Stable rings generated by their units

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PII. S0161171201005798 http://ijmms.hindawi.com © Hindawi Publishing Corp.

STABLE RINGS GENERATED BY THEIR UNITS

HUANYIN CHEN

(Received 12 September 2000)

Abstract.We introduce the class of rings satisfying(m,1)-stable range and investigate equivalent characterizations of such rings. These give generalizations of the corresponding results by Badawi (1994), Ehrlich (1976), and Fisher and Snider (1976).

2000 Mathematics Subject Classiﬁcation. 19B10, 16E50.

LetRbe an associative ring with identity. A ringRis said to have stable range one provided thataR+bR=Rimplies thata+by∈U (R)fory∈R. It is well known that MRcancels from direct sums if EndRMhas stable range one. For further properties of stable range one condition, we refer the reader to [1,2,5,7,9,10,13,14].

Many authors have studied rings generated by their units (see [3,4,7,8,10,12]). It was shown that every unit-regular ring in which 2 is invertible is generated by its unit (see [7, Theorem 5]) and every stronglyπ-regular ring in which 2 is invertible is generated by its units (see [8, Theorem 3]). So far one always investigate such rings under stable range one condition.

In this paper, we generalize stable range one condition and introduce rings satis-fying(m,1)-stable range so as to investigate rings generated by their units. Also we give generalizations of the corresponding results in [3,7,8].

Throughout, rings are associative with identity and modules are right modules. GLn(R)denotes the general linear group ofR,U (R)denotes the set of units ofR, and thatUm(R)= {x∈R| ∃u1, . . . , um∈U (R)such thatx=u1+···+um}. LetBij(x)=

I2+xeij (i≠j, 1≤i, j≤2),[α, β]=αe11+βe22, whereeij (1≤i, j≤2)are matrix units (1 in thei, jposition and 0 elsewhere).

Deﬁnition₁. _{The ring}_R_{is said to satisfy}_(m,₁₎_{-stable range provided that}_aR+

bR=Rimplies thata+by∈U (R)fory∈Um(R).

Proposition₂. _{The following are equivalent:} (1)The ringRsatisﬁes(m,1)-stable range.

(2)Wheneverax+b=1, there existsy∈Um(R)such thata+by∈U (R).

Proof. ₍₁₎⇒_{(2). The proof is obvious.}

(2)⇒(1). GivenaR+bR=R, thenax+by=1 for somex, y∈R. So we can ﬁnd z∈Um(R)such thataxz+b=u∈U (R), and thenaxzu−1+bu−1=1. Hence we havew∈Um(R)such thata+bu−1w∈U (R). Clearly,u−1w∈Um(R), as desired.

Proposition₃. _{The following are equivalent:} (1)The ringRsatisﬁes(m,1)-stable range.

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Proof. ₍₁₎⇒_{(2). Given ¯}_a_x_¯+¯_b=_{1 in}¯ _R/J(R)_{, then}_ax+_(b+_{r )}=_{1 for some}_r∈

J(R). SinceRsatisﬁes(m,1)-stable range, we havey∈Um(R)such thata+(b+r )y∈

U (R). Therefore ¯a+_b¯_y_¯_∈_{U (R/J(R))}_{with ¯}_y_∈_U_m_(R/J(R))_{, hence}_R/J(R)_satisﬁes (m,1)-stable range byProposition 2.

(2)⇒(1). Givenax+b=1 inR, then ¯ax¯+b¯=¯1 inR/J(R). So there is ¯y∈Um(R/J(R)) such that ¯a+_b¯_y_¯₌_u_¯_∈_{U (R/J(R))}_{. Assume that} _y₌_w₁₊_w₂_{+ ··· +}_w_m _{with all} wi∈U (R/J(R)). Since units lift moduloJ(R), we may assume that allwi∈U (R)and

u∈U (R), and thata+b(w1+w2+ ··· +wm)=u+r for somer∈J(R). Obviously,

u+r∈U (R)andw1+w2+···+wm∈Um(R). HenceRsatisﬁes(m,1)-stable range, as asserted.

Theorem₄. _Let_R_{be an associative ring with identity,}_K_{a set of some elements of} R. Then the following are equivalent:

(1)Wheneverax+b=1, there existsy∈Ksuch thata+by∈U (R).

(2)Wheneverax+b=1, there existsz∈Ksuch thatx+zb∈U (R).

Proof. (1)⇒(2). Since ax+b = 1, we see that a₁−_xb−1 = ₋x₁1−_axa ∈GL₂(R). Clearly,xa+(1−xa)=1. So there existsz∈Ksuch thatx+(1−xa)z=u∈U (R). Hence a−b

1 x −11 0

z1

=u∗

∗ ∗∈GL2(R). Thus we know that a1−xb −1

=u∗ ∗ ∗−1 0z1

. Therefore a−b

1 x

= 1 0 z1

_u_∗

∗ ∗−1. Since there is w ∈ U (R) such that u∗ ∗∗ = _{1 0}

∗1 _u ₀

0w ₁_∗

0 1

, we have v =w−1 _∈ _{U (R)}_{such that} u∗

∗ ∗−1 = 10 1∗ _u−1₀

0 v _{1 0}

∗1

. Hence a−b

1 x

= 1 0 z1

_{∗ ∗} ∗v

, v ∈ U (R). So 1 0 −z1

_a₋_b 1 x

= ∗ ∗∗v

. Thus, we see that x+zb=v∈U (R), as required.

(2)⇒(1). Applying (1)⇒(2) to the opposite ringRop_{, we complete the proof.}

Theorem 4is a general result for symmetry of stable range conditions. As appli-cations, we see that stable range one conditions, unit 1-stable range conditions and rings having many unit-regular elements are symmetric. The following result shows that(m,1)-stable range condition is right-left symmetric.

Corollary₅. _{The following are equivalent:} (1)The ringRsatisﬁes(m,1)-stable range.

(2)Wheneverax+b=1, there exists somez∈Um(R)such thatx+zb∈U (R). (3)WheneverRa+Rb=R, there exists somez∈Um(R)such thata+zb∈U (R).

Proof. (1)(2). Set_K=_U_m_(R)_{. Then the equivalence follows by}_{Theorem 4}_. (3)⇒(2). The proof is trivial.

(2)⇒(3). Given Ra+Rb = R, then xa+yb= 1 for some x, y ∈ R. So we have s ∈ Um(R) such that sxa+b = u ∈U (R), hence u−1sxa+u−1b =1. Therefore

a+vu−1_b_∈_{U (R)}_{for some}_v_∈_U

m(R), as required.

Proposition₆. _{The following are equivalent:} (1)The ringRsatisﬁes(m,1)-stable range.

(2) For any A ∈ GL2(R), there exists some w ∈ Um(R) such that A =

[∗,∗]B21(w)B12(∗)B21(∗).

(3) For any A ∈ GL2(R), there exists some w ∈ Um(R) such that A =

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Proof. ₍₁₎⇒_{(2). Let}_A∈_GL₂_(R)_{, and let}_A−1=_(b_ij₎_{. Since} _b₁₁_R+_b₁₂_R=_R_{, we} can ﬁnd some y ∈Um(R)such that b11+b12y =u∈U (R). We easily check that

A−1₌_B

21(b21+b22u−1)[u, b22−(b21+b22y)u−1b12]B12(u−1b12)B21(−y). ThusA=

[∗,∗]B21(w)B12(∗)B21(∗)for somew∈Um(R). (2)⇒(1). Given ax+b=1 in R, then we have a b

−1x

∈GL2(R). Thus we have a

w∈Um(R)such that _{a b}

−1x −1

=[∗,∗]B21(w)B12(∗)B21(∗).Therefore we see that _{a b}

−1x

=[∗,∗]B21(∗)B12(∗)B21(−y)for somey∈Um(R).Consequently, a+by∈

U (R)withy∈Um(R), as desired.

(1)(3). Applying (1)(2) to the opposite ringRop_{, we complete the proof by the} symmetry of(m,1)-stable range conditions.

LetRbe generated bymunits. IfR has stable range one, then it satisﬁes(m,1) -stable range. Conversely, we easily check that every ring satisfying(m,1)-stable range is generated by m+1 units. Now we show that (m,1)-stable range condition is inherited by matrix rings.

Lemma7. The following are equivalent: (1)The ringRsatisﬁes(m,1)-stable range.

(2)Givenax+b =1in R, then there existsy ∈R such that a+by∈U (R)and

1−xy∈Um(R).

(3) Givenax+b =1 inR, then there exists z∈R such that x+zb∈U (R) and

1−za∈Um(R).

Proof. (1)⇒(2). Given ax+b = 1 in R, then ₋a b₁_x ∈ GL₂(R). In view of

Proposition 6, we have aw∈Um(R)such that _{a b}

−1x

=[∗,∗]B21(w)B12(∗)B21(∗). So we can ﬁnd some−y∈Rsuch thata b

−1x)=[∗,∗]B21(w)B12(∗)B21(−y). There-forea+by∈U (R)and 1−xy= −(−1+xy)∈Um(R), as required.

(2)⇒(1). Givenax+b=1 inR, then there exists somey∈Rsuch thata+by=u∈

U (R) and 1− xy = v ∈ Um(R). So we know that ₋a b1x 1 0

y1

= u b −v x

=

[∗,∗]B21(w)B12(∗)for somew∈Um(R). Thus _{a b}

−1x

=[∗,∗]B21(w)B12(∗)B21(−y). So we can ﬁndz∈Um(R)such that

_{1 0} z1

_{a b} −1x

=[∗,∗]B12(∗)B21(∗). Consequently, we show thatx+zb∈U (R)for somez∈Um(R). ThereforeRsatisﬁes(m,1)-stable range byCorollary 5.

(1)(3). Applying (1)(2) to the opposite ringRop_{, we complete the proof.}

In [6], the author shows that every matrix ring over a ring satisfying unit 1-stable range also satisﬁes unit 1-stable range. Now we extend [6, Theorem 2.2] to(m,1) -stable range conditions by a similar route.

Theorem8. IfRsatisﬁes(m,1)-stable range, then so doesM_n(R)for anyn≥1.

Proof. GivenBC+D=I_ninM_n(R), thenA=₋B_I D nC

∈GL2n(R). SetA=(Aij) (1≤

i, j≤2)with allAij=(aijst)∈Mn(R) (1≤s, t≤n). Then there existx1, . . . , xn, y1, . . . ,

yn∈R such that a1111x1+ ··· +a111nxn+a1211y1+ ··· +a121nyn =1, . . . , a11n1x1+ ··· +

a11

nnxn+a12n1y1+ ··· +a12nnyn=0,a2111x1+ ··· +a211nxn+a2211y1+ ··· +a221nyn=0, . . . ,

a21n1x1+ ··· +a21nnxn+a22n1y1+ ··· +a22nnyn=0.In view ofLemma 7, there isz1∈R such thata11

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[∗,∗]A[∗,∗]B21(∗)=                     

u1 a1112 ··· a111n a1211 ··· a121n

0 b1122 ··· b2n11 b1221 ··· b122n

..

. ... . .. ... ... . .. ...

0 b11

n2 ··· bnn11 bn112 ··· bnn12

a2111v1 a2112 ··· a211n a2211 ··· a221n

..

. ... . .. ... ... . .. ...

a21n1v1 a21n2 ··· a21nn a22n1 ··· a22nn                      . (1)

Likewise, we haveu2, u3, . . . , un∈U (R)andv2, v3, . . . , vn∈Um(R)such that

[∗,∗]A[∗,∗]B21(∗)=                        

u1 ∗ ∗ ··· ∗ a1211 ··· a121n

0 u2 ∗ ··· ∗ b2112 ··· b2n12

0 0 u3 ··· ∗ c3112 ··· c123n

..

. ... ... . .. ... ... . .. ...

0 0 0 ··· un d12n1 ··· d12nn

a2111v1 a2112v2 a2113v3 ··· a211nvn a2211 ··· a221n

..

. ... ... . .. ... ... . .. ...

a21

n1v1 a21n2v2 a21n3v3 ··· a21nnvn a22n1 ··· a22nn                         . (2)

Similar to the consideration in [6, Theorem 2.2] , we can ﬁnd someE∈GLn(R)such that[∗,∗]A[∗,∗]B21(∗)=[∗,∗]B21(−E−1diag(v1, . . . , vn))B12(∗). Consequently,A=

[∗,∗]B21(W )B12(∗)B21(∗)withW∈Um(Mn(R)). So there isW∈Um(Mn(R))such that

B21

W B D

−In C

=[∗,∗]B12(∗)B21(∗), soC+WD∈GLn(R). (3)

It follows fromCorollary 5thatMn(R)satisﬁes(m,1)-stable range.

Corollary₉. _Let_R_satisfy_(m,₁₎_{-stable range, then every}_n×_n_{matrix over}_R_is

the sum ofm+1invertible matrices.

Proof. _Let_A∈_M_n_(R)_{. Since}_R_satisﬁes_(m,₁₎_{-stable range, so does}_M_n_(R)_from

Theorem 8. AsAMn(R)+InMn(R)=Mn(R), we can ﬁnd someU∈Um(Mn(R))such thatA+In×U=V∈GLn(R). ThusA=(−U )+V, as desired.

Recall that a ringRis said to be an exchange ring if for every rightR-moduleAand any two decompositionsA=M⊕N=i∈IAi, whereMR RR and the index setI is ﬁnite, then there exist submodulesA_i⊆Aisuch thatA=M⊕(

i∈IAi). A ringR is said to be stronglyπ-regular provided that for anyx∈R, there exists a positive integernsuch thatxn₌_xn+1_y_{for some}_y_∈_R_.

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of M. Henriksen [11], we claim that the ringRhas stable range one if and only if the ring M2(R)satisﬁes(3,1)-stable range. For exchange rings, we now derive the following.

Lemma₁₀._Let_R_{be an exchange ring with}₁_/₂∈_R_{. Then the following are equivalent:} (1)The exchange ringRhas stable range one.

(2)The exchange ringRsatisﬁes(7,1)-stable range.

Proof. ₍₂₎⇒_{(1). The proof is clear.}

(1)⇒(2). Givenax+b=1 inR, thena+by∈U (R)fory∈R. SinceRis an exchange ring, there exists an idempotente∈Rsuch thate=ysand 1−e=(1−y)t. Obviously, ey and(1−e)(1−y)are both regular. Thusey=f u, (1−e)(1−y)=gv for some f =f2_{, g}₌_g2_∈_R _and_{u, v}_∈_{U (R)}_{. Hence}_y₌_ey₋₍₁₋_e)(₁₋_y)₊₁₋_e₌_{f u}₋

gv+1−e. As 2∈U (R), we see thatf=2−1₊₂−1₍₂_f₋₁₎_,_g₌₂−1₊₂−1₍₂_g₋₁₎_and

e=2−1₊₂−1₍₂_e₋₁₎_{. Clearly, 2}−1₍₂_f₋₁_),₂−1₍₂_g₋₁_),₂−1₍₂_e₋₁₎_∈_{U (R)}_{. Therefore}

y∈U7(R), as required.

Theorem11. LetRbe a stronglyπ-regular ring. If2is a nonnilpotent ofR, then

there exists some nonzero idempotente∈R such thatMn(eRe)satisﬁes (7,1)-stable

range.

Proof. _Since_R_{is a strongly}_π_{-regular ring, there exists}_n≥_{1 such that 2}n=_eu for somee=e2_{, u}_∈_{U (R)}_{. Since 2 is a nonnilpotent of}_R_{, we see that}_e_≠_{0. Assume} thatuv=1 forv∈R. We easily check that(eue)(eve)=2n_eve₌_euve₌_e_{. Likewise,} we have(eve)(eue)=e. Thus 2e∈U (eRe). On the other hand, we know thateReis a stronglyπ-regular ring. By virtue of [1, Theorem 4],Rhas stable range one. Thus we complete the proof byTheorem 8andLemma 10.

Proposition12. The following are equivalent: (1)The ringRsatisﬁes(m,1)-stable range.

(2)WheneveraR+bR=dR, there existy∈Um(R),u∈U (R)such thata+by=du. (3)WheneverRa+Rb=dR, there existz∈Um(R),u∈U (R)such thata+zb=ud.

Proof. (1)⇒(2). GivenaR+bR=dR, then(a, b)M₂(R)=(d,0)M₂(R). Assume that (d,0)A=(a, b)and (a, b)B=(d,0). FromAB+(I2−AB)=I2, we have Y ∈M2(R) such thatA+(I2−AB)Y=W∈GL2(R). Thus(a, b)=(d,0)A=(d,0)(A+(I2−AB))=

(d,0)W. Assume thatW=(wij). Thenw11R+w12R=R, whencew11+w12y=u∈

U (R)fory∈Um(R). Thereforea+by=du, as desired. (2)⇒(1). The proof is trivial.

(1)(3). Applying (1)(2) to the opposite ringRop_{, we complete the proof by the} symmetry of(m,1)-stable range property.

Corollary₁₃. _Let_R_{be a ring which is quasi-injective as a right}_R_{-module. Then}

the following are equivalent:

(1)The ringRsatisﬁes(m,1)-stable range.

(2)Wheneverr·ann(a)∩r·ann(b)=r·ann(d), there existsz∈Um(R)such that

r·ann(a)∩r·ann(b)=r·ann(a+zb).

(3)Wheneverl·ann(a)∩l·ann(b)=l·ann(d), there existsy∈Um(R)such that

l·ann(a)∩l·ann(b)=l·ann(a+by).

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thata+zb=dufor someu∈U (R). Thereforer·ann(a)∩r·ann(b)=r·ann(d)=

r·ann(a+zb), as desired.

(2)⇒(1). Assume thatRa+Rb=R. Thenr·ann(a)∩r·ann(b)=r·ann(1). Thus, we claim that r·ann(a)∩r·ann(b)=r·ann(a+zb)for a z∈Um(R). Therefore

r·ann(1)=r·ann(a+zb). By [5, Proposition 3.4], we show thatR=R(a+zb), and then a+zb=u is left invertible inR. Assume thatvu=1 for some v∈R. From Rv+R(1−uv)=R, we also havew ∈Um(R)such thatv+w(1−uv)=tis left invertible inR. Clearly, we havetu=(v+w(1−uv))u=1. Hencetis a unit ofR. Thereforea+zb=uis a unit ofR, as desired.

(1)(2). By the symmetry of(m,1)-stable range condition, we complete the proof.

Acknowledgements. This work was supported by grant no. 19801012 of the National Natural Science Foundation of China, and the Ministry of Education of China. The author would like to thank the referee for many helpful suggestions.

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Huanyin Chen: Department of Mathematics, Hunan Normal University, Changsha 410006, China