Research Article Minor Prime Factorization for n-d Polynomial Matrices over Arbitrary Coefficient Field

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1 Complexity, Article ID , 9 pages Research Article Minor Prime Factorization for n-d Polynomial Matrices over Arbitrary Coefficient Field Jinwang Liu, Dongmei Li, and Licui Zheng School of Mathematics and Computing Sciences, Hunan University of Science and Technology, Xiangtan, Hunan, China Correspondence should be addressed to Dongmei Li; dmli@hnust.edu.cn Received 11 January 2018; Revised 4 September 2018; Accepted 10 September 2018; Published 8 October 2018 Academic Editor: Eulalia Martínez Copyright 2018 Jinwang Liu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In this paper, we investigate two classes of multivariate (n-d) polynomial matrices whose coefficient field is arbitrary and the greatest common divisor of maximal order minors satisfy certain condition. Two tractable criterions are presented for the existence of minor prime factorization, which can be realized by programming and complexity computations. On the theory and application, we shall obtain some new and interesting results, giving some constructive computational methods for carrying out the minor prime factorization. 1. Introduction In the polynomial approach, pioneered by Rosenbrock [1], matrices over R s, s d/dt are used to represent linear systems of ordinary differential equations. For more general linear functional systems, e.g., partial differential systems or delay-differential systems, the resulting system matrices are multivariate. The problem of multivariate (n-d) polynomial matrix has attracted much attention over the past decades because of its diverse applications in n-d systems, network theory, Wiener-Hopf equations, controls, and signal processing ([2 8]); its factorizations have attracted much attention over the past decades because of their wide applications ([9 25]). Lin-Bose conjecture is one of the important problems of n-d polynomial matrix factorizations, which has been proved in [15 20]. It gave a sufficient condition for a matrix to have zero prime factorization or minor prime factorization. After that, Lin et al. proposed a tractable criterion for the existence of minor prime factorizations for a class of n-d polynomial matrices whose greatest common divisor of maximal order minors is z 1 f z 2,, z n. They have presented a constructive method for carrying out the minor prime factorization when it exists [14], where the coefficient field is the complex number field. In this paper, we investigate the n-d polynomial matrices whose coefficient field is arbitrary and minor prime factorizations for two classes of n-d polynomial matrices whose the greatest common divisor of maximal order minors is z 1 f z 2,, z n p or k=1 z 1 z 1k. In this case, because the coefficient field is an arbitrary field, the many previous results (lemmas) cannot be used directly. By giving some new key lemmas, two tractable criterions for the existence of minor prime factorization are obtained. These criterions can be judged by using programming of reduced Gröbner basis. Furthermore, we give constructive computational methods for carrying out the minor prime factorization. 2. Notations and Preliminaries For an arbitrary field K, let K z 1, z 2,, z n denote the set of polynomial ring in n variables z 1, z 2,, z n with coefficients in K and K denote an algebraic closed field of K. For the convenience, K z 1, z 2,, z n is denoted simply by K z ; when K = C is the complex number field, C z 1, z 2,, z n is denoted simply by C z. Let K l m z denote the set of l m matrices with entries in K z and 0 r t denote the r t zero matrices. Throughout the paper, the argument z is omitted whenever its omission does not cause confusion. Definition 2.1 (see [21]). Let F z K l m z be of normal rank r and a 1 z, a 2 z,, a β z denote all the r r minors

2 2 Complexity of the matrix F z. Extracting the greatest common divisor (g.c.d.), denoted by d F or d,ofa 1 z, a 2 z,, a β z gives a i z = db i z, i =1,2,, β 1 Then, b 1 z, b 2 z,, b β z minors of F z. are called the reduced Definition 2.2. Let F z K l m z be of normal full rank, then F z is said to be zero left prime (zero right prime) if the l l m m minors of F z generate the unit ideal K z, which is simplified as that F z is ZLP (ZRP). First, the following two lemmas and a well-known result (Theorem 1 in [14]) are required. Lemma 2.1. Let g z K z and f z K z 2,, z n. Supposing that g f, z 2,, z n =0, then z 1 f z 2,, z n is a divisor of g z. The proof is simple and omitted. Lemma 2.2. Let f 1 z, f 2 z,, f s z K z be polynomials, then they have no common zeros in K n (are zero coprime) if and only if there are u 1 z, u 2 z,, u s z K z such that u 1 z f 1 z + u 2 z f 2 z + + u s z f s z =1, 2 i.e., f 1 z, f 2 z,, f s z is a ZLP row vector in K n z,or f 1 z,, f s z generate the unit ideal K z. Proof 1. Necessity. It is clear that f i z K z, i =1,2,, s. Let I K and I K denote the ideal generated by f 1 z, f 2 z,, f s z in K z and in K z, respectively. From the assumption and Hilbert Zero Theorem, I K = K z. Then the unit element 1 (in K) may be represented by a Gröbner basis G of I K, i.e., 1 may be reduced to 0 by G. According to computational methods of G, we have that G K z, then 1 can be represented by G in K z, i.e., there are u 1 z, u 2 z,, u s z K z such that u 1 z f 1 z + u 2 z f 2 z + + u s z f s z =1 3 Sufficiency. We assume that f 1 z, f 2 z,, f s z have common zeros in K n and z 0 = z 11, z 21,, z n1 is one of them. From (2), we have u 1 z 0 f 1 z 0 + u 2 z 0 f 2 z u s z 0 f s z 0 =0 1 4 This is a contradiction. So f 1 z, f 2 z,, f s z no common zeros in K n. have 3. Main Results For the complex number field C and an n-d polynomial matrix F z 1, z 2,, z n C z, from Lemma 2.2, we see that the reduced minors of F z 1, z 2,, z n are zero coprime if and only if the reduced minors of F z 1, z 2,, z n generate the unit ideal C z. In the following, we investigate the case in which F z 1, z 2,, z n K z, where K is an arbitrary field (may be a finite field); we shall extend the range of applicability of Theorem 2.1 and present our main results. First, several key lemmas are presented. Lemma 3.1. Let Q K l 1 l z be of normal full rank, if the reduced minors of Q generate K z, then there is a ZLP matrix w K l 1 z such that Q w =0 l 1 1. Proof 2. Since the reduced minors of Q generate the unit ideal K z, by Lin-Bose Theorem (Theorem 3.3 in [18]), Q can be l 1 l 1 factorized as Q = Q 1 B 1 such that Q 1 R is of normal l 1 l full rank, and B 1 K is ZRP. By Quillen-Suslin Theorem [26], there is a unimodular U 1 K l l z such that B 1 is its first l 1 rows, i.e., U 1 = B T 1 VT T 1, where V1 K 1 l. Let U 1 1 = V 2 w, where V 2 R l l 1, w R l 1, then B 1 w =0 l 1 1. Obviously, w is ZLP. We have that Q w = Q 1 B 1 w = Q 1 0 l 1 l =0 l 1 1. Lemma 3.2. Let F z K m l z l m and d F = q s=1 z 1 f s z 2,, z n, where f i f j 0 for i j. Then rank F f p z 2,, z n, z 2,, z n is l 1, where p =1,2,, q. Proof 3. Let F z = f ij, for p 1, 2,, q, we may m l write f ij as f ij = q ij z 1 f p z 2,, z n + b ij, 5 where q ij K z, b ij K z 2,, z n, i =1,2,, m, j =1,2,, l. Setting A a q ij m l ij z 1 f p z 2,, z, B n m l b ij, we have that F = A + B. Since d F = q m l s=1 z 1 f s z 2,, z n, then there is an l l minor Δ of F z such that z 1 f p z 2,, z n is a simple divisor of Δ. Let F j 1, j 2,, j l denote an l l submatrix consisting of the j 1, j 2,, j l rows of F such that z 1 f p z 2,, z n is a simple divisor of det F j 1, j 2,, j l. Then Theorem 2.1 (see [14]). Let F z C m l z, d F = z 1 f z 2,, z n. Then a necessary and sufficient condition for F z to admit a minor prime factorization F z = F 0 z G 0 z with F 0 z K m l z, G 0 z K l l z, and det G 0 z = d F is that there exists a full rank l 1 l submatrix F 1 z 2,, z n F 1 f z 2,, z n, z 2,, z n of F f z 2,, z n, z 2,, z n such that the reduced minors of F 1 z 2,, z n are zero coprime. Fj 1, j 2,, j l = a j1 1 + b j1 1 a j1 2 + b j1 2 a j1 l + b j1 l a j2 1 + b j2 1 a j2 2 + b j2 2 a j2 l + b j2 l a jl 1 + b jl 1 a jl 2 + b jl 2 a jl l + b jl l 6

3 Complexity 3 Since z 1 f p z 2,, z n a ij, where i =1,2,, m, j =1, 2,, l, b ij is independent of z 1, and z 1 f p z 2,, z n is a simple divisor of det F j 1, j 2,, j l, by the properties of determinant, some k 1 k l exists such that of det F z 1, z 20,, z n0, by the properties of determinant, there is some k 1 k l such that b 11 b 1k 1 a 1k b 1k+1 b 1l b j1 1 b j1 k 1 a j1 k b j1 k+1 b j1 l b j2 1 b j2 k 1 a j2 k b j2 k+1 b j2 l 0 7 b 21 b 2k 1 a 2k b 2k+1 b 2l b l1 b lk 1 a lk b lk+1 b ll 0 11 b jl 1 b jl k 1 a jl k b jl k+1 b jl l From (7) and Laplace Theorem, there is a permutation i 1 i 2 i l 1 of j 1 j 2 j l such that From (11) and Laplace Theorem, there is a permutation i 1 i 2 i l 1 of 12 l such that b i1 1 b i1 k 1 b i1 k+1 b i1 l Δ k b i1 1 b i1 k 1 b i1 k+1 b i1 l b i2 1 b i2 k 1 b i2 k+1 b i2 l 0 8 Δ k b i2 1 b i2 k 1 b i2 k+1 b i2 l b il 1 1 b il 1 k 1 b il 1 k+1 b il b il 1 1 b il 1 k 1 b il 1 k+1 b il 1 1 It is obvious that F f p z 2,, z n, z 2,, z n = B, then Δ k is an l 1 l 1 minor of F f p z 2,, z n, z 2,, z n. Since d F f p z 2,, z n, z 2,, z n =0, combined with (8), we have that rank F f p z 2,, z n, z 2,, z n is l 1. Lemma 3.3. Let F z K l l z and det F = q s=1 z 1 f s z 2,, z n, where 0 f i f j K for i j. Then rank F f p z 2,, z n, z 2,, z n = l 1 for every z 2,, z n K n 1, and rank F f p z 2,, z n, z 2,, z n = l 1; furthermore, the l 1 l 1 minors of F f p z 2,, z n, z 2,, z n generate the unit ideal K z, where p =1,2,, q. Proof 4. For arbitrary but fixed z 2 = z 20,, z n = z n0, where z 20,, z n0 K n 1, let z 1p = f p z 20,, z n0, then z 11, z 12,, z 1q are distinguished. Assuming that F z 1, z 20,, z n0 = f ij l l, for p 1, 2,, q, we may write f ij as f ij = q ij z 1 z 1p + b ij, 9 where q ij K z 1, b ij K, i, j =1,2,, l. Setting A a q ij l l ij z 1 z, B b 1p l l ij, then we l l have that F z 1, z 20,, z n0 = A + B, det Fz 1, z 20,, z n0 = a 11 + b 11 a 12 + b 12 a 1l + b 1l a 21 + b 21 a 22 + b 22 a 2l + b 2l a l1 + b l1 a l2 + b l2 a ll + b ll 10 Since det F z 1, z 20,, z n0 = q s=1 z z 1s, z 1 z 1p a ij, where i, j =1,2,, l, and b ij K, z 1 z 1p is a simple divisor It is obvious that F f p z 20,, z n0, z 20,, z n0 = B, then Δ k is an l 1 l 1 minor of F f p z 20,, z n0, z 20,, z n0.sincedet F f p z 20,, z n0, z 20,, z n0 =0,combined with (12), we have that rank F f p z 2,, z n, z 2,, z n is l 1 for every z 2,, z n K n 1, i.e., the l 1 l 1 minor of F f p z 2,, z n, z 2,, z n have no common zeros in K n 1. Noting that det F f p z 2,, z n, z 2,, z n =0, then rank F f p z 2,, z n, z 2,, z n = l 1. Furthermore, by Lemma 2.2, we obtain that the l 1 l 1 minors of F f p z 2,, z n, z 2,, z n generate the unit ideal K z. Corollary 3.1. Let F z K l l z with det F z = q k=1 z 1 z 1k, where z 11, z 12,, z 1q K are distinguished. Then rank F z 1k, z 2,, z n = l, the l 1 l 1 minors of F z 1k, z 2,, z n generate the unit ideal K z, where k =1,2,, q. Lemma 3.4. Let G z K l l z, det G z = q s=1 z 1 f s z 2,, z n, where 0 f i f j K for i j. Then, for p 1, 2,, q, G f p z 2,, z n, z 2,, z n = Gz diag 0, 1,,1 V p, where G z K l l z 2,, z n, V p K l l z 2,, z n is an invertible matrix. Proof 5. Since det G z = q s=1 z 1 f s z 2,, z n, for p 1, 2,, q, by Lemma 3.3, rank G f p z 2,, z n, z 2,, z n = l 1 and the l 1 l 1 minors of G f p z 2,, z n, z 2,, z n generate K z. By Lemma 3.1, there is a ZLP vector w p K l 1 z 2,, z n such that G F f p z 2,, z n, z 2,, z n w p z 2,, z n = 0,,0 T. According to Quillen- 13

4 4 Complexity Suslin Theorem [26], a unimodular matrix U 1 K l l z 2,, z n can be constructed such that w p z 2,, z n is its first column, then G f p z 2,, z n, z 2,, z n U 1 = Gz diag 0, 1,,1, where G z K l l z 2,, z n. Setting V p = U 1 1, then the desired results are obtained. Lemma 3.5. Let F z K l 1 l z,iff z is of normal full rank and F z has the following factorization: Fz = Az diag 0, 1,,1 Vz, 15 where A z K l 1 l z, V z K l l z is a unimodular matrix. Then the reduced minors of F z generate the unit ideal K z. Proof 6. We partition A z, V z respectively, 14 as the following form, Az = A 1 z A 2 z, 16 Vz = V T 1 z VT 2 z T, where A 1 z K l 1 1 z, A 2 z K l 1 l 1 z, V 1 z K 1 l z, V 2 z K l 1 l z. Then V 2 z is ZLP, i.e., the reduced minors of V 2 z generate the unit ideal K z, and Fz = 0 A 2 z V T 1 z VT 2 z T = A 2 z V 2 z So the reduced minors of F z are the same as that of V 2 z, which generate the unit ideal K z. Theorem 3.1. Let K be an arbitrary field, F z K m l z be of normal full rank l m, d F = z 1 f z 2,, z n d z. Then a necessary and sufficient condition for F z to admit a minor prime factorization F z = F 0 z G 0 z with F 0 z K m l z, G 0 z K l l z, and det G 0 z = d z is that there exists a normal full rank l 1 l submatrix F 1 z 2,, z n F 1 f z 2,, z n, z 2,, z n of F f z 2,, z n, z 2,, z n such that the reduced minors of F 1 z 2,, z n generate the unit ideal K z. Proof 7. Necessity. Supposing that F z prime factorization 17 admits a minor Fz = F 0 z G 0 z, 18 with F 0 z K m l z, G 0 z K l l z, and det G 0 z = z 1 f z 2,, z n = d z. From Lemma 3.4, G 0 f z 2,, z n, z 2,, z n = G 0 z diag 0, 1,,1 V, wherev K z 2,, z n is unimodular. Combined with (18), we have F f z 2,, z n, z 2,, z n = F 0 f z 2,, z n, z 2,, z n G 0 f z 2,, z n, z 2,, z n = F 0 f z 2,, z n, z 2,, z n G 0 z diag 0, 1,,1 V By Lemma 3.2, we see that the normal rank of F f z 2,, z n, z 2,, z n is l 1. Hence, from (19), there exists a normal full rank l 1 l submatrix F 1 z 2,, z n F 1 f z 2,, z n, z 2,, z n of F f z 2,, z n, z 2,, z n such that F 1 z 2,, z n = Az diag 0, 1,,1 V, 20 where A z is a l 1 l submatrix of F 0 f z 2,, z n, z 2,, z n G 0 z. By Lemma 3.5, the reduced minors of F 1 z 2,, z n generate the unit ideal K z. Sufficieny. Assuming that there exists a normal full rank l 1 l submatrix F 1 z 2,, z n F 1 f z 2,, z n, z 2,, z n of F f z 2,, z n, z 2,, z n such that the reduced minors of F 1 z 2,, z n generate the unit ideal K z. According to Lemma 3.1, a ZRP vector ω K l 1 z 2,, z n can be constructed such that F 1 z 2,, z n ω z 2,, z n =0,,0 T 21 Since F f z 2,, z n, z 2,, z n is of normal rank l 1, by (21) and the fact that F 1 z 2,, z n is a normal full rank submatrix of F f z 2,, z n, z 2,, z n, we have F f z 2,, z n, z 2,, z n ω z 2,, z n =0,,0 T 22 By Quillen-Suslin Theorem [26], a unimodular square matrix V K l l z 2,, z n can be constructed such that ω z 2,, z n is its first column; combined with Lemma 2.1, we have Fz Vz 2,, z n = F 0 z diag z 1 f z 2,, z n,1,,1, where F 0 z K m l z. It follows that Fz = F 0 z G 0 z, 24 where G 0 z = diag z 1 f z 2,, z n,1,,1 V 1 K l l z with det G 0 z = z 1 f z 2,, z n = d F. Remark 3.1. When K = C is the complex number field, according to Hilbert Zero Theorem, we see that the condition in Theorem 3.1 is the same as that of Theorem 2.1, and Theorem 3.1 is a generalization of Theorem 2.1.

5 Complexity 5 According to Theorem 3.1, in order to check that F z satisfies the conditions of Theorem 3.1, we first find all the l 1 l submatrices of F f z 2,, z n, z 2,, z n, denoted by F 1, F 2,, F s, where s = C l 1 m = m m 1 m l / l 1. For every F i 1 i s, we compute the l 1 l 1 minors of F i, saying as a i1, a i2,, a il. Let the ideal I i = a i1, a i2,, a il, we compute the reduced Gröbner basis G i of I i.ifg i contains only a polynomial d i z, by the following result (Theorem 3.2), the reduced minors of F i generate the unit ideal K z. Theorem 3.2. Let d i z a i1, a i2,, a il, i.e., be the greatest common divisor of a ij = b j d i z, j =1,2,, l, 25 and G i be the reduced Gröbner basis of the ideal I i = a i1, a i2,, a il. Then G i contains only a polynomial c i z if and only if b 1, b 2,, b l generate the unit K z. Proof 8. Necessity. Assuming that G i = c i z, it is obvious that c i z a ij, where j =1,2,, l. Then c i z d i z. Setting d i z = p i z c i z, since c i z I i, we have that there are u 1, u 2,, u l such that c i z = u 1 a i1 + u 2 a i2 + + u l a il = u 1 b 1 d i z + u 2 b 2 d i z + + u l b l d i z, d i z = p i z c i z = p i u 1 b 1 d i z + p i u 2 b 2 d i z + + p i u l b l d i z So 1=p i u 1 b 1 + p i u 2 b p i u l b l. Thus, b 1, b 2,, b l generate the unit ideal K z. Sufficiency. Assuming that b 1, b 2,, b l generate the unit ideal K z, then the reduced Gröbner basis of the ideal b 1, b 2,, b l is 1. Furthermore, the reduced Gröbner basis G i of I i = b 1 d i z, b 2 d i z,, b l d i z is d i z. The reduced Gröbner basis algorithm of an ideal can be found in [16]. Remark 3.2. From the discussion above, we see that judging whether F z satisfies the conditions of Theorem 3.1 or not, we only need to compute the reduced Gröbner basis of I i, where i =1,2,, l. Lin et al. studied the matrix factorization for a class of matrix whose g.c.d. of maximal order minors has divisor p k=1 z 1 z 1k ([13]); in the following, we study the minor prime factorization for this class of matrices and propose a tractable criterion. Theorem 3.3. Let K be an arbitrary field, F z K m l l m be of normal full rank, d F = p k=1 z 1 z 1k, where z 11, z 12,, z 1p K are distinguished. Then F z admits a minor prime factorization F z = F 0 z G 0 z with F 0 z K m l z, 26 G 0 z K l l z, and det G 0 z = d F if and only if there are normal full rank l 1 l submatrices F k z 2,, z n F k z 1k, z 2,, z n of F z 1k, z 2,, z n such that the reduced minors of F k z 2,, z n generate the unit ideal K z, where k =1,2,, p. Proof 9. Necessity. Supposing that F z prime factorization as follows: admits a minor Fz = F 0 z G 0 z, 27 where F 0 z K m l z, G 0 z K l l z, and det G 0 z = d F. For arbitrary but fixed k 1 k p, from (27) and Lemma 3.4, we have Fz 1k, z 2,, z n = F 0 z 1k, z 2,, z n G 0 z 1k, z 2,, z n = F 0 z 1k, z 2,, z n Gz 1k, z 2,, z n diag 0, 1,,1 V 28 By Lemma 3.2, we obtain that rank F z 1k, z 2,, z n is l 1. Hence, from (28), there is a normal full rank l 1 l submatrix F k z 2,, z n F k z 1k, z 2,, z n of F z 1k, z 2,, z n such that F k z 2,, z n = Az diag 0, 1,,1 V, 29 where A z is an l 1 l submatrix of F 0 z 1k, z 2,, z n G z 1k, z 2,, z n. By Lemma 3.5, the reduced minors of F k z 2,, z n also generate the unit ideal K z. Sufficiency. If there exists a normal full rank l 1 l submatrix F 1 z 2,, z n F 1 z 11, z 2,, z n of F z 11, z 2,, z n such that the reduced minors of F 1 z 2,, z n generate the unit ideal K z, according to Lemma 3.1, we can construct a ZRP vector w 1 K l l z 2,, z n such that F 1 z 2,, z n w 1 z 2,, z n =0,,0 T 30 By Lemma 3.2, F z 11, z 2,, z n is of normal rank l 1, from (30), and the fact that F 1 z 2,, z n is a normal full rank submatrix of F z 11, z 2,, z n, we have Fz 11, z 2,, z n w 1 z 2,, z n =0,,0 T 31 By Quillen-Suslin Theorem [26], a unimodular square matrix V 1 K l l z 2,, z n can be constructed such that w 1 is its first columns; combined with Lemma 2.1, we have Fz V 1 z 2,, z n = A 1 z diag z 1 z 11,1,,1, 32 where A 1 z K m l z. It follows that Fz = A 1 z G 1 z, 33 where G 1 z = diag z 1 z 11,1,,1 V 1 1.

6 6 Complexity It is obvious that det G 1 z = z 1 z 11, d A 1 = p k=2 z 1 z 1k, and from (33), we have Fz 12, z 2,, z n F 2 z 12, z 2,, z n = A 1 z 12, z 2,, z n diag z 12 z 11,1,,1 V 1 1, 34 = A 12 z 12, z 2,, z n diag z 12 z 11,1,,1 V 1 1, 35 where A 12 z 12, z 2,, z n is an l 1 l submatrix of A 1 z 12, z 2,, z n. By Lemma 3.2, rank F z 12, z 2,, z n is l 1 and from (34), rank A 1 z 12, z 2,, z n is also l 1. By the assumption, the reduced minors of F 2 z 12, z 2,, z n generate the unit ideal K z. Since G 1 z 12,1,,1 = diag z 12 z 11, 1,,1 V 1 1 is unimodular, by (35) and Cauchy-Binet formula, we obtain that the reduced minors of A 12 z 12, z 2,, z n also generate the unit ideal K z. Repeating the procedure above for A 1 z with respect to z 1 z 12, we have or A 1 z = A 2 z G 2 z 36 Fz = A 2 z G 2 z G 1 z, 37 such that A 2 z K m l z, G 2 z = diag z 1 z 12,1,,1 V 1 2 with det G 2 z = z 1 z 12, V 2 is a unimodular square matrix, d A 2 = p k=3 z 1 z 1k. And F 3 z 13, z 2,, z n = A 23 z 13, z 2,, z n diag z 13 z 12,1,,1 V 1 2 diag z 13 z 11,1,,1 V 1 1, 38 where A 23 z 13, z 2,, z n is an l 1 l submatrix of A 2 z 13, z 2,, z n. Furthermore, since F 3 z 13, z 2,, z n is of normal rank l 1 and the reduced minors of F 3 z 13, z 2,, z n generate the unit ideal K z, by (38), then A 23 z 13, z 2,, z n is of normal rank l 1, and the reduced minors of A 23 z 13, z 2,, z n also generate the unit ideal K z. Repeating the same procedure continuously with respect to z 1 z 1k, k =3,, p, we obtain that or Fz = A p z G p z G 2 z G 1 z 39 Fz = F 0 z G 0 z, 40 where F 0 z = A p z K m l z, G 0 z = p k=1 G p z, det G 0 z = p k=1 z 1 z 1k. Remark 3.3. Similar to Remark 3.2, judging whether F z satisfies the conditions of Theorem 3.1 or not, we only need to compute the reduced Gröbner bases of some ideals. Example 3.1. Let K Z/ 5 be the finite field, F z 1, z 2, z 3 K z 1, z 2, z 3, and Fz 1, z 2, z 3 = 2z z3 1 z2 2 +2z3 1 z 3 +4z 1 z 3 +3z 2 2 z 3 +4z z 2z 3 3z 2 +4z 3 3z z 1z z 1z z 1z 2 +4z z2 2 z 3 3z 1 4z 2 1 z 2 + z 1 z 2 2 z 3 +3z 1 z z 1z 3 +4z 2 2 z 3 + z 2 z 3 3z z 3 41 It is easy to check that the greatest common divisor of the 2 2 minors of F z is d F = z 1 +2z z 3 = z 1 3z z 3. By computing, we have F 3z z 3, z 2, z 3 = 4z z 2z 3 + z 2 3 3z 2 +4z 3 2z z2 2 z 3 + z 2 z 3 +3z 2 3 4z z 3 4z z4 2 z 3 +4z 3 2 z 3 +3z 2 2 z3 2 + z 2z z 2z 3 2z z2 2 z 3 +3z z 3 42 Obviously, F 3z z 3, z 2, z 3 is of normal rank 1, the reduced minors of the first row of F 3z z 3, z 2, z 3 (a 1 2 submatrix of F 3z z 3, z 2, z 3 ) are 3z 2 +4z 3 and 1 (since 4z z 2z 3 + z 2 3 = 3z 2 +4z 3 3z 2 +4z 3 ), which generate the unit ideal K z. By Theorem 3.1, there is a minor prime factorization for F z 1, z 2, z 3.

7 Complexity 7 To construct such a factorization, let Let 3 0 U = z 2 +3z 3 2, 43 we have 0 z 2 +3z 3 F 3z z 3, z 2, z 3 U = 0 3z z 3, 0 4z z2 2 z 3 + z z 3 44 we have z z 3 z 2 +3z 3 F 0 z = 4z 1 + z 2 2 z 1, 4z 1 z 3 + z 3 z z 3 G 0 z = z 1 +2z z , 4z 2 +2z 3 3 Fz 1, z 2, z 3 = F 0 z 1, z 2, z 3 G 0 z 1, z 2, z 3 47 which gives Fz 1, z 2, z 3 U = z z 3 z 2 +3z 3 4z 1 + z 2 2 z 1 4z 1 z 3 + z 3 z z 3 z 1 +2z z Remark 3.4. In our two main theorems (Theorem 3.1 and 3.3), F z K m l l m are of normal full rank. In fact, our methods can also deal with some matrices not of full normal rank. When the normal rank of F z is r and r < l, if the reduce minors of F z generate the unit ideal K z, one has that F z = Q 1 z B 1 z such that Q 1 K m r of normal full rank, and B 1 K r l is ZRP. In order to consider the factorization of F z, one needs to consider the factorization of Q 1 z. See the following example. Example 3.2. Let K be the rational number field, A z K 3 3 z 1, z 2, z 3, and Az = 3z 2 1 4z 1z 2 4z 2 3z 2 1 z 2 4z 2 1 z 2 + z 1 z 2 1 3z 2 1 z 2 +4z 1 z 2 z 2 0 z z 2 1 z 2z 3 z 1 z 2 z 3 z 3 1 z 2z 3 z 2 1 z 2z 3 +3z 3 z 2 1 z 2z 3 +2z 1 z 3 By computing the minors of A z, we have that rank A z =2, d A = z 1 1 z 1 3, the reduced minors of A z generate the unit ideal K z. Using Theorem 3.2 in [18], we have thata z = F z B z, where Fz = Bz =, 3z 3 4z 1 z 2 4z 2 z 1 1 3z 1 z 2 9z 2 z 1 3 z 2 1 z 2z 3 + z 1 z 2 z 3 1 z 1 1 3z 2 1+3z 1 z 2 z 3z 2, 49 and 2, which generate the unit ideal K z. F 3, z 2, z 3 is also of normal rank 1, the reduced minors of the first row of F 3, z 2, z 3 are 8z 2 and 8, which also generate the unit ideal K z. By Theorem 3.3, there is a minor prime factorization for F z 1, z 2, z 3. To construct such a factorization, let we have U 1 = 1 0 3z 2 1, 50 and B z is ZLP. It is easy to check that the greatest common divisor of the 2 2 minors of F z is d F = z z 1 +3= z 1 1 z 1 3. Obviously, F 1, z 2, z 3 is of normal rank 1, the reduced minors of the second row of F 1, z 2, z 3 (a 1 2 submatrix of F 1, z 2, z 3 ) are 6z 2 F 1, z 2, z 3 = 0 0 6z 2 2, 9z 2 z 3 3z 3

8 8 Complexity which gives Fz 1, z 2, z 3 U 1 = Thus, Fz 1, z 2, z 3 = Let F 1, z 2, z 3 U 1 = z 1 z 2 + z 2 z z 1 3 z 1 z 2 z 3 3z 3 3z 1 z 2 + z 2 z z 1 3 z 1 z 2 z 3 3z 3, 51 z z z Thus, F 1 z 1, z 2, z 3 = Fz 1, z 2, z 3 = Let F 0 z = z 1 z 2 z z 2 z 1 3 z 2 z 3 3z 3 z 1 z 2 z z 2 z 1 3 z 2 z 3 3z 3 z z 2 1 z 1 z 2 z z 2 z 1 3, z 2 z 3 3z 3 G 0 z = z z 2 1 z z 2 1, z z 2 1 z z 2 1, we have which gives F 1 z 1, z 2, z 3 = U 2 = F 1 3, z 2, z 3 = F 1 3, z 2, z 3 U 2 = F 1 z 1, z 2, z 3 U 2 = 3z 1 z 2 + z 2 z z 1 3, z 1 z 2 z 3 3z z 2 1, 8z , 3z 2 z 3 3z , 0 3z 3 z 1 z 2 z z 2 z 1 3 z 2 z 3 3z 3 z it is clear that det G 0 z = z 2 1 4z 1 +3=d A. We have Fz 1, z 2, z 3 = F 0 z 1, z 2, z 3 G 0 z 1, z 2, z 3, Az 1, z 2, z 3 = F 0 z 1, z 2, z 3 G 0 z 1, z 2, z 3 Bz 1, z 2, z 3, where B z 1, z 2, z 3 is ZLP. The desired minor prime factorization for A z 1, z 2, z 3 is obtained. 4. Conclusions In this paper, we have investigated the minor prime factorization for two classes of n-d polynomial matrices whose p g.c.d. of maximal order minors is z 1 f z 2,, z n or k=1 z 1 z 1k with their entries in K z, where K is an arbitrary field. We have presented some key lemmas and two tractable criterions for the existence of minor prime factorizations. These criterions are easily checked by computing reduced Gröbner basis of the ideal generated by lower order minors, and the minor prime factorization can be carried out by Lin s constructive methods. Data Availability No data were used to support this study. Conflicts of Interest The authors declare that they have no conflicts of interest. 59

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