Minimal Non-Commutative n-insertive Rings

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1 Acta Mathematica Sinica, English Series Jan., 003, Vol.19, No.1, pp Minimal Non-Commutative n-insertive Rings Li Qiong XU Wei Min XUE Department of Mathematics, Fujian Normal University, Fuzhou , P. R. China Abstract Let n be an integer with n > 1. If p is the smallest prime factor of n, weprovethat a minimal non-commutative n-insertive ring contains p 4 elements and these rings have five (p +4) isomorphic classes for p =(p ). Keywords Finite rings, n-insertive rings, n-commutative rings MR(000) Subject Classification 16P10, 16U80 All rings in this paper are finite associative rings with identity. If n Z is an integer, we call a ring R an n-insertive (n-commutative) ring in case for a, b R and ab = n then arb = nr for each r R (then ba = n). The structure of minimal non-commutative zero-insertive (zerocommutative) rings was given in our recent paper [1]. In this sequel we study minimal noncommutative n-insertive (n-commutative) rings, where R is called a minimal non-commutative n-insertive (n-commutative) ring if R has the smallest order R among the non-commutative n-insertive (n-commutative) rings. If n > 1andp is the smallest prime factor of n, ourmain theorem states that a minimal non-commutative n-insertive ring contains p 4 elements and these rings have five (p + 4) isomorphic classes for p =(p ). In particular, a minimal noncommutative p-insertive ring contains p 4 elements and these rings have five (p + 4) isomorphic classes for p =(p ). Finally, we show that there exists a unique minimal non-commutative n-commutative ring, which contains 8 (16) elements if n is odd (even). Let p be a fixed prime number. The field with p m elements is denoted by GF (p m ), and Z t = Z/tZ = {0, 1,...,t 1}. Throughout this paper, R is a finite associative ring with identity. We denote by X(R), R, andj the characteristic, the order, and the Jacobson radical of R, respectively. It is well known that if R = p α1 1 pαs s (p 1,...,p s are distinct prime numbers), then R = R 1 R s,whereeach R i = p αi i. If R = R 1 R s one shows that R is n-insertive (n-commutative) if and only if each R i is n-insertive (n-commutative). Hence minimal non-commutative n-insertive (n-commutative) rings are indecomposable, so they have prime power orders. It was proved in [] that if R does not contain a cube factor then R is Received February 5, 000, Accepted June 1, 000 This research is supported by the National Natural Science Foundation of China, and the Scientific Research Foundation for Bai-Qian-Wan Project, Fujian Province of China

2 14 Xu L. Q. and Xue W. M. commutative; and if R is non-commutative and R = P 3 then R ( ) Zp Z p = 0 Z p. This ring is not zero-insertive by [1, Lemma 3]. In fact, it is not n-insertive for each integer n. However, we need a lemma first. We know that every finite ring is 1-commutative, i.e., every one-sided invertible element is two-sided invertible. But a finite ring need not be 1-insertive, unless it is commutative. Lemma 1 The following are equivalent for the finite ring R: (1) R is commutative; () R is 1-insertive; (3) Every invertible element of R lies in the center; (4) R is n-insertive for each integer n with n 1 R; (5) R is n-insertive for some integer n with n 1 R. Proof One checks that (1) (3) () (4) (5) (). It remains to prove (3) (1). If r is a nilpotent element of R, then1 r is an invertible element, and 1 r lies in the center by (3). Hence r lies in the center, too. Now the commutativity of R follows from an old theorem of Herstein [3]. Note that the equivalence () (3) (4) (5) of Lemma 1 holds for any ring which need not be finite. But there does exist an infinite 1-insertive ring which is not commutative. In fact, we let Z t [x, y] denote the polynomial ring over Z t with non-commutative indeterminants x and y. Then the non-commutative infinite ring Z [x, y] is 1-insertive, since the identity 1 is the only invertible element of Z [x, y]. ( ) Zp Z Lemma The ring p 0 Z p is not n-insertive for each integer n. ( ) Zp Z P 0 Z p Proof We already know that R = is not zero-insertive by [1, Lemma 3]. Since X(R) = p, R is not n-insertive if p n. Now letting p n, then n 1 R. Since R is not commutative, it is not n-insertive by Lemma 1. Next we give a non-commutative zero-insertive (p-insertive) ring with R = p 4. The ideal of Z t [x, y] generated by {f 1,...,f n } Z t [x, y] is denoted by f 1,...,f n. Example 3 Let R = Z p [x, y]/ x 3,y,yx,x xy, x p. Then R is a non-commutative ring with R = p 4,X(R) =p,andpx = py = 0, where we identify x = x and y =ȳ. Toshow R is p-insertive we let a = a 0 + a 1 x + a y and b = b 0 + b 1 x + b y R such that p = ab = a 0 b 0 + pa 1 (b 1 + b )+(a 0 b 1 + a 1 b 0 )x +(a 0 b + a b 0 )y. Then a 0 b 0 + pa 1 (b 1 + b ) p (mod p ), (1) a 0 b 1 + a 1 b 0 0(modp), () a 0 b + a b 0 0(modp). (3) Now axb = p(a 0 b 1 + a 1 b 0 + a 0 b )+a 0 b 0 x = pa 0 b, where the last equality holds by () and (1). If p a 0 then axb = pa 0 b =0=px. Ifp a 0 then p b 0 by (1), and so p b by (3). We also have

3 Minimal Non-Commutative n-insertive Rings 143 axb = pa 0 b =0=px. Similarly, ayb =0=py. Therefore, for every r = c 0 + c 1 x + c y R we must have arb = pr. Consequently,R is a p-insertive ring. Since J = p 3 and J 3 =0,R is a zero-insertive ring by the following lemma: Lemma 4 Let R be a ring with R = p 4.If J = p 3 and J 3 =0,thenR is zero-insertive. Proof Since R/J = p, R/J = Z p and R is a local ring. Let a, b R such that ab =0. If a/ J or b/ J, thena or b is invertible and then b =0ora =0. Inthiscase,arb =0for each r R. Now suppose a, b J and r = i + j R, where0 i p 1andj J. Then arb = aib + ajb = ajb =0,sinceJ 3 =0. Because of Lemma, we consider non-commutative rings of order p 4. The above lemma gives a class of rings which are zero-insertive. On the other hand, the next lemma gives some rings which are neither zero-insertive (zero-commutative) nor p-insertive (p-commutative). Lemma 5 Let R be a non-commutative ring. If R = p 4 and X(R) =p = J, then R is neither p-insertive (p-commutative) nor zero-insertive (zero-commutative). Proof Using the results in [4, p. 115] we have t, e R such that (1) t =0=pt = pe, e = e, te =0andet = t or, () t =0=pt = pe, e = e, te = t and et =0. If (1) holds, we have (p t + e)(1 + t e) =p but (p t + e)t(1 + t e) =t pt, andsor is not p-insertive. In this case, (1 e)(p + t) =p (p + t)(1 e), so R is not p-commutative, either. If () holds, we have (1 + t e)(p t + e) =p but (1 + t e)e(p t + e) =t pe, andso R is not p-insertive. In this case, (p+t)(1 e) =p (1 e)(p+t), and R is not p-commutative. From (1) or () we see that e is a non-central idempotent element, so R is not zero-insertive by [1, Lemma 3]. Clearly, R is not zero-commutative in either case. In [1, Theorem 8], we proved that every minimal non-commutative zero-insertive ring contains 16 elements, and up to an isomorphism there are five such rings as follows: R 1 = Z [x, y]/ x 3,y 3,yx,x xy, y xy, R = Z 4 [x, y]/ x 3,y 3,yx,x xy, x,y, {( ) a b R 3 = 0 a a, b GF (4)}, R 4 = Z [x, y]/ x 3,y,yx,x xy, R 5 = Z 4 [x, y]/ x 3,y,yx,x xy, x, where X(R 1 )=X(R 3 )=X(R 4 )=,X(R )=X(R 5 )=4. HenceR 1,R 3 and R 4 are also -insertive rings. By Example 3, R 5 is -insertive. Similarly, R is also -insertive. One notes that every zero-commutative ring is zero-insertive, but the converse is false. e.g., the ring R 3 is zero-commutative, but none of the the zero-insertive rings R 1,R,R 4 and R 5 is zero-commutative. In the following notations, we suppose the prime number p :

4 144 Xu L. Q. and Xue W. M. Let {[ a b S 0 = 0 a p ] a, b GF (p )}, S i = Z p [x, y]/ x 3,y 3,yx,x xy, y (i 1)xy,i=1,...,p, S p+1 = Z p [x, y]/ x,y,xy+ yx, S p+ = Z p [x, y]/ x,y,xy+ yx, xy p, S p++j = Z p [x, y]/ x 3,y 3,yx,x xy, x p, y (j 1)xy,j =1,...,p, ( ) p +1 S p+3 = Z p / x 3,y 3,yx,x xy, x p, y xy, where Z p and is not a square. It is easy to see that each of the rings S 0,S 1,...,and S p+3 is a non-commutative ring with order p 4. These rings are not isomorphic to each other by [4]. It is easy to see that S 0 is zero-commutative, so it is also zero-insertive. The rings S 1,..., and S p+3 are also zero-insertive by Lemma 4. Since the zero-insertive rings S 0,S 1,...,and S p+1 have characteristic p, theyarep-insertive rings, too. According to Example 3, S p+3 is p- insertive. Similarly, S p+,...,ands p+3 are also p-insertive. Hence each of the rings S 0,S 1,..., and S p+3 is a non-commutative zero-insertive (p-insertive) ring with order p 4. Now we consider minimal non-commutative n-insertive (n-commutative) rings for an arbitrary integer n. Lemma 6 If m is an integer with m 1 R, thenr is n-insertive (n-commutative) if and only if R is nm-insertive (nm-commutative). Proof Straightforward. Lemma 7 (1) If p, every R {S 0,S 1,...,S p+3 } is n-insertive for each integer n with p n. () Every R {R 1,R,R 4,R 5 } is n-insertive for each even integer n. Proof (1) We know that R is zero-insertive (p-insertive) and X(R) =p or p. If p n, n =0 in R, thenr is n-insertive in this case. Let p n and n = pm. Then p m and m 1 R, so R is pm-insertive by Lemma 6. Similarly, we can prove (). According to Lemma 1, every finite 1-insertive ring is commutative. On the other hand, the structure of the minimal non-commutative zero-insertive rings was given in [1, Theorem 8]. Hence we assume n > 1 in the following main theorem, which gives the structure of minimal non-commutative n-insertive rings: Theorem 8 Let n be an integer with n > 1. Ifp is the smallest prime factor of n, then a minimal non-commutative n-insertive ring contains p 4 elements, and up to isomophism there are : (1) Five such rings for p =,namelyr 1,R,R 3,R 4 and R 5 ; and () p +4 such rings for p,namelys 0,S 1,..., and S p+3.

5 Minimal Non-Commutative n-insertive Rings 145 Proof We prove only (), since (1) is similar. Let R be a minimal non-commutative n-insertive ring. Then there exists a prime number q such that q t = R p 4 by Lemma 7. Since R is non-commutative, t 4 by Lemma, and so q p. Ifq<p,thenq n, son 1 R and R is commutative by Lemma 1, a contradiction. Hence q = p, andt = 4, i.e., R = p 4. Therefore by Lemma 7 again, the rings S 0,S 1,...,andS p+3 are minimal non-commutative n-insertive rings. Since R is non-commutative, X(R) =p or p by [4, Proposition 1]. Let n = p i m,where i 1andp m. Sincem 1 R, R is p i -insertive by Lemma 6. Hence R is either p-insertive or zero-insertive, since X(R) =p or p. (i) X(R) =p. If J = p 3, by [4, 3.1] we have R = S 1,..., or S p+1. If J = p, by [4, 3.] we have R = S 0. By [4, 3.3, 3.4] we know J p or 0. (ii) X(R) =p. By [4, Lemma 3] and Lemma 5 we have J = p 3. Then by [4, pp ] we know that R = S p+,...,ors p+3. Corollary 9 A minimal non-commutative p-insertive ring contains p 4 elements, and up to isomorphism there are : (1) Five such rings for p =,namelyr 1,R,R 3,R 4 and R 5 ; and () p +4 such rings for p,namelys 0,S 1,...,andS p+3. We call the ring R a pz-insertive ring in case R is pm-insertive for each integer m. Corollary 10 A minimal non-commutative pz-insertive ring contains p 4 elements, and up to isomorphism there are : (1) Five such rings for p =,namelyr 1,R,R 3,R 4 and R 5 ; and () p +4 such rings for p,namelys 0,S 1,...,andS p+3. Proof By Lemma 7 and Corollary 9. Finally, we give the structure of minimal non-commutative n-commutative rings. It was shown in [1, Corollary 9] that the ring R 3 is the unique (up to isomorphism) minimal noncommutative zero-commutative ring. [ ] Z Z Proposition 11 (1) If n is an odd integer, then up to isomorphism, the ring 0 Z is the unique minimal non-commutative n-commutative ring. () If n is an even integer, then up to isomorphism, the ring R 3 is the unique minimal non-commutative n-commutative ring. Proof Since every finite ring R is 1-commutative, by Lemma 6, R is also m-commutative for each integer m with m 1 R. [ Z Z (1) Each odd integer n is invertible in 0 Z ], which is the unique minimal non-commutative ring, so it is also [ the ] unique minimal non-commutative n-commutative ring. Z Z () Since 0 Z has characteristic and is not zero-commutative, it is not n-commutative, either. Now R 3 is zero-commutative and X(R 3 )=,sor 3 is n-commutative. Since R 3 = 16, R 3 is a minimal non-commutative n-commutative ring. Suppose R is also a minimal noncommutative n-commutative ring. Then R = 16. We need to prove R = R 3. By [4, Proposition 1], X(R) =or4. IfX(R) =or4 n, R is also zero-commutative, so R = R 3 by [1, Corollary

6 146 Xu L. Q. and Xue W. M. 9]. So let X(R) =4and4 n. Write n =m with m odd. Since m 1 R, R is -commutative by Lemma 6. By [4, Lemma 3] and Lemma 5, J = 8. Hence, it follows from [4, pp ] that R = R or R 5. This is a contradiction, since either of R and R 5 is not -commutative. If S Z, we call the ring R an S-commutative ring in case R is n-commutative for each integer n S. Let(1+Z) be the set of all odd integers. [ ] Z Z Corollary 1 (1) Up to isomorphism, the ring 0 Z is the unique minimal non-commutative (1 + Z)-commutative ring. () Up to isomorphism, the ring R 3 is the unique minimal non-commutative Z-commutative (Z-commutative) ring. Proof Since X(R 3 )=,R 3 is n-commutative for each odd integer n. Hence the parenthetical version of () also holds. References [1] Xu, L. Q., Xue, W. M.: Structure of minimal non-commutative zero-insertive rings. Math. J. Okayama Univ. 40, (1998) [] Eldridge, K. E.: Orders for finite noncommutative rings with unity. Amer. Math. Monthly, 75, (1968) [3] Herstein, I. N.: A note on rings with central nilpotent elements. Proc.A.M.S., 5, 60 (1954) [4] Derr, J. B., Orr, G. F., Peck, P. S.: Noncommutative rings of order p 4. J. Pure Appl. Algebra, 97, (1994)