ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.82415 J. Algebra Comb. Discrete Appl. 3(2) • 91–96 Received: 18 November 2015 Accepted: 2 Feburary 2016 Journal of Algebra Combinatorics Discrete Structures and Applications Matrix rings over a principal ideal domain in which elements are nil-clean Research Article Somayeh Hadjirezaei, Somayeh Karimzadeh Abstract: An element of a ring R is called nil-clean if it is the sum of an idempotent and a nilpotent element. A ring is called nil-clean if each of its elements is nil-clean. S. Breaz et al. in [1] proved their main result that the matrix ring Mn(F) over a field F is nil-clean if and only if F ∼= F2, where F2 is the field of two elements. M. T. Koşan et al. generalized this result to a division ring. In this paper, we show that the n × n matrix ring over a principal ideal domain R is a nil-clean ring if and only if R is isomorphic to F2. Also, we show that the same result is true for the 2×2 matrix ring over an integral domain R. As a consequence, we show that for a commutative ring R, if M2(R) is a nil-clean ring, then dimR = 0 and charR/J(R) = 2. 2010 MSC: 15A23, 15B33, 16S50 Keywords: Nil-clean matrix, Idempotent matrix, Nilpotent matrix, Principal ideal domain 1. Introduction Throughout this paper, all rings are associative with identity. An element in a ring R is said to be (strongly) clean if it is the sum of an idempotent and a unit element(and these commute ). A (strongly) clean ring is one in which every element is (strongly) clean. Local rings are obviously strongly clean. Strongly clean rings were introduced by Nicholson [8]. An element in a ring R is said to be (strongly) nil-clean if it is the sum of an idempotent and a nilpotent element(and these commute). A (strongly) nil-clean ring is one in which every element is (strongly) nil-clean. It is easy to see that every strongly nil-clean element is strongly clean and that every nil-clean ring is clean ([3, Proposition 3.1.3]). Nil-clean rings were extensively investigated by Diesl in [3] and [4]. S. Breaz et al. in [1] proved their main result that the matrix ring Mn(F) over a field F is nil-clean if and only if F ∼= F2, where F2 is the field of two elements. M. T. Koşan et al. in [6], generalized this result to a division ring. That is, the matrix ring Mn(D) over a division ring D is nil-clean if and only if D ∼= F2. We show that this is true for a principal ideal domain (PID). Somayeh Hadjirezaei (Corresponding Author), Somayeh Karimzadeh; Vali-e-Asr University of Rafsanjan (email: s.hajirezaei@vru.ac.ir, karimzadeh@vru.ac.ir). 91 S. Hadjirezaei, S. Karimzadeh / J. Algebra Comb. Discrete Appl. 3(2) (2016) 91–96 Throughout this paper an integral domain is a commutative ring without zero divisors and the Jacobson radical of a ring is denoted by J(R). We write Mn(R) for the n×n matrix ring over R, In for the n×n identity matrix. 2. Main results First, we recall from [5, Proposition VII.2.11 ], the following Proposition. Proposition 2.1. If A is an n × m matrix of rank r > 0 over a principal ideal domain R, then A is equivalent to a matrix of the form ( Lr 0 0 0 ) , where Lr is an r×r diagonal matrix with nonzero diagonal entries d1, ..., dr such that d1 | ... | dr. The ideals (d1), ..., (dr) in R are uniquely determined by the equivalence class of A. Further, we use the following lemmas. Lemma 2.2. (See [4, Proposition 3.14]) Let R be a nil-clean ring. Then the element 2 is (central) nilpotent and, as such, is always contained in J(R). Lemma 2.3. (See [9, Corollary 5]) Let A be an n×n idempotent matrix over a ring R. If A is equivalent to a diagonal matrix, then A is similar to a diagonal matrix. Next Lemmas are the main results of [1] and [6]. Lemma 2.4. (See [1, Theorem 3]) Let F be a field and let n ≥ 1. Then Mn(F) is a nil-clean ring if and only if F ∼= F2. Lemma 2.5. (See [6, Theorem 3]) Let D be a division ring and let n ≥ 1. Then Mn(D) is a nil-clean ring if and only if D ∼= F2. Theorem 2.6. Let R be a principal ideal domain and let n ≥ 1. Then Mn(R) is a nil-clean ring if and only if R ∼= F2. Proof. If R ∼= F2, then by Lemma 2.4, Mn(R) is a nil-clean ring. Now, assume that Mn(R) is a nil-clean ring. By Lemma 2.2, 2In is a nilpotent element. Thus 2 = 0 in R, because R is an integral domain. Proof in the case n = 1 is obvious, so assume that n > 1. Take a ∈ R \{0, 1} and put A =   a 0 . . . 0 0 0 . . . 0 ... ... ... ... 0 0 . . . 0   = E + N, where E is an idempotent element and N is a nilpotent element of Mn(R). By Proposition 2.1, E is equivalent to a diagonal matrix. Thus by Lemma 2.3, E is similar to a diagonal matrix where it’s entries are 0 and 1. Hence U−1EU = ( Ik 0 0 0 ) , for some invertible matrix U = (uij) ∈ Mn(R). Therefore U−1AU = ( Ik 0 0 0 ) + N′, (1) where N′ = U−1NU is a nilpotent element. Since a is not nilpotent, hence U−1AU is not nilpotent, so k ≥ 1. If k = n, then A = In +N is invertible, a contradiction because det A = 0. Thus 1 ≤ k < n. Since In + N ′ is invertible, U(In + N′) is invertible. We have U(In + N ′) = U ( Ik 0 0 0 ) + UN′ + U ( 0 0 0 In−k ) = AU + U ( 0 0 0 In−k ) 92 S. Hadjirezaei, S. Karimzadeh / J. Algebra Comb. Discrete Appl. 3(2) (2016) 91–96 =   au11 . . . au1n 0 . . . 0 ... ... ... 0 . . . 0   +   0 . . . 0 u1(k+1) . . . u1n 0 . . . 0 u2(k+1) . . . u2n ... ... ... ... ... 0 . . . 0 un(k+1) . . . unn   =   au11 . . . au1k (1 + a)u1(k+1) . . . (1 + a)u1n 0 . . . 0 u2(k+1) . . . u2n ... ... ... ... ... ... 0 . . . 0 un(k+1) . . . unn   . We imply that k = 1 and u11 6= 0. Thus U(In + N ′) =   au11 (1 + a)u12 . . . (1 + a)u1n 0 u22 . . . u2n ... ... ... ... 0 un2 . . . unn   . Put U1 :=   u22 . . . u2n... ... ... un2 . . . unn   . Since det(U(In + N′)) = au11 det U1, U1 is invertible in Mn−1(R) and u11 is invertible in R, hence (1) implies that ( a 0 0 0 ) U = U ( 1 0 0 0 ) + UN′. This implies that ( u−111 0 0 U−11 )( a 0 0 0 )( u11 0 0 U1 )( u−111 0 0 U−11 ) U = ( u−111 0 0 U−11 ) U ( 1 0 0 0 ) + ( u−111 0 0 U−11 ) UN′, i.e., ( a 0 0 0 ) V = V ( 1 0 0 0 ) + V N′, (2) where V = ( u−111 0 0 U−11 ) U = ( 1 X Y In−1 ) . Let V −1 = ( c X′ Y ′ C1 ) . From V V −1 = V −1V = In, it follows that 1 = c + XY ′ = c + X′Y In−1 = Y X ′ + C1 = Y ′X + C1 0 = X′ + XC1 = cX + X ′ 93 S. Hadjirezaei, S. Karimzadeh / J. Algebra Comb. Discrete Appl. 3(2) (2016) 91–96 0 = cY + Y ′ = Y ′ + C1Y. Since 2 = 0 in R (by Lemma 2.2, and since R is an integral domain) hence, we have 1 = −1 in R, so c = −c. Therefore X′ = −cX = cX, Y ′ = cY and C1 = In−1 + Y X′ = In−1 + cY X. Also, 1 = c + XY ′ = c + cXY = c(1 + XY ), so c is a unit element of R and XY = 1 + c−1. (3) Hence V −1 = ( c cX cY In−1 + cY X ) . If XY = 0, then c = 1 and V −1 = ( 1 X Y In−1 + Y X ) . Then by (2), N ′′ := V NV −1 = ( a 0 0 0 ) + V ( 1 0 0 0 ) V −1 = ( 1 + a X Y Y X ) , and, for k ≥ 1, N ′′ k+1 = ( (1 + a)k+1 (1 + a)k+1X (1 + a)k+1Y (1 + a)k+1Y X ) 6= 0 (as (1 + a) 6= 0). This is a contradiction because N ′′ is a nilpotent matrix. Therefore XY 6= 0. From (2) it follows that( 1 X 0 In−1 )( a 0 0 0 )( 1 X 0 In−1 )( 1 X 0 In−1 ) V = ( 1 X 0 In−1 ) V ( 1 0 0 0 ) + ( 1 X 0 In−1 ) V N′, i.e., ( a X 0 0 ) P = P ( 1 0 0 0 ) + PN′, (4) where P = ( 1 X 0 In−1 ) V = ( 1 X 0 In−1 )( 1 X Y In−1 ) = ( 1 + XY X + X Y In−1 ) . Since 2 = 0 in R, hence X + X = 2X = 0. Also by (3), we have XY − 1 = XY + 1 = c−1. Hence P = ( c−1 0 Y In−1 ) and P−1 = ( c 0 cY In−1 ) . It follows from (4) that 4 := PNP−1 = ( a aX 0 0 ) + P ( 1 0 0 0 ) P−1 = ( 1 + a aX cY 0 ) . If Q is an n×n matrix, then we will write Q in block form Q = ( Q11 Q12 Q21 Q22 ) , where Q11, Q12, Q21, Q22 have size 1×1, 1× (n−1), (n−1)×1 and (n−1)× (n−1), respectively. For k ≥ 1 we have 4k+1 = 4k4 = ( (4k)11 (4)k12 (4k)21 (4k)22 )( 1 + a aX cY 0 ) = ( (4k)11(1 + a) + (4)k12cY a(4k)11X (4k)21(1 + a) + (4k)22cY a(4k)21X ) . (5) 94 S. Hadjirezaei, S. Karimzadeh / J. Algebra Comb. Discrete Appl. 3(2) (2016) 91–96 An easy induction shows that there exist ak, bk, ck ∈ R such that for k ≥ 1 we have (4k)12 = bkX, (4k)21 = ckY, (4k)22 = akY X. (6) Since 4 is a nilpotent matrix and 421 = cY 6= 0, there exists a positive integer s such that (4s+1)21 = 0 but (4s)21 6= 0. Then by (5) and (6), 4s+1 = ( (4s+1)11 (4s+1)12 0 csaY X ) , where csa 6= 0. For r ∈ R, it is easily seen that rY X = 0 if and only if r = 0. We have (4s+1)k22 = (csa) k(XY )k−1Y X. Since csa 6= 0 and XY 6= 0, hence (4s+1)k22 6= 0, for k ≥ 2. It is a contradiction because 4 is nilpotent. Theorem 2.7. Let R be an integral domain . If Mn(R) is a nil-clean ring, then R is a field. Proof. Let Q be the field of fractions of R and 0 6= a ∈ R. We know that aIn is nil-clean. So, aIn = E + N with E idempotent and N nilpotent. We have In = a−1E + a−1N, in Mn(Q) . Thus a−1E ( and consequently E) is invertible in Mn(Q). Since E is idempotent, so E = In. Therefore aIn is invertible, hence R is a field. Lemma 2.8. Let R be an integral domain and 0, I2 6= A ∈ M2(R). Then A is idempotent if and only if rank(A) = 1 and tr(A) = 1. Proof. By [2, Lemma 1.5]. Lemma 2.9. Let R be an integral domain. If A ∈ Mn(R) be a nilpotent matrix, then det(A) = 0. Proof. Let A be a nonzero nilpotent matrix. Thus there exists some k ∈ N such that Ak = 0. Thus adj(A)Ak = 0. Hence det(A)Ak−1 = 0. So det(A) adj(A)Ak−1 = 0. Therefore (det(A))2Ak−2 = 0. Continuing this process we have (det(A))k−1A = 0. Since R is an integral domain and A 6= 0, hence det(A) = 0 Theorem 2.10. Let R be an integral domain. Then M2(R) is a nil-clean ring if and only if R ∼= F2. Proof. ⇐=)This is by Theorem 2.6. =⇒) Assume that R is not isomorphic to F2. So, there exists a ∈ R\{0, 1}. Put A = ( a 0 0 0 ) = E + N, where E is idempotent and N is a nilpotent matrix. If E = I2, then A is invertible, a contradiction. If E = 0, then A is nilpotent. Hence a = 0, a contradiction. So by Lemma 2.8, E = ( e b c 1−e ) , where e, b, c ∈ R and e(1 − e) = bc. Hence N = ( n −b −c −(1−e) ) , for some n ∈ R. By Lemma 2.9, −n(1 − e) = bc. Therefore e(1 − e) = −n(1 − e). If e 6= 1, then e = −n. So N = −E, a contradiction. Thus e = 1 and bc = 0. Hence b = 0 or c = 0. We consider two cases. Case 1) Let b = 0. So N = ( n 0 0 0 ) . Since N is nilpotent, hence there exists a positive integer k such that nk = 0. So n = 0. Therefore a = 1. Case 2) Let c = 0. Thus N = ( n −b 0 0 ) . Since N is nilpotent, hence there exists a positive integer k such that nk = 0. So n = 0. Therefore a = 1. Let R be a commutative ring with identity. By a chain of prime ideals of R we mean a finite strictly increasing sequence of prime ideals of R of the type Po $ P1 $ P2 $ ... $ Pn. The integer n is called the length of the chain. 95 S. Hadjirezaei, S. Karimzadeh / J. Algebra Comb. Discrete Appl. 3(2) (2016) 91–96 Definition 2.11. The Krull dimension of R is the supremum of all lengths of chains of prime ideals of R. Krull dimension of R is denoted by dimR. Corollary 2.12. Let R be a commutative ring. If M2(R) is a nil-clean ring, then dimR = 0 and charR/J(R) = 2. Proof. Let P be a prime ideal of R. We have M2(R/P) = M2(R)/M2(P) is nil-clean. Hence by Theorem 2.10, R/P ∼= F2. So P is a maximal ideal of R and 2 ∈ J(R). Therefore charR/J(R) = 2. Remark 2.13. Note that all of these results can also be obtained as some consequences of [7, Theorem 6.1]. Acknowledgment: The authors are grateful to the referees’ invaluable comments, which helped to improve our study. References [1] S. Breaz, G. Călugăreanu, P. Danchev, T. Micu, Nil-clean matrix rings, Linear Algebra Appl. 439(10) (2013) 3115-3119. [2] J. Chen, X. Yang, Y. Zhou, On strongly clean matrix and triangular matrix rings, Comm. Algebra. 34(10) (2006) 3659–3674. [3] A. J. Diesl, Classes of strongly clean rings, Ph. D. thesis, University of California, Berkeley, 2006. [4] A. J. Diesl, Nil clean rings, J. Algebra. 383 (2013) 197–211. [5] T. W. Hungerford, Algebra, Springer-Verlag, 1980. [6] M.T. Koşan, T. K. Lee, Y. Zhou, When is every matrix over a division ring a sum of an idempotent and a nilpotent?, Linear Algebra Appl. 450 (2014) 7–12. [7] T. Koşan, Z. Wang, Y. Zhou, Nil-clean and strongly nil-clean rings, J. Pure Appl. Algebra. 220(2) (2016) 633–646. [8] W. K. Nicholson, Strongly clean rings and Fitting’s lemma, Comm. Algebra. 27(8) (1999) 3583–3592. [9] G. Song, X. 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