ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.451229 J. Algebra Comb. Discrete Appl. 5(3) • 117–127 Received: 16 October 2016 Accepted: 25 April 2018 Journal of Algebra Combinatorics Discrete Structures and Applications Characterization of 2×2 nil-clean matrices over integral domains Research Article Kota Nagalakshmi Rajeswari, Umesh Gupta Abstract: Let R be any ring with identity. An element a ∈ R is called nil-clean, if a = e + n where e is an idempotent element and n is a nil-potent element. In this paper we give necessary and sufficient conditions for a 2×2 matrix over an integral domain R to be nil-clean. 2010 MSC: 15A24 Keywords: Nil-clean matrix, Idempotent matrix, Nil-potent matrix, Diophantine equation 1. Introduction Let R be an associative ring with identity. An element a ∈ R is called clean if a = e + u, where e, u ∈ R, e is an idempotent and u is a unit. We shall call such a representation a clean representation of a. Further, a is called strongly clean (uniquely clean) if it has a clean representation in which the idempotent and the unit commute (it has only one clean representation). The ring R is said to be clean (strongly clean, uniquely clean) if each of its element is. The notion of clean rings was introduced by W. K. Nicholson [10] while studying rings with exchange property and observed that these rings have the exchange property. The notions of strongly clean and uniquely clean ring were introduced subsequently. Ever since, many mathematicians have studied these rings extensively. In an attempt to unify several variants of strongly clean property A. J. Diesl [7] has introduced the notion of nil-clean rings. A ring R is nil-clean if each of its elements is the sum of an idempotent and a nil-potent. The notion of strongly nil-clean (uniquely nil-clean) were defined analogous to those of strongly clean (uniquely clean). These rings have been studied in great detail in [7]. Among other things, the author has shown [[7], Prop 3.4] that a nil-clean ring is clean and immediately posed the question: "Whether or not a nil-clean element in a ring is clean?" K. N. Rajeswari (Corresponding Author); School of Mathematics, Vigyan Bhawan, Khandwa Road, Devi Ahilya University, Indore-452 001, India (email: knr k@yahoo.co.in). Umesh Gupta; Sanghvi Institute of Management and Science, Behind IIM, Pigdamber, Indore-453 331, India (email: ugupta19@gmail.com). 117 https://orcid.org/0000-0002-7722-2491 https://orcid.org/0000-0001-7732-5768 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 This question was answered in negative by Andrica Dorin and Calugareanu Grigore in [3]. They have given an example of a matrix in M2(ZZ) which is nil-clean but not clean. In [2] they have also given example of a unit regular matrix (matrix which is the product of an idempotent matrix and a unit) which is not nil-clean and that of a nil-clean matrix which is not unit regular. To tackle questions as above, characterization of elements having a prescribed decomposition, nil- clean decomposition in the present context, are essential. As matrices are good sources of examples, it is nice to have characterization of matrices having nil-clean decomposition. Another motivation to study nil-clean matrices stems from Lie Theory, as strongly nil-clean decomposition of matrices over fields is the classical Jordan-Chevalley decomposition which is of central importance in Lie Theory. The nil-clean (strongly nil-clean) ring theory has application in various branches of mathematics. For instance in [6] nil-cleanness of endomorphism ring of a group of finite rank and in [12] nil-cleanness of group rings have been studied. In [4], the authors have introduced the notion of nil-clean graph of a ring R for which R is the set of vertices and a, b ∈ R are adjacent if a + b is nil-clean. The characterization of nil-clean elements will certainly help understand such applications. In this paper, we give a characterization of nil-clean 2 × 2 matrices over an integral domain. The idempotents, unipotents (element of the form 1 + n, where n is nil-potent) and nil-potents in any ring R are nil-clean. By nil-clean we mean, nil-clean elements other than these. It is well known that when R is an integral domain, an idempotent 2 × 2 matrix other than 0 and I is of the form [ x y w 1−x ] with x(1 − x) = yw. For the sake of completeness, a proof is given [see: Lemma 2.1]. Similarly, 2 × 2 nil-potent matrices are of the form [ a b c −a ] with a2 + bc = 0 [see: Lemma 2.2]. Thus a nil-clean matrix A = [ a b c d ] satisfies the conditions (i) Tr(A) = 1. (ii) det [ a−x b−y c−w d−1 + x ] = 0, where x(1−x) = yw. Expanding the determinant and simplifying gives (2a−1)x + cy + bw = a2 + bc. (1) Thus, if [ a b c d ] is nil-clean as above, then it has trace 1 and the equation (1) has a solution x, y, w with x(1−x) = yw. We shall show these two conditions characterize a nil-clean matrix [see: Theorem 3.1]. In [9], the authors have characterized nil-clean matrices over UFD. But, our approach is different from their approach. The authors of [9] have used characterization of rank one 2 × 2 matrices over a UFD R to arrive at the characterization while we have used solutions of certain Diophantine equation to get the characterization. When the entries are integers, the equation (1) is a linear Diophantine equation in 3 variables which has a solution if and only if gcd(2a − 1, c, b) divides (a2 + bc) and this forces gcd(2a − 1, b, c) = 1. The condition that equation (1) has a solution satisfying x(1 − x) = yw yields a second degree Diophantine equation in two variables. This equation has been analyzed for its solutions in section 4. Discussion in the sections 3 and 4 has been summarized in section 5 along with illustrations. It has been shown [[7], Cor. 5.4] that central idempotents and central nil-potents in a ring are uniquely nil-clean. In section 6, we let R be a Principal Ideal Domain (PID) and show using the characterization of section 3 that if an idempotent in M2(R) is not central then it is not uniquely nil-clean. Thus in M2(R), the only uniquely nil-clean idempotents are central idempotents i.e., the zero matrix and the identity matrix. We observe that nilpotents in M2(R) are uniquely nil-clean. 118 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 Throughout the paper R is an integral domain, Mn(R) is the ring of n × n matrices over R. For A ∈ Mn(R), Tr(A) denotes the trace of A, R(A) denotes its rank and |A| its determinant. Let Rn be the free R-module of n tuples of elements of R. 2. Preliminaries Let R be an integral domain. In this section we recall some results and facts. Lemma 2.1. E = [ x y w z ] ∈ M2(R) (E 6= I, E 6= 0) is idempotent if and only if it is of the form[ x y w 1−x ] with x(1−x) = yw. Proof. Let E = [ x y w z ] (E 6= I, E 6= 0) be an idempotent matrix. As idempotent matrix, which is not identity is singular, we have xz = yw and as Tr(E) = R(E) = 1, x + z = 1 i.e., z = 1 − x. Hence the matrix is [ x y w 1−x ] with x(1−x) = yw. Converse can be checked by direct calculation. Lemma 2.2. N = [ a b c d ] ∈ M2(R) is nil-potent if and only if it is of the form [ a b c −a ] with a2+bc = 0. Proof. Let N = [ a b c d ] be a nil-potent matrix. Since trace and determinant of a nil-potent matrix are always zero, a + d = 0 and bc − ad = 0 i.e., d = −a and bc + a2 = 0. Hence the matrix is [ a b c −a ] with bc + a2 = 0. Converse is easy to check. Proposition 2.3. The equation x2 −y2 = c admits integer solutions if and only if c belongs to 4ZZ or it is odd. Proof. (See [[13], Property 1]). Proposition 2.4. The equation x2−dy2 = c2 (where d is not a perfect square) admits an infinite number of solutions in IN. Proof. (See [[13], Property 2]). Theorem 2.5. Let R be a PID and A ∈ Mn(R) be an idempotent matrix. Then A is similar to the matrix [ Ir 0 0 0 ] where Ir is the identity matrix of order r and r = R(A). Proof. Let B be a basis of Rn, the free R-module of rank n. Then A determines a module homomor- phism T : Rn → Rn which is idempotent and whose matrix with respect to B is A. Let I be the identity transformation, N(T) and N(I − T) be the null spaces of T and I − T respectively. As R is a PID, N(T) and N(I − T) are free. As T is idempotent, for x ∈ Rn, Tx ∈ N(I − T) and (I − T)x ∈ N(T). Hence Rn = N(I − T) + N(T). Clearly, N(I − T) ∩ N(T) = 0. Thus Rn = N(I − T) ⊕ N(T). Let {v1, ....vr} be a basis of N(I−T) and {vr+1, ....vn} be a basis of N(T). Then B′ = {v1, ....vr, vr+1, ....vn} 119 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 is a basis of Rn and the matrix of T with respect to B′ is [ Ir 0 0 0 ] . Therefore, A is similar to [ Ir 0 0 0 ] and r = R(A). It is well known that a polynomial equation with integral coefficients for which integral solutions are desired is called a Diophantine equation. In the later part we will use two types of Diophantine equations, one linear Diophantine equation in three variables and other is quadratic Diophantine equation in two variables. A linear Diophantine equation with three variables is of the form ax + by + cz = d (2) where a, b, c and d are integers. This equation has a solution if and only if gcd(a, b, c) divides d (see [11]). It is also that once it has a solution, it has infinite number of solutions. Solution to the equation (2) is obtained as follows (see [8]): Let gcd(a, b) = p, a′ = a/p and b′ = b/p. Let (u0, v0) be any one solution of a′u + b′v = c, (z0, t0) be any one solution of cz + pt = d and (x0, y0) be any one solution of a′x + b′y = t0, then the general solution of ax + by + cz = d is x = x0 + b ′k −u0m y = y0 −a′k −v0m z = z0 + pm k, m ∈ ZZ We shall denote quadratic Diophantine equation in two variables by Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (3) where A, B, C, D, E and F are integers. We shall use the notations of [3] and follow the method of solution given there. Let D = B2 − 4AC, g = gcd(B2 − 4AC, 2AE − BD) and 4 = 4ACF + BDE − AE2 −CD2 −FB2. Then the equation reduces to gX2 − D g Y 2 = −4A 4 g (4) where X = D g y + 2AE−BD g and Y = 2Ax + By + D. Now, we have the following cases: If D > 0 and perfect square, the equation (4) admits finite number of integer solutions provided right hand side of (4) is non zero [see: Proposition 2.3]. If D > 0 and not a perfect square, the equation (4) admits an infinite number of solutions in IN[see: Proposition 2.4] If D < 0 i.e., B2 − 4AC < 0. (see [1]) Since the ellipse is a closed figure, the number of solutions will be finite(if a solution exists). The equation is Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (5) ⇒ Cy2 + (Bx + E)y + (Ax2 + Dx + F) = 0 ⇒ y = −(Bx + E)± √ (Bx + E)2 −4C(Ax2 + Dx + F) 2C (6) 120 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 for a value of x there will be two values of y except at the left and right extremes of the ellipse. In this case there will be only one value of y. To determine the location of the left and right extremes we should equal the square root to zero. i.e., (Bx + E)2 −4C(Ax2 + Dx + F) = 0 ⇒ (B2 −4AC)x2 + 2(BE −2CD)x + (E2 −4CF) = 0 so the value of x should be between the roots of this equation. If the roots are not real, there will be no solution to the equation (5), all integer values of x should be replaced in equation (6) in order to find an integer value of y. 3. Main result Let R be an integral domain. It is clear that idempotent matrices, nil-potent matrices and unit matrices of the form I + N, where I is the identity matrix and N is a nil-potent matrix are nil-clean. We shall call these trivially nil-clean. So from now onwards, nil-clean means non-trivially nil-clean. In this section, we shall discuss necessary and sufficient conditions for a matrix A ∈ M2(R) to be nil-clean. Theorem 3.1. Let A = [ a b c d ] be any matrix in M2(R). Then A is nil-clean if and only if Tr(A) = 1 and (2a−1)x + cy + bw = a2 + bc has a solution satisfying x(1−x) = yw. Proof. Let A be nil-clean. Say A = E + N, where E 6= 0, E 6= Id, is an idempotent matrix and N is a nil-potent matrix. Then A−E = [ a−x b−y c−w d−1 + x ] is nil-potent. Hence (A−E)2 = [ a−x b−y c−w d−1 + x ]2 = 0. Since, the trace of a nil-potent matrix is zero, a − x + d − 1 + x = 0 ⇒ a + d = 1 i.e. Tr(A) = 1. Also, determinant of a nil-potent matrix is zero, Hence (a−x)(d−1 + x)− (b−y)(c−w) = 0 ⇒ (a−x)(x−a)− (b−y)(c−w) = 0 ⇒ (a−x)2 + (b−y)(c−w) = 0 ⇒ x2 −2ax + a2 + bc−yc− bw + yw = 0 ⇒ x2 −2ax + a2 + bc−yc− bw + x−x2 = 0 ⇒ x(1−2a)−yc−wb + a2 + bc = 0 ⇒ (2a−1)x + cy + bw = a2 + bc. Conversely, let Tr(A) = 1 and (2a−1)x + cy + bw = a2 + bc have a solution x, y, w with x(1−x) = yw. Let E = [ x y w 1−x ] . Then A−E = [ a−x b−y c−w d−1 + x ] . Now, Tr(A−E) = (a−x) + (d−1 + x) = a + d−1 = 0 (∵ Tr(A) = 1) 121 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 And |A−E| = ∣∣∣∣∣ a−x b−yc−w d−1 + x ∣∣∣∣∣ = (a−x)(d−1 + x)− (b−y)(c−w) = (a−x)(x−a)− (b−y)(c−w) = −x2 −a2 + 2ax− bc + yc + bw −yw = −x2 −a2 + 2ax− bc + yc + bw − (x−x2) (∵ yw = x(1−x)) = (2a−1)x + cy + bw − (a2 + bc) = 0 [∵ (2a−1)x + cy + bw = a2 + bc] Hence A−E is nil-potent, which proves A is nil-clean. 4. Analysis on main result when R = ZZ In this section, we shall analyze the linear Diophantine equation (2a − 1)x + cy + bw = a2 + bc as to when it has a solution satisfying x(1−x) = yw. Let a1 = 2a−1 and d1 = a2 + bc, then the equation becomes a1x + cy + bw = d1. Let gcd(a1, c) = p, a′1 = a1/p and c ′ = c/p. Let (u0, v0) be any one solution of a′1u + c ′v = b, (w0, t0) be any one solution of bw + pt = d1 and (x0, y0) be any one solution of a′1x + c ′y = t0, then the general solution of a1x + cy + bw = d1 is x = x0 + c ′k −u0m y = y0 −a′1k −v0m w = w0 + pm The condition x−x2 = yw ⇒(x0 + c′k −u0m)− (x0 + c′k −u0m)2 = (y0 −a′1k −v0m)(w0 + pm) ⇒(x0 + c′k −u0m)− (x20 + c ′2k2 + u20m 2 + 2x0c ′k −2x0u0m− 2c′ku0m) = y0w0 −a′1kw0 −v0mw0 + y0pm−a ′ 1kpm−v0m 2p ⇒c′2k2 − (2c′u0 + a′1p)km + (u 2 0 −v0p)m 2 + (2x0c ′ − c′ −a′1w0)k + (u0 −2x0u0 −v0w0 + y0p)m + (x20 −x0 + y0w0) = 0 ⇒Ak2 + Bkm + Cm2 + Dk + Em + F = 0 (7) where A = c′2, B = −(2c′u0 + a′1p), C = (u 2 0 −v0p), D = (2x0c ′ − c′ −a′1w0), E = (u0 −2x0u0 −v0w0 + y0p), and F = (x 2 0 −x0 + y0w0) Now, D = B2 −4AC = (2c′u0 + a ′ 1p) 2 −4c′2(u20 −v0p) = 4c′2u20 + a ′2 1 p 2 + 4c′u0a ′ 1p−4c′2u20 + 4c′2v0p = a21 + 4c ′u0a ′ 1p + 4c ′2v0p = a21 + 4c ′u0a ′ 1p + 4c ′p(b−a′1u0) (∵ a′1u0 + c′v0 = b) = a21 + 4cb. 2AE −BD = 2c′2(u0 −2x0u0 −v0w0 + y0p) + (2c′u0 + a′1p)(2x0c′ − c′ −a′1w0) 122 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 = 2c′2u0 −4c′2x0u0 −2c′2v0w0 + 2c′2y0p + 4c′2u0x0 −2c′2u0 −2c′u0a′1w0 + 2a′1px0c′ −a′1pc′ −a′21 pw0 = −2c′w0(a′1u0 + c′v0) + 2c′p(a′1x0 + c′y0)−a′1pc′ −a′21 pw0 = −2c′w0b + 2c′pt0 −a′1pc′ −a′21 pw0 ( ∵ a′1x0 + c ′y0 = t0 a′1u0 + c ′v0 = b ) = −2c′w0b + 2c′(d1 − bw0)−a′1pc′ −a′21 pw0 = 2c′d1 −4c′w0b−a′1pc′ −a′21 pw0 = 1 p (2cd1 −4cw0b−a1c−a21w0) Let g = gcd(D, 2AE −BD). 4 = 4ACF + BDE −AE2 −CD2 −FB2 = 4c′2(u20 −v0p)(x20 −x0 + y0w0)− (2c′u0 + a′1p)(2x0c′ − c′ −a′1w0) (u0 −2x0u0 −v0w0 + y0p)− c′2(u0 −2x0u0 −v0w0 + y0p)2 − (u20 −v0p)(2x0c′ − c′ −a′1w0)2 − (x20 −x0 + y0w0)(2c′u0 + a′1p)2 = 4c′2u20x 2 0 −4c′2u20x0 + 4c′2u20y0w0 −4c′2v0px20 + 4c′2v0px0 −4c′2v0py0w0 −4c′2x0u20 + 2c′2u20 + 2c′a′1w0u20 + 8c′2u20x20 −4c′2u20x0 −4c′u20a′1w0x0 +4c′2u0v0w0x0 −2c′2u0v0w0 −2c′u0a′1w20v0 −4c′2u0y0px0 + 2c′2u0y0p +2c′u0a ′ 1y0pw0 −2a′1px0c′u0 + a′1pc′u0 + a′21 pw0u0 + 4a′1x20c′u0p −2a′1pc′x0u0 −2a′21 pw0x0u0 + 2a′1pv0w0x0c′ −a′1pc′v0w0 −a′21 w20v0p −2a′1p2y0x0c′ + a′1p2c′y0 + a′21 p2y0w0 − c′2u20 −4c′2x20u20 − c′2v20w20 −c′2y20p2 + 4c′2x0u20 + 2c′2u0v0w0 −2c′2u0y0p−4c′2x0u0v0w0 +4c′2x0u0y0p + 2c ′2v0w0y0p−4u20x20c′2 −u20c′2 −u20a′21 w20 + 4u20x0c′2 +4u20x0c ′a′1w0 −2u20c′a′1w0 + 4v0px20c′2 + c′2v0p + a′21 w20pv0 −4c′2x0v0p −4c′a′1w0x0v0p + 2c′a′1w0v0p−4x20c′2u20 −x20a′21 p2 −4c′a′1x20u0p + 4c′2u20x0 +a′21 p 2x0 + 4c ′a′1u0x0p−4c′2u20y0w0 −a′21 p2y0w0 −4c′a′1u0y0w0p = (4c′2u20x 2 0 + 8c ′2u20x 2 0 −4c′2u20x20 −4c′2u20x20 −4c′2u20x20) + (−4c′2u20x0 −4c′2u20x0 −4c′2u20x0 + 4c′2u20x0 + 4c′2u20x0 + 4c′2u20x0) + (4c′2u20y0w0 −4c′2u20y0w0) + (4c′2v0px20 −4c′2v0px20) + (4c′2v0px0 −4c′2v0px0) +(2c′2v0py0w0 −4c′2v0py0w0) + (2c′2u20 − c′2u20 − c′2u20) + (2c′a′1w0u20 −2c′a′1w0u20) + (4c′u20a′1w0x0 −4c′u20a′1w0x0) + (4c′2u0v0w0x0 −4c′2u0v0w0x0) + (2c′2u0v0w0 −2c′2u0v0w0)−2c′u0a′1w20v0 + (4c′2u0y0px0 −4c′2u0y0px0) + (2c′2u0y0p−2c′2u0y0p) + (2c′u0a′1y0pw0 −4c′u0a′1y0pw0) +(4a′1px0c ′u0 −2a′1px0c′u0 −2a′1px0c′u0) + a′1pc′u0 + a′21 pw0u0 +(4a′1x 2 0c ′u0p−4a′1x20c′u0p)−2a′21 pw0x0u0 + (2a′1pv0w0x0c′− 4a′1pv0w0x0c ′)−a′1pc′v0w0 + (a′21 w20v0p−a′21 w20v0p)−2a′1p2y0x0c′ 123 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 +a′1p 2c′y0 + (a ′2 1 p 2y0w0 −a′21 p2y0w0)− c′2v20w20 − c′2y20p2 −u20a′21 w20 +c′2v0p + 2c ′a′1w0v0p−x20p2a′21 + a′21 p2x0 = −2c′2v0py0w0 −2c′u0a′1w20v0 −2c′a′1u0y0w0p + a′1pc′u0 + a′21 pw0u0 −2a′21 pw0x0u0 −2c′a′1pw0v0x0 −a′1pc′v0w0 −2a′1p2y0x0c′ + a′1p2c′y0 −c′2v20w20 − c′2y20p2 −u20a′21 w20 + c′2v0p + 2c′a′1w0v0p−x20p2a′21 + a′21 p2x0 = −(c′2y20p2 + c′2v20w20 + 2c′2v0py0w0)− (x20p2a′21 + u20a′21 w20 + 2a′21 px0u0w0) −2(a′1p2y0x0c′ + c′a′1y0u0w0p + c′v0w0x0pa′1 + c′u0w20v0a′1) + a′1pc′u0 +a′21 pw0u0 + a ′ 1p 2c′y0 + c ′2v0p + c ′a′1w0v0p + a ′2 1 p 2x0. = −(c′y0p + c′v0w0 + x0pa′1 + u0a′1w0)2 + c′p(a′1u0 + c′v0) + a′1pw0 (a′1u0 + c ′v0) + a ′ 1p 2(a′1x0 + c ′y0) = −[p(a′1x0 + c′y0) + w0(c′v0 + a′1u0)]2 + c′pb + a′1pw0b +a′1p 2t0 [ ∵ a′1u0 + c ′v0 = b a′1x0 + c ′y0 = t0 ] = −(pt0 + bw0)2 + cb + a′1p(bw0 + pt0) = −d21 + cb + a1d1 (∵ pt0 + bw0 = d1) = −(a2 + bc)2 + (2a−1)(a2 + bc) + bc = −(a2 + bc)2 + 2a(a2 + bc)−a2 − bc + bc = −(a2 + bc−a)2 = −(d1 −a)2 The equation (7) reduces to gX2 − D g Y 2 = −4A4 g Or X21 − DY 2 = [2c′(d1 −a)]2 (8) Where X1 = gX = Dm + (2AE − BD) = (a21 + 4bc)m + 1 p (2cd1 − 4cw0b − a1c − a21w0) and Y = 2Ak + Bm + D = 2c′2k − (2c′u0 + a1)m + (2x0c′ − c′ − a′1w0). The general methods of solutions of (8) have already been discussed in section 2. 5. Examples In this section, we shall briefly summarize the discussion in section 3 and 4. For a given matrix A = [ a b c d ] to be nil-clean the following conditions are necessary and sufficient : (1) Tr(A) = 1 (2) gcd(2a−1, b, c) = 1 (3) The equation (8) of section 4 has a solution. The existence of solutions and the number of solutions is determined by the discriminant D of the equation (8) which is (2a−1)2 + bc. (i) If D > 0, a perfect square and right hand side of (8) is non zero, then it has finitely many solutions (see Example 5.1 below). (ii) If D > 0 and not a perfect square (8) has infinitely many solutions (see Example 5.2 below). (iii) If D < 0, equation (8) may or may not have a solution. If (8) has a solution then it has finitely many (see Example 5.3 below). Finally, Example 5.4 illustrates the case where (8) does not have a solution. 124 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 Example 5.1. Consider the matrix A = [ 1 3 2 0 ] . The linear Diophantine equation (2a−1)x+cy+bw = a2 + bc for A is x + 2y + 3w = 7. The solutions are x = 1 + 2k − m, y = 3 − k − m, w = m where k and m are integers. Using condition x(1 − x) = yw, we get the quadratic Diophantine equation 4k2−5km + 2k + 2m = 0. Here D = 25 which is square of 5 and solutions are (4, 4), (-2, -1), (0, 0) and (1, 2). For the solution (4, 4), we get the solution of linear Diophantine equation x = 5, y = −5, w = 4. Let E = [ 5 −5 4 −4 ] . Then A−E is nil-potent. Example 5.2. Consider the matrix A = [ 2 3 2 −1 ] . The linear Diophantine equation (2a − 1)x + cy + bw = a2 + bc for A is 3x + 2y + 3w = 10. The solutions are x = 2k − m, y = 5 − 3k, w = m where k and m are integers. Using condition x(1 − x) = yw, we get the quadratic Diophantine equation 4k2 −7km + m2 −2k + 6m = 0. Here D = 33 which is not a perfect square and this equation has infinite number of solutions, one of the solution is (2, 2) for which, we get the solution of linear Diophantine equation x = 2, y = −1, w = 2. Let E = [ 2 −1 2 −1 ] . Then A−E is nil-potent. Example 5.3. Consider the matrix A = [ 2 3 −1 −1 ] The linear Diophantine equation (2a−1)x+cy+bw = a2 + bc for A is 3x − y + 3w = 1. The solutions are x = −k, y = −1 − 3k + 3m, w = m where k and m are integers. Using condition x(1 − x) = yw, we get the quadratic Diophantine equation k2 − 3km + 3m2 + k − m = 0. Here D = −3 < 0 and solutions are (-2, -1), (-1, 0) and (0, 0). For the solution (-2, -1), we get the solution of linear Diophantine equation x = 2, y = 2, w = −1. Let E = [ 2 2 −1 −1 ] . Then A−E is nil-potent. Example 5.4. Consider the matrix A = [ 2 3 −5 −1 ] . The linear Diophantine equation (2a−1)x + cy + bw = a2 + bc for A is 3x − 5y + 3w = −11. The solutions are x = 3 − 5k − m, y = 4 − 3k, w = m where k and m are integers. Using condition x(1 − x) = yw, we get the quadratic Diophantine equation 25k2 + 7km + m2 −25k −m + 6 = 0. Here D = −51 < 0. This equation has no solution as the equation (B2 −4AC)k2 + 2(BE −2CD)k + (E2 −4CF) = 0 i.e., 51k2 −86k + 23 = 0 has roots 0.33 and 1.35 and the integer k = 1 does not yield integer value for m. Hence 3x−5y + 3w = −11 does not have a solution satisfying x(1−x) = yw. 6. Uniquely nil-clean matrices In [7], it has been shown that central idempotents and central nilpotents are uniquely nil-clean. When R is a PID, we show that only uniquely nil-clean idempotent elements of M2(R) are central idempotents and observe that all nilpotents in M2(R) are uniquely nil-clean. Remark 6.1. Nil-clean property is preserved under similarity. Proof. Let A be nil-clean and A = E + N, where E, an idempotent matrix and N, a nilpotent matrix. Say Nk = 0 . Let B be similar to A. Then B = CAC−1 for some invertible matrix C. Therefore, B = CEC−1 + CNC−1, where (CEC−1)2 = CE2C−1 = CEC−1 and (CNC−1)k = CNkC−1 = 0. Hence B is nil-clean. Lemma 6.2. Let R be a PID and E ∈ M2(R) be an idempotent which is not central. Then E is not uniquely nil-clean. 125 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 Proof. Since E is not central, E 6= 0 and E 6= I. Then E = [ x y w 1−x ] , where yw = x(1 − x). As every idempotent matrix of rank r over a PID is similar to the matrix of the form [ Ir 0 0 0 ] (By Theorem 2.5), E is similar to [ 1 0 0 0 ] . Let CEC−1 = [ 1 0 0 0 ] . The equation (2a − 1)x + cy + bw = a2 + bc of section 3, for [ 1 0 0 0 ] reduces to x = 1. As we are interested in the solution x, y, w with yw = x(1−x), we have yw = 0. Hence either y = 0 and w is arbitrary or y is arbitrary and w = 0. Hence all the nil-clean decomposition of [ 1 0 0 0 ] are[ 1 0 0 0 ] = [ 1 0 w 0 ] + [ 0 0 −w 0 ] , w is arbitrary and [ 1 0 0 0 ] = [ 1 y 0 0 ] + [ 0 −y 0 0 ] , y is arbitrary. Therefore, E = C−1 [ 1 0 w 0 ] C + C−1 [ 0 0 −w 0 ] C, w is arbitrary and E = C−1 [ 1 y 0 0 ] C + C−1 [ 0 −y 0 0 ] C, y is arbitrary. Hence E is not uniquely nil-clean. Example 6.3. Let R = ZZ and let A = [ 9 −6 12 −8 ] and C = [ 3 −2 −4 3 ] then C−1AC = [ 1 0 0 0 ] . The nil-clean decomposition of [ 1 0 0 0 ] is [ 1 0 0 0 ] = [ 1 0 w 0 ] + [ 0 0 −w 0 ] where w is arbitrary. Therefore, [ 9 −6 12 −8 ] = [ 3 2 4 3 ]([ 1 0 w 0 ] + [ 0 0 −w 0 ])[ 3 −2 −4 3 ] = ([ 3 + 2w 0 4 + 3w 0 ] + [ −2w 0 −3w 0 ])[ 3 −2 −4 3 ] = [ 9 + 6w −6−4w 12 + 9w −8−6w ] + [ −6w 4w −9w 6w ] where w ∈ ZZis arbitrary Hence A is not uniquely nil-clean. Lemma 6.4. If A ∈ M2(R) is nilpotent, then it is uniquely nil-clean. Proof. Let A ∈ M2(R) be nilpotent. Suppose A = E + N , where E, an idempotent matrix and N, a nilpotent matrix, is a nil-clean representation of A. Comparing the traces on both sides, we have Tr(E) = 0. Therefore, E = 0 and A = N. In [5], the notion of nil-clean elements has been generalized to ideals of a ring R. An ideal I of R is nil-clean if I = E +N, where E and N are ideals of R, E2 = E and N is a nil ideal(i.e., all elements of N 126 K. N. Rajeswari, U. Gupta / J. Algebra Comb. Discrete Appl. 5(3) (2018) 117–127 are nil-potent). While the ideals generated by idempotents are idempotent, those generated by nilpotent elements need not be nil or nil-potent when the ring R is non-commutative. In view of the results of the paper, we ask the following: Question 6.4. If I is the principal ideal generated by a nil-clean element in M2(R), R an integral domain, when is I nil-clean? Acknowledgment: The authors thank the referees immensely for their suggestions which have improved the earlier version of the paper. References [1] D. Alpern, Generic Two Integer Variable Equation Solver, 2018, available at www.alpertron.com.ar/QUAD.HTM. [2] T. Andreescu, D. Andrica, Quadratic Diophantine Equations, Springer, New York, 2015. [3] D. Andrica, G. Călugăreanu, A nil–clean 2×2 matrix over the integers which is not clean, J. Algebra Appl. 13(6) (2014) 1450009. [4] D. K. Basnet, J. Bhattacharyya, Nil clean graph of rings, arXiv:1701.07630 [math.RA], https://arxiv.org/abs/1701.07630. [5] A. T. Block Gorman, Generalizations of Nil Clean to Ideals, Wellesley College, Honors Thesis Col- lection, (388) 2016. [6] S. Breaz, G. Călugăreanu, P. Danchev, T. Micu, Nil–clean matrix rings, Linear Algebra Appl. 439(10) (2013) 3115–3119. [7] A. J. Diesl, Nil–clean rings, J. Algebra 383(1) (2013) 197–211. [8] Diophantine Equation ax + by + cz = d Solver, www.mathafou.free.fr/ex e_en/exedioph3.html. [9] S. Hadjirezaei, S. Karimzadeh, On the nil–clean matrix over a UFD, J. Alg. Struc. Appl. 2(2) (2015) 49–55. [10] W. K. Nicholson, Lifting idempotents and exchange rings, Trans. Amer. Math. Soc. 229 (1977) 269–278. [11] I. Niven, H. S. Zuckerman, An Introduction to the Theory of Numbers, John Wiley–Sons, 3rd edition, 1972. [12] S. Sahinkaya, G. Tang, Y. Zhou, Nil–clean group rings, J. Algebra Appl. 16(7) (2017) 1750135. [13] F. Smarandache, Existence and number of solutions of Diophantine quadratic equations with two unkowns in ZZ and IN, arXiv:0 704.3716 [math.GM], http://arxiv.org/abs/0704.3716. 127 https://www.alpertron.com.ar/QUAD.HTM https://www.alpertron.com.ar/QUAD.HTM https://doi.org/10.1007/978-0-387-54109-9 http://dx.doi.org/10.1142/S0219498814500091 http://dx.doi.org/10.1142/S0219498814500091 https://repository.wellesley.edu/thesiscollection/388/ https://repository.wellesley.edu/thesiscollection/388/ https://doi.org/10.1016/j.laa.2013.08.027 https://doi.org/10.1016/j.laa.2013.08.027 https://doi.org/10.1016/j.jalgebra.2013.02.020 https://doi.org/10.1090/S0002-9947-1977-0439876-2 https://doi.org/10.1090/S0002-9947-1977-0439876-2 https://mathscinet.ams.org/mathscinet-getitem?mr=344181 https://mathscinet.ams.org/mathscinet-getitem?mr=344181 https://doi.org/10.1142/S0219498817501353 Introduction Preliminaries Main result Analysis on main result when R= ZZ Examples Uniquely nil-clean matrices References