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When is every matrix over a division ring a sum of an idempotent and a nilpotent?
Linear Algebra and its Applications 450 (2014) 7–12
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
When is every matrix over a division ring a sum of an idempotent and a nilpotent? M. Tamer Koşan a , Tsiu-Kwen Lee b,1 , Yiqiang Zhou c,∗ a Department of Mathematics, Gebze Institute of Technology, Gebze/Kocaeli, Turkey b Department of Mathematics, National Taiwan University, Taipei 106, Taiwan c Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, NL A1C 5S7, Canada
a r t i c l e
i n f o
Article history: Received 28 October 2013 Accepted 27 February 2014 Available online 17 March 2014 Submitted by P. Semrl MSC: 15A23 15B33 16S50 16U60
a b s t r a c t A ring is called nil-clean if each of its elements is a sum of an idempotent and a nilpotent. In response to a question of S. Breaz et al. in [1], we prove that the n × n matrix ring over a division ring D is a nil-clean ring if and only if D ∼ = F2 . As consequences, it is shown that the n × n matrix ring over a strongly regular ring R is a nil-clean ring if and only if R is a Boolean ring, and that a semilocal ring R is nil-clean if and only if its Jacobson radical J(R) is nil and R/J(R) is a direct product of matrix rings over F2 . © 2014 Elsevier Inc. All rights reserved.
Keywords: Idempotent matrix Nilpotent matrix Nil-clean matrix Matrix ring Semilocal ring Division ring Strongly regular ring
* Corresponding author. E-mail addresses: [email protected] (M.T. Koşan), [email protected] (T.-K. Lee), [email protected] (Y. Zhou). 1 Member of Mathematics Division (Taipei Office), National Center for Theoretical Sciences. http://dx.doi.org/10.1016/j.laa.2014.02.047 0024-3795/© 2014 Elsevier Inc. All rights reserved.
8
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
1. Introduction A ring is called nil-clean if each of its elements is a sum of an idempotent and a nilpotent. Nil-clean rings were extensively investigated by Diesl in [2] and [3]. Motivated by Diesl’s question whether the matrix ring over a nil-clean ring is again nil-clean, S. Breaz et al. in [1] proved their main result that the matrix ring Mn (F ) over a field F is nilclean if and only if F ∼ = F2 . This result has several interesting consequences, including a complete characterization of the finite rank Abelian groups with nil-clean endomorphism ring (see [1]). The proof of this result heavily relies on the commutativity of a field, and it was asked in [1] whether the result can be proved for a division ring instead of a field. As a response to this question, we prove that the matrix ring Mn (D) over a division ring D is a nil-clean ring if and only if D ∼ = F2 . As consequences, it is shown that the matrix ring Mn (R) over a strongly regular ring R is a nil-clean ring if and only if R is a Boolean ring, and that a semilocal ring R is nil-clean if and only if its Jacobson radical J(R) is nil and R/J(R) is a direct product of matrix rings over F2 (as suggested in [1]). Throughout, rings are associative with nonzero identity. The Jacobson radical of a ring R is denoted by J(R). We write Mn (R) for the n × n matrix ring over R, In for the n × n identity matrix, and F2 for the field of two elements. 2. The results Our first lemma was proved in [3], which clearly implies that a nil-clean ring R with J(R) = 0 has characteristic 2. Lemma 1. (See [3, Proposition 3.14].) Let R be a nil-clean ring. Then the element 2 is a (central) nilpotent and, as such, is always contained in J(R). Our second lemma is the main result in [1]. Lemma 2. (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 . We are now ready to prove our main result. 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 . Proof. (⇐=). This is by Lemma 2. (=⇒). First note that 2 = 0 in D by Lemma 1. Assume on the contrary that D is not isomorphic to F2 . To get a contradiction, take a ∈ D\{0, 1}. Then a, 1 − a are not nilpotents of D, and this implies that a ∈ D is not a sum of an idempotent and a nilpotent. Hence the claim holds for n = 1. Let us assume that n 2. By hypothesis,
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
9
⎛a 0 ··· 0⎞ ⎜0 0 ··· 0⎟ ⎟ A := ⎜ ⎝ ... ... . . . ... ⎠ = E + B, 0 0 ··· 0 where E 2 = E ∈ Mn (D) and B ∈ Mn (D) is a nilpotent. It is well known that every matrix over a division ring is equivalent to a diagonal matrix. Particularly, E is equivalent to a diagonal matrix. So, by [4, Corollary 5], E is similar to a diagonal matrix whose diagonal entries are 0 or 1. Hence, there exists an invertible matrix U = (uij ) ∈ Mn (D) such that U −1 EU = I0k 00 , and so U
−1
AU =
0 0
Ik 0
+ B,
(2.1)
where B = U −1 BU is a nilpotent. Since U −1 AU is not a nilpotent, we have k 1. If k = n, then A = In + B is invertible, a contradiction. So 1 k < n. Since In + B is invertible, U (In + B ) is invertible. Note that U In + B = U
0 0
Ik 0
= AU + U ⎛ au
11
⎜ 0 =⎜ ⎝ ... 0 ⎛ au11 ⎜ 0 ⎜ =⎜ . ⎝ ..
+ U B + U
0 0
0
0 0
0
In−k
In−k
⎛ 0 · · · 0 u1,k+1 · · · au1n ⎞ 0 · · · 0 u2,k+1 ··· 0 ⎟ ⎜ ⎜ +⎜. .. ⎟ .. .. ⎠ ⎝ .. . . . ··· 0 0 · · · 0 un,k+1 · · · au1k ··· 0 .. . ···
0
(1 + a)u1,k+1 u2,k+1 .. .
0
We deduce that k = 1, u11 = 0 and ⎛
···
u22 ⎜ .. U1 := ⎝ .
⎞ u2n .. ⎟ . ⎠
· · · unn
un2
is invertible in Mn−1 (D). Thus, (2.1) becomes
and it follows that
a 0
0 0
U =U
1 0
0 0
⎞ u1n u2n ⎟ ⎟ .. ⎟ . ⎠
· · · unn ⎞ · · · (1 + a)u1n ⎟ ··· u2n ⎟ ⎟. .. ⎠ . ···
un,k+1
··· ···
+ U B,
unn
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
10
−1 u−1 0 a 0 u11 0 u11 0 11 U 0 U1−1 0 0 0 U1 0 U1−1
−1
−1 1 0 u11 0 0 u11 U + U B, = 0 0 0 U1−1 0 U1−1
i.e.,
b 0
0 0
where b = u−1 11 au11 and V = VV
−1
−1
=V
V =V
u−1 11
0 U1−1
0
1 0
0 0
1 U = Y
+ V B,
X In−1
(2.2)
. Write V −1 = Yc
X C1
. From
V = In , it follows that 1 = c + XY = c + X Y In−1 = Y X + C1 = Y X + C1 0 = X + XC1 = cX + X 0 = Y c + Y = Y + C1 Y.
Thus, we obtain that X = cX, Y = Y c and C1 = In−1 + Y X = In−1 + Y cX. Moreover, 1 = c + XY = c + XY c = (1 + XY )c, so c = 0 and XY = 1 + c−1 . Therefore, c cX V −1 = Y c In−1 +Y cX with XY = 1 + c−1 . 1 X If XY = 0, then c = 1 and V −1 = Y In−1 +Y X . Then, by (2.2),
B := V B V
−1
=
b 0
and, for all k 1, (B )k+1 =
0 0
+V
1 0
0 0
V
(1+b)k+1
(1+b)k X Y (1+b)k Y (1+b)k−1 X
−1
=
1+b Y
X YX
,
= 0 (as 1 + b = 0). This is a
contradiction because B is a nilpotent. Therefore, we can assume that XY = 0. Now by (2.2), we have
1 X b 0 1 X 1 X V 0 In−1 0 0 0 In−1 0 In−1
1 0 1 X 1 X V + V B, = 0 0 0 In−1 0 In−1 i.e.,
where P =
1
X 0 In−1
V =
b bX 0 0
c−1 Y
:= P B P −1 =
0 In−1
P =P
1 0
and P −1 =
b bX 0 0
+P
1 0
0 0
+ P B,
c 0 Y c In−1
0 0
(2.3)
. It follows from (2.3) that
P −1 =
1+b Yc
bX 0
.
(2.4)
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
11
For our next computation, it will be convenient to introduce the following notation. π (Q) π12 (Q) , If Q is an n × n matrix over D, we will write Q in block form Q = π11 21 (Q) π22 (Q) where π11 (Q), π12 (Q), π21 (Q) and π22 (Q) have sizes 1 × 1, 1 × (n − 1), (n − 1) × 1 and (n − 1) × (n − 1), respectively. We calculate that, for k 1,
π11 (k+1 ) π21 (k+1 )
π12 (k+1 ) π22 (k+1 )
= k+1 = k
=
=
π11 (k ) π21 (k )
π12 (k ) π22 (k )
1+b Yc
π11 (k )(1 + b) + π12 (k )Y c π21 (k )(1 + b) + π22 (k )Y c
bX 0
π11 (k )bX π21 (k )bX
.
(2.5)
Note that π12 () = bX and π21 () = Y c and π22 () = 0 (= Y 0X). By (2.5), an easy inductive argument shows that there exist ak , bk , ck ∈ D such that π12 k = bk X,
π21 k = Y ck ,
π22 k = Y ak X
for k = 1, 2, . . . . (2.6)
Because is a nilpotent and π21 () = Y c = 0, there exists a positive integer s such that π21 (s+1 ) = 0 but π21 (s ) = 0. Then, by (2.5) and (2.6),
s+1
=
π11 (s+1 ) π12 (s+1 ) 0 Y cs bX
,
where cs b = 0. Because Y has size (n − 1) × 1 and X has size 1 × (n − 1), we see that, for d ∈ D, Y dX = 0 if and only if d = 0; so for d = 0, (Y dX)2 = Y (dXY d)X = 0 as dXY d = d(XY )d = 0 and, similarly (Y dX)k = 0 for all k > 0. In particular, Y cs bX ∈ Mn−1 (D) is not a nilpotent, and hence s+1 is not a nilpotent. Therefore, is not a nilpotent, a contradiction. 2 Theorem 3 can be extended from a division ring to a strongly regular ring. Recall that a ring is called strongly regular if it is a von Neumann regular ring with all idempotents central. Every strongly regular ring contains no nonzero nilpotents. It is well known that a ring R is strongly regular if and only if each element of R is a product of a unit and a central idempotent. Corollary 4. Let R be a strongly regular ring and let n 1. Then Mn (R) is a nil-clean ring if and only if R is Boolean. Proof. (⇐=). This is by [1, Corollary 6]. (=⇒). Assume on the contrary that R is not Boolean. As R is strongly regular, R contains a unit u = 1. Then 1 − u = ve where v is a unit of R and e is a nonzero central idempotent. Thus (1 − e)R is a proper ideal of R, so there exists a maximal ideal I of R such that (1 − e)R ⊆ I. As every quotient ring of a strongly regular ring is again
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M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
strongly regular, R/I is strongly regular, and hence is a division ring (being a simple ring). Since a quotient ring of a nil-clean ring is nil-clean, Mn (R/I) ∼ = Mn (R)/Mn (I) is nil-clean. Therefore, R/I ∼ = F2 by Theorem 3. It follows that ve = 1 − u ∈ I, so e ∈ I. Hence 1 = e + (1 − e) ∈ I, a contradiction. 2 The assumption that R is strongly regular in Corollary 4 cannot be weakened to R being unit-regular. Here a ring R is unit-regular if, for each a ∈ R, a = aua for a unit u of R. Indeed, for R = Mn (F2 ) (n 2), Mk (R) ∼ = Mnk (F2 ) is nil-clean, R is unit-regular but not Boolean. To give another application of Theorem 3, we need the following result from [3]: Lemma 5. (See [3, Corollary 3.17].) A ring R is nil-clean if and only if J(R) is nil and R/J(R) is nil-clean. A ring R is called semilocal if R/J(R) is semisimple Artinian. Because a finite direct product of rings is nil-clean if and only if each direct summand is nil-clean, the next corollary is a quick consequence of Theorem 3, Lemma 5 and the Wedderburn–Artin Theorem. Corollary 6. A semilocal ring R is nil-clean if and only if J(R) is nil and R/J(R) is a direct product of matrix rings over F2 . Acknowledgements We are grateful to the referee for his invaluable comments. Yiqiang Zhou acknowledges gratefully the support from NCTS of Taipei and from TUBITAK of Turkey for his visits to the National Taiwan University and the Gebze Institute of Technology, and the hospitality from the two universities. The research of the second author was supported by NSC of Taiwan and by NCTS of Taipei, and that of the third author by a Discovery Grant from NSERC of Canada. References [1] S. Breaz, G. Călugăreanu, P. Danchev, T. Micu, Nil-clean matrix rings, Linear Algebra Appl. 439 (2013) 3115–3119. [2] A.J. Diesl, Classes of strongly clean rings, PhD thesis, University of California, Berkeley, 2006. [3] A.J. Diesl, Nil clean rings, J. Algebra 383 (2013) 197–211. [4] G. Song, X. Guo, Diagonability of idempotent matrices over noncommutative rings, Linear Algebra Appl. 297 (1999) 1–7.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
When is every matrix over a division ring a sum of an idempotent and a nilpotent? M. Tamer Koşan a , Tsiu-Kwen Lee b,1 , Yiqiang Zhou c,∗ a Department of Mathematics, Gebze Institute of Technology, Gebze/Kocaeli, Turkey b Department of Mathematics, National Taiwan University, Taipei 106, Taiwan c Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, NL A1C 5S7, Canada
a r t i c l e
i n f o
Article history: Received 28 October 2013 Accepted 27 February 2014 Available online 17 March 2014 Submitted by P. Semrl MSC: 15A23 15B33 16S50 16U60
a b s t r a c t A ring is called nil-clean if each of its elements is a sum of an idempotent and a nilpotent. In response to a question of S. Breaz et al. in [1], we prove that the n × n matrix ring over a division ring D is a nil-clean ring if and only if D ∼ = F2 . As consequences, it is shown that the n × n matrix ring over a strongly regular ring R is a nil-clean ring if and only if R is a Boolean ring, and that a semilocal ring R is nil-clean if and only if its Jacobson radical J(R) is nil and R/J(R) is a direct product of matrix rings over F2 . © 2014 Elsevier Inc. All rights reserved.
Keywords: Idempotent matrix Nilpotent matrix Nil-clean matrix Matrix ring Semilocal ring Division ring Strongly regular ring
* Corresponding author. E-mail addresses: [email protected] (M.T. Koşan), [email protected] (T.-K. Lee), [email protected] (Y. Zhou). 1 Member of Mathematics Division (Taipei Office), National Center for Theoretical Sciences. http://dx.doi.org/10.1016/j.laa.2014.02.047 0024-3795/© 2014 Elsevier Inc. All rights reserved.
8
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
1. Introduction A ring is called nil-clean if each of its elements is a sum of an idempotent and a nilpotent. Nil-clean rings were extensively investigated by Diesl in [2] and [3]. Motivated by Diesl’s question whether the matrix ring over a nil-clean ring is again nil-clean, S. Breaz et al. in [1] proved their main result that the matrix ring Mn (F ) over a field F is nilclean if and only if F ∼ = F2 . This result has several interesting consequences, including a complete characterization of the finite rank Abelian groups with nil-clean endomorphism ring (see [1]). The proof of this result heavily relies on the commutativity of a field, and it was asked in [1] whether the result can be proved for a division ring instead of a field. As a response to this question, we prove that the matrix ring Mn (D) over a division ring D is a nil-clean ring if and only if D ∼ = F2 . As consequences, it is shown that the matrix ring Mn (R) over a strongly regular ring R is a nil-clean ring if and only if R is a Boolean ring, and that a semilocal ring R is nil-clean if and only if its Jacobson radical J(R) is nil and R/J(R) is a direct product of matrix rings over F2 (as suggested in [1]). Throughout, rings are associative with nonzero identity. The Jacobson radical of a ring R is denoted by J(R). We write Mn (R) for the n × n matrix ring over R, In for the n × n identity matrix, and F2 for the field of two elements. 2. The results Our first lemma was proved in [3], which clearly implies that a nil-clean ring R with J(R) = 0 has characteristic 2. Lemma 1. (See [3, Proposition 3.14].) Let R be a nil-clean ring. Then the element 2 is a (central) nilpotent and, as such, is always contained in J(R). Our second lemma is the main result in [1]. Lemma 2. (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 . We are now ready to prove our main result. 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 . Proof. (⇐=). This is by Lemma 2. (=⇒). First note that 2 = 0 in D by Lemma 1. Assume on the contrary that D is not isomorphic to F2 . To get a contradiction, take a ∈ D\{0, 1}. Then a, 1 − a are not nilpotents of D, and this implies that a ∈ D is not a sum of an idempotent and a nilpotent. Hence the claim holds for n = 1. Let us assume that n 2. By hypothesis,
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
9
⎛a 0 ··· 0⎞ ⎜0 0 ··· 0⎟ ⎟ A := ⎜ ⎝ ... ... . . . ... ⎠ = E + B, 0 0 ··· 0 where E 2 = E ∈ Mn (D) and B ∈ Mn (D) is a nilpotent. It is well known that every matrix over a division ring is equivalent to a diagonal matrix. Particularly, E is equivalent to a diagonal matrix. So, by [4, Corollary 5], E is similar to a diagonal matrix whose diagonal entries are 0 or 1. Hence, there exists an invertible matrix U = (uij ) ∈ Mn (D) such that U −1 EU = I0k 00 , and so U
−1
AU =
0 0
Ik 0
+ B,
(2.1)
where B = U −1 BU is a nilpotent. Since U −1 AU is not a nilpotent, we have k 1. If k = n, then A = In + B is invertible, a contradiction. So 1 k < n. Since In + B is invertible, U (In + B ) is invertible. Note that U In + B = U
0 0
Ik 0
= AU + U ⎛ au
11
⎜ 0 =⎜ ⎝ ... 0 ⎛ au11 ⎜ 0 ⎜ =⎜ . ⎝ ..
+ U B + U
0 0
0
0 0
0
In−k
In−k
⎛ 0 · · · 0 u1,k+1 · · · au1n ⎞ 0 · · · 0 u2,k+1 ··· 0 ⎟ ⎜ ⎜ +⎜. .. ⎟ .. .. ⎠ ⎝ .. . . . ··· 0 0 · · · 0 un,k+1 · · · au1k ··· 0 .. . ···
0
(1 + a)u1,k+1 u2,k+1 .. .
0
We deduce that k = 1, u11 = 0 and ⎛
···
u22 ⎜ .. U1 := ⎝ .
⎞ u2n .. ⎟ . ⎠
· · · unn
un2
is invertible in Mn−1 (D). Thus, (2.1) becomes
and it follows that
a 0
0 0
U =U
1 0
0 0
⎞ u1n u2n ⎟ ⎟ .. ⎟ . ⎠
· · · unn ⎞ · · · (1 + a)u1n ⎟ ··· u2n ⎟ ⎟. .. ⎠ . ···
un,k+1
··· ···
+ U B,
unn
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
10
−1 u−1 0 a 0 u11 0 u11 0 11 U 0 U1−1 0 0 0 U1 0 U1−1
−1
−1 1 0 u11 0 0 u11 U + U B, = 0 0 0 U1−1 0 U1−1
i.e.,
b 0
0 0
where b = u−1 11 au11 and V = VV
−1
−1
=V
V =V
u−1 11
0 U1−1
0
1 0
0 0
1 U = Y
+ V B,
X In−1
(2.2)
. Write V −1 = Yc
X C1
. From
V = In , it follows that 1 = c + XY = c + X Y In−1 = Y X + C1 = Y X + C1 0 = X + XC1 = cX + X 0 = Y c + Y = Y + C1 Y.
Thus, we obtain that X = cX, Y = Y c and C1 = In−1 + Y X = In−1 + Y cX. Moreover, 1 = c + XY = c + XY c = (1 + XY )c, so c = 0 and XY = 1 + c−1 . Therefore, c cX V −1 = Y c In−1 +Y cX with XY = 1 + c−1 . 1 X If XY = 0, then c = 1 and V −1 = Y In−1 +Y X . Then, by (2.2),
B := V B V
−1
=
b 0
and, for all k 1, (B )k+1 =
0 0
+V
1 0
0 0
V
(1+b)k+1
(1+b)k X Y (1+b)k Y (1+b)k−1 X
−1
=
1+b Y
X YX
,
= 0 (as 1 + b = 0). This is a
contradiction because B is a nilpotent. Therefore, we can assume that XY = 0. Now by (2.2), we have
1 X b 0 1 X 1 X V 0 In−1 0 0 0 In−1 0 In−1
1 0 1 X 1 X V + V B, = 0 0 0 In−1 0 In−1 i.e.,
where P =
1
X 0 In−1
V =
b bX 0 0
c−1 Y
:= P B P −1 =
0 In−1
P =P
1 0
and P −1 =
b bX 0 0
+P
1 0
0 0
+ P B,
c 0 Y c In−1
0 0
(2.3)
. It follows from (2.3) that
P −1 =
1+b Yc
bX 0
.
(2.4)
M.T. Koşan et al. / Linear Algebra and its Applications 450 (2014) 7–12
11
For our next computation, it will be convenient to introduce the following notation. π (Q) π12 (Q) , If Q is an n × n matrix over D, we will write Q in block form Q = π11 21 (Q) π22 (Q) where π11 (Q), π12 (Q), π21 (Q) and π22 (Q) have sizes 1 × 1, 1 × (n − 1), (n − 1) × 1 and (n − 1) × (n − 1), respectively. We calculate that, for k 1,
π11 (k+1 ) π21 (k+1 )
π12 (k+1 ) π22 (k+1 )
= k+1 = k
=
=
π11 (k ) π21 (k )
π12 (k ) π22 (k )
1+b Yc
π11 (k )(1 + b) + π12 (k )Y c π21 (k )(1 + b) + π22 (k )Y c
bX 0
π11 (k )bX π21 (k )bX
.
(2.5)
Note that π12 () = bX and π21 () = Y c and π22 () = 0 (= Y 0X). By (2.5), an easy inductive argument shows that there exist ak , bk , ck ∈ D such that π12 k = bk X,
π21 k = Y ck ,
π22 k = Y ak X
for k = 1, 2, . . . . (2.6)
Because is a nilpotent and π21 () = Y c = 0, there exists a positive integer s such that π21 (s+1 ) = 0 but π21 (s ) = 0. Then, by (2.5) and (2.6),
s+1
=
π11 (s+1 ) π12 (s+1 ) 0 Y cs bX
,
where cs b = 0. Because Y has size (n − 1) × 1 and X has size 1 × (n − 1), we see that, for d ∈ D, Y dX = 0 if and only if d = 0; so for d = 0, (Y dX)2 = Y (dXY d)X = 0 as dXY d = d(XY )d = 0 and, similarly (Y dX)k = 0 for all k > 0. In particular, Y cs bX ∈ Mn−1 (D) is not a nilpotent, and hence s+1 is not a nilpotent. Therefore, is not a nilpotent, a contradiction. 2 Theorem 3 can be extended from a division ring to a strongly regular ring. Recall that a ring is called strongly regular if it is a von Neumann regular ring with all idempotents central. Every strongly regular ring contains no nonzero nilpotents. It is well known that a ring R is strongly regular if and only if each element of R is a product of a unit and a central idempotent. Corollary 4. Let R be a strongly regular ring and let n 1. Then Mn (R) is a nil-clean ring if and only if R is Boolean. Proof. (⇐=). This is by [1, Corollary 6]. (=⇒). Assume on the contrary that R is not Boolean. As R is strongly regular, R contains a unit u = 1. Then 1 − u = ve where v is a unit of R and e is a nonzero central idempotent. Thus (1 − e)R is a proper ideal of R, so there exists a maximal ideal I of R such that (1 − e)R ⊆ I. As every quotient ring of a strongly regular ring is again
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strongly regular, R/I is strongly regular, and hence is a division ring (being a simple ring). Since a quotient ring of a nil-clean ring is nil-clean, Mn (R/I) ∼ = Mn (R)/Mn (I) is nil-clean. Therefore, R/I ∼ = F2 by Theorem 3. It follows that ve = 1 − u ∈ I, so e ∈ I. Hence 1 = e + (1 − e) ∈ I, a contradiction. 2 The assumption that R is strongly regular in Corollary 4 cannot be weakened to R being unit-regular. Here a ring R is unit-regular if, for each a ∈ R, a = aua for a unit u of R. Indeed, for R = Mn (F2 ) (n 2), Mk (R) ∼ = Mnk (F2 ) is nil-clean, R is unit-regular but not Boolean. To give another application of Theorem 3, we need the following result from [3]: Lemma 5. (See [3, Corollary 3.17].) A ring R is nil-clean if and only if J(R) is nil and R/J(R) is nil-clean. A ring R is called semilocal if R/J(R) is semisimple Artinian. Because a finite direct product of rings is nil-clean if and only if each direct summand is nil-clean, the next corollary is a quick consequence of Theorem 3, Lemma 5 and the Wedderburn–Artin Theorem. Corollary 6. A semilocal ring R is nil-clean if and only if J(R) is nil and R/J(R) is a direct product of matrix rings over F2 . Acknowledgements We are grateful to the referee for his invaluable comments. Yiqiang Zhou acknowledges gratefully the support from NCTS of Taipei and from TUBITAK of Turkey for his visits to the National Taiwan University and the Gebze Institute of Technology, and the hospitality from the two universities. The research of the second author was supported by NSC of Taiwan and by NCTS of Taipei, and that of the third author by a Discovery Grant from NSERC of Canada. References [1] S. Breaz, G. Călugăreanu, P. Danchev, T. Micu, Nil-clean matrix rings, Linear Algebra Appl. 439 (2013) 3115–3119. [2] A.J. Diesl, Classes of strongly clean rings, PhD thesis, University of California, Berkeley, 2006. [3] A.J. Diesl, Nil clean rings, J. Algebra 383 (2013) 197–211. [4] G. Song, X. Guo, Diagonability of idempotent matrices over noncommutative rings, Linear Algebra Appl. 297 (1999) 1–7.