Strongly Clean Matrices over Commutative Domains

Algebra Colloquium 21 : 2 (2014) 257–266 DOI: 10.1142/S1005386714000212

Algebra Colloquium c 2014 AMSS CAS ° & SUZHOU UNIV

Strongly Clean Matrices over Commutative Domains∗ Huanyin Chen

Algebra Colloq. 2014.21:257-266. Downloaded from www.worldscientific.com by HANGZHOU NORMAL UNIVERSITY on 06/19/14. For personal use only.

Department of Mathematics, Hangzhou Normal University Hangzhou 310036, China E-mail: [email protected] Received 6 October 2010 Revised 11 December 2010 Communicated by L.A. Bokut Abstract. We get criteria of strong cleanness for several classes of 2 × 2 matrices over integers. For commutative local domains, we establish ones in terms of solvability of quadratic equations. Strongly clean matrices over power series are also studied. 2010 Mathematics Subject Classification: 16S50, 16E50 Keywords: strong cleanness, ring of integers, local domain

1 Introduction Throughout, all rings are associative rings with identity. We say that an element a ∈ R is strongly clean provided that there exists an idempotent e ∈ R and a unit u ∈ R such that a = e + u and eu = ue. A ring R is strongly clean in case every element in R is strongly clean. As is well known, every strongly π-regular ring is strongly clean (cf. [8]). If R is strongly clean, then so is each of its corners (cf. [6, Theorem 2.4]). But strong cleanness is not Morita invariant. In fact, it is hard to determine when a 2 × 2 matrix is strongly clean. In [5, Lemma 1.6], Chen et al. observed a necessary condition of strong cleanness for commutative domains. Using this fact, they discussed strongly clean 2 × 2 matrices over some special classes of commutative local domains. In Section 2, we extend [5, Lemma 1.6] and observe that strong cleanness of 2 × 2 matrices over commutative domains can deduce a system of quadratic equations. From these, we get criteria of strong cleanness for several classes of 2 × 2 matrices over integers. In [4, Theorem 8], Chen et al. studied strong cleanness of 2 × 2 matrices over commutative local rings by means of solvability. Recently, these results are extensively improved by many authors (cf. [1], [7], [9], [11]). In Section 3, we discuss the strong cleanness of 2 × 2 matrices over commutative local domains, and get a ∗

The research of the author was supported by the Natural Science Foundation of Zhejiang Province (LY13A010019) and the Fund of Hangzhou Normal University, China.

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new criterion in terms of solvability of the preceding quadratic equations in R for M2 (R) to be strongly clean. Strong cleanness of matrices over the rings of power series are also discussed. 2 Matrices over the Ring Z of Integers

³

In what follows, we always use ha, bi to stand for the matrix following simple fact can be verified straightforwardly:

a 0 b 0

´ ∈ M2 (R). The


Lemma 2.1. [5, Lemma 1.5] Let R be a commutative³domain.´Then A ∈ M2 (R) is a b an idempotent if and only if A = 0 or A = I2 or A = c 1 − a , where bc = a − a2 in R. Theorem 2.2. Let R be a commutative domain. Then ha, bi ∈ M2 (R) is strongly clean if and only if either a ∈ U (R) or 1 − a ∈ U (R). Proof. If a ∈ U (R), then µ ¶ µ ¶ µ ¶ a 0 0 0 a 0 = + . b 0 −a−1 b 1 (1 + a−1 )b −1 If 1 − a ∈ U (R), then µ

a 0 b 0

¶

µ =

1 0 0 1

¶

µ +

a−1 0 b −1

¶ .

In these cases, ha, bi ∈ M2 (R) is strongly clean. Conversely, assume that A := ha, bi is strongly clean. If I2 − A is invertible, then 1 − a ∈ GL2 (R). Thus, we may assume that I2 − A is a nonunit. Obviously, A is a nonunit. Write³ A = E +´ U with x z E = E 2 , U ∈ GL2 (R) and EU = U E. In view of Lemma 2.1, E = y 1 − x , where ³ ´ ∗ −z zy = x−x2 . As U = A−E ∈ GL2 (R), we see that U = ∗ x − 1 . From EA = AE, we deduce az = 0, thus we may assume that z = 0; otherwise, I2 − A ∈ GL2 (R). This implies x(1 − x) ³ = 0, ´ so x = 0 or x = 1. As U ∈ GL ³ 2 (R), x ´6= 1. This implies a 0 0 0 x = 0, hence E = y 1 . It follows from A − E = b − y −1 ∈ GL2 (R) that a ∈ U (R), and we are done. ¤ Corollary 2.3. Let a, b ∈ Z. Then ha, bi ∈ M2 (Z) if and only if a = 0, ±1, 2. Proof. Clearly, a ∈ U (Z) or 1 − a ∈ U (Z) if and only if a = 0, ±1, 2. Therefore, the proof is true by Theorem 2.2. ¤ According to Corollary 2.3, h7, 7i ∈³ M2 (Z) clean. But it is clean ´ is ³not strongly ´ 5 −5 4 5 from the clean representation h7, 7i = 4 −4 + 3 4 . Thus, every strongly clean ring is clean, but the converse is not true. Let A = (aij ) ∈ M2 (R), sA = a11 − a22 and tA = (trA)2 − 4 det A. By a similar route, we can extend [5, Lemma 1.6] as follows.

Strongly Clean Matrices over Commutative Domains

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Lemma 2.4. Let R be a commutative domain and A ∈ M2 (R). If A is strongly clean in M2 (R), then A is invertible or I2 − A is invertible or there exists q ∈ R with q ≡ 1 (mod 2) such that s2A = tA q 2 . Proof. Suppose that A and I2 − A are nonunits. In view of Lemma 2.1, ³ ´ there a b exist a, b, c ∈ R such that A = E + (A − E), where E = E 2 = c 1 − a , bc =


a − a2 , EA = AE, A − E ∈ GL2 (R). As EA = AE, we get sA b = a12 (2a − 1) and sA c = a21 (2a − 1). Hence, s2A (a − a2 ) = s2A bc = a12 a21 (2a − 1)2 . That is, (s2A + 4a12 a21 )(a − a2 ) = a12 a21 . Obviously, tA = s2A + 4a12 a21 . Therefore, s2A = tA − 4a12 a21 = tA − 4tA (a − a2 ) = tA (1 − 2a)2 . Set q = 1 − 2a. Then s2A = tA q 2 and q ≡ 1 (mod 2), as required. ¤ In [10], Rajeswari and Aziz studied strong cleanness of 2 × 2 matrices over the ring of integers by means of a kind of the Diophantine equation. Now we completely determine the strong cleanness for several classes of 2 × 2 matrices over integers. ³ ´ x+1 y Theorem 2.5. A matrix x x ∈ M2 (Z) is strongly clean if and only if x = −1, y = −3, 0, ±1; x = 0; or x = 1, y = 0, ±1, 3. ³ ´ ³ ´ ³ ´ ³ ´ x+1 y 1 y 0 −y 1 2y Proof. Let A = + 0 −1 , thus A x x . If x = 0, then 0 0 = 0 1 is strongly clean. If x = 1, y = ±1, 3 or x = −1, y = ±1, −3, then A ∈ GL2 (Z) or I2 − A ∈ GL2 (Z), hence A is strongly clean. In addition, we have strongly clean expresses µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ 2 0 1 0 1 0 0 0 1 0 −1 0 = + , = + . 1 1 1 0 0 1 −1 −1 −1 0 0 −1 Thus, A is strongly clean provided that´ x = ±1, y = 0, as required. ³ x+1 y Conversely, assume that x x ∈ M2 (Z) is strongly clean. If x 6= ±1, then A and I2 − A are nonunits. In view of Lemma 2.4, we have q ∈ Z such that 2 1 = (1 + 4xy)q . This ³ ´ implies 1 + 4xy = 1, hence x = 0 or y = 0. If y = 0, then we a c have E = b 1 − a ∈ M2 (Z) and U ∈ GL2 (Z) such that µ ¶ x+1 0 = E + U, E = E 2 , EU = U E. x x 2 From³EA = AE, we ´ deduce c = 0, so a = a. This implies a = 0, 1. If a = 0, then x+1 0 U= ∈ GL2 (Z), and thus x2 − 1 = ±1. Hence, x = 0. If a = 1, then ³ ∗ ´x − 1 x 0 U = ∗ x ∈ GL2 (Z), and thus x = ±1. This gives a contradiction. Thus, we consider the case x = ±1. If x = 1 and y 6= ³0, ±1,´3, then ³ A and ´ I2 − A are nonunits, so we have some a c 2 y a, b, c ∈ R such that 1 1 = b 1 − a + U is a strongly clean expression. It ³ ´³ ´ ³ ´³ ´ a c a c 2 y 2 y follows from 1 1 1 1 that c = yb, 2a − b = 1 and b 1−a = b 1−a

c + y = 2ay. As a − a2 = bc, we see that (4y + 1)a2 − (4y + 1)a + y = 0. Thus, 1 a = 12 ± 2√4y+1 6∈ Z for all y ∈ Z − {0}, a contradiction.

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If x = −1 and y 6=³ −3, 0, ±1, ´ then ³ A and ´ I2 − A are nonunits, so we have some a c 0 y a, b, c ∈ R such that −1 −1 = b 1 − a + U is a strongly clean expression. It ³ ´³ ´ ³ ´³ ´ a c a c 0 y 0 y follows from −1 −1 −1 −1 that c = −yb, 2a + b = 1 b 1−a = b 1−a and c + y = 2ay. In addition, a − a2 = bc. As a result, we can get (4y − 1)a2 − 1 (4y − 1)a + y = 0. Thus, a = 12 ± 2√4y−1 6∈ Z for all y ∈ Z − {0}, a contradiction. Therefore, we complete the proof. ¤ ³ ´ x x Corollary 2.6. A matrix y x + 1 ∈ M2 (Z) is strongly clean if and only if x = −1, y = −3, 0, ±1; x = 0; or x = 1, y = 0, ±1, 3. ´ ´ ³ ´ ³ ´³ ´³ ´ ³ ³ 0 1 0 1 −1 0 1 x+1 y 0 1 x x . = and Proof. Clearly, y x + 1 = 1 0 1 0³ 1 0 x x ´1 0 ´ ³ x+1 y x x Thus, y x + 1 ∈ M2 (Z) is strongly clean if and only if so is x x ∈ Z. Therefore, we obtain the result from Theorem 2.5. ¤ ´ ³ x−1 y Theorem 2.7. A matrix x x ∈ M2 (Z) is strongly clean if and only if x = −3, y = −7; x = −1, y = −7, −5, −3, −1; x = 0; x = 1, y = 0, ±1; x = 2, y = 0; or x = 3, y = 3. Proof. It suffices to consider (i) x = 0; (ii) x = 1, y = 0; (iii) x =³ 2, y = ´0. x−1 y Otherwise, A or I2 − A is invertible, then A is strongly clean. Let A = x x . ³ ´ ³ ´ ³ ´ −1 y 0 y −1 0 If x = 0, then 0 0 = 0 1 + 0 −1 , thus A is strongly clean. In addition, we have strongly clean expresses µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ 0 0 1 0 −1 0 1 0 0 0 1 0 = + , = + . 1 1 −1 0 2 1 2 2 2 1 0 1 Hence, A is strongly clean in³ the cases ´ (ii) and (iii), as required. x−1 y Conversely, assume that x x ∈ M2 (Z) is strongly clean. If A is invertible, then x2 − (y + 1)x ± 1 = 0. That is, y + 1 = x ± x1 . Hence, x = 1, y = ±1 or x = −1, y = −1, −3. If I2 −A is invertible, then x2 −(y +3)x+1 = 0 or x2 −(y +3)x+3 = 0. That is, y + 3 = x + x1 or y + 3 = x + x3 . Hence, x = 1, y = ±1; x = −1, y = −7, −5; x = 3, y = 3 or x = −3, y = −7. Otherwise, it follows from Lemma 2.4 that we have q ∈ Z such that 1 = (1³+ 4xy)q 2´. We infer 1 + 4xy = 1, hence x = 0 or y = 0. a c If y = 0, then we have E = b 1 − a ∈ M2 (Z) and U ∈ GL2 (Z) such that µ

x−1 0 x x

¶ = E + U,

E = E2,

EU = U E.

As in³the proof of´ Theorem 2.5, we deduce c = 0, then a = 0, 1. If ³a = 0, then ´ x−1 0 x−2 0 U = ∈ GL (Z), and thus x = 0, 2. If a = 1, then U = 2 ∗ x−1 ∗ x ∈ GL2 (Z), and thus x = 1. As a result, we get x = 0; x = 2, y = 0; or x = 1, y = 0, as required. ¤


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´ ³ x x Corollary 2.8. A matrix y x − 1 ∈ M2 (Z) is strongly clean if and only if x = −3, y = −7; x = −1, y = −7, −5, −3, −1; x = 0; x = 1, y = 0, ±1; x = 2, y = 0; or x = 3, y = 3. ³ ´ ³ ´³ ´³ ´ ³ ´ x x 0 1 x−1 y 0 1 x x Proof. Clearly, y x − 1 = 1 0 , so x´ x 1 0 y x − 1 ∈ M2 (Z) is ³ x−1 y strongly clean if and only if so is x x ∈ Z. Therefore, the result follows from Theorem 2.7. ¤


3 Commutative Local Domains ³ ´ x+1 x Let A = f (x) x ∈ Z[x]. Then sA = 1. If A is strongly clean, then 1 = tA q 2 for some q ∈ Z. This implies tA = 1. It follows from tA = s2A + txf (x) that xf (x) = 0. Thus, we claim that A is not strongly clean in Z[x] for any 0 6= f (x) ∈ Z[x]. Let R be a commutative domain, and let P be a prime ideal of R. Then RP is a commutative local domain. Theorem 3.1. Let R be a commutative local domain of characteristic not equal to 2. Then the following are equivalent: (1) A ∈ M2 (R) is strongly clean. ³ ´ 0 λ (2) A is invertible or I2 − A is invertible or A is similar to a matrix B = 1 µ with λ ∈ J(R), µ ∈ 1 + J(R) such that s2B = tB q 2 , q ≡ 1 (mod 2). Proof. (1)⇒(2) In view of [2, Lemma A is invertible or I2 − A is invertible ³ 16.4.11], ´ 0 λ or A is similar to a matrix B = 1 µ with λ ∈ J(R), µ ∈ 1 + J(R). As A is strongly clean, so is B. Thus, s2B = tB q 2 , q ≡ 1 (mod 2) by Lemma 2.4. (2)⇒(1) If A is invertible or I2 − A is invertible, then A ∈ M2³(R) is ´strongly 0 b , where clean. So we may assume that A is similar to a matrix B = (bij ) = 1 b12 22 b12 ∈ J(R), b22 ∈ 1 + J(R), b11 = 0 and b21 = 1. In addition, s2B = tB q 2 , q ≡ 1 (mod 2). Obviously, sB = −b22 ∈ U (R) and tB = s2B + 4b12 . Write 1 − q = 2x for some x ∈ R. Let a = 1 − x. Then q = 2a − 1. ³ ´ a b −1 If a ∈ U (R), we choose b = s−1 B b12 q and c = sB b21 q. Let E = c 1 − a . Then −2 2 2 2 2 2 2 4bc = 4s−2 B b12 b21 q = sB (tB − sB )q = 1 − q = 1 − (2a − 1) = 4(a − a ), hence 2 bc = a − a because the characteristic of R is not 2 . This implies E = E 2 . As b21 b = b12 c, we have b11 a+b21 b = b11 a+b12 c and b12 c+b22 (1−a) = b21 b+b22 (1−a). Since sB b = b12 (2a − 1) and sB c = b21 (2a − 1), we get b12 a + b22 b = b11 b + b12 (1 − a) and b11 c + b21 (1 − a) = b21 a + b22 c. Thus, we have EB = BE. It suffices to show that B − E ∈ GL2 (R). Observe that

det(B − E) = (b11 − a)(b22 + a − 1) − (b12 − b)(b21 − c) = det B + sB a − b11 + 2b12 c ∈ U (R). If a 6∈ U (R), then a ∈ J(R), hence x ∈ U (R). Thus, q ³ = 1 − 2x´with x ∈ U (R). x b −1 −1 Choose b = −sB b12 q and c = −sB b21 q. Let E = c 1 − x . Then 4bc =

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−2 2 2 2 2 2 2 2 4s−2 B b12 b21 q = sB (tB −sB )q = 1−q = 1−(1−2x) = 4(x−x ), hence bc = x−x . 2 This implies E = E . As b21 b = b12 c, we see that b11 x + b21 b = b11 x + b12 c and b12 c + b22 (1 − x) = b21 b + b22 (1 − x). As sB b = b12 (2x − 1) and sB c = b21 (2x − 1), we get b12 x + b22 b = b11 b + b12 (1 − x) and b11 c + b21 (1 − x) = b21 x + b22 c. Thus, we have EB = BE. It suffices to show that B − E ∈ GL2 (R). Observe that

det(B − E) = (b11 − x)(b22 + x − 1) − (b12 − b)(b21 − c) = det B + sB x − b11 + 2b12 c ∈ U (R).


Thus, we conclude that B ∈ M2 (R) is strongly clean, then so is A, as required. ¤ Corollary 3.2. Let R be a commutative local domain of characteristic not equal to 2. Then the following are equivalent: (1) M2 (R) is strongly clean. (2) For any A ∈ M2 (R), A is invertible or I2 − A is invertible or s2A = tA q 2 , q ≡ 1 (mod 2). Proof. (1)⇒(2) It is obvious by Lemma 2.4. (2)⇒(1) Let A ∈ M2 (R). In view of [2, Lemma 16.4.11], A is invertible or I2 − A ³ ´ 0 λ is invertible or A is similar to a matrix B = 1 µ with λ ∈ J(R), µ ∈ 1 + J(R). If A is invertible or I2 − A is invertible, then A is strongly clean. Otherwise, B, I2 −B 6∈ GL2 (R). By hypothesis, s2B = tB q 2 , q ≡ 1 (mod 2). According to Theorem 3.1, B is strongly clean, then so is A. ¤ Corollary 3.3. Let R be a commutative local domain. If 12 ∈ R, then the following are equivalent: (1) M2 (R) is strongly clean. (2) For any A ∈ M2 (R), A is invertible or I2 − A is invertible or there exists q ∈ R such that s2A = tA q 2 . Proof. It is immediate from Corollary 3.2.

¤

Proposition 3.4. Let R be a commutative local domain with 21 ∈ R. If there exists q ∈ U (R) such that s2A = tA q 2 , then A ∈ M2 (R) is strongly clean. Proof. If A or I2 − A is invertible, then A is strongly clean. Now we assume that A and I2 − A are nonunits. Thus, det A, det(I2 − A) ∈ J(R). As 1 − trA + det A = det(I2 − A) ∈ J(R), tr(A) ∈ 1 + J(R). Hence, tA = (trA)2 − 4 det A ∈ 1 + J(R), so tA ∈ U (R). Clearly, tA = (sA (−q −1 ))2 , where sA (−q −1 ) ∈ U (R). Set u = sA (−q −1 ). If u ∈ −1 + J(R), then tA = v 2 , where v = −u ∈ 1 + J(R). In light of [5, Lemma 2.3], A ∈ M2 (R) is strongly clean. As R is local, we may assume that u ∈ −1 + U (R). Since 21 ∈ R, there exists some d ∈ R³ such that ´ sA − u = 2d. Let a = −du−1 , b = −a12 u−1 , c = −a21 u−1 , and E =

a b c 1−a .

Clearly, tA = s2A + 4a12 a21 . It is easy to verify 4bc = 4a12 a21 u−2 = (tA − s2A )u−2 = 1 − s2A u−2 = 1 − (sA u−1 )2 = 1 − (1 − 2a)2 = 4(a − a2 ), hence bc = a − a2 . This implies E = E 2 . As a21 b = a12 c, we see that a11 a + a21 b = a11 a + a12 c and a12 c + a22 (1 − a) = a21 b + a22 (1 − a). As sA b = a12 (2a − 1) and sA c = a21 (2a − 1),

263


we get a12 a + a22 b = a11 b + a12 (1 − a) and a11 c + a21 (1 − a) = a21 a + a22 c. Thus, we have EA = AE. It suffices to show that A − E ∈ GL2 (R). Observe that 2 det(A − E) = 2(a11 − a)(a22 + a − 1) − 2(a12 − b)(a21 − c) = 2 det A + 2sA a − 2a11 + 4a12 c = 2 det A + s(1 − sA u−1 ) − 2a11 − 4a12 a21 u−1 = 2 det A − (s2A + 4a12 a21 )u−1 + (s − 2a11 ) = 2 det A − tA u−1 − trA = 2 det A − u − trA ∈ −(1 + u) + J(R).


As u ∈ −1 + U (R), we see that det(A − E) ∈ U (R), hence A − E ∈ GL2 (R). Therefore, A is strongly clean. ¤ Corollary 3.5. Let R be a commutative local domain with ∈ R. Then the ´following are equivalent: ³ x+1 y (1) z x ∈ M2 (R) is strongly clean. ³ ´ x−1 y (2) z x ∈ M2 (R) is strongly clean.

1 2

∈ R, and let x, y, z

(3) There exists q ∈ R such that (1 + 4yz)q 2 = 1. ´ ³ x+1 y Proof. (1)⇒(3) Let A = z x . Then sA = 1 and tA = 1 + 4yz. In light of Lemma 2.4, there exists q ∈ R such that (1 + 4yz)q 2 = 1. (3)⇒(1) By hypothesis, s2A = tA q 2 with q ∈ U (R). According to Proposition 3.4, we conclude that A is strongly clean, as asserted. (2)⇔(3) It is proved in the similar manner. ¤ Analogously, we could derive the following: Corollary 3.6. Let R be a commutative local domain with ∈ R. Then the ´following are equivalent: ³ x y (1) z x + 1 ∈ M2 (R) is strongly clean. ³ ´ x y (2) z x − 1 ∈ M2 (R) is strongly clean.

1 2

∈ R, and let x, y, z

(3) There exists q ∈ R such that (1 + 4yz)q 2 = 1. Let R be³a commutative local domain in which 2 is invertible, and let x, y ∈ R. ´ x±1 y Claim that 0 x is strongly clean. Choose q = 1. Then we are through by 1 Corollaries ³ ´3.5 and 3.6. Note that 2 ∈ R is necessary in the preceding corollaries, 3 0 e.g., 2 2 ∈ M2 (Z) is not strongly clean by Theorem 2.5. But (1 + 4 · 0 · 2)q 2 = 1 for©q =¯ 1 ∈ R. Let p ∈ªZ be a prime number. The local ring Z(p) of the ring Z at p ¯ is m n m, n ∈ N, p - n . Obviously, Z(p) is a commutative local domain. ³ ´ −2 3 Example 3.7. Let A = −4 5 ∈ M2 (Z(5) ). Then sA = −7 and tA = 1. Choose

q = 7. Then q ∈ U (Z(5) ) and s2A = tA q 2 . Clearly, Z(5) is a commutative local domain with 12 ∈ Z(5) . In light of Proposition 3.4, A is strongly ³ clean´ in M ³ 2 (Z´(5) ). This can also be seen from the strongly clean expression A =

−3 3 −4 4

+

1 0 0 1 .

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4 Matrices over Power Series The main purpose of this section is to investigate strongly clean matrices over power series for commutative local domains.


Theorem 4.1. Let R be a commutative local domain with M2 (R[[x]]). Then the following are equivalent: (1) A(x) ∈ M2 (R[[x]]) is strongly clean. (2) A(0) ∈ M2 (R) is strongly clean.

1 2

∈ R, and let A(x) ∈

Proof. (1)⇒(2) Since A(x) ∈ M2 (R[[x]]) is strongly clean, there exists an idempotent E(x) ∈ M2 (R[[x]]) and U (x) ∈ GL2 (R[[x]]) such that A(x) = E(x) + U (x) and E(x)U (x) = U (x)E(x). Therefore, A(0) = E(0) + U (0), E(0)2 = E(0) and E(0)U (0) = U (0)E(0). This implies that A(0) ∈ M2 (R) is strongly clean. (2)⇒(1) It is easy to verify that R[[x]] is a commutative local domain with 12 ∈ R. If A(0) ∈ GL2 (R), then A(x) ∈ GL2 (R[[x]]) is strongly clean. If I2 −A(0) ∈ GL2 (R), then I2 − A(x) ∈ GL2 (R[[x]]) is strongly clean, and so A(x) ∈ M2 (R[[x]]) is strongly clean. Thus, we may assume ³ that´ A(0), I2 − A(0) 6∈ GL2 (R). In view of Theorem 0 λ 3.1, A(0) is similar to B = 1 µ , where λ ∈ J(R), µ ∈ 1 + J(R), and s2B = tB q 2 for some q ∈ R. ³Obviously, ´ q ∈ U (R). By virtue of [3, Lemma 3.1], A(x) is similar to B(x) =

0 λ(x) 1 µ(x) , where λ(x) ∈ J(R[[x]]), µ(x) ∈ 1 + J([[x]]). Clearly,

P∞ i tB(x) ∈ U (R[[x]]). Write s2B(x) t−1 i=0 ci x , where each ci ∈ R. Clearly, B(x) = c0 = q 2 . Construct q0 , q1 , q2 , q3 , . . . by the following equalities: q0 = q, 2q q1 = c1 , P0 ∞ 2q0 q2 +q12 = c2 , 2q0 q3 +2q1 q2 = c3 , 2q0 q4 +2q1 q3 +q22 = c4 , . . . . Set q(x) = i=0 qi xi . Then s2B(x) = tB(x) q(x)2 . As 21 ∈ R, we see that q(x) ≡ 1 (mod 2). According to Theorem 3.1, A(x) ∈ R[[x]] is strongly clean. ¤ Let Ã A(x) =

! P∞ P∞ n 3 + n=1 2n−1 −2 + n=1 xn 3n x ∈ M2 (Z(5) [[x]]). P∞ P∞ −4 + n=1 (−1)n xn 5 + n=1 21n xn

In view of Example 3.7, A(0) ∈ M2 (Z(5) ) is strongly clean. Therefore, we conclude that A(x) ∈ M2 (Z(5) [[x]]) is strongly clean from Theorem 4.1. Corollary 4.2. Let R be a commutative local domain with following are equivalent: (1) M2 (R) is strongly clean. (2) M2 (R[[x]]) is strongly clean. Proof. It is obvious from Theorem 4.1.

1 2

∈ R. Then the

¤

Proposition 4.3. Let R be a commutative local domain with 12 ∈ R, and let A(x) ∈ M2 (R[[x]]/(xn+1 )) (n ≥ 0). Then the following are equivalent: (1) A(x) is strongly clean in M2 (R[[x]]). (2) A(0) is strongly clean in M2 (R).

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Proof. (1)⇒(2) It is proved analogously to Theorem 4.1. (2)⇒(1) Clearly,


R[[x]]/(xn+1 ) = {a0 + a1 x + · · · + an xn | a0 , a1 , . . . , an ∈ R, xn+1 = 0}, J(R[[x]]/(xn+1 )) = {a0 + a1 x + · · · + an xn | a0 ∈ J(R), a1 , . . . , an ∈ R, xn+1 = 0}. Let ϕ : R[[x]] → R[[x]]/(xn+1 ) be the canonical map. It deduces a surjective ring morphism ψ : M2 (R[[x]]) → M2 (R[[x]]/(xn+1 )). Since A(x) ∈ M2 (R[[x]]/(xn+1 )), we can find B(x) ∈ M2 (R[[x]]) such that ψ(B(x)) = A(x) with B(0) = A(0). Thus, B(0) is strongly clean. By Theorem 4.1, B(x) ∈ M2 (R[[x]]) is strongly clean. So we have E(x) = E 2 (x) ∈ M2 (R[[x]]) and U (x) ∈ GL2 (R[[x]]) such that B(x) = E(x) + U (x), E(x)W (x) = W (x)E(x). So A(x) = ψ(E(x)) + ψ(U (x)) and ψ(E(x))ψ(U (x)) = ψ(U (x))ψ(E(x)). Obviously, ψ(E(x)) ∈ M2 (R[[x]]/(xn+1 )) is an idempotent. Further, ψ(U (x)) ∈ U (R[[x]]/(xn+1 )), and we are done. ¤ Corollary 4.4. Let R be a commutative local domain with following are equivalent: (1) M2 (R) is strongly clean. (2) M2 (R[[x]]/(xn+1 )) is strongly clean. (3) M2 (R[[x]]/(xn+1 )) is strongly clean.

1 2

∈ R. Then the

Proof. Since R[[x]]/(xn+1 ) = R[x]/(xn+1 ), it is obvious by Proposition 4.3. ¤ µ ¶ x 2 Example 4.5. Let R = Z3 [x]/(x2 ) and A(x) = ∈ M2 (R). Then A(0) = 2x 1 µ ¶ 0 2 . Obviously, Z3 is a commutative local domain in which 2 is invertible. 0 1

Clearly, sA(0) = −1 and tA(0) = 1. Then we can find q = 1 ∈ U (Z3 ) such that s2A(0) = tA(0) q. In light of Proposition 3.4, A(0) ∈ M2 (Z3 ) is strongly clean. According to Proposition 4.3, A(x) ∈ M2 (R) is strongly clean. In fact, ¶ we have the ³ ´ µ 2 + 2x 1 + x strongly clean decomposition A(x) = 1 +x 2x 1 +x 2x + in M2 (R). x

1 + 2x

Acknowledgement. The author is grateful to the referee for his/her suggestions which made the new version clearer and simpler.

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