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Morphic rings and unit regular rings
Journal of Pure and Applied Algebra 210 (2007) 501–510 www.elsevier.com/locate/jpaa
Morphic rings and unit regular rings Tsiu-Kwen Lee a,1 , Yiqiang Zhou b,∗ a Department of Mathematics, National Taiwan University, Taipei 106, Taiwan b Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, Newfoundland A1C 5S7, Canada
Received 9 July 2006; received in revised form 4 October 2006 Available online 17 November 2006 Communicated by R. Parimala
Abstract A ring R is called left morphic if R/Ra ∼ = l(a) for every a ∈ R. A left and right morphic ring is called a morphic ring. If Mn (R) is morphic for all n ≥ 1 then R is called a strongly morphic ring. A well-known result of Erlich says that a ring R is unit regular iff it is both (von Neumann) regular and left morphic. A new connection between morphic rings and unit regular rings is proved here: a ring R is unit regular iff R[x]/(x n ) is strongly morphic for all n ≥ 1 iff R[x]/(x 2 ) is morphic. Various new families of left morphic or strongly morphic rings are constructed as extensions of unit regular rings and of principal ideal domains. This places some known examples in a broader context and answers some existing questions. c 2006 Elsevier B.V. All rights reserved.
MSC: Primary: 16E50; 16U99; secondary: 16S70; 16S35
0. Introduction By the fundamental homomorphism theorem of modules, for any element a in a ring R, R/l(a) ∼ = Ra as left R-modules where l(a) = {r ∈ R : ra = 0} is the left annihilator of a in R. Dually, if R/Ra ∼ = l(a), then a is called a left morphic element of R. The ring R is called left morphic if every element of R is left morphic. Right morphic rings are defined analogously. A left and right morphic ring is simply called a morphic ring. A ring R is called strongly morphic if the matrix ring Mn (R) is morphic for all n ≥ 1. An element a ∈ R is called unit regular if a = aua for some unit u of R and the ring R is called unit regular if all elements of R are unit regular. A well-known result of Erlich [7] states that an element a in a ring R is unit regular if and only if a is both left morphic and regular in R. This result is crucial in proving several important results of unit regular rings (see [2,8]). The study of morphic rings was started by Nicholson and S´anchez Campos in [9–11] and was continued by others in [4–6]. If R is a ring and σ : R → R is a ring endomorphism, let R[x; σ ] denote the ring of skew polynomials over R, that is, all formal polynomials in x with coefficients from R with multiplication defined by xr = σ (r )x. The first two sections deal with the question when R[x; σ ]/(x n ) is left morphic. If D is a division ring and σ is an endomorphism ∗ Corresponding author. Tel.: +1 705 737 8797; fax: +1 709 737 3010.
E-mail addresses: [email protected] (T.-K. Lee), [email protected] (Y. Zhou). 1 Member of Mathematics Division (Taipei Office), National Center for Theoretical Sciences. c 2006 Elsevier B.V. All rights reserved. 0022-4049/$ - see front matter doi:10.1016/j.jpaa.2006.10.005
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of D with σ (1) = 1, then D[x; σ ]/(x 2 ) is a left morphic ring by [9]. This ring turns out to be a prototype for all local, left morphic rings with nilpotent Jacobson radical, and the latter plays a significant role in building up the structures of more general morphic rings (see [9]). Later in [6] it was proved that R[x; σ ]/(x 2 ) is left morphic for any strongly regular ring (i.e., any unit regular ring whose idempotents are central) R with an endomorphism σ satisfying σ (e) = e for all e2 = e ∈ R and that R[x]/(x 2 ) is strongly morphic for any semisimple ring R. A question left open in [6] is whether R[x]/(x 2 ) is morphic for any unit regular ring R. This question is settled here. In fact, it is proved that if R is a unit regular ring and σ is an onto endomorphism of R with σ (e) = e for all e2 = e ∈ R, then R[x; σ ]/(x n ) is left morphic for all n ≥ 1. It is further proved that a ring R is unit regular iff R[x]/(x n ) is strongly morphic for all n ≥ 1 iff R[x]/(x 2 ) is morphic. This result is used to provide a large class of examples of semiprimitive, strongly morphic rings that are not regular. The morphic property of the trivial extension R ∝ M of a ring R with a bimodule M over R was discussed in [6]. One of the results obtained in [6] says that if R is a PID (principal ideal domain) with the field Q(R) of quotients, then R ∝ Q(R) R is morphic. This result cannot be extended to a UFD (unique factorization domain) because, as noticed in Section 3, for a UFD R, R ∝ Q(R) R is morphic iff R is a PID. If R is any factor ring of a PID and if M is a bimodule over R such that r x = xr for all r ∈ R and x ∈ M, a necessary and sufficient condition for R ∝ M to be morphic is obtained. It is also proved here that if I is an ideal of a unit regular ring then R ∝ I is strongly morphic. Section 4 presents an example of a commutative strongly morphic chain ring whose Jacobson radical is nil but not T -nilpotent. This answers in the negative a question of Camillo and Nicholson [3] on quasi-morphic rings. It is observed that a ring R is a finite direct product of matrix rings over morphic chain rings iff R is a morphic ring with a complete set {e1 , . . . , en } of orthogonal idempotents such that ei Rei is a chain ring for all i. All rings R are associative with identity. The Jacobson radical and the group of units of a ring R are denoted by J (R) and U (R) respectively. As usual, we write Z, Q, and Zn for the ring of integers, the field of rational numbers, and the ring of integers modulo n. The field of quotients of an integral domain R is denoted Q(R). By a subring of a ring R, we shall always mean a subring containing the identity of R. For a ∈ R, the left annihilator of a in R is denoted by l(a) or l R (a), and the right annihilator of a in R is denoted by r(a) or r R (a). 1. The ring R[x; σ ]/(x n ) R
For a, b ∈ R, we write a ∼ b (or simply a ∼ b) to mean that Ra = l(b) and Rb = l(a). By [9, Lemma 1], a ∈ R R
is left morphic iff a ∼ b for some b ∈ R. Lemma 1 ([9]). Let a, b ∈ R and let u ∈ U (R). The following are equivalent: (1) a ∼ b. (2) ua ∼ bu −1 . (3) au ∼ u −1 b. It was proved in [6] that for a strongly regular ring R with an endomorphism σ : R → R such that σ (e) = e for all e2 = e ∈ R, R[x; σ ]/(x 2 ) is a left morphic ring. This is a special case of the next result. Theorem 2. Let R be a unit regular ring and let σ : R → R be a ring endomorphism such that σ (e) = e for all e2 = e ∈ R. Then R[x; σ ]/(x 2 ) and R[x]/(x n+1 ) are left morphic rings for all n ≥ 0. Proof. Let S = R[x; σ ]/(x n+1 ). Then S = {r0 + r1 x + · · · + rn x n : ri ∈ R, i = 0, 1, . . . , n} with x n+1 = 0 and xr = σ (r )x for all r ∈ R. Let α = r0 + r1 x + · · · + rn x n ∈ S. We show that α is left morphic in S. Since R is unit regular, every element of R is the product of a unit and an idempotent. Thus, by multiplying α by a suitable unit of R we can assume that r0 = e0 is an idempotent by Lemma 1. Because [1 − (1 − e0 )r1 x]α(1 − r1 x) = e0 + (1 − e0 )r1 (1 − e0 )x + · · ·
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where both 1 − (1 − e0 )r1 x and 1 − r1 x are units of S, we can further assume that r1 ∈ (1 − e0 )R(1 − e0 ) because of Lemma 1. Now [1 − (1 − e0 )r2 x 2 ]α(1 − r2 x 2 ) = e0 + r1 x + (1 − e0 )r2 (1 − e0 )x + · · · where both 1 − (1 − e0 )r2 x 2 and 1 − r2 x 2 are units of S, so we can assume that r2 ∈ (1 − e0 )R(1 − e0 ) by Lemma 1. A simple induction shows that we can assume that α = e0 + r 1 x + r 2 x 2 + · · · + r n x n ,
ri ∈ (1 − e0 )R(1 − e0 )
for i = 1, . . . , n.
Because (1 − e0 )R(1 − e0 ) is unit regular by [8, Corollary 4.7], write r1 = ue1 where e1 is an idempotent of (1 − e0 )R(1 − e0 ) and u is a unit of (1 − e0 )R(1 − e0 ) with inverse v. Then e0 + v is a unit of R with inverse e0 + u. Since (e0 + v)α = e0 + e1 x + vr2 x 2 + · · · + vrn x n , by Lemma 1 we can assume that α = e0 + e1 x + r 2 x 2 + · · · + r n x n ,
ri ∈ (1 − e0 )R(1 − e0 )
for i = 2, . . . , n.
Now [1 − (1 − e1 )r2 x]α(1 − r2 x) = e0 + e1 x + (1 − e1 )r2 (1 − e1 )x 2 + · · · . Since 1 − (1 − e1 )r2 x and 1 − r2 x are units of S, by Lemma 1 we can assume that α = e0 + e1 x + r2 x 2 + · · · + rn x n where r2 ∈ (1 − e1 )(1 − e0 )R(1 − e0 )(1 − e1 ) and ri ∈ (1 − e0 )R(1 − e0 ) for i = 3, . . . , n. As above, a simple induction shows that we can assume that α = e 0 + e1 x + r 2 x 2 + · · · + r n x n , e02 = e0 ,
e12 = e1 ∈ (1 − e0 )R(1 − e0 ),
ri ∈ (1 − e1 )(1 − e0 )R(1 − e0 )(1 − e1 )
for i = 2, . . . , n.
Repeating the process, by induction we can assume that α = e0 + e1 x + e2 x 2 + · · · + en x n , e02 = e0 , ei2 = ei ∈ (1 − ei−1 ) · · · (1 − e0 )R(1 − e0 ) · · · (1 − ei−1 )
for i = 1, . . . , n.
Let β = b0 + b1 x + · · · + bn x n where b0 = (1 − e0 )(1 − e1 ) · · · (1 − en−1 )(1 − en ) b1 = (1 − e0 )(1 − e1 ) · · · (1 − en−1 ) ··· bn = 1 − e0 . Thus, we have Sα = Re0 + R(e0 + e1 )x + · · · + R(e0 + · · · + en )x n = l(β), l(α) = Rb0 + Rb1 x + · · · + Rbn x n = Sβ. So α ∼ β. The proof is complete.
The assumption that σ (e) = e for all e2 = e ∈ R in Theorem 2 can not be removed by the next example. Example 3. There exists a Boolean ring R and an endomorphism σ : R → R with σ (1) = 1, but R[x; σ ]/(x 2 ) is not left morphic. Proof. Consider the direct product R = Z2 × Z2 and let σ : R → R be given by (a1 , a2 ) 7→ (a2 , a1 ). Then σ is an endomorphism of R with σ (1) = 1. Let b = (1, 0) ∈ R and S = R[x; σ ]/(x 2 ). We next show that there do not exist c, d ∈ R such that l(bx) = S(c + d x)
and
S(bx) = l(c + d x).
(1.1)
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Suppose that c, d ∈ R satisfy (1.1). Then x ∈ l(bx) = S(c + d x) = {r c + [r d + sσ (c)]x : r, s ∈ R}. So r c = 0 and 1 = r d + sσ (c) for some r, s ∈ R. Thus c = c(r d + sσ (c)) = scσ (c). Since bx 6= 0 and bx ∼ c + d x, c + d x cannot be a unit of S. So c 6= 1. Thus cσ (c) = 0 and hence c = 0. Thus, r d = 1, showing that d = 1. But l(x) = Sx 6= S(bx). Hence S is not left morphic. It is possible that the ring in Theorem 2 is neither right morphic nor strongly left morphic as shown by the next example. Example 4. By [9], there exists a field F and an endomorphism σ of F with σ (1) = 1 such that F[x; σ ]/(x 2 ) is left morphic but not right morphic and that the 2 × 2 matrix ring over F[x; σ ]/(x 2 ) is not left morphic. The next corollary extends a result in [6] that R[x]/(x 2 ) is strongly morphic for any semisimple ring R and gives an affirmative answer to the question whether R[x]/(x 2 ) is morphic for any unit regular ring R (see [6, p. 148]). Corollary 5. If R is a unit regular ring, then R[x]/(x n ) is strongly morphic for all n ≥ 1. Proof. If R is unit regular then so is Mk (R) for every k ≥ 1 by [8, Corollary 4.7]. Thus, the claim follows by Theorem 2 because Mk (R[x]/(x n )) ∼ = Mk (R)[x]/(x n ). The next example is a contrast to Corollary 5. Example 6. The ring Z2 [x, y]/(x 2 , y 2 ) is not left morphic. Proof. Let S = Z2 [x, y]/(x 2 , y 2 ). Then S = {a + bx + cy + d x y : a, b, c, d ∈ Z2 }. Let α = x y ∈ S and we show that α is not left morphic in S. Otherwise there exists β = a0 + b0 x + c0 y + d0 x y ∈ S such that Sα = l(β) and Sβ = l(α). Notice that Sα = Z2 x y and l(α) = Z2 x + Z2 y + Z2 x y. It follows from Sβ = l(α) that a0 = 0, and so β = b0 x + c0 y + d0 x y. We have that Sβ = {ab0 x + ac0 y + (ad0 + bc0 + cb0 )x y : a, b, c ∈ Z2 } and that l(β) = {a + bx + cy + d x y ∈ S : ab0 = 0, ac0 = 0, ad0 + bc0 + cb0 = 0}. So it follows from Sβ = l(α) that Z2 b0 = Z2 and Z2 c0 = Z2 . Hence b0 = c0 = 1. Thus l(β) = {b(x − y) + d x y : b, d ∈ Z2 }. Hence l(β) 6= Sα. This is a contradiction. Let D be a ring and C be a subring of D. The set R[D, C] = {(d1 , . . . , dn , c, c, . . .) : di ∈ D, c ∈ C, n ≥ 1} is a ring where addition and multiplication are defined componentwise. In response to a question of Nicholson and morphic ring is regular, it is shown in [4, Example 0.1] that if S´anchez Campos [9] whether n every semiprimitive o x y D = M2 (Z2 ) and C = : x, y ∈ Z then R[D, C] is a semiprimitive, strongly morphic ring that is not 2 0 x regular. As a consequence of Theorem 2, we can now provide a large class of examples of semiprimitive, strongly morphic rings that are not regular. Lemma 7 ([4]). Let C be a subring of D. Then R[D, C] is left morphic iff D is left morphic and, for any x ∈ C, C
D
there exists y ∈ C such that x ∼ y and x ∼ y. There is a canonical way to identify R[x]/(x n ) with a subring of Mn (R). Theorem 8. Let A be a unit regular ring and n ≥ 1. Then R[Mn (A), that is not regular.
A[x] (x n ) ]
is a semiprimitive, strongly morphic ring
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Proof. For all k ≥ 1, Mk (A) is unit regular and we have A[x] A[x] ∼ Mk R Mn (A), n = R Mk (Mn (A)), Mk (x ) (x n ) Mk (A)[x] ∼ , = R Mn (Mk (A)), (x n ) A[x] so it suffices to show that R[Mn (A), (x n ) ] is a semiprimitive, morphic ring that is not regular. The proof uses an idea similar to the proof of Theorem 2. For simplicity, we only give the proof for the case where n = 2. Let D = M2 (A), C
D
and S = R[D, C]. Let α = a + bx ∈ C. Next we show that there exists β ∈ C such that α ∼ β and α ∼ β, n o so S is left morphic by Lemma 7. We identify A[x]/(x 2 ) with r0 rs : r, s ∈ A and α with a0 ab . Write a = ue −1 −1 a b 0 1 −u −1 b e (1 − e)u −1 b(1 − e) = , we where u ∈ U (A) and e2 = e ∈ A. Since 10 −(1 − 1e)u b u 0 u −1 0 a 0 1 e 0 e can assume that α = 0e ef where e2 = e ∈ A and f 2 = f ∈ (1 − e)A(1 − e). Let β = 1 − e0 − f 1 −1 − e − f ∈ C. Then r s Cα = : r ∈ Ae, s ∈ A(e + f ) = lC (β) 0 r r s lC (α) = : r ∈ A(1 − e − f ), s ∈ A(1 − e) = Cβ 0 r Re R(e + f ) Dα = = l D (β) Re R(e + f ) R(1 − e − f ) R(1 − e) l D (α) = = Dβ. R(1 − e − f ) R(1 − e)
C=
A[x] , (x 2 )
C
D
So α ∼ β and α ∼ β. Hence S is left morphic by Lemma 7. Symmetrically, S is right morphic. Since D is semiprimitive, so is S. Since C is an image of S, S is not regular. 2. The converse of Corollary 5 If R is unit regular then R[x]/(x 2 ) is strongly morphic by Corollary 5. But Z4 [x]/(x 2 ) is not morphic (see [6, p. 144]), even though Z4 is. By [6, Corollary 21], if R[x]/(x 2 ) is left morphic, then so is R. So far, all known examples of rings R for which R[x]/(x 2 ) is morphic are unit regular. Is it true that R must be unit regular if R[x]/(x 2 ) is morphic? This turns out to be true. Recall that a ring R is called right P-injective if, for every a ∈ R, every R-homomorphism a R → R R can be extended to an R-homomorphism R R → R R . Equivalently, R is right P-injective iff lr(a) = Ra. By [9, Theorem 24(1)], every left morphic ring is right P-injective. Theorem 9. The following are equivalent for a ring R: (1) R is unit regular. (2) R[x]/(x n ) is strongly morphic for all n ≥ 1. (3) R[x]/(x 2 ) is morphic. Proof. (1) ⇒ (2) is by Corollary 5. (2) ⇒ (3) is obvious. (3) ⇒ (1). Suppose that S := R[x]/(x 2 ) is morphic. Then R is morphic by [6, Corollary 21]. Let a ∈ R. We next show that R is regular by proving that a is regular in R. Thus R is unit regular by Erlich [7]. Since a is morphic in R, there exists b ∈ R such that Ra = l(b) and l(a) = Rb. Since R is left P-injective by [9, Theorem 24], one obtains that a R = rl(a) = r(Rb) = r(b) and r(a) = r(Ra) = rl(b) = b R. Now let α = bx ∈ S. Since S is left morphic, there exists β = c + d x ∈ S such that Sα = l(β) and l(α) = Sβ. Since S is right morphic, it is left P-injective
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by [9, Theorem 24]. Thus, as above, we have αS = rl(α) = r(Sβ) = r(β) and r(α) = r(Sα) = r(l(β)) = β S. By computation, one has l(α) = {r + sx : r ∈ l(b), s ∈ R}, Sβ = {r c + (r d + sc)x : r, s ∈ R}, r(α) = {r + sx : r ∈ r(b), s ∈ R}, β S = {cr + (dr + cs)x : r, s ∈ R}. Thus, x ∈ l(α) = Sβ and x ∈ r(α) = β S. Hence there exist r, s, r 0 , s 0 ∈ R such that 0 = r c,
1 = r d + sc,
0 = cr 0 ,
1 = dr 0 + cs 0 .
So r = r (dr 0 + cs 0 ) = r dr 0 = (r d + sc)r 0 = r 0 . Thus, c = c(r d + sc) = cr d + csc = cr 0 d + csc = csc. Therefore, c is regular in R. It follows from l(α) = Sβ that Rc = l(b). Since l(b) = Ra, we have Ra = Rc. Therefore a is regular in R since c is regular in R. We have been unable to answer whether R[x]/(x 2 ) being left morphic implies that R is unit regular. But we do have the following. Proposition 10. If R[x]/(x 2 ) is left morphic, then R is left morphic and semiprimitive. Proof. Suppose that S := R[x]/(x 2 ) is left morphic. Then R is left morphic by [6, Corollary 21]. Let a ∈ J (R). Since a is left morphic in R, there exists b ∈ R such that Ra = l(b) and Rb = l(a). Let α = bx ∈ S. Since S is left morphic, there exists β := c + d x ∈ S where c, d ∈ R such that l(α) = Sβ. One easily obtains l(α) = {r + sx : r ∈ l(b), s ∈ R}, Sβ = {r c + (r d + sc)x : r, s ∈ R}. Thus, x ∈ l(α) = Sβ. Hence there exist r, s ∈ R such that 0 = r c,
and
1 = r d + sc.
It follows from l(α) = Sβ that Rc = l(b). Since l(b) = Ra, we have Ra = Rc. So c ∈ J (R). Hence r d = 1 − sc is a unit of R. Thus r is a unit because R is directly finite (i.e., r s = 1 in R implies sr = 1) by [9, Proposition 6]. So, c = r −1 (r c) = 0. Hence a = 0. 3. The trivial extension R ∝ M Let R be a ring and M a bimodule over R. Recall that the trivial extension of R and M is R ∝ M = {(a, x) : a ∈ R, x ∈ M} with addition defined componentwise and multiplication defined by (a, x)(b, y) = (ab, ay + xb). n o In fact, R ∝ M is isomorphic to the subring a0 ax : a ∈ R, x ∈ M of the formal 2 × 2 matrix ring R0 MR , and ∼ R[x]/(x 2 ). For convenience, we let I ∝ X = {(a, x) : a ∈ I, x ∈ X } where I is a subset of R and X is a R∝R= subset of M. If σ : R → R is a ring endomorphism of R, let R(σ ) be the (R, R)-bimodule defined by R R(σ ) = R R and m ◦ r = mσ (r ) for all m ∈ R(σ ) and r ∈ R. Then R[x; σ ]/(x 2 ) ∼ = R ∝ R(σ ). Proposition 11. Let D be a division ring and V be a bimodule over D. Then D ∝ V is left morphic iff dim( D V ) ≤ 1. Proof. Let S = D ∝ V . Suppose that D ∝ V is left morphic and that V 6= 0. Take 0 6= x ∈ V . Then (0, x) ∼ (b, y) where b ∈ D and y ∈ V . Since S(0, x) = 0 ∝ Dx and l(0, x) = 0 ∝ V , it must be that b = 0. So (0, x) ∼ (0, y). Thus, 0 ∝ Dx = S(0, x) = l(0, y) = 0 ∝ V , showing that V = Dx. So dim( D V ) = 1. Conversely, if V = 0 then D ∝ V ∼ = D is clearly left morphic. If dim( D V ) = 1 and if 0 6= (a, x) ∈ S, then (a, x) ∼ (0, x) when a = 0 and (a, x) ∼ (0, 0) when a 6= 0. So S is left morphic.
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Next we consider when R ∝ M is left morphic for a unit regular ring R. By Corollary 5, R ∝ R is strongly morphic if R is unit regular. Let I be an ideal of a unit regular ring R. Is R ∝ I strongly morphic? Theorem 12. If I is an ideal of a unit regular ring R, then R ∝ I is strongly morphic. Proof. Let S = R ∝ I and let α = (a, x) ∈ S. We first show that S is left morphic by proving that α is left morphic in S. Since R is unit regular, a = ue where u is a unit of R and e2 = e ∈ R. Then (u −1 , 0) is a unit of S and (u −1 , 0)α = (e, u −1 x). Thus, by Lemma 1, we may assume that a = e and α = (e, x). Because (1, −(1 − e)x)α(1, −x) = (e, (1 − e)x(1 − e)) and because both (1, −(1 − e)x) and (1, −x) are units of S, by Lemma 1 it suffices to show that (e, (1 − e)x(1 − e)) is left morphic in S. Since (1 − e)R(1 − e) is again unit regular by [8, Corollary 4.7], as in the proof of Theorem 2, write (1 − e)x(1 − e) = v f where f ∈ I is an idempotent of (1 − e)R(1 − e) and v is a unit of (1 − e)R(1 − e) with inverse w. Then e + w is a unit of R with inverse e + v. Since (e + w, 0)(e, v f ) = (e, f ), by Lemma 1, we can finally assume that α = (e, f ) where e2 = e ∈ R and f 2 = f ∈ I with e f = f e = 0. Let β = (1 − e − f, f ) ∈ S. We verify that α ∼ β. We have Sα = S(e, f ) = Re ∝ (R f + I e), l(α) = l(e, f ) = R(1 − e − f ) ∝ I (1 − e), Sβ = S(1 − e − f, f ) = R(1 − e − f ) ∝ (R f + I (1 − e − f )), l(β) = l(1 − e − f, f ) = Re ∝ I (e + f ). For any c ∈ I , ce = (ce)(e + f ) and c f = (c f )(e + f ). It follows that I (e + f ) = I e + I f . Because R f = (R f ) f ⊆ I f , we obtain that R f + I e = I (e + f ). So Sα = l(β). Moreover, for any c ∈ I and any r ∈ R, c(1 − e) = c f + c(1 − e − f ) ∈ R f + I (1 − e − f ) and r f + c(1 − e − f ) = [r f + c(1 − f )](1 − e) with r f + c(1 − f ) ∈ I . It follows that I (1 − e) = R f + I (1 − e − f ). So Sβ = l(α). Hence α ∼ β. Thus S is left morphic. It is similar to show that S is right morphic. For any n > 0, Mn (S) ∼ = Mn (R) ∝ Mn (I ). Since Mn (R) is again unit regular, as above S is strongly morphic. For an ideal I of a unit regular ring R, if R I is finitely generated then I is a ring direct summand of R. Hence, by Theorem 12, R ∝ RI is strongly morphic. But, in general, R ∝ RI need not be left morphic. Example 13. There exists a unit regular ring R and an ideal I of R such that R ∝
R I
is not left morphic.
∞ F be a direct product of rings where each F = Z and let R be the subring of Q generated Proof. Let Q = Πi=1 i i 2 L∞ by I := i=1 Fi and 1 Q . Then R is unit regular (Boolean indeed) and I is an ideal of R. Set S = R ∝ RI . We show α ∼ β where that S is not left morphic. L Consider α = (a, x) ∈ S where a = 1 F1 and x = 1 + I . Suppose thatL β = (b, y) ∈ S. Then ( i≥2 Fi ) ∝ RI = l(α) = Sβ = {(r b, r y + zb) : r ∈ R, z ∈ RI }. This gives that i≥2 Fi = Rb is finitely generated. This is a contradiction. So α is not left morphic in S.
Let R be a PID that is not a field. Two elements a, b ∈ R are equivalent if a = ub for some u ∈ U (R). The primes in R are those non-invertible elements divisible only by elements equivalent to themselves and invertible elements. L Let P be the set consisting of one element from each equivalence class of primes in R. If Q = Q(R), then Q = P R p∞ R is the minimal cogenerator in R-Mod, where R p∞ = { pan + R : a ∈ R, n ∈ Z} (see [1, p. 215]). In particular, Q ∼L = {Z p∞ : p is a prime number} is the minimal cogenerator in Z-Mod. Z
Theorem 14. Let R be a PID that is not a field with Q = Q(R) and let M be a bimodule over R such that r x = xr for all x ∈ M and all r ∈ R. Then (1) R ∝ M is morphic iff M = (2) R ∝ Q R is strongly morphic.
Q R.
Proof. (1) The implication “⇐” is [6, Theorem 13]. The proofs for “(1) ⇒” and for (2) are the same as those for [6, Theorem 14 and Corollary 15]. Theorem 14 does not hold for a UFD. The greatest common divisor of two elements a, b in a UFD R is denoted by gcd(a, b).
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Theorem 15. Let R be a UFD with the quotient field Q. Then R ∝
Q R
is morphic iff R is a PID.
Q Proof. If R is a PID then R ∝ Q R is morphic by [6, Theorem 13]. Now assume that T = R ∝ R is morphic. For any −1 nonzero elements s, t in R with gcd(s, t) = 1, (0, x) ∈ T is morphic where x = st + R ∈ Q R . Let s (t R) denote the −1 set {r ∈ R : sr ∈ t R}. Then, by [6, Lemma 12], there exists k ∈ R such that s (t R) = k R and, for any c, d ∈ R, ck ∈ d R implies that ct ∈ d(s R + t R). Since gcd(s, t) = 1, s −1 (t R) = t R. So t R = k R. Take c = 1 and d = t. Then ck ∈ d R, thus ct ∈ d(s R + t R). This shows that c ∈ s R + t R, i.e., R = s R + t R. So, for any nonzero elements s, t s t in R, R = gcd(s,t) R + gcd(s,t) R. This clearly shows that gcd(s, t)R = s R + t R. Hence R is a PID by a well-known result of PID’s.
Theorem 16. Let R be a PID and let a ∈ R be such that a = p1 · · · pk q1s1 · · · qlsl where pi , q j are distinct primes of R, k ≥ 1, l ≥ 0, and all s j ≥ 2. Let T = R/a R and M be a bimodule over T such that xt = t x for all t ∈ T and x ∈ M. The following are equivalent: (1) T ∝ M is morphic. (2) T ∝ M is strongly morphic. (3) M ∼ = R/b R where b is a divisor of p1 · · · pk . Proof. (3) ⇒ (2). Suppose (3) holds and suppose b = p1 · · · pm with m ≤ k. Then T ∝ M ∼ = ( p1RR ⊕ R R R R R R R R ∼ ( · · · ⊕ pk R ⊕ s1 ⊕ · · · ⊕ sl ) ∝ ( p1 R ⊕ · · · ⊕ pm R ⊕ (0) ⊕ · · · ⊕ (0)) = p1 R ∝ p1 R ) ⊕ · · · ⊕ ( pm R ∝ pm R ) ⊕ R
q1 R
R pm+1 R
⊕···⊕
R pk R
ql R R s ql l R
⊕
⊕···⊕
R . s ql l R
The last isomorphism is by [6, Lemma 4]. Each of the summands of
the latter is a local morphic ring with nilpotent Jacobson radical by [9, Theorem 9]. Thus, by [9, Theorem 17], T ∝ M is strongly morphic. (1) ⇒ (3). Write T = T1 ⊕· · ·⊕Tk ⊕S1 ⊕· · ·⊕Sl where Ti ∼ = sRj for i = 1, . . . , k and j = 1, . . . , l. = pRi R and S j ∼ qj R
Then M = M1 ⊕ · · · ⊕ Mk ⊕ N1 ⊕ · · · ⊕ Nl where Mi ∼ = S j M for i = 1, . . . , k and j = 1, . . . , l. = Ti M and N j ∼ Thus, by [6, Lemma 4], (T1 ∝ M1 ) ⊕ · · · ⊕ (Tk ∝ Mk ) ⊕ (S1 ∝ N1 ) ⊕ · · · ⊕ (Sl ∝ Nl ) ∼ = T ∝ M is morphic. So, for any 1 ≤ i ≤ k, Ti ∝ Mi is morphic, and hence either Mi = 0 or Mi ∼ = pRi R by Proposition 11. Moreover, for any 1 ≤ j ≤ l, S j ∝ N j is morphic. We next show that N j = 0. We let S = S j , N = N j , q = q j , s = s j . If N 6= 0, S∝N
take 0 6= x ∈ N . We claim l S (x) 6= 0. Since (0, x) is morphic in S ∝ N , (0, x) ∼ (r, y) where r ∈ S and y ∈ N . If l S (x) = 0, then r = 0. Thus, 0 ∝ Sx = (S ∝ N )(0, x) = l(0, y) = l S (y) ∝ N , showing that N = Sx. So N ∼ = S. Thus S ∝ S ∼ = S ∝ N is morphic. Hence S is regular by Theorem 9, a contradiction. So the claim holds and it follows that q s−1 N = 0 and that l S (x) = q n S with 1 ≤ n < s. Now direct computation shows that 0 ∝ Sx = (0, x)(S ∝ N ) = rl((0, x)) = r(l S (x) ∝ N ) = r(q n S ∝ N ) ⊇ q s−n S ∝ 0. This contradiction shows that N = 0. Hence M = M1 ⊕ · · · Mk ⊕ N1 ⊕ · · · ⊕ Nl ∼ = M1 ⊕ · · · ⊕ Mk . We can assume that Mi ∼ = R/ pi R for i = 1, . . . , k1 ≤ k and Mi = 0 for i = k1 + 1, . . . , k. Then M ∼ = R/b R with b = p1 · · · pk1 . This proves (3). Corollary 17 ([6]). Let d, n be positive integers with d > 1. Then Zdn ∝ Zd is morphic iff gcd(n, d) = 1 and d is a product of distinct primes. 4. An example of morphic chain rings A ring R is called right chain if the right ideals of R form a chain under usual inclusion, or equivalently the principal right ideals of R form a chain. A left and right chain ring is called a chain ring. Any one-sided chain ring is local. Examples of local left morphic rings with nilpotent Jacobson radical have been given in [9,10].
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Example 18. There is a commutative chain ring R satisfying: (1) R is strongly morphic; (2) J (R) is nil but not T -nilpotent. L Proof. Let F be a field and R = {Fαr : 0 ≤ r < 1} be a vector space over F with a basis {αr : 0 ≤ r < 1}, with multiplication defined by α , if r + s < 1; αr αs = r +s 0, if r + s ≥ 1. Then R is a commutative ring with J (R) = Σ {Fαr : 0 < r < 1}. Clearly J (R) is nil. Each principal ideal of R is generated by some αr . So the principal ideals form a chain. Hence R is a chain ring. If let ai = α2−i ∈ J (R) for i = 1, 2, . . ., then for any n > 0, a1 a2 · · · an = αr 6= 0 where r = 1 − 21n . So J (R) is not T -nilpotent and thus R is not perfect. For any 0 6= x ∈ R, it can easily be seen that x = uαr for some unit u of R and some 0 ≤ r < 1. Since R
αr ∼ α1−r , x is morphic in R by Lemma 1. It can be easily proved as [9, Theorem 17] that R is strongly morphic.
Remark 19. Following Camillo and Nicholson [3], a ring R is called left quasi-morphic if {Ra : a ∈ R} = {l(b) : b ∈ R}. A left and right quasi-morphic ring is called a quasi-morphic ring. These rings include regular rings and morphic rings. A question raised in [3] asks whether a quasi-morphic ring containing no infinite sets of orthogonal idempotents is necessarily perfect. The answer is “No” by Example 18. A ring R is said to satisfy (∗) if it contains a set of orthogonal idempotents e1 , . . . , en with 1 = e1 + · · · + en such that ei Rei is a chain ring for all i. Theorem 20. The following are equivalent for a ring R: (1) R is a morphic ring satisfying (∗). (2) R is a finite direct product of matrix rings over morphic chain rings. Proof. The proof of “(1) ⇒ (2)” is the same as that of [9, Theorem 29]. As in the proof of [9, Theorem 17], it can be shown that every morphic chain ring is strongly morphic. Thus, the implication “(2) ⇒ (1)” follows. To give our concluding example, we need a lemma. Lemma 21. Let R be a ring and e be a central idempotent. Then the element a ∈ R is left morphic in R iff ea is left morphic in e Re and (1 − e)a is left morphic in (1 − e)R(1 − e). R
eR
Proof. This is because a ∼ b iff ea ∼ eb and (1 − e)a
(1−e)R
∼ (1 − e)b.
Note that the assumption that e is central can not be removed from Lemma 21 by the next remark (see [9, Lemma 14]). Remark 22. Let R, σ, b be as in Example 3. Let S = R[x; σ ]/(x 2 ) and let e = b. Then e is an idempotent of S, eSe
e(bx)e = 0, and (1 − e)(bx)(1 − e) = 0. Hence we have that e(bx)e ∼ 0 and (1 − e)bx(1 − e) is not left morphic in S by Example 3.
(1−e)S(1−e)
∼
0. But bx
Example 23. If R is a strongly regular ring and if C2 is the cyclic group of order 2, then the group ring RC2 is morphic. Proof. Write C2 = {1, g}. Let S = RC2 and let α = a + bg ∈ S. We want to show that α is left morphic in S. Since −1 R is strongly regular, write a = u 1 e1 where u 1 ∈ U (R) and e1 is a central idempotent. Then u −1 1 α = e1 + u 1 bg. By Lemma 1, we may assume that a = e1 is an idempotent. So α = e1 + bg. Because e1 α = e1 + e1 bg ∈ e1 S = (e1 R)C2 and (1 − e1 )α = (1 − e1 )bg ∈ (1 − e1 )S = ((1 − e1 )R)C2 and because of Lemma 21, α is left morphic in S iff e1 + e1 bg is left morphic in (e1 R)C2 and (1 − e1 )bg is left morphic in ((1 − e1 )R)C2 . Since (1 − e1 )bg is a strongly regular element in ((1 − e1 )R)C2 (i.e., a product of a unit and a central idempotent), (1 − e1 )bg is left morphic in ((1 − e1 )R)C2 by Lemma 1. Thus we only need to show that e1 + e1 bg is left morphic in (e1 R)C2 . Therefore, without
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loss of generality, we can assume that α = 1 + bg. Write b = u 2 e2 where u 2 ∈ U (R) and e2 is a central idempotent of R. Because (1 − e2 )α = 1 − e2 is left morphic in (1 − e2 )S, by Lemma 21 it suffices to show that e2 α = e2 + (u 2 e2 )g is left morphic in e2 S = (e2 R)C2 . Therefore, we may further assume that α = 1 + vg where v ∈ U (R). Now write 1 − v 2 = eu where u ∈ U (R) and e is a central idempotent. Let β = −(1 − e)v + (1 − e)g = (1 − e)(g − v) ∈ S. Noting that (1 − e)v 2 = 1 − e, one obtains Sα = {(a + bg)(1 + vg) : a, b ∈ R} = {(a + bv) + (av + b)g : a, b ∈ R}, l(α) = {a + bg : a + bv = av + b = 0} = {a + bg : a = −bv, be = 0} = R(1 − e)(g − v), Sβ = = = = =
{(a + bg)(1 − e)(g − v) : a, b ∈ R} {(1 − e)(−av + b) + (1 − e)(a − bv)g : a, b ∈ R} {−(1 − e)(a − bv)v + (1 − e)(a − bv)g : a, b ∈ R} {(1 − e)(a − bv)(g − v) : a, b ∈ R} R(1 − e)(g − v),
l(β) = {a + bg : (a + bg)(1 − e)(g − v) = 0} = {a + bg : (1 − e)(a − bv) = 0}. Thus, Sα = l(β), and it is easy to see that Sβ ⊆ l(β). For any r +sg ∈ l(β) with r, s ∈ R, r +sg = (a+bv)+(av+b)g where a = r − (s − r v)u −1 v and b = (s − r v)u −1 . So r + sg ∈ Sα. Therefore, α ∼ β. Acknowledgments Part of the work was done during a visit, sponsored by NCTS of Taipei, by the second author to the National Taiwan University. He gratefully acknowledges the financial support from NCTS and kind hospitality from the institute. The research of the first author was supported by NSC of Taiwan, and that of the second author by NSERC of Canada. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
F.W. Anderson, K.R. Fuller, Rings and Categories of Modules, Springer-Verlag, New York, 1973. V.P. Camillo, D. Khurana, A characterization of unit-regular rings, Comm. Algebra 29 (2001) 2293–2295. V.P. Camillo, W.K. Nicholson, Quasi-morphic rings, 2006. preprint. J. Chen, Y. Li, Y. Zhou, Constructing morphic rings, in: Advances in Ring Theory, World Sci. Publ., Hackensack, NJ, 2005, pp. 26–32. J. Chen, Y. Li, Y. Zhou, Morphic group rings, J. Pure Appl. Algebra 205 (3) (2006) 621–639. J. Chen, Y. Zhou, Morphic rings as trivial extensions, Glasgow Math. J. 47 (2005) 139–148. G. Erlich, Units and one-sided units in regular rings, Trans. Amer. Math. Soc. 216 (1976) 81–90. K.R. Goodearl, Von Neumann regular rings, in: Monographs and Studies in Mathematics, Pitman, Boston, Mass, London, 1979. W.K. Nicholson, E. S´anchez Campos, Rings with the dual of the isomorphism theorem, J. Algebra 271 (2004) 391–406. W.K. Nicholson, E. S´anchez Campos, Principal rings with the dual of the isomorphism theorem, Glasgow Math. J. 46 (2004) 181–191. W.K. Nicholson, E. S´anchez Campos, Morphic modules, Comm. Algebra 33 (2005) 2629–2647.
Morphic rings and unit regular rings Tsiu-Kwen Lee a,1 , Yiqiang Zhou b,∗ a Department of Mathematics, National Taiwan University, Taipei 106, Taiwan b Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, Newfoundland A1C 5S7, Canada
Received 9 July 2006; received in revised form 4 October 2006 Available online 17 November 2006 Communicated by R. Parimala
Abstract A ring R is called left morphic if R/Ra ∼ = l(a) for every a ∈ R. A left and right morphic ring is called a morphic ring. If Mn (R) is morphic for all n ≥ 1 then R is called a strongly morphic ring. A well-known result of Erlich says that a ring R is unit regular iff it is both (von Neumann) regular and left morphic. A new connection between morphic rings and unit regular rings is proved here: a ring R is unit regular iff R[x]/(x n ) is strongly morphic for all n ≥ 1 iff R[x]/(x 2 ) is morphic. Various new families of left morphic or strongly morphic rings are constructed as extensions of unit regular rings and of principal ideal domains. This places some known examples in a broader context and answers some existing questions. c 2006 Elsevier B.V. All rights reserved.
MSC: Primary: 16E50; 16U99; secondary: 16S70; 16S35
0. Introduction By the fundamental homomorphism theorem of modules, for any element a in a ring R, R/l(a) ∼ = Ra as left R-modules where l(a) = {r ∈ R : ra = 0} is the left annihilator of a in R. Dually, if R/Ra ∼ = l(a), then a is called a left morphic element of R. The ring R is called left morphic if every element of R is left morphic. Right morphic rings are defined analogously. A left and right morphic ring is simply called a morphic ring. A ring R is called strongly morphic if the matrix ring Mn (R) is morphic for all n ≥ 1. An element a ∈ R is called unit regular if a = aua for some unit u of R and the ring R is called unit regular if all elements of R are unit regular. A well-known result of Erlich [7] states that an element a in a ring R is unit regular if and only if a is both left morphic and regular in R. This result is crucial in proving several important results of unit regular rings (see [2,8]). The study of morphic rings was started by Nicholson and S´anchez Campos in [9–11] and was continued by others in [4–6]. If R is a ring and σ : R → R is a ring endomorphism, let R[x; σ ] denote the ring of skew polynomials over R, that is, all formal polynomials in x with coefficients from R with multiplication defined by xr = σ (r )x. The first two sections deal with the question when R[x; σ ]/(x n ) is left morphic. If D is a division ring and σ is an endomorphism ∗ Corresponding author. Tel.: +1 705 737 8797; fax: +1 709 737 3010.
E-mail addresses: [email protected] (T.-K. Lee), [email protected] (Y. Zhou). 1 Member of Mathematics Division (Taipei Office), National Center for Theoretical Sciences. c 2006 Elsevier B.V. All rights reserved. 0022-4049/$ - see front matter doi:10.1016/j.jpaa.2006.10.005
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of D with σ (1) = 1, then D[x; σ ]/(x 2 ) is a left morphic ring by [9]. This ring turns out to be a prototype for all local, left morphic rings with nilpotent Jacobson radical, and the latter plays a significant role in building up the structures of more general morphic rings (see [9]). Later in [6] it was proved that R[x; σ ]/(x 2 ) is left morphic for any strongly regular ring (i.e., any unit regular ring whose idempotents are central) R with an endomorphism σ satisfying σ (e) = e for all e2 = e ∈ R and that R[x]/(x 2 ) is strongly morphic for any semisimple ring R. A question left open in [6] is whether R[x]/(x 2 ) is morphic for any unit regular ring R. This question is settled here. In fact, it is proved that if R is a unit regular ring and σ is an onto endomorphism of R with σ (e) = e for all e2 = e ∈ R, then R[x; σ ]/(x n ) is left morphic for all n ≥ 1. It is further proved that a ring R is unit regular iff R[x]/(x n ) is strongly morphic for all n ≥ 1 iff R[x]/(x 2 ) is morphic. This result is used to provide a large class of examples of semiprimitive, strongly morphic rings that are not regular. The morphic property of the trivial extension R ∝ M of a ring R with a bimodule M over R was discussed in [6]. One of the results obtained in [6] says that if R is a PID (principal ideal domain) with the field Q(R) of quotients, then R ∝ Q(R) R is morphic. This result cannot be extended to a UFD (unique factorization domain) because, as noticed in Section 3, for a UFD R, R ∝ Q(R) R is morphic iff R is a PID. If R is any factor ring of a PID and if M is a bimodule over R such that r x = xr for all r ∈ R and x ∈ M, a necessary and sufficient condition for R ∝ M to be morphic is obtained. It is also proved here that if I is an ideal of a unit regular ring then R ∝ I is strongly morphic. Section 4 presents an example of a commutative strongly morphic chain ring whose Jacobson radical is nil but not T -nilpotent. This answers in the negative a question of Camillo and Nicholson [3] on quasi-morphic rings. It is observed that a ring R is a finite direct product of matrix rings over morphic chain rings iff R is a morphic ring with a complete set {e1 , . . . , en } of orthogonal idempotents such that ei Rei is a chain ring for all i. All rings R are associative with identity. The Jacobson radical and the group of units of a ring R are denoted by J (R) and U (R) respectively. As usual, we write Z, Q, and Zn for the ring of integers, the field of rational numbers, and the ring of integers modulo n. The field of quotients of an integral domain R is denoted Q(R). By a subring of a ring R, we shall always mean a subring containing the identity of R. For a ∈ R, the left annihilator of a in R is denoted by l(a) or l R (a), and the right annihilator of a in R is denoted by r(a) or r R (a). 1. The ring R[x; σ ]/(x n ) R
For a, b ∈ R, we write a ∼ b (or simply a ∼ b) to mean that Ra = l(b) and Rb = l(a). By [9, Lemma 1], a ∈ R R
is left morphic iff a ∼ b for some b ∈ R. Lemma 1 ([9]). Let a, b ∈ R and let u ∈ U (R). The following are equivalent: (1) a ∼ b. (2) ua ∼ bu −1 . (3) au ∼ u −1 b. It was proved in [6] that for a strongly regular ring R with an endomorphism σ : R → R such that σ (e) = e for all e2 = e ∈ R, R[x; σ ]/(x 2 ) is a left morphic ring. This is a special case of the next result. Theorem 2. Let R be a unit regular ring and let σ : R → R be a ring endomorphism such that σ (e) = e for all e2 = e ∈ R. Then R[x; σ ]/(x 2 ) and R[x]/(x n+1 ) are left morphic rings for all n ≥ 0. Proof. Let S = R[x; σ ]/(x n+1 ). Then S = {r0 + r1 x + · · · + rn x n : ri ∈ R, i = 0, 1, . . . , n} with x n+1 = 0 and xr = σ (r )x for all r ∈ R. Let α = r0 + r1 x + · · · + rn x n ∈ S. We show that α is left morphic in S. Since R is unit regular, every element of R is the product of a unit and an idempotent. Thus, by multiplying α by a suitable unit of R we can assume that r0 = e0 is an idempotent by Lemma 1. Because [1 − (1 − e0 )r1 x]α(1 − r1 x) = e0 + (1 − e0 )r1 (1 − e0 )x + · · ·
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503
where both 1 − (1 − e0 )r1 x and 1 − r1 x are units of S, we can further assume that r1 ∈ (1 − e0 )R(1 − e0 ) because of Lemma 1. Now [1 − (1 − e0 )r2 x 2 ]α(1 − r2 x 2 ) = e0 + r1 x + (1 − e0 )r2 (1 − e0 )x + · · · where both 1 − (1 − e0 )r2 x 2 and 1 − r2 x 2 are units of S, so we can assume that r2 ∈ (1 − e0 )R(1 − e0 ) by Lemma 1. A simple induction shows that we can assume that α = e0 + r 1 x + r 2 x 2 + · · · + r n x n ,
ri ∈ (1 − e0 )R(1 − e0 )
for i = 1, . . . , n.
Because (1 − e0 )R(1 − e0 ) is unit regular by [8, Corollary 4.7], write r1 = ue1 where e1 is an idempotent of (1 − e0 )R(1 − e0 ) and u is a unit of (1 − e0 )R(1 − e0 ) with inverse v. Then e0 + v is a unit of R with inverse e0 + u. Since (e0 + v)α = e0 + e1 x + vr2 x 2 + · · · + vrn x n , by Lemma 1 we can assume that α = e0 + e1 x + r 2 x 2 + · · · + r n x n ,
ri ∈ (1 − e0 )R(1 − e0 )
for i = 2, . . . , n.
Now [1 − (1 − e1 )r2 x]α(1 − r2 x) = e0 + e1 x + (1 − e1 )r2 (1 − e1 )x 2 + · · · . Since 1 − (1 − e1 )r2 x and 1 − r2 x are units of S, by Lemma 1 we can assume that α = e0 + e1 x + r2 x 2 + · · · + rn x n where r2 ∈ (1 − e1 )(1 − e0 )R(1 − e0 )(1 − e1 ) and ri ∈ (1 − e0 )R(1 − e0 ) for i = 3, . . . , n. As above, a simple induction shows that we can assume that α = e 0 + e1 x + r 2 x 2 + · · · + r n x n , e02 = e0 ,
e12 = e1 ∈ (1 − e0 )R(1 − e0 ),
ri ∈ (1 − e1 )(1 − e0 )R(1 − e0 )(1 − e1 )
for i = 2, . . . , n.
Repeating the process, by induction we can assume that α = e0 + e1 x + e2 x 2 + · · · + en x n , e02 = e0 , ei2 = ei ∈ (1 − ei−1 ) · · · (1 − e0 )R(1 − e0 ) · · · (1 − ei−1 )
for i = 1, . . . , n.
Let β = b0 + b1 x + · · · + bn x n where b0 = (1 − e0 )(1 − e1 ) · · · (1 − en−1 )(1 − en ) b1 = (1 − e0 )(1 − e1 ) · · · (1 − en−1 ) ··· bn = 1 − e0 . Thus, we have Sα = Re0 + R(e0 + e1 )x + · · · + R(e0 + · · · + en )x n = l(β), l(α) = Rb0 + Rb1 x + · · · + Rbn x n = Sβ. So α ∼ β. The proof is complete.
The assumption that σ (e) = e for all e2 = e ∈ R in Theorem 2 can not be removed by the next example. Example 3. There exists a Boolean ring R and an endomorphism σ : R → R with σ (1) = 1, but R[x; σ ]/(x 2 ) is not left morphic. Proof. Consider the direct product R = Z2 × Z2 and let σ : R → R be given by (a1 , a2 ) 7→ (a2 , a1 ). Then σ is an endomorphism of R with σ (1) = 1. Let b = (1, 0) ∈ R and S = R[x; σ ]/(x 2 ). We next show that there do not exist c, d ∈ R such that l(bx) = S(c + d x)
and
S(bx) = l(c + d x).
(1.1)
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Suppose that c, d ∈ R satisfy (1.1). Then x ∈ l(bx) = S(c + d x) = {r c + [r d + sσ (c)]x : r, s ∈ R}. So r c = 0 and 1 = r d + sσ (c) for some r, s ∈ R. Thus c = c(r d + sσ (c)) = scσ (c). Since bx 6= 0 and bx ∼ c + d x, c + d x cannot be a unit of S. So c 6= 1. Thus cσ (c) = 0 and hence c = 0. Thus, r d = 1, showing that d = 1. But l(x) = Sx 6= S(bx). Hence S is not left morphic. It is possible that the ring in Theorem 2 is neither right morphic nor strongly left morphic as shown by the next example. Example 4. By [9], there exists a field F and an endomorphism σ of F with σ (1) = 1 such that F[x; σ ]/(x 2 ) is left morphic but not right morphic and that the 2 × 2 matrix ring over F[x; σ ]/(x 2 ) is not left morphic. The next corollary extends a result in [6] that R[x]/(x 2 ) is strongly morphic for any semisimple ring R and gives an affirmative answer to the question whether R[x]/(x 2 ) is morphic for any unit regular ring R (see [6, p. 148]). Corollary 5. If R is a unit regular ring, then R[x]/(x n ) is strongly morphic for all n ≥ 1. Proof. If R is unit regular then so is Mk (R) for every k ≥ 1 by [8, Corollary 4.7]. Thus, the claim follows by Theorem 2 because Mk (R[x]/(x n )) ∼ = Mk (R)[x]/(x n ). The next example is a contrast to Corollary 5. Example 6. The ring Z2 [x, y]/(x 2 , y 2 ) is not left morphic. Proof. Let S = Z2 [x, y]/(x 2 , y 2 ). Then S = {a + bx + cy + d x y : a, b, c, d ∈ Z2 }. Let α = x y ∈ S and we show that α is not left morphic in S. Otherwise there exists β = a0 + b0 x + c0 y + d0 x y ∈ S such that Sα = l(β) and Sβ = l(α). Notice that Sα = Z2 x y and l(α) = Z2 x + Z2 y + Z2 x y. It follows from Sβ = l(α) that a0 = 0, and so β = b0 x + c0 y + d0 x y. We have that Sβ = {ab0 x + ac0 y + (ad0 + bc0 + cb0 )x y : a, b, c ∈ Z2 } and that l(β) = {a + bx + cy + d x y ∈ S : ab0 = 0, ac0 = 0, ad0 + bc0 + cb0 = 0}. So it follows from Sβ = l(α) that Z2 b0 = Z2 and Z2 c0 = Z2 . Hence b0 = c0 = 1. Thus l(β) = {b(x − y) + d x y : b, d ∈ Z2 }. Hence l(β) 6= Sα. This is a contradiction. Let D be a ring and C be a subring of D. The set R[D, C] = {(d1 , . . . , dn , c, c, . . .) : di ∈ D, c ∈ C, n ≥ 1} is a ring where addition and multiplication are defined componentwise. In response to a question of Nicholson and morphic ring is regular, it is shown in [4, Example 0.1] that if S´anchez Campos [9] whether n every semiprimitive o x y D = M2 (Z2 ) and C = : x, y ∈ Z then R[D, C] is a semiprimitive, strongly morphic ring that is not 2 0 x regular. As a consequence of Theorem 2, we can now provide a large class of examples of semiprimitive, strongly morphic rings that are not regular. Lemma 7 ([4]). Let C be a subring of D. Then R[D, C] is left morphic iff D is left morphic and, for any x ∈ C, C
D
there exists y ∈ C such that x ∼ y and x ∼ y. There is a canonical way to identify R[x]/(x n ) with a subring of Mn (R). Theorem 8. Let A be a unit regular ring and n ≥ 1. Then R[Mn (A), that is not regular.
A[x] (x n ) ]
is a semiprimitive, strongly morphic ring
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Proof. For all k ≥ 1, Mk (A) is unit regular and we have A[x] A[x] ∼ Mk R Mn (A), n = R Mk (Mn (A)), Mk (x ) (x n ) Mk (A)[x] ∼ , = R Mn (Mk (A)), (x n ) A[x] so it suffices to show that R[Mn (A), (x n ) ] is a semiprimitive, morphic ring that is not regular. The proof uses an idea similar to the proof of Theorem 2. For simplicity, we only give the proof for the case where n = 2. Let D = M2 (A), C
D
and S = R[D, C]. Let α = a + bx ∈ C. Next we show that there exists β ∈ C such that α ∼ β and α ∼ β, n o so S is left morphic by Lemma 7. We identify A[x]/(x 2 ) with r0 rs : r, s ∈ A and α with a0 ab . Write a = ue −1 −1 a b 0 1 −u −1 b e (1 − e)u −1 b(1 − e) = , we where u ∈ U (A) and e2 = e ∈ A. Since 10 −(1 − 1e)u b u 0 u −1 0 a 0 1 e 0 e can assume that α = 0e ef where e2 = e ∈ A and f 2 = f ∈ (1 − e)A(1 − e). Let β = 1 − e0 − f 1 −1 − e − f ∈ C. Then r s Cα = : r ∈ Ae, s ∈ A(e + f ) = lC (β) 0 r r s lC (α) = : r ∈ A(1 − e − f ), s ∈ A(1 − e) = Cβ 0 r Re R(e + f ) Dα = = l D (β) Re R(e + f ) R(1 − e − f ) R(1 − e) l D (α) = = Dβ. R(1 − e − f ) R(1 − e)
C=
A[x] , (x 2 )
C
D
So α ∼ β and α ∼ β. Hence S is left morphic by Lemma 7. Symmetrically, S is right morphic. Since D is semiprimitive, so is S. Since C is an image of S, S is not regular. 2. The converse of Corollary 5 If R is unit regular then R[x]/(x 2 ) is strongly morphic by Corollary 5. But Z4 [x]/(x 2 ) is not morphic (see [6, p. 144]), even though Z4 is. By [6, Corollary 21], if R[x]/(x 2 ) is left morphic, then so is R. So far, all known examples of rings R for which R[x]/(x 2 ) is morphic are unit regular. Is it true that R must be unit regular if R[x]/(x 2 ) is morphic? This turns out to be true. Recall that a ring R is called right P-injective if, for every a ∈ R, every R-homomorphism a R → R R can be extended to an R-homomorphism R R → R R . Equivalently, R is right P-injective iff lr(a) = Ra. By [9, Theorem 24(1)], every left morphic ring is right P-injective. Theorem 9. The following are equivalent for a ring R: (1) R is unit regular. (2) R[x]/(x n ) is strongly morphic for all n ≥ 1. (3) R[x]/(x 2 ) is morphic. Proof. (1) ⇒ (2) is by Corollary 5. (2) ⇒ (3) is obvious. (3) ⇒ (1). Suppose that S := R[x]/(x 2 ) is morphic. Then R is morphic by [6, Corollary 21]. Let a ∈ R. We next show that R is regular by proving that a is regular in R. Thus R is unit regular by Erlich [7]. Since a is morphic in R, there exists b ∈ R such that Ra = l(b) and l(a) = Rb. Since R is left P-injective by [9, Theorem 24], one obtains that a R = rl(a) = r(Rb) = r(b) and r(a) = r(Ra) = rl(b) = b R. Now let α = bx ∈ S. Since S is left morphic, there exists β = c + d x ∈ S such that Sα = l(β) and l(α) = Sβ. Since S is right morphic, it is left P-injective
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by [9, Theorem 24]. Thus, as above, we have αS = rl(α) = r(Sβ) = r(β) and r(α) = r(Sα) = r(l(β)) = β S. By computation, one has l(α) = {r + sx : r ∈ l(b), s ∈ R}, Sβ = {r c + (r d + sc)x : r, s ∈ R}, r(α) = {r + sx : r ∈ r(b), s ∈ R}, β S = {cr + (dr + cs)x : r, s ∈ R}. Thus, x ∈ l(α) = Sβ and x ∈ r(α) = β S. Hence there exist r, s, r 0 , s 0 ∈ R such that 0 = r c,
1 = r d + sc,
0 = cr 0 ,
1 = dr 0 + cs 0 .
So r = r (dr 0 + cs 0 ) = r dr 0 = (r d + sc)r 0 = r 0 . Thus, c = c(r d + sc) = cr d + csc = cr 0 d + csc = csc. Therefore, c is regular in R. It follows from l(α) = Sβ that Rc = l(b). Since l(b) = Ra, we have Ra = Rc. Therefore a is regular in R since c is regular in R. We have been unable to answer whether R[x]/(x 2 ) being left morphic implies that R is unit regular. But we do have the following. Proposition 10. If R[x]/(x 2 ) is left morphic, then R is left morphic and semiprimitive. Proof. Suppose that S := R[x]/(x 2 ) is left morphic. Then R is left morphic by [6, Corollary 21]. Let a ∈ J (R). Since a is left morphic in R, there exists b ∈ R such that Ra = l(b) and Rb = l(a). Let α = bx ∈ S. Since S is left morphic, there exists β := c + d x ∈ S where c, d ∈ R such that l(α) = Sβ. One easily obtains l(α) = {r + sx : r ∈ l(b), s ∈ R}, Sβ = {r c + (r d + sc)x : r, s ∈ R}. Thus, x ∈ l(α) = Sβ. Hence there exist r, s ∈ R such that 0 = r c,
and
1 = r d + sc.
It follows from l(α) = Sβ that Rc = l(b). Since l(b) = Ra, we have Ra = Rc. So c ∈ J (R). Hence r d = 1 − sc is a unit of R. Thus r is a unit because R is directly finite (i.e., r s = 1 in R implies sr = 1) by [9, Proposition 6]. So, c = r −1 (r c) = 0. Hence a = 0. 3. The trivial extension R ∝ M Let R be a ring and M a bimodule over R. Recall that the trivial extension of R and M is R ∝ M = {(a, x) : a ∈ R, x ∈ M} with addition defined componentwise and multiplication defined by (a, x)(b, y) = (ab, ay + xb). n o In fact, R ∝ M is isomorphic to the subring a0 ax : a ∈ R, x ∈ M of the formal 2 × 2 matrix ring R0 MR , and ∼ R[x]/(x 2 ). For convenience, we let I ∝ X = {(a, x) : a ∈ I, x ∈ X } where I is a subset of R and X is a R∝R= subset of M. If σ : R → R is a ring endomorphism of R, let R(σ ) be the (R, R)-bimodule defined by R R(σ ) = R R and m ◦ r = mσ (r ) for all m ∈ R(σ ) and r ∈ R. Then R[x; σ ]/(x 2 ) ∼ = R ∝ R(σ ). Proposition 11. Let D be a division ring and V be a bimodule over D. Then D ∝ V is left morphic iff dim( D V ) ≤ 1. Proof. Let S = D ∝ V . Suppose that D ∝ V is left morphic and that V 6= 0. Take 0 6= x ∈ V . Then (0, x) ∼ (b, y) where b ∈ D and y ∈ V . Since S(0, x) = 0 ∝ Dx and l(0, x) = 0 ∝ V , it must be that b = 0. So (0, x) ∼ (0, y). Thus, 0 ∝ Dx = S(0, x) = l(0, y) = 0 ∝ V , showing that V = Dx. So dim( D V ) = 1. Conversely, if V = 0 then D ∝ V ∼ = D is clearly left morphic. If dim( D V ) = 1 and if 0 6= (a, x) ∈ S, then (a, x) ∼ (0, x) when a = 0 and (a, x) ∼ (0, 0) when a 6= 0. So S is left morphic.
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Next we consider when R ∝ M is left morphic for a unit regular ring R. By Corollary 5, R ∝ R is strongly morphic if R is unit regular. Let I be an ideal of a unit regular ring R. Is R ∝ I strongly morphic? Theorem 12. If I is an ideal of a unit regular ring R, then R ∝ I is strongly morphic. Proof. Let S = R ∝ I and let α = (a, x) ∈ S. We first show that S is left morphic by proving that α is left morphic in S. Since R is unit regular, a = ue where u is a unit of R and e2 = e ∈ R. Then (u −1 , 0) is a unit of S and (u −1 , 0)α = (e, u −1 x). Thus, by Lemma 1, we may assume that a = e and α = (e, x). Because (1, −(1 − e)x)α(1, −x) = (e, (1 − e)x(1 − e)) and because both (1, −(1 − e)x) and (1, −x) are units of S, by Lemma 1 it suffices to show that (e, (1 − e)x(1 − e)) is left morphic in S. Since (1 − e)R(1 − e) is again unit regular by [8, Corollary 4.7], as in the proof of Theorem 2, write (1 − e)x(1 − e) = v f where f ∈ I is an idempotent of (1 − e)R(1 − e) and v is a unit of (1 − e)R(1 − e) with inverse w. Then e + w is a unit of R with inverse e + v. Since (e + w, 0)(e, v f ) = (e, f ), by Lemma 1, we can finally assume that α = (e, f ) where e2 = e ∈ R and f 2 = f ∈ I with e f = f e = 0. Let β = (1 − e − f, f ) ∈ S. We verify that α ∼ β. We have Sα = S(e, f ) = Re ∝ (R f + I e), l(α) = l(e, f ) = R(1 − e − f ) ∝ I (1 − e), Sβ = S(1 − e − f, f ) = R(1 − e − f ) ∝ (R f + I (1 − e − f )), l(β) = l(1 − e − f, f ) = Re ∝ I (e + f ). For any c ∈ I , ce = (ce)(e + f ) and c f = (c f )(e + f ). It follows that I (e + f ) = I e + I f . Because R f = (R f ) f ⊆ I f , we obtain that R f + I e = I (e + f ). So Sα = l(β). Moreover, for any c ∈ I and any r ∈ R, c(1 − e) = c f + c(1 − e − f ) ∈ R f + I (1 − e − f ) and r f + c(1 − e − f ) = [r f + c(1 − f )](1 − e) with r f + c(1 − f ) ∈ I . It follows that I (1 − e) = R f + I (1 − e − f ). So Sβ = l(α). Hence α ∼ β. Thus S is left morphic. It is similar to show that S is right morphic. For any n > 0, Mn (S) ∼ = Mn (R) ∝ Mn (I ). Since Mn (R) is again unit regular, as above S is strongly morphic. For an ideal I of a unit regular ring R, if R I is finitely generated then I is a ring direct summand of R. Hence, by Theorem 12, R ∝ RI is strongly morphic. But, in general, R ∝ RI need not be left morphic. Example 13. There exists a unit regular ring R and an ideal I of R such that R ∝
R I
is not left morphic.
∞ F be a direct product of rings where each F = Z and let R be the subring of Q generated Proof. Let Q = Πi=1 i i 2 L∞ by I := i=1 Fi and 1 Q . Then R is unit regular (Boolean indeed) and I is an ideal of R. Set S = R ∝ RI . We show α ∼ β where that S is not left morphic. L Consider α = (a, x) ∈ S where a = 1 F1 and x = 1 + I . Suppose thatL β = (b, y) ∈ S. Then ( i≥2 Fi ) ∝ RI = l(α) = Sβ = {(r b, r y + zb) : r ∈ R, z ∈ RI }. This gives that i≥2 Fi = Rb is finitely generated. This is a contradiction. So α is not left morphic in S.
Let R be a PID that is not a field. Two elements a, b ∈ R are equivalent if a = ub for some u ∈ U (R). The primes in R are those non-invertible elements divisible only by elements equivalent to themselves and invertible elements. L Let P be the set consisting of one element from each equivalence class of primes in R. If Q = Q(R), then Q = P R p∞ R is the minimal cogenerator in R-Mod, where R p∞ = { pan + R : a ∈ R, n ∈ Z} (see [1, p. 215]). In particular, Q ∼L = {Z p∞ : p is a prime number} is the minimal cogenerator in Z-Mod. Z
Theorem 14. Let R be a PID that is not a field with Q = Q(R) and let M be a bimodule over R such that r x = xr for all x ∈ M and all r ∈ R. Then (1) R ∝ M is morphic iff M = (2) R ∝ Q R is strongly morphic.
Q R.
Proof. (1) The implication “⇐” is [6, Theorem 13]. The proofs for “(1) ⇒” and for (2) are the same as those for [6, Theorem 14 and Corollary 15]. Theorem 14 does not hold for a UFD. The greatest common divisor of two elements a, b in a UFD R is denoted by gcd(a, b).
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Theorem 15. Let R be a UFD with the quotient field Q. Then R ∝
Q R
is morphic iff R is a PID.
Q Proof. If R is a PID then R ∝ Q R is morphic by [6, Theorem 13]. Now assume that T = R ∝ R is morphic. For any −1 nonzero elements s, t in R with gcd(s, t) = 1, (0, x) ∈ T is morphic where x = st + R ∈ Q R . Let s (t R) denote the −1 set {r ∈ R : sr ∈ t R}. Then, by [6, Lemma 12], there exists k ∈ R such that s (t R) = k R and, for any c, d ∈ R, ck ∈ d R implies that ct ∈ d(s R + t R). Since gcd(s, t) = 1, s −1 (t R) = t R. So t R = k R. Take c = 1 and d = t. Then ck ∈ d R, thus ct ∈ d(s R + t R). This shows that c ∈ s R + t R, i.e., R = s R + t R. So, for any nonzero elements s, t s t in R, R = gcd(s,t) R + gcd(s,t) R. This clearly shows that gcd(s, t)R = s R + t R. Hence R is a PID by a well-known result of PID’s.
Theorem 16. Let R be a PID and let a ∈ R be such that a = p1 · · · pk q1s1 · · · qlsl where pi , q j are distinct primes of R, k ≥ 1, l ≥ 0, and all s j ≥ 2. Let T = R/a R and M be a bimodule over T such that xt = t x for all t ∈ T and x ∈ M. The following are equivalent: (1) T ∝ M is morphic. (2) T ∝ M is strongly morphic. (3) M ∼ = R/b R where b is a divisor of p1 · · · pk . Proof. (3) ⇒ (2). Suppose (3) holds and suppose b = p1 · · · pm with m ≤ k. Then T ∝ M ∼ = ( p1RR ⊕ R R R R R R R R ∼ ( · · · ⊕ pk R ⊕ s1 ⊕ · · · ⊕ sl ) ∝ ( p1 R ⊕ · · · ⊕ pm R ⊕ (0) ⊕ · · · ⊕ (0)) = p1 R ∝ p1 R ) ⊕ · · · ⊕ ( pm R ∝ pm R ) ⊕ R
q1 R
R pm+1 R
⊕···⊕
R pk R
ql R R s ql l R
⊕
⊕···⊕
R . s ql l R
The last isomorphism is by [6, Lemma 4]. Each of the summands of
the latter is a local morphic ring with nilpotent Jacobson radical by [9, Theorem 9]. Thus, by [9, Theorem 17], T ∝ M is strongly morphic. (1) ⇒ (3). Write T = T1 ⊕· · ·⊕Tk ⊕S1 ⊕· · ·⊕Sl where Ti ∼ = sRj for i = 1, . . . , k and j = 1, . . . , l. = pRi R and S j ∼ qj R
Then M = M1 ⊕ · · · ⊕ Mk ⊕ N1 ⊕ · · · ⊕ Nl where Mi ∼ = S j M for i = 1, . . . , k and j = 1, . . . , l. = Ti M and N j ∼ Thus, by [6, Lemma 4], (T1 ∝ M1 ) ⊕ · · · ⊕ (Tk ∝ Mk ) ⊕ (S1 ∝ N1 ) ⊕ · · · ⊕ (Sl ∝ Nl ) ∼ = T ∝ M is morphic. So, for any 1 ≤ i ≤ k, Ti ∝ Mi is morphic, and hence either Mi = 0 or Mi ∼ = pRi R by Proposition 11. Moreover, for any 1 ≤ j ≤ l, S j ∝ N j is morphic. We next show that N j = 0. We let S = S j , N = N j , q = q j , s = s j . If N 6= 0, S∝N
take 0 6= x ∈ N . We claim l S (x) 6= 0. Since (0, x) is morphic in S ∝ N , (0, x) ∼ (r, y) where r ∈ S and y ∈ N . If l S (x) = 0, then r = 0. Thus, 0 ∝ Sx = (S ∝ N )(0, x) = l(0, y) = l S (y) ∝ N , showing that N = Sx. So N ∼ = S. Thus S ∝ S ∼ = S ∝ N is morphic. Hence S is regular by Theorem 9, a contradiction. So the claim holds and it follows that q s−1 N = 0 and that l S (x) = q n S with 1 ≤ n < s. Now direct computation shows that 0 ∝ Sx = (0, x)(S ∝ N ) = rl((0, x)) = r(l S (x) ∝ N ) = r(q n S ∝ N ) ⊇ q s−n S ∝ 0. This contradiction shows that N = 0. Hence M = M1 ⊕ · · · Mk ⊕ N1 ⊕ · · · ⊕ Nl ∼ = M1 ⊕ · · · ⊕ Mk . We can assume that Mi ∼ = R/ pi R for i = 1, . . . , k1 ≤ k and Mi = 0 for i = k1 + 1, . . . , k. Then M ∼ = R/b R with b = p1 · · · pk1 . This proves (3). Corollary 17 ([6]). Let d, n be positive integers with d > 1. Then Zdn ∝ Zd is morphic iff gcd(n, d) = 1 and d is a product of distinct primes. 4. An example of morphic chain rings A ring R is called right chain if the right ideals of R form a chain under usual inclusion, or equivalently the principal right ideals of R form a chain. A left and right chain ring is called a chain ring. Any one-sided chain ring is local. Examples of local left morphic rings with nilpotent Jacobson radical have been given in [9,10].
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Example 18. There is a commutative chain ring R satisfying: (1) R is strongly morphic; (2) J (R) is nil but not T -nilpotent. L Proof. Let F be a field and R = {Fαr : 0 ≤ r < 1} be a vector space over F with a basis {αr : 0 ≤ r < 1}, with multiplication defined by α , if r + s < 1; αr αs = r +s 0, if r + s ≥ 1. Then R is a commutative ring with J (R) = Σ {Fαr : 0 < r < 1}. Clearly J (R) is nil. Each principal ideal of R is generated by some αr . So the principal ideals form a chain. Hence R is a chain ring. If let ai = α2−i ∈ J (R) for i = 1, 2, . . ., then for any n > 0, a1 a2 · · · an = αr 6= 0 where r = 1 − 21n . So J (R) is not T -nilpotent and thus R is not perfect. For any 0 6= x ∈ R, it can easily be seen that x = uαr for some unit u of R and some 0 ≤ r < 1. Since R
αr ∼ α1−r , x is morphic in R by Lemma 1. It can be easily proved as [9, Theorem 17] that R is strongly morphic.
Remark 19. Following Camillo and Nicholson [3], a ring R is called left quasi-morphic if {Ra : a ∈ R} = {l(b) : b ∈ R}. A left and right quasi-morphic ring is called a quasi-morphic ring. These rings include regular rings and morphic rings. A question raised in [3] asks whether a quasi-morphic ring containing no infinite sets of orthogonal idempotents is necessarily perfect. The answer is “No” by Example 18. A ring R is said to satisfy (∗) if it contains a set of orthogonal idempotents e1 , . . . , en with 1 = e1 + · · · + en such that ei Rei is a chain ring for all i. Theorem 20. The following are equivalent for a ring R: (1) R is a morphic ring satisfying (∗). (2) R is a finite direct product of matrix rings over morphic chain rings. Proof. The proof of “(1) ⇒ (2)” is the same as that of [9, Theorem 29]. As in the proof of [9, Theorem 17], it can be shown that every morphic chain ring is strongly morphic. Thus, the implication “(2) ⇒ (1)” follows. To give our concluding example, we need a lemma. Lemma 21. Let R be a ring and e be a central idempotent. Then the element a ∈ R is left morphic in R iff ea is left morphic in e Re and (1 − e)a is left morphic in (1 − e)R(1 − e). R
eR
Proof. This is because a ∼ b iff ea ∼ eb and (1 − e)a
(1−e)R
∼ (1 − e)b.
Note that the assumption that e is central can not be removed from Lemma 21 by the next remark (see [9, Lemma 14]). Remark 22. Let R, σ, b be as in Example 3. Let S = R[x; σ ]/(x 2 ) and let e = b. Then e is an idempotent of S, eSe
e(bx)e = 0, and (1 − e)(bx)(1 − e) = 0. Hence we have that e(bx)e ∼ 0 and (1 − e)bx(1 − e) is not left morphic in S by Example 3.
(1−e)S(1−e)
∼
0. But bx
Example 23. If R is a strongly regular ring and if C2 is the cyclic group of order 2, then the group ring RC2 is morphic. Proof. Write C2 = {1, g}. Let S = RC2 and let α = a + bg ∈ S. We want to show that α is left morphic in S. Since −1 R is strongly regular, write a = u 1 e1 where u 1 ∈ U (R) and e1 is a central idempotent. Then u −1 1 α = e1 + u 1 bg. By Lemma 1, we may assume that a = e1 is an idempotent. So α = e1 + bg. Because e1 α = e1 + e1 bg ∈ e1 S = (e1 R)C2 and (1 − e1 )α = (1 − e1 )bg ∈ (1 − e1 )S = ((1 − e1 )R)C2 and because of Lemma 21, α is left morphic in S iff e1 + e1 bg is left morphic in (e1 R)C2 and (1 − e1 )bg is left morphic in ((1 − e1 )R)C2 . Since (1 − e1 )bg is a strongly regular element in ((1 − e1 )R)C2 (i.e., a product of a unit and a central idempotent), (1 − e1 )bg is left morphic in ((1 − e1 )R)C2 by Lemma 1. Thus we only need to show that e1 + e1 bg is left morphic in (e1 R)C2 . Therefore, without
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loss of generality, we can assume that α = 1 + bg. Write b = u 2 e2 where u 2 ∈ U (R) and e2 is a central idempotent of R. Because (1 − e2 )α = 1 − e2 is left morphic in (1 − e2 )S, by Lemma 21 it suffices to show that e2 α = e2 + (u 2 e2 )g is left morphic in e2 S = (e2 R)C2 . Therefore, we may further assume that α = 1 + vg where v ∈ U (R). Now write 1 − v 2 = eu where u ∈ U (R) and e is a central idempotent. Let β = −(1 − e)v + (1 − e)g = (1 − e)(g − v) ∈ S. Noting that (1 − e)v 2 = 1 − e, one obtains Sα = {(a + bg)(1 + vg) : a, b ∈ R} = {(a + bv) + (av + b)g : a, b ∈ R}, l(α) = {a + bg : a + bv = av + b = 0} = {a + bg : a = −bv, be = 0} = R(1 − e)(g − v), Sβ = = = = =
{(a + bg)(1 − e)(g − v) : a, b ∈ R} {(1 − e)(−av + b) + (1 − e)(a − bv)g : a, b ∈ R} {−(1 − e)(a − bv)v + (1 − e)(a − bv)g : a, b ∈ R} {(1 − e)(a − bv)(g − v) : a, b ∈ R} R(1 − e)(g − v),
l(β) = {a + bg : (a + bg)(1 − e)(g − v) = 0} = {a + bg : (1 − e)(a − bv) = 0}. Thus, Sα = l(β), and it is easy to see that Sβ ⊆ l(β). For any r +sg ∈ l(β) with r, s ∈ R, r +sg = (a+bv)+(av+b)g where a = r − (s − r v)u −1 v and b = (s − r v)u −1 . So r + sg ∈ Sα. Therefore, α ∼ β. Acknowledgments Part of the work was done during a visit, sponsored by NCTS of Taipei, by the second author to the National Taiwan University. He gratefully acknowledges the financial support from NCTS and kind hospitality from the institute. The research of the first author was supported by NSC of Taiwan, and that of the second author by NSERC of Canada. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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