reduction Formula for Subresultants Some proofs in the commutative subresultant theory are tacitly based on the following fact: if A and B are two univariate commutative polynomials in the indeterminate, say x, then xprem(a, B) = prem(xa, xb). However, if A and B are in R[X], then in general Xprem(A, B) prem(xa, XB). The next lemma describes a relation between Xprem(A, B) and prem(xa, XB). Lemma 4.1 For i N, ( ) X i prem(a, B) prem(x i A, X i B) is an R-linear combination of X i 1 A,..., A, X m n+i B,..., B.

6 96 Z. M. Li Proof Set b = lc(b) [m n+1]. By (??) we get and, since lc(x i B) = σ i (lc(b)), for all i N, X i ba = X i QB + X i prem(a, B) σ i (b)x i A = Q i X i B + prem(x i A, X i B), where Q, Q i R[X] both with degree m n. Lemma?? implies that X i ba σ i (b)x i A is an R-linear combination of X i 1 A,..., A. The lemma is proved by subtracting the above two equations. We are ready to present the row-reduction formula for subresultants, by which the techniques for proving the commutative subresultant theorem will be extended to Ore polynomials. Lemma 4.2 Let A, B R[X] with respective degrees m and n, where m n 0. If there exist non-zero elements q, r, s R and F, G R[X] such that qb = rf and sres n 1 (A, B) = s G, then q [m j] lc(b) [m n+1][n j] sres j (A, B) = r [m j] s [n j] X n j 1 F,..., F, X m j 1 G,..., G. (6) Proof Let C = prem(a, B), C k = prem(x k A, X k B), for k N, and S j = sres j (A, B), for j = n 1, n 2,..., 0. By Definition?? we have S j = X n j 1 A,..., XA, A, }{{} n j X m j 1 B,..., XB, B. }{{} m j The pseudo-remainder formula for X n j 1 A and X n j 1 B implies that the polynomial σ n j 1 (lc(b)) [m n+1] X n j 1 A C n j 1 is an R-linear combination of the polynomials Thus, Lemma?? implies that the polynomial X m j 1 B,..., X n j B, X n j 1 B. σ n j 1 (lc(b)) [m n+1] X n j 1 A X n j 1 C is an R-linear combination of the polynomials Hence, we have X n j 2 A,..., XA, A, X m j 1 B,..., XB, B. σ n j 1 (lc(b)) [m n+1] S j = X n j 1 C, X n j 2 A,..., A, X m j 1 B,..., B. (7)

7 Ore Polynomials 97 The same reasoning allows us to replace X i A by X i C on the right-hand of (??), while at the same time multiply the power σ i (lc(b)) [m n+1] on the left-hand of (??), for i = n j 2, n j 3,..., 0. We eventually arrive at Lemma?? then asserts that lc(b) [n j][m n+1] S j = X n j 1 C, X n j 2 C,..., C, X m j 1 B,..., B. lc(b) [n j][m n+1] S j = X m j 1 B,..., B, X n j 1 S n 1,..., S n 1. Thus, the lemma follows form Lemma??. 5. Subresultant Theorem Notation To avoid endlessly repeating the assumptions, in the rest of this article, we let A and B be in R[X] with respective degrees m and n, where m n 0. Let S n = B and S j = sres j (A, B), for j = n 1, n 2,..., 0. The subresultant sequence S(A, B) consists of the polynomials A, S n,, S 1, S 0. Definition 5.1 The jth subresultant S j is said to be regular if deg S j = j and otherwise S j is defective. In particular, the nth subresultant S n is always regular. First, we demonstrate the relation between the members of S(A, B) and subresultants of the two consecutive non-zero members of S(A, B) in the next lemma. Lemma 5.1 Let α i = lc(s i ), for i = 0, 1,..., n; β n = σlc(s n ) [m n] ; and β i = σlc(s i ), for i = 0, 1,..., n 1. If S j+1 is regular, for some j with n 1 j 0, then the following hold: 1. If S j = 0, then S i = 0, for j 1 i If S j 0 and deg S j = r, then S i = 0, (j 1 i r + 1), (8) and β [j r] j+1 S r = β [j r] j S j, (9) α [r i] j+1 β[j i] j+1 S i = sres i (S j+1, S j ), (r 1 i 0). (10) Proof The proof will be done by induction on the sequence of the regular subresultants in S(A, B). As S n is the first regular subresultant in S(A, B), we start with the case in which j = n 1. Let i be an integer such that n 2 i 0. By the definition of subresultants, we have S i = X n 1 i A,..., A, X m 1 i S n,..., S n. It then follows from the row-reduction formula (??) that α [m n+1][n i] n S i = R i, (11)

8 98 Z. M. Li where R i = X m 1 i S n,..., S n, X n 1 i S n 1,..., S n 1. If S n 1 = 0, then R i = 0, and hence, S i = 0 by (??). Assume that r 0. For n 2 i r + 1, deg S n > deg ( X n 1 i ) S n 1. Consequently, R i = 0 and S i = 0 by (??). If i = r, then R r = α n [m r] β [n 1 r] S n 1. Hence, (??) can be rewritten as Since n 1 α [m n+1][n r] n α [m n+1][n r] n S r = α [m r] n β [n 1 r] n 1 S n 1. = α n [m r] β n [n r 1] by Lemma??, (??) holds for j = n 1. For r 1 i 0, R i = (σ r i α n ) [m r] sres i (S n, S n 1 ), which, together with (??), implies that Since α [m n+1][n i] n S i = (σ r i α n ) [m r] sres i (S n, S n 1 ). α [m n+1][n i] n = α n [r i] (σ r i α n ) [m r] β n [n 1 i] by Lemma??, (??) holds for j = n 1. The proof of the base case is done. We assume that the lemma holds for the regular subresultant S j+1 and that deg S j = r. In other words, the first assertion of the theorem, (??), (??) and (??) hold. If S j = 0 then there is no regular subresultant that follows S j. So, there is nothing to prove. Suppose S j 0. Then the regular subresultant next to S j+1 must be S r by our induction hypothesis. Let deg(s r 1 ) = t. We have to prove that if S r 1 = 0, S i = 0 for r 2 i 0, and that if S r 1 0, S i = 0, (r 2 i t + 1), (12) and β r [r 1 t] S t = β [r 1 t] r 1 S r 1, (13) α r [t i] β r [r 1 i] S i = sres i (S r, S r 1 ), (t 1 i 0). (14) Before going to induction, we point out two important relations hiding in (??). Equating the leading coefficients of both sides of (??) yields Applying σ to both sides of this equality, we get Based on the induction hypothesis we claim that β [j r] j+1 α r = β [j r] j α j. (15) (σβ j+1 ) [j r] β r = β [j r+1] j. (16) α r [j i+1] β r [r i 1] S i = T i, (r 2 i 0), (17)

9 Ore Polynomials 99 where T i = X j i S r,..., S r, X r i 1 S r 1,..., S r 1. (18) Proof of the Claim. Observe that (??) and (??) (setting i = r 1) imply that and β [j r] j+1 S r = β [j r] j S j (19) α j+1 β [j r+1] j+1 S r 1 = sres r 1 (S j+1, S j ). (20) The induction hypothesis (??) asserts that, for t i 0, α [r i] j+1 β[j i] j+1 S i = X r i 1 S j+1,..., S j+1, X j i S j,... S j. It then follows from (??), (??), and Lemma?? that ( β [j r][j i+1] j = β [j r][j i+1] j+1 α [j r+2][r i] ) ( j α [r i] ) j+1 β[j i] j+1 S i ( α [r i] ) j+1 β[j r+1][r i] j+1 T i. Note that, in the above deduction, S j+1, S j, r, S r and S r 1 play the roles of A, B, n, F, and G in the statement of Lemma??, respectively. Simplifying the σ-expressions in the above equality by Lemma??, (??) and (??), we prove the claim (??) (for the details of this simplification, please see [?, page 27]). If S r 1 is zero, then T i is zero, so is S i, for r 2 i 0. If r 2 i t + 1, then T i = 0 since deg(s r ) > deg(x r i 1 S r 1 ). It follows from (??) that S i = 0. If i = t, then T i = α r [j t+1] β [r t 1] S r 1 whence (??) implies Equation (??) follows. If t 1 i 0, then Hence (??) implies Observe that α [j t+1] r β r [r t 1] r 1 S t = α r [j t+1] β [r t 1] r 1 S r 1. T i = (σ t i α r ) [j t+1] sres i (S r, S r 1 ). α r [j i+1] β r [r i 1] S i = (σ t i α r ) [j t+1] sres i (S r, S r 1 ). (21) α [j i+1] r = α [(t i)+(j t+1)] r = α [t i] r (σ t i α r ) [j t+1]. Using this relation to remove the like σ-factorial expressions from both sides of (??), we find (??).

10 100 Z. M. Li A B... Sj+1 is regular. S j is defective of order r.. S i = 0 (j > i > r). S r is regular.... a regular subresultant a defective subresultant. zero subresultants a regular subresultant... Fig. 1. The gap structure of S(A, B) Theorem 5.2 (Subresultant Theorem) Let α i = lc(s i ), for i = n, n 1,..., 0; β n = σlc(s n ) [m n], and β i = σlc(s i ), for i = n 1,..., 0. If S j+1 is regular, for some j with n 1 j 0, and deg S j = r, then 1. If S j = 0, then 2. If S j 0, then and S i = 0, (j 1 i 0). (22) S i = 0, (j 1 i r + 1), (23) β [j r] j+1 S r = β [j r] j S j, (24) α j+1 β [j r+1] j+1 S r 1 = ( 1) j r prem(s j+1, S j ). (25) Proof Equations (??), (??), and (??) were proved in Lemma??. If i = r 1, (??) in Lemma?? becomes α j+1 (β j+1 ) [j r+1] S r 1 = sres r 1 (S n, S n 1 ) Equation (??) then follows from Lemma??. A formula-free version of Theorem?? reads: Corollary 5.3 Let n 1 j 0. If S j+1 is regular and deg S j = r, then, for all i with j 1 i r + 1, the subresultant S i is zero. If, moreover, S j 0 then S j and S r are R-linearly dependent. In particular, S r is regular. This corollary can be illustrated by the gap structure of S(A, B) in Figure 1.

11 Ore Polynomials Subresultant Algorithm We shall now extend the subresultant sequences of the first and second kinds (see Rubald [?]), and describe the subresultant algorithm. Definition 6.1 The subresultant sequence of A and B of the first kind is the subsequence of S(A, B) that consists of the following polynomials: A, B, and S j if S j+1 is regular and S j is nonzero. The subresultant sequence of A and B of the second kind is the subsequence of S(A, B) that consists of A, B and the other regular subresultants of S(A, B). We denote the subresultant sequences of A and B of the first and second kinds by S 1 (A, B) and S 2 (A, B), respectively. The next lemma describes the relation between S 1 (A, B) and S 2 (A, B). Lemma 6.1 Let S 2 (A, B) consist of A, S n, S jk, S jk 1,, S j1, S j0, with n > j k > j k 1 > > j 0. Then S 1 (A, B) consists of Proof The sequence H: A, S n, S n 1, S jk 1, S jk 1 1,, S j1 1, A, S n, S n 1, S jk 1, S jk 1 1,, S j1 1, is a subsequence of S 1 (A, B) by Definition??. Notice that S j0 1 is zero, for otherwise there would be a regular subresultant S r with r = deg S j0 1, but S j0 is the last regular subresultant in S(A, B), a contradiction. Hence all the subresultants following S j0 are zero because of (??) in Theorem??. H is S 1 (A, B). Let the subresultants S ji+1 and S ji be two consecutive members of S 2 (A, B). Then Corollary?? asserts that the subresultants between S ji+1 1 and S ji are all zero. Hence, all the non-zero subresultants are contained in either S 2 (A, B) or S 1 (A, B). Accordingly, all the defective subresultants are contained in S 1 (A, B). The members of S 1 (A, B) and S 2 (A, B) are R-linearly dependent in order. If there is no defective subresultant in S(A, B), then both S 1 (A, B) and S 2 (A, B) coincide with S(A, B). We present the subresultant algorithm in the next theorem, which is an extension to the commutative subresultant algorithm in [?]. The algorithm computes S 1 (A, B) without expanding determinants directly. It proceeds as the Euclidean algorithm but removes an extraneous factor from the coefficients in the pseudo-remainder after each pseudo-division. A byproduct of this algorithm is the leading coefficients of the members of S 2 (A, B), that is, the leading coefficients of all regular subresultants of A and B. Theorem 6.2 (Subresultant Algorithm) 1. If S n 1 is nonzero, the third member of S 1 (A, B) is S n 1 = ( 1) m n+1 prem(a, B).

12 102 Z. M. Li 2. If S k is the fourth member of S 1 (A, B), and let l be (n deg S n 1 ), then where e = ( 1) l+1 σ(lc(b)) [m n][l] lc(b). S k = prem(b, S n 1 )/e, (26) 3. If S i, S j and S k are three consecutive members in S 1 (A, B), where n > i > j > k, and let l be (deg S i deg S j ), then where e = ( 1) l+1 σ(lc(s j+1 )) [l] lc(s i ). S k = prem(s i, S j )/e, (27) Proof The first assertion is due to Lemma??. The second assertion follows from (??) in Theorem?? (set j = n 1). To prove the last assertion, we let deg S i = t and deg S j = r. Since both S i+1 and S j+1 are regular by the definition of S 1 (A, B), t = j +1 by Corollary??. It then follows from (??) in Theorem?? that σ(lc(s i+1 )) [i t] S j+1 = σ(lc(s i )) [i t] S i. (28) From this equation we see that (lc(s i )) [i t+1] prem(s i, S j ) = lc(s i )prem(σ(lc(s i )) [i t] S i, S j ) = lc(s i )prem(σ(lc(s i+1 )) [i t] S j+1, S j ) = lc(s i )σ(lc(s i+1 )) [i t] prem(s j+1, S j ) = ( 1) j r+2 σ(lc(s i+1 )) [i t] lc(s j+1 )lc(s i ) σ(lc(s j+1 )) [j+1 r] S r 1 (by (??)). Note that S r 1 = S k since S r is regular. Thus, the theorem will be proved if lc(s i ) [i t+1] = σ(lc(s i+1 )) [i t] lc(s j+1 ). (29) holds. Equating the leading coefficients on both sides of (??) yields (??). By Theorem?? we can design the subresultant algorithm for computing S 1 (A, B). In order to compute S k by S i and S j according to the third assertion of Theorem??, we need to compute lc(s j+1 ). This leading coefficient can be obtained recursively from (??). Corollary 6.3 Both S 1 (A, B) and S 2 (A, B) are p.r.s. s. Proof S 1 (A, B) is a p.r.s. by Theorem??. S 2 (A, B) is a p.r.s. because each member of S 2 (A, B) is R-linearly dependent on one (and only one) member of S 1 (A, B) by Lemma??.

13 Ore Polynomials Applications First, we extend the resultant of two differential operators given in [?,?]. Definition 7.1 The subresultant S 0 is called Sylvester s resultant of A and B and denoted by res (A, B). Proposition 7.1 Let d be the degree of the gcrd of A and B. Then the following hold. 1. S d is a gcrd of A and B. 2. d = 0 if and only if res (A, B) = UA is an lclm of A and B, where U is the determinant of order (m + n 2d + 2) whose first (m + n 2d + 1) columns are the first (m + n 2d + 1) columns of mat(x n d A,..., XA, A, X m d B,..., XB, B) and whose last column is the transpose of the vector (X n d,... X, 1, 0, 0,..., 0). }{{} m d+1 Proof Since S 2 (A, B) is a p.r.s., the last member of S 2 (A, B) is a gcrd of A and B. This member is a regular subresultant, so, it must be S d. In particular, d = 0 if and only if S 0 0. The first and second assertions are proved. To prove the last assertion, we first show that UA is right divisible by B. Let M be the matrix whose first (m + n 2d + 1) columns are the same as those of U, and its last column is the transpose of the vector (X n d A,..., XA, A, X m d B,..., XB, B). Then M is zero because M = S d 1 when d > 0, and the last column of M can be reduced to zero by the previous columns when d = 0. Expanding M according to its last column yields UA + V B = 0, for some V R[X], that is, UA is right divisible by B. By [?, p. 486] the degree of lclms of A and B is equal to (m + n d). Hence, it suffices to prove that the degree of U is equal to (n d), that is, the cofactor of X n d in the determinant U is nonzero. From the definition of U one sees that this cofactor is the product of ±σ m d (lc(b)) and the coefficient of X d in S d. As S d is a gcrd of A and B, the degree of S d is d. Hence, the cofactor is nonzero Example 4 Let D = d dt, and let and A = (t 1)D 3 + ( t 2 + t)d 2 + ( 2t + 3)D t B = 3D 2 + (t 3 3t)D t 4 3 be in Z[t][D]. By Proposition?? a gcrd of A and B is t 1 t 2 + t A 3 3t + t 3 DB 0 3 B = (t7 t t 3 + 9t 2 )(D t),

14 104 Z. M. Li and that an lclm of A and B is t 1 t 2 + t + 1 4t + 4 t 2 D 0 t 1 t 2 + t 2t t + t 3 9 t 4 + 6t 2 8t 3 + 6t t + t 3 6 t 4 + 3t t + t 3 0 A. Remark 1 The determinant formula for gcrds is a generalization of the corresponding formula for gcds in [?]. The determinant formula for lclms is possibly new. With these two formulas, one can estimate the coefficient and degree bounds for gcrds lclms. Remark 2 If R is a unique factorization domain, then σ m d (lc(b))σ n d (lc(a))lc(s d ) is a multiple of the leading coefficient of the primitive lclm of A and B, because UA is an lclm of A and B and the leading coefficient of U is σ m d (lc(b))lc(s d ). In the rest of this section we study algorithms for computing lclms. Computing lclms is not as simple as computing lcms in the commutative case because of the non-commutativity of multiplication. One method for computing lclms is the extended (right) Euclidean algorithm [?]. Of course, it is sufficient to use the half-extended Euclidean algorithm, which computes only one co-sequence. If one uses the polynomial division over the fraction field of R to carry out the half-extended Euclidean algorithm, the algorithm is inefficient because there are too many gcd-computations among coefficients. We present a fraction-free version of the half-extend Euclidean algorithm for computing lclms, which reduces coefficient growth by exact division over R. For a later convenience we modify our notation. Let A 1 = A, A 2 = B and S 1 (A, B) be the sequence: A 1, A 2, A 3,..., A k. Let U 1 = 1 and U 2 = 0. If A i is the jth subresultant of A and B, then we let U i be the (m + n 2j) (m + n 2j) determinant whose first (m + n 2j 1) column is the same as those in mat(x n j 1 A,..., A, X m j 1 B,..., B). and whose last row is the transpose of (X n j 1,... X, 1, 0, 0,..., 0). }{{} m j Note that the index j decreases as the index i increases. The sequence U 1, U 2, U 3,..., U k is called the first co-sequence associated with S 1 (A, B), because U i A 1 A i mod A 2 for i = 1, 2,..., k, i.e., U i A 1 A i is right-divisible by A 2. Another property of the first cosequence is that deg U i < (n deg A i ), for i = 3,..., k, because deg A i j if A i is the jth subresultant of A and B. By Theorem?? we have, for i = 3, 4,..., k, l i A i 2 = Q i A i 1 + e i A i,

15 Ore Polynomials 105 where l i = lc(a i 1 ) [deg A i 2 deg A i 1 +1], e 3 is ( 1) m n+1 and e i is the extraneous factor e removed in (??) or (??) when we use the subresultant algorithm to compute A i. The next proposition provides a fraction-free version of the half-extended Euclidean algorithm. Proposition 7.2 With the notation just introduced, we have for i = 3,..., k. Furthermore is an lclm of A 1 and A 2. Proof Let V 1 = U 1 and V 2 = U 2, and let U i = (l i 1 U i 2 Q i U i 1 )/e i (30) L = (l k U k 1 Q k U k )A 1 (31) V i = (l i 1 V i 2 Q i V i 1 )/e i, (32) for i = 3, 4,..., k. Since l i A i 2 = Q i A i 1 + e i A i, V i A 1 A i mod A 2, for i = 3,..., k. Now, we prove U i = V i, for i = 3,..., k. Notice that (U i V i )A 1 is right divisible by A 2. Observe that (??) implies deg V i = (deg A 2 deg A i 1 ) and recall that deg U i < (deg A 2 deg A i ). We conclude deg(u i V i ) < (deg A 2 deg A k ). It then follows that U i V i = 0, for otherwise (U i V i )A 1 would be a non-zero common left multiple of A 1 and A 2 with degree less than (deg A 1 + deg A 2 deg A k ), the degree of lclms of A 1 and A 2, a contradiction. The formula (??) holds. Since U i is in R[X], the division in (??) is exact. In particular, we have U k 1 A 1 A k 1 mod A 2, and U k A 1 A k mod A 2, and l k A k 1 = Q k A k. Hence L defined in (??) is right-divisible by A 2. The degree of L is (deg A 2 deg A k ), since deg U k 1 = (deg A 2 deg A k 2 ), deg U k = (deg A 2 deg A k 1 ), and deg Q k = (deg A k 1 deg A k ). We present an experimental comparison among three algorithms for computing lclms of two Ore polynomials A and B in R[X], where deg A = m and deg B = n. The first algorithm (labeled L) has two time-consuming costs: compute the solution space V of the linear homogeneous system whose matrix is the transpose of M = mat(x n A,..., A, X m B,..., B) adjust the parameters in V to find a linear relation (over R) among the polynomials X n r A,..., A, X m r B,..., B with r as large as possible.

16 106 Z. M. Li The second algorithm (labeled S) is given in Proposition??, which has two time-consuming costs: compute S 1 (A, B) : A 1, A 2,..., A k compute the 1st co-sequence associated with S 1 (A, B). The third algorithm (labeled M) is based on Proposition??, which has two costs: compute the degree of gcrds of A and B expand the determinant U given in Proposition??. If the gcrd of A and B is trivial, then both Algorithms L and S compute some triangular forms of the matrix M. However, Algorithm S makes use of the special structure of M and avoids finding solutions of the linear homogeneous system defined by the transpose of M. Furthermore, Algorithm S avoids any gcd-computation in R and controls the growth of coefficients by dividing out extraneous factors. But Algorithm S has the additional cost for computing the first co-sequence. If the gcrd of A and B is of degree d > 0, then Algorithm S only needs to compute a triangular form of the matrix mat(x n d A,..., A, X m d B,..., B). But Algorithm L still needs to compute a triangular form of the transpose of the matrix mat(x n A,..., A, X m B,..., B). Moreover the higher the degree d is, the higher the second cost of Algorithm L may be, because the solution space V contains more parameters. If the degree of the gcrd of A and B can be computed efficiently, then Algorithm M needs only to expand one determinant of order (m + n 2d + 2). For example, if R = Z[t], then the gcrd of A and B can be computed efficiently by the modular method in [?]. We compared the function LCLM in the Maple package diffop by Marc van Hoeji, which appeared to use the idea of Algorithm L, with our Maple implementations of Algorithm S and Algorithm M. We generated three random polynomials in Z[t][D], A, B and C, with respective degrees d A, d B, and d C in X, regarded A, B and C as linear differential operators, and computed the primitive lclm of AC and BC. In Algorithm M the degree of the gcrd of AC and BC was computed by the modular algorithm in [?]. Our experiment was carried out in Maple V (Release 3) on an Alpha-workstation. The random polynomials in the experiment were generated by the Maple function randpoly. When d C = 1, we set C = 1. Some of the timings are summarized in Figure 2, in which the column labeled l gives the average maximal length of the integral coefficients of AC and AC, and the column labeled t gives the average degree of AC and BC in the variable t. All the timings are Maple CPU time and given in seconds. The timings show that Algorithms S and M tend to be faster than Algorithm L when d C increases. But I think that the difference between Algorithm L and Algorithms M and

17 Ore Polynomials 107 d A + d C d B + d C d C t l L S M Fig. 2. Times for computing lclms S should be smaller when d C is zero, because, in this case, the most time-consuming computation in the three algorithms is to triangularize the matrix M. For the time being, my explanation on this difference is that there is about one-third of computing time of LCLM spent on other costs, since LCLM works for various coefficient domains. I wish that better understandings about the function LCLM in the the package diffop, and functions solve and linsolve in Maple would result a clear explanation of the timings, and lead to a better way of designing experimental data. Acknowledgment The author thanks Dr. Chardin for the discussion on differential subresultants, and thanks Dr. van Hoeji for putting his package diffop on the internet. References [1] L.M. Berkovich and V.G. Tsirulik. Differential Resultants and Some of Their Applications. In Differential nye Uravneniya, 22, No. 5, , [2] M. Bronstein and M. Petkovšek. On Ore Rings, Linear Operators and Factorisation. Programming and Comput. Software, 20, pp , [3] M. Bronstein and M. Petkovšek An introduction to pseudo-linear algebra, Theoretical Computer Science, 157, pp. 3 33, [4] W.S. Brown and J.F. Traub. On the Euclid s Algorithm and the Theory of Subresultants. JACM 18, pp , [5] M. Chardin. Differential Resultants and Subresultants. Proceedings of Fundamentals of Computation Theory, Budach L. (ed.) Lecture Notes in Computer Science, No. 529, pp , [6] G.E. Collins. Subresultant and Reduced Polynomial Remainder Sequences. JACM 16, pp , 1967.

18 108 Z. M. Li [7] Z. Li. A Subresultant Theory for Linear Differential, Linear Difference, and Ore Polynomials with Applications. PhD Thesis, Technical Report 96-14, Research Institute for Symbolic Computation, Johannes Kepler University, Linz, A-4040, Austria, [8] Z. Li and I. Nemes. A Modular Algorithm for Computing Greatest Common Right Divisors of Ore Polynomials, in Proceedings of the 1997 International Symposium on Symbolic and Algebraic Computation, , ACM Press, [9] R. Loos. Generalized Polynomial Remainder Sequence. Computer Algebra, Symbolic and Algebraic Computation, B. Buchberger, G. E. Collins, and R. Loos (eds.), Springer-Verlag, Wien-New York, pp [10] B. Mishra. Algorithmic Algebra, Texts and Monographs in Computer Science, Gries D. and Schneider F.B. (eds.) Springer-Verlag [11] O. Ore. Theory of Non-Commutative Polynomials, Annals of Math 34 pp , [12] O. Perron. Algebra I die Grundlagen, Berlin: de Gruyter & Co., [13] C.M. Rubald. Algorithms for Polynomials over a Real Algebraic Number Field. PhD Thesis, Department of Computer Science, University of Wisconsin