A NOTE ON RINGS OF WEAKLY STABLE RANGE ONE

A NOTE ON RINGS OF WEAKLY STABLE RANGE ONE

Huanyin Chen and Miaosen Chen

(Received July 2004)

Abstract. It is shown that if R and S are Morita equivalent rings then R has weakly stable range 1 (written as wsr(R) = 1) if and only if S has. Let T be the ring of a Morita context (R, S, M, N, ψ, φ) with zero pairings. If wsr(R) = wsr(S) = 1, we prove that T is a weakly stable ring.

A ring R is said to have weakly stable range one if aR + bR = R implies that there exists a y ∈ R such that a + by ∈ R is right or left invertible. We denote this by wsr(R) = 1. By [2, Proposition 6], it is known that a regular ring R is one–sided unit–regular if and only if wsr(R) = 1. Many authors have studied rings of weakly stable range one, for example [2 –5] and [7–8].

In this note, we investigate equivalent characterizations of weakly stable range one. We prove that if R and S are Morita equivalent rings then wsr(R) = 1 if and only if wsr(S) = 1. This generalizes a corresponding result for one-sided unit–regular rings (cf. [Corollary 7]3). Furthermore, we study weakly stable range one over trivial extensions of rings, power series rings and the ring of a Morita context (R, S, M, N, ψ, φ). In addition, we prove that if T is the ring of a Morita context (R, S, M, N, ψ, φ) with zero pairings and wsr(R) = wsr(S) = 1, then T is a weakly stable ring where here sr(R) = 1 indicates that R has stable range one, i.e., aR + bR = R implies that there exists a y ∈ R such that a + by ∈ R is invertible
.

Throughout, all rings are associative with identity. We use M n (R) to denote the ring of all n × n matrices over the ring R. We use N to denote the set of all natural numbers. The notation A . ^⊕ B means that A is isomorphic to a direct summand of B. We write R ≈ S to denote that the rings R and S are Morita equivalent. For any n ≥ 1 and any module A, we let nA denote the direct sum of n copies of A.

Lemma 1. Let A be a right R–module such that wsr End R (A)

= 1. Then wsr End _R (nA)
= 1 for all n ∈ N.

Proof. Given M = A 1 ⊕ B = A 2 ⊕ C with A 1 ∼ = nA ∼ = A 2 , we have M = A 11 ⊕ · · · ⊕ A 1n ⊕ B = A 21 ⊕ · · · ⊕ A 2n ⊕ C with A 1i ∼ = A ∼ = A 2i for all i. As wsr End R (A)
= 1, by [3, Proposition 2], we can find some D 1 , E 1 ⊆ M such that M = D 1 ⊕ E 1 ⊕ (A 12 ⊕ · · · ⊕ A 1n ⊕ B) = D 1 ⊕ (A 22 ⊕ · · · ⊕ A 2n ⊕ C) or M = D 1 ⊕ (A 12 ⊕ · · · ⊕ A 1n ⊕ B) = D 1 ⊕ E 1 ⊕ (A 22 ⊕ · · · ⊕ A 2n ⊕ C). Thus we get M = (E 1 ⊕ A 12 ) ⊕ (A 13 ⊕ · · · ⊕ A 1n ⊕ B ⊕ D 1 ) = A 22 ⊕ (A 23 ⊕ · · · ⊕ A 2n ⊕ C ⊕ D 1 ) or M = A 12 ⊕(A 13 ⊕· · ·⊕A 1n ⊕B ⊕D 1 ) = (E 1 ⊕A 22 )⊕(A 23 ⊕· · ·⊕A 2n ⊕C ⊕D 1 ). As a result, we get M = A ⁰ ₁₂ ⊕(A 13 ⊕· · ·⊕A 1n ⊕B⊕D 1 ) = A ⁰ ₂₂ ⊕(A 23 ⊕· · ·⊕A 2n ⊕C⊕D 1 ), where A ⁰ ₁₂ = E ₁ ⊕ A 12 or A ⁰ ₁₂ = A ₁₂ and A ⁰ ₂₂ = A ₂₂ or A ⁰ ₂₂ = E ₁ ⊕ A 22 . Clearly,

1991 Mathematics Subject Classification 16U99.

Key words and phrases: Weakly stable range one; exchange ring.

A ⁰ ₁₂ ∼ = A ∼ = A ⁰ ₂₂ . By [3, Proposition 2] again, we can find D 2 ⊆ M such that M = A ⁰ ₁₃ ⊕ (A 14 ⊕ · · · ⊕ A 1n ⊕ B ⊕ D 1 ⊕ D 2 ) = A ⁰ ₂₃ ⊕ (A 24 ⊕ · · · ⊕ A 2n ⊕ C ⊕ D 1 ⊕ D 2 ) with A ⁰ ₁₃ ∼ = A ∼ = A ⁰ ₂₃ . Continuing in this way, we get D 3 , · · · , D n−1 ⊆ M such that M = A ⁰ _1n ⊕ (D 1 ⊕ D 2 ⊕ · · · ⊕ D n−1 ⊕ B) = A ⁰ _2n ⊕ (D 1 ⊕ D 2 ⊕ · · · ⊕ D n−1 ⊕ C) with A ⁰ _1n ∼ = A ∼ = A ⁰ _2n . Thus we can find D n , E ⊆ M such that M = (D 1 ⊕ D 2 ⊕ D n ) ⊕ E ⊕ B = (D 1 ⊕ D 2 ⊕ · · · ⊕ D n ) ⊕ C or M = (D 1 ⊕ D 2 ⊕ · · · ⊕ D n ) ⊕ B = (D 1 ⊕ D 2 ⊕ · · · ⊕ D n ) ⊕ E ⊕ C. By [3, Proposition 2] again, we prove that

wsr End R (nA)
= 1. 

Theorem 2. If S ≈ R then wsr(R) = 1 if and only if wsr(S) = 1.

Proof. Suppose that wsr(R) = 1. As S ≈ R, there exist n ∈ N and E = E ² ∈ M _n (R) such that S ∼ = EM n (R)E. In view of Lemma 1, we have wsr M _n (R)
= 1.

Therefore wsr(S) = 1 by [3, Proposition 3]. The converse is symmetric.  It follows from Theorem 2 that if wsr(R) = 1, then wsr M n (R)
=1 for all n ∈ N.

Lemma 3. Let A and B be right R–modules. If sr(End _R A) = 1, then wsr End _R (A ⊕ B)
= 1 if and only if wsr(End R B) = 1.

Proof. Suppose that wsr(End R B) = 1. Consider two right R–module decomposi- tions M = A 1 ⊕B 1 = A 2 ⊕B 2 with A 1 ∼ = A⊕B ∼ = A 2 . Then we have A 1 = A 11 ⊕A 12

and A 2 = A 21 ⊕ A 22 with A 11 ∼ = A ∼ = A 21 and A 12 ∼ = B ∼ = A 22 . It follows that M = A ₁₁ ⊕(A 12 ⊕B) = A 21 ⊕(A 22 ⊕C) with A 11 ∼ = A ∼ = A 21 . Since sr(End _R A) = 1, we can find a right R–module D such that M = D ⊕ (A ₁₂ ⊕ B) = D ⊕ (A 22 ⊕ C).

This means that M = A ₁₂ ⊕ (D ⊕ B) = A ₂₂ ⊕ (D ⊕ C) with A ₁₂ ∼ = B ∼ = A 22 . According to [3, Proposition 2], there exist right R–modules E and F such that M = E ⊕ F ⊕ (D ⊕ B) = E ⊕ (D ⊕ C) or M = E ⊕ (D ⊕ B) = E ⊕ F ⊕ (D ⊕ C).

Let G = E ⊕ D. Then M = G ⊕ F ⊕ B = G ⊕ C or M = G ⊕ B = G ⊕ F ⊕ C. By [3, Proposition 2] again, it follows that wsr End R (A ⊕ B)
= 1.

Conversely, assume that wsr End _R (A ⊕ B)

= 1. Let e : A ⊕ B → A ⊕ B given by e(x + y) = x for any x ∈ A, y ∈ B. Then e ∈ End _R (A ⊕ B) is an idempotent. Clearly, e End _R (A ⊕ B)
e ∼ = End R A and so the proof is complete by

[3, Proposition 3]. 

Let A be an artinian right R–module. Then End R (A) is semilocal; hence, sr(End R A) = 1. It follows by Lemma 3 that wsr End R (A ⊕ B)

= 1 if and only if wsr(End R B) = 1 for any right R–module B.

Recall that a Morita context denoted by (R, S, M, N, ψ, φ) consists of two rings R, S, two bimodules _R N _S , _S M _R and a pair of bimodule homomorphisms ψ : N N

S M → R and φ : M N

R N → S (called the pairings) which satisfy the following associativity: ψ(n N m)n ⁰ = nφ(m N n ⁰ ), φ(m N n)m ⁰ = mψ(n N m ⁰ ).

These conditions insure that the set T of generalized matrices ( _{m s} ^{r n} ), where r ∈ R, s ∈ S, m ∈ M, n ∈ N, forms a ring, called the ring of the context. We let diag(r, s) denote the matrix for which m = n = 0.

Theorem 4. Let T be the ring of the Morita context (R, S, M, N, ψ, φ). If sr(R) = 1, then wsr(T ) = 1 if and only if wsr(S) = 1.

Proof. Let e = diag(1, 0). Then e = e ² ∈ T . Clearly, sr End T (eT )
= sr(eT e) =

sr(R) = 1. Furthermore, we see that wsr End _T ((1−e)T )
= wsr (1−e)T (1−e)
=

wsr(S). Then, from Lemma 3, we deduce that wsr End T (eT ⊕ (1 − e)T )
= 1 if

and only if wsr(S) = 1, as required. 

Corollary 5. Let T be the ring of the Morita context (R, S, M, N, ψ, φ) with sr(R) = wsr(S) = 1, and let A be a finitely generated projective right T –module.

If B and C are any right T –modules such that A ⊕ B ∼ = A ⊕ C, then B . ^⊕ C or C . ^⊕ B.

Proof. Since A is a finitely generated projective right T –module, we can find n ∈ N and E = E ² ∈ M n (T ) such that End _T (A) ∼ = EM n (T )E. By virtue of Theorem 4, wsr(T ) = 1. Using Theorem 2, we have wsr M _n (T )
= 1. In view of [3, Proposition 3], it follows that wsr End R (A)
= 1. Since there is an isomorphism ψ : A ⊕ B ∼ = A ⊕ C, we get A ⊕ C = ψ(A) ⊕ ψ(B). By [3, Proposition 2], we can find right T –modules D and E such that A ⊕ C = D ⊕ E ⊕ C = D ⊕ ψ(B) or A ⊕ C = D ⊕ C = D ⊕ E ⊕ ψ(B). Thus either B . ^⊕ C or C . ^⊕ B.  Lemma 6. Let K be an ideal of a ring T , and let R be a subring of T which contains K. If sr(R/K) = 1 and wsr(T ) = 1, then wsr(R) = 1.

Proof. Given ax + b = 1 in R, then ax + b = 1 in R/K. As sr(R/K) = 1, we can find a z ∈ R such that a + bz ∈ U (R/K), the group of units of R/K. So we have w ∈ R and k ∈ K such that (a + bz)w + k = 1 in R. As a result, we deduce that (a + bz)xk + b(1 − zx)k = k in R. So we get (a + bz)w + k = (a + bz)w + (a + bz)xk + b(1 − zx)k = 1; hence, (a + bz)(w + xk) + b(1 − zx)k = 1 in T . Since wsr(T ) = 1, we can find a y ∈ R such that a + bz + b(1 − zx)ky is a right or left invertible in T . Since a + bz ∈ U (R/K), we have u ∈ R such that u(a + bz) = 1 = (a + bz)u in R/K. If a + bz + b(1 − zx)ky
v = 1 in T , then u(a + bz)v = u in T /K. This infers that v − u ∈ K; hence v ∈ R. If v a+bz+b(1−zx)ky
= 1 in T , then v(a + bz)u = u in T /K. Thus we get v−u ∈ K;

hence again v ∈ R. Consequently, we have shown that a + b z + (1 − zx)ky
∈ R is right or left invertible in R. Therefore wsr(R) = 1, as asserted.  Theorem 7. Let T be the subdirect product of two rings R and S. If sr(R) = 1, then wsr(T ) = 1 if and only if wsr(S) = 1.

Proof. Since T is the subdirect product of the rings R and S, there exist two epimorphisms ϕ : T → R and ψ : T → S such that Kerϕ ∩ Kerψ = 0. If wsr(T ) = 1, then wsr(T /Kerψ) = 1; hence, wsr(S) = 1. Assume now that wsr(S) = 1. Then wsr(R ⊕ S) = 1. Construct a map φ : T → T /Kerϕ ⊕ T /Kerψ given by φ(x) = (x + Kerϕ, x + Kerψ) for any x ∈ R. Clearly, =m φ is a subring of T /Kerϕ ⊕ T /Kerψ. In addition, wsr(T /Kerϕ ⊕ T /Kerψ) = wsr(R ⊕ S) = 1. Fur- thermore, Im(φ| _Kerϕ ) is an ideal of T /Kerϕ⊕T /Kerψ. Clearly, =m φ/=m(φ| _Kerϕ ) ∼ = T /Kerϕ. So sr =m φ/=m(φ| Kerϕ )
= 1. Since T ∼ = =mφ, by Lemma 6, we see that

wsr(T ) = 1, as asserted. 

Let I be an ideal of a ring R. We use l.ann(I) to denote the annihilator ideal {x ∈ R|xI = 0}.

Corollary 8. Let I be an ideal of a ring R. If sr(R/I) = 1, then wsr(R) = 1 if

and only if wsr R/l.ann(I)
= 1.

Proof. If wsr(R) = 1, then one easily checks that wsr R/l.ann(I)

= 1. Now assume that wsr R/l.ann(I)
= 1 and set J = I ∩ l.ann(I). Since (R/J)/(I/J) ∼ = R/I and (R/J )/(l.ann(I)/J ) ∼ = R/l.ann(I), it follows that sr (R/J )/(I/J )

= 1 and wsr (R/J )/(l.ann(I)/J )

= 1. Clearly, R/J is the subdirect product of (R/J )/(I/J ) and (R/J )/(l.ann(I)/J ). Thus wsr(R/J ) = 1 by Theorem 7. Given ax + b = 1 in R. Then ax + b = 1 in R/J . Hence we can find a y ∈ R such that a + by ∈ R/J is right or left invertible. If (a + by)u = 1 for a v ∈ R, then (a + by)u − 1 ∈ J . This infers that 1 − (a + by)u
²

= 0, and then (a + by)u 2 − (a + by)
= 1 and so a + by ∈ R is right invertible. If u(a + by) = 1 for a v ∈ R, we can similarly prove that a + by ∈ R is left invertible. Therefore wsr(R) = 1, as

required. 

If M is a R–R–bimodule, then the trivial extension of R by M is the ring T (R, M ) = R × M with addition defined as usual and multiplication defined by (r 1 , m 1 )(r 2 , m 2 ) = (r 1 r 2 , r 1 m 2 + m 1 r 2 ) for r 1 , r 2 ∈ R and m 1 , m 2 ∈ M . Next we show that trivial extensions provide a large class of rings with weakly stable range one.

Theorem 9. Let M be a R–R–bimodule. Then wsr T (R, M )
= 1 if and only if wsr(R) = 1.

Proof. Suppose that wsr(R) = 1. Given (a, m)(x, n) + (b, p) = (1, 0) in T (R, M ), then ax + b = 1 in R. So we can find a y ∈ R such that a + by = u ∈ R is right or left invertible in R. Hence (a, m) + (b, p)(y, 0) = (u, m + py). If there exists a v ∈ R such that uv = 1, then (u, m + py) v, −v(m + py)v
= (1, 0). If there exists a v ∈ R such that vu = 1, then v, −v(m + py)v
(u, m + py) = (1, 0). Thus (a, m) + (b, p)(y, 0) ∈ T (R, M ) is right or left invertible. Thus wsr T (R, M )
= 1, as required.

Suppose that wsr T (R, M )
= 1. Given ax + b = 1 in R, then (a, 0)(x, 0) + (b, 0) = (1, 0) in T (R, M ). So we can find a (y, q) ∈ T (R, M ) such that (a, 0) + (b, 0)(y, q) = (a + by, bq) ∈ T (R, M ) is right or left invertible. Therefore a + by ∈ R

is right or left invertible, as required. 

Corollary 10. For any ring R, wsr T (R, R)
= 1 if and only if wsr(R) = 1.

Proof. This is an immediate consequence of Theorem9. 

Theorem 11. wsr R[[x 1 , · · · , x n ]]
= 1 if and only if wsr(R) = 1, for any ring R.

Proof. It suffices to show that wsr R[[x]]
= 1 if and only if wsr(R) = 1. Suppose

that wsr(R) = 1. If f (x)g(x) + h(x) = 1 in R[[x]], then f (0)g(0) + h(0) = 1 in R

and so we can find y ∈ R such that u := f (0) + h(0)y ∈ R is right or left invertible.

First assume that uv = 1 for a v ∈ R. Choose b ₀ = v,

b 1 = −v(a 1 b 0 ), b ₂ = −v(a ₁ b ₁ + a ₂ b ₀ ),

.. .

b n = −v

n

X

i=1

a i b n−i , .. . .

Then it is easy to verify that f (x) + g(x)y
P ∞ i=0 b _i x ⁱ

= 1. This infers that f (x) + g(x)y ∈ R[[x]] is right invertible. A similar argument shows that if vu = 1 for some v ∈ R then f (x) + g(x)y ∈ R[[x]] is left invertible. Consequently, we show that wsr R[[x]]
= 1, as asserted.

Conversely, if wsr R[[x]]

= 1 then wsr(R) = 1 since R is a factor ring of

R[[x]]. 

Corollary 12. Let V be a vector space over a division ring. Then wsr End _D (V ) [[x 1 , · · · , x n ]]
= 1.

Proof. As wsr End D (V )
= 1, we get the result by Theorem 11.  We say that x, y ∈ R are pseudo–orthogonal, in symbols x\y, if (xRy) ² = 0 and (yRx) ² = 0. Let R ⁻¹ _p denote {u ∈ R | (1 − uv)\(1 − vu) for some v ∈ R}.

Then, if u ∈ R, we claim that u ∈ R ⁻¹ _p if and only if there exist a, b ∈ R such that (1 − ua)\(1 − bu). To see this, suppose that a, b ∈ R are such that (1 − ua)\(1 − bu);

then (1 − ua)R(1 − bu)
²

= 0 = (1 − bu)R(1 − ua)
²

. If v = a + b − aub, then 1 − uv = (1 − ua)(1 − ub) and 1 − vu = (1 − au)(1 − bu) and so (1 − uv) R(1 − vu)
²

= (1 − ua)(1 − ub)R(1 − au)(1 − bu)
²

⊆ (1 − ua)R(1 − bu)
²

= 0;

hence, (1 − uv)R(1 − vu)
2

= 0. Likewise, we get (1 − vu)R(1 − uv)
2

= 0. This gives (1 − uv)\(1 − vu), as required.

We say that R is a weakly stable ring provided that aR + bR = R implies that there exists a y ∈ R such that a + by ∈ R _p ⁻¹ .

Theorem 13. Let T be the ring of a Morita context (R, S, M, N, ψ, φ) where ψ, φ are both zero. If wsr(R) = wsr(S) = 1, then T is a weakly stable ring.

Proof. Given

 a 11 a 12

a 21 a 22

  x 11 x 12

x 21 x 22

 +

 b 11 b 12

b 21 b 22



=

 1 0 0 1



in T , then we get a 11 x 11 + b 11 = 1 in R and a 22 x 22 + b 22 = 1 in S. Since wsr(R) =

wsr(S) = 1, there exist y 1 ∈ R and y 2 ∈ S such that a 11 + b 11 y 1 = u 1 is right or

left invertible in R and a 22 + b 22 y 2 = u 2 is right or left invertible in S. So we have

v ₁ ∈ R such that u 1 v ₁ = 1 or v ₁ u ₁ = 1 and v ₂ ∈ S such that u 2 v ₂ = 1 or v ₂ u ₂ = 1.

(1) If u 1 v 1 = 1 and u 2 v 2 = 1, then

 1 0

0 1
− ^u _{∗∗ u}

¹

^∗

₂

 _v

1

0 0 v

₂



T 

1 0 0 1
−  _v

1

0 0 v

₂

 _u

1

∗

∗∗ u

2



= _{∗∗ 0} ^{0 ∗}
∗ ∗

∗∗ ∗
= _{∗∗ 0} ^{0 ∗}
and so, since

 0 ∗

∗∗ 0

 ²

= 0, we see that

 a ₁₁ a ₁₂ a 21 a 22

 +

 b ₁₁ b ₁₂ b 21 b 22

  y ₁ 0 0 y 2



∈ T _p ⁻¹ . (2) If u 1 v 1 = 1 and v 2 u 2 = 1, then we check similarly that

 _{1 0}

0 1
− ^u _{∗∗ u}

¹

^∗

₂

 _v

1

0 0 v

₂

 T  _{1 0}

0 1
−  _v

1

0 0 v

₂

 _u

1

∗

∗∗ u

2



= _{∗∗ 0} ^{0 0}
and then that

 a 11 a 12

a 21 a 22

 +

 b 11 b 12

b 21 b 22

  y 1 0 0 y 2



∈ T _p ⁻¹ . (3) Similarly, if v 1 u 1 = 1 and u 2 v 2 = 1, then

 1 0

0 1
− ^u _{∗∗ u}

¹

^∗

₂

 _v

1

0 0 v

2



T 

1 0 0 1
−  _v

1

0 0 v

2

 _u

1

∗

∗∗ u

2



= ^{0 ∗} _{0 0}
and so

 a 11 a 12

a 21 a 22

 +

 b 11 b 12

b 21 b 22

  y 1 0 0 y 2



∈ T _p ⁻¹ . (4) Finally, if u 1 v 1 = 1 and v 2 u 2 = 1, then

 1 0

0 1
− ^u _{∗∗ u}

¹

^∗

₂

 _v

1

0 0 v

2



T 

1 0 0 1
−  _v

1

0 0 v

2

 _u

1

∗

∗∗ u

2



= ( _{∗∗ 0} ^{0 ∗} ) and it follows that

 a 11 a 12

a 21 a 22

 +

 b 11 b 12

b 21 b 22

  y 1 0 0 y 2



∈ T _p ⁻¹ . In all cases,

 a 11 a 12

a 21 a 22

 +

 b 11 b 12

b 21 b 22

  y 1 0 0 y 2



∈ T _p ⁻¹ .

Therefore T is a weakly stable ring. 

For any ring R, let T M ₂ (R) denote the ring of all 2 × 2 lower triangular matrices over R.

Corollary 14. If wsr(R) = 1, then every regular element in T M ₂ (R) is a product of an idempotent in T M ₂ (R) and an element in T M ₂ (R) ⁻¹ _p .

Proof. By virtue of Theorem 13, T M 2 (R) is a weakly stable ring.

Given any (von Neumann) regular element x ∈ T M 2 (R), there exists a

y ∈ T M 2 (R) such that x = xyx and y = yxy. It follows from xy + (1 − xy) =

diag(1, 1) that there exists a z ∈ T M 2 (R) such that x + (1 − xy)z = u ∈ T M 2 (R) ⁻¹ _p .

Therefore x = xy x + (1 − xy)z
= (xy)u, as asserted. 

In [1], an element u in a ring R is called quasi–invertible provided there exist a, b ∈ R such that (1 − ua)R(1 − bu) = (1 − bu)R(1 − ua) = 0. (Such a u is von Neumann regular, and a and b may be chosen so that a = b and uau = u.) The ring R is then defined to be a QB–ring provided that whenever xa + b = 1 in R, there exists y ∈ R such that a + yb is quasi–invertible.

Let V be a vector space over a division ring D. Then T M 2 (End D (V )) is a weakly stable ring. We end this note by asking a natural problem: Is T M 2 End D (V )
a QB–ring?

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Huanyin Chen

Department of Mathematics Zhejiang Normal University Jinhua

Zhejiang 321004 P.R. CHINA [email protected]

Miaosen Chen

Department of Mathematics Zhejiang Normal University Jinhua

Zhejiang 321004 P.R. CHINA

[email protected]