An element a ∈ R is strongly clean provided that it is the sum of an idempotent and a unit that commutes. A ring R is strongly clean provided that every element in R is strongly clean. A ring R is local if it has only one maximal right ideal. As is well known, a ring R is local if and only if for any x∈ R, x or 1-x is invertible. Strongly clean matrices over commutative local rings was extensively studied by many authors from very different view points (see [1, 2, 3, 5, 6, 7, 8, 9, 10, 11]). Recently, a related cleanness of triangular matrix rings over abelian rings was studied by Diesl et al. (see [9]).

Following Diesl, we say that a ∈ R is strongly rad-clean provided that there exists an idempotent e ∈ R such that a−e∈U(R), ae=ea and eae∈J(eRe) (see [9]). A ring R is strongly rad-clean provided that every element in R is strongly rad-clean. Strongly rad-clean rings form a natural subclass of strongly clean rings which have stable range one (see [4]). Let M be a right R-module, and let φ∈endR(M). Then we include a relevant diagram to reinforce the theme of direct sum decompositions:

If such diagram holds we call this is an AB-decomposition for φ. It turns out by [2, Lemma 40] that φ is strongly π-regular if and only if there is an AB-decomposition with φ|B∈N(end(B)) (the set of nilpotent elements).

Further, φ is strongly rad-clean if and only if there is an AB-decomposition with φ|B∈J(end(B)) (the Jacobson radical of end(B)). Thus, strong rad-cleanness can be seen as a natural extension of strong π-regularity. In [2, Theorem 12], the authors gave a criterion to characterize when a square matrix over a commutative local ring is strongly clean. We extend this result to strongly rad-clean matrices over a commutative local ring. We completely determine when a 2×2 matrix over a commutative local ring has such clean decomposition related to its Jacobson radical. Application to the matrices over power-series is also studied.

Throughout, all rings are commutative with an identity and all modules are unitary left modules. Let M be a left R-module. We denote the endomorphism ring of M by end(M) and the automorphism ring of M by aut(M), respectively. The characteristic polynomial of A is the polynomial χ(A)=det(tIn−A). We always use J(R) to denote the Jacobson radical and U(R) is the set of invertible elements of a ring R. M2(R) stands for the ring of all 2×2 matrices over R, and GL2(R) denotes the 2-dimensional general linear group of R.

In this section, we study the structure of strongly rad-clean elements in various situations related to ordinary ring extensions which have roles in ring theory. We start with a well known characterization of strongly rad-clean element in the endomorphism ring of a module M.

Lemma 2.1. Let E=end(RM), and let α∈E. Then the following are equivalent:

• (1) α∈E is strongly rad-clean.

• (2) There exists a direct sum decomposition M=P⊕Q where P and Q are α-invariant, and α|P∈aut(P) and α|Q∈J(end(Q)).

Proof. See [4, Proposition 4.1.2]. Lemma 2.2. Let R be a ring, let M be a left R-module. Suppose that x,y,a,b∈end(RM) such that xa+yb=1M,xy=yx=0,ay=ya and xb=bx. Then M=ker(x)⊕ker(y) as left R-modules.

Proof. See [2, Lemma 11].

A commutative ring R is projective-free if every finitely generated projective R-module is free. Evidently, every commutative local ring is projective-free. We now derive

Lemma 2.3. Let R be projective-free. Then A∈M2(R) is strongly rad-clean if and only if A∈GL2(R), or A∈J(M2(R)), or A is similar to diag(α,β) with α∈J(R) and β∈U(R).

Proof. ⇒ Write A=E+U,E2=E,U∈GL2(R),EA=AE∈J(M2(R)). Since R is projective-free, there exists P∈GLn(R) such that PEP−1=diag(0,0),diag(1,1) or diag(1,0). Then (i) PAP−1=PUP−1; hence, A∈GL2(R), (ii) (PAP−1)diag(1,1)=diag(1,1)(PAP−1)∈J(M2(R), and so A∈J(M2(R)). (3) (PAP−1)diag(1,0)=diag(1,0)(PAP−1)∈J(M2(R) and PAP−1−diag(1,0)∈GL2(R). Hence, PAP−1=abcd with a∈J(R),b=c=0 and d∈UR). Therefore A is similar to diag(α,β) with α∈J(R) and β∈U(R).

⇐ If A∈GL2(R) or A∈J(M2(R)), then A is strongly rad-clean. We now assume that A is similar to diag(α,β) with α∈J(R) and β∈U(R). Then A is similar to 1000+α−100β where

Therefore A∈M2(R) is strongly rad-clean.

Theorem 2.4. Let R be projective-free. Then A∈M2(R) is strongly rad-clean if and only if

• (1) A∈GL2(R)), or

• (2) A∈J(M2(R)), or

• (3) x2=tr(A)x−detA has roots α∈U(R),β∈J(R).

Proof. ⇒ By Lemma 2.3, A∈GL2(R), or A∈J(M2(R)), or A is similar to a matrix α00β, where α∈J(R) and β∈U(R). Then χ(A)=(x−α)(x−β) has roots α∈U(R),β∈J(R).

⇐ If (1) or (2) holds, then A∈M2(R) is strongly rad-clean. If (3) holds, we assume that χ(A)=(t−α)(t−β). Choose X=A−αI2 and Y=A−βI2. Then

By virtue of Lemma 2.2, we have 2R=ker(X)⊕ker(Y). For any x∈ker(X), we have (x)AX=(x)XA=0, and so (x)A∈ker(X). Then ker(X) is A-invariant. Similarly, ker(Y) is A-invariant. For any x∈ker(X), we have 0=(x)X=(x)(A−αI2); hence, (x)A=(x)αI2. By hypothesis, we have A|ker(X)∈J(end(ker(X))). For any y∈ker(Y), we prove that

This implies that (y)A=(y)(βI2). Obviously, A|ker(Y)∈aut(ker(Y)). Therefore A∈M2(R) is strongly rad-clean by Lemma 2.1.We have accumulated all the information necessary to prove the following.

Theorem 2.5. Let R be a commutative local ring, and let A∈M2(R). Then the following are equivalent:

• (1) A∈M2(R) is strongly rad-clean.

• (2) A∈GL2(R) or A∈J(M2(R)), or trA∈U(R) and the quadratic equation x2+x=−detAtr2A has a root in J(R).

• (3) A∈GL2(R) or A∈J(M2(R)), or trA∈U(R),detA∈J(R) and the quadratic equation x2+x=detAtr2A−4detA is solvable.

Proof. (1)⇒(2) Assume that A∈GL2(R) and A∈J(M2(R)). By virtue of Theorem 2.4, trA∈U(R) and the characteristic polynomial χ(A) has a root in J(R) and a root in U(R). According to Lemma 2.3, A is similar to λ00μ, where λ∈J(R),μ∈U(R). Clearly, y2−(λ+μ)y+λμ=0 has a root λ in J(R).

Hence so does the equation

Set z=(λ+μ)−1y. Then

That is, z2−z=−(λ+μ)−2λμ. Consequently, z2−z=−detAtr2A has a root in J(R). Let x=-z. Then x2+x=−detAtr2A has a root in J(R).

(2)⇒(3) By hypothesis, we prove that the equation y2−y=−detAtr2A has a root a∈J(R). Assume that trA∈ U(R). Then (a(2a−1)−1)2−(a(2a−1)−1)=detAtr2A⋅(4(a2−a)+1)=detAtr2A⋅(−4(trA)−2detA+1)=detAtr2A−4detA. Therefore the equation y2−y=detAtr2A−4detA is solvable. Let x=-y. Then x2+x=detAtr2A−4detA is solvable.

(3)⇒(1) Suppose A∈GL2(R) and A∈J(M2(R)). Then trA∈U(R),detA∈J(R) and the equation x2+x=detAtr2A−4detA has a root. Let y=-x. Then y2−ydetAtr2A−4detA has a root a∈R. Clearly, b:=1−a∈R is a root of this equation. As a2−a∈J(R), we see that either a∈J(R) or 1−a∈J(R). Thus, 2a−1=1−2(1−a)∈U(R). It is easy to verify that (a(2a−1)−1trA)2−trA⋅(a(2a−1)−1trA)+detA=−tr2A⋅(a2−a)4(a2−a)+1+detA=0. Thus the equation y2−trA⋅y+detA=0 has roots a(2a−1)−1trA and b(2b−1)−1trA. Since ab∈ J(R), we see that a+b=1 and either a∈J(R) or b∈J(R). Therefore y2−trA⋅y+detA=0 has a root in U(R) and a root in J(R). Since R is a commutative local ring, it is projective-free. By virtue of Theorem 2.4, we obtain the result.

Corollary 2.6. Let R be a commutative local ring, and let A∈M2(R). Then the following are equivalent:

• (1) A∈M2(R) is strongly clean.

• (2) I2−A∈GL2(R) or A∈M2(R) is strongly rad-clean.

Proof. (2)⇒(1) is trivial.

(1)⇒(2) In view of [3, Corollary 16.4.33], A∈GL2(R), or I2−A∈GL2(R) or trA∈U(R),detA∈J(R) and the quadratic equation

x2−x=detAtr2A−4detA is solvable. Hence x2+x=detAtr2A−4detA is solvable. According to Theorem 2.5, we complete the proof.

Corollary 2.7. Let R be a commutative local ring. If 12∈R, then the following are equivalent:

• (1) A∈M2(R) is strongly rad-clean.

• (2) A∈GL2(R) or A∈J(M2(R)), or trA∈U(R),detA∈J(R) and tr2A−4detA is square.

Proof. (1)⇒(2) According to Theorem 2.5, A∈GL2(R) or A∈J(M2(R)), or trA∈U(R),detA∈J(R) and the quadratic equation x2−x=detAtr2A−4detA is solvable. If a∈R is the root of the equation, then (2a−1)2=4(a2−a)+1=tr2Atr2A−4detA∈U(R). As in the proof of Theorem 2.5 , 2a−1∈U(R). Therefore tr2A−4detA=(trA⋅(2a−1)−1)2.

(2)⇒(1) If trA∈U(R),detA∈J(R) and tr2A−4detA=u2 for some u∈R, then u∈U(R) and the equation x2+x=detAtr2A−4detA has a root −12u−1(trA+u). By virtue of Theorem 2.5, A∈M2(R) is strongly rad-clean.

Every strongly rad-clean matrix over a ring is strongly clean. But there exist strongly clean matrices over a commutative local ring which is not strongly rad-clean as the following shows.

Example 2.8. Let R=ℤ4, and let A=2302∈M2(R). R is a commutative local ring. Then A=1001+1301 is a strongly clean decomposition. Thus A∈M2(R) is strongly clean. If A∈M2(R) is strongly rad-clean, there exist an idempotent E∈M2(R) and an invertible U∈M2(R) such that A=E+U, EA=AE and EAE∈J(M2(R)). Hence, AU=A(A−E)=(A−E)A=UA, and then E=A−U∈GL2(R) as A4=0. This implies that E=I2, and so EAE=A∈J(M2(R)), as J(R)=2R. This gives a contradiction. Therefore A∈M2(R) is not strongly rad-clean.

Following Cui and Chen, an element a∈R is quaspolar if there exists an idempotent e∈comm(a) such that a+e∈U(R) and ae∈Rqnil (see [6]). Obviously, A is strongly J-clean⇒A is strongly rad-clean⇒A is quasipolar. But the converses are not true, as the following shows:

Example 2.9. (1) Let R be a commutative local ring and A=1110 be in M₂(R). Since A∈GL2(R), by Lemma 2.3, it is strongly rad-clean but is not strongly J-clean, as I2−A∈J(M2(R)).

(2) Let R=ℤ(3) and A=21−11. Then trA=3∈J(R) and detA=3∈J(R). Hence A is quasipolar by [5, Theorem 2.6]. Note that trA∉U(R), A∉GL2(R) and A∉J(M2(R)). Thus, A is not strongly rad-clean, in terms of Corollary 2.7.

Set B12(a)=1a01 and B21(a)=10a1. We now derive

Theorem 2.10. Let R be a commutative local ring. Then the following are equivalent:

• (1) Every A∈M2(R) with invertible trace is strongly rad-clean.

• (2) For any λ∈J(R),μ∈U(R), the quadratic equationx2=μx+λ is solvable.

Proof. (1)⇒(2) Let λ∈J(R),μ∈U(R). Choose A=0−λ1μ. Then A∈M2(R) is strongly rad clean. Obviously, A∈GL2(R) and A∈J(M2(R)). In view of Theorem 2.4, we see that the quadratic equation x2=μx+λ is solvable.

(2)⇒(1) Let A=abcd with tr(A)∈U(R).

Case I. c∈U(R). Then

for some λ,μ∈R. If λ∈U(R), then A∈GL2(R), and so it is strongly rad-clean. If λ∈J(R), then μ∈U(R). Then A is strongly rad-clean by Theorem 2.5.

Case II. b∈U(R). Then

and the result follows from Case I.

Case III. c,b∈J(R),a−d∈U(R). Then

where a−d+b−c∈U(R); hence the result follows from Case I.

Case IV. c,b∈J(R),a,d∈U(R). Then

hence, A∈GL2(R).

Case V. c,b,a,d∈J(R). Then A∈J(M2(R)), and so tr(A)∈J(R), a contradiction.

Therefore A∈M2(R) with invertible trace is strongly rad-clean.

Example 2.11. Let R=ℤ4. Then R is a commutative local ring. For any λ∈J(R),μ∈U(R), we directly check that the quadratic equation x2=μx+λ is solvable. Applying Theorem 2.10, every 2×2 matrix over R with invertible trace is strongly rad-clean. In this case, M2(R) is not strongly rad-clean.

Example 2.12. Let R=ℤ^2 be the ring of 2-adic integers. Then every 2×2 matrix with invertible trace is strongly rad-clean.

Proof. Obviously, R is a commutative local ring. Let λ∈J(R),μ∈U(R). Then 0λ1μ∈M2(R) is strongly clean, by [5, Theorem 3.3]. Clearly, det(A)=−λ∈J(R). As R/J(R)≅ℤ2, we see that μ∈1+J(R), and then det(A−I2)=1−λ−μ∈J(R). In light of [5, Lemma 3.1], the equation x2=μx+λ is solvable. This completes the proof, by Theorem 2.10.

We note that matrix with non-invertible trace over commutative local rings maybe not strongly rad-clean. For instance, A=1111∈M2(ℤ^2) is not strongly rad-clean.

We now apply our preceding results and investigate strongly rad-clean matrices over power series over commutative local rings.

Lemma 3.1. Let R be a commutative ring, and let A(x1,⋯,xn)∈M2(R[[x1,⋯,xn]]). Then the following hold:

• (1) A(x1,⋯,xn)∈GL2(R[[x1,⋯,xn]]) if and only if A(0,⋯,0)∈GL2(R).

• (2) A(x1,⋯,xn)∈J(M2(R[[x1,⋯,xn]])) if and only if A(0,⋯,0)∈J(M2(R)).

Proof. (1) We suffice to prove for n=1. If A(x1)∈GL2(R[[x1]]), it is easy to verify that A(0)∈GL2(R). Conversely, assume that A(0)∈GL2(R). Write

where A(0)=a0b0c0d0. We note that the determinant of A(x1) is a0d0−c0b0+x1f(x1), which is a unit plus an element of the radical of R[[x1]]. Thus, A(x1)∈GL2(R[[x1]]), as required.

(2) It is immediate from (1).

Theorem 3.2. Let R be a commutative local ring, and let A(x1,⋯,xn)∈M2(R[[x1,⋯,xn]]). Then the following are equivalent:

• (1) A(x1,⋯,xn)∈M2(R[[x1,⋯,xn]]) is strongly rad-clean.

• (2) A(x1,⋯,xn)∈M2(R[[x1,⋯,xn]]/(x1m1⋯xnmn)) is strongly rad-clean.

• (3) A(0,⋯,0)∈M2(R) is strongly rad-clean.

Proof. (1)⇒(2) and (2)⇒(3) are obvious.

(3)⇒(1) It will suffice to prove for n=1. Set x=x₁. Clearly, R[[x]] is a commutative local ring. Since A(0) is strongly clean in M2(R), it follows from Theorem 2.4 that A(0)∈GL2(R), or A(0)∈J(M2(R)), or χ(A(0)) has a root α∈J(R) and a root β∈U(R). If A(0)∈GL2(R) or A(0)∈J(M2(R)), in view of Lemma 3.1, A(x)∈GL2(R[[x]]) or A(x)∈J(M2(R[[x]])). Hence, A(x)∈M2(R[[x]]) is strongly rad-clean. Thus, we may assume that χ(A(0))=t2+μt+λ has a root α∈J(R) and a root β∈U(R).

Write χ(A(x))=t2+μ(x)t+λ(x) where μ(x)=∑ i=0∞μixi,λ(x)=∑ i=0∞λixi∈R[[x]] and μ0=μ,λ0=λ. Let b0=α. It is easy to verify that μ0=α+β∈U(R). Hence, 2b0+μ0∈U(R). Choose

Then y=∑ i=0∞bixi∈R[[x]] is a root of χ(A(x)). In addition, y∈J(R[[x]]) as b0∈J(R). Since y2+μ(x)y+λ(x)=0, we have χ(A(x))=(t−y)(t+y)+μ(t−y)=(t−y)(t+y+μ). Set z=−y−μ. Then z∈U(R[[x]]) as μ∈U(R[[x]]). Therefore χ(A(x)) has a root in J(R[[x]]) and a root in U(R[[x]]). According to Theorem 2.4, A(x)∈M2(R[[x]]) is strongly rad-clean, as asserted.

Corollary 3.3. Let R be a commutative local ring. Then the following are equivalent:

• (1) Every A∈M2(R) with invertible trace is strongly rad-clean.

• (2) Every A(x1,⋯,xn)∈M2(R[[x1,⋯,xn]]) with invertible trace is strongly rad-clean.

Proof. (1)⇒(2) Let A(x1,⋯,xn)∈M2(R[[x1,⋯,xn]]) with invertible trace. Then trA(0,⋯,0)∈U(R). By hypothesis, A(0,⋯,0)∈M2(R) is strongly rad-clean. In light of Theorem 2.4, A(x1,⋯,xn)∈M2(R[[x1,⋯,xn]]) is strongly rad-clean.

(2)⇒(1) is obvious.