Neumann Near-rings and Neumann Centers

We study near-rings N with the property (∗) : xS = Sx for each x ∈ N and each n-subset S of N . By localizing (∗), we obtain center-like subsets, which we study for the case of rings. In response to a question of B.H. Neumann, we defined a Qn–ring to be a ring R with the property that xS = Sx for all x ∈ R and all n-subsets S of R. In [3] we investigated basic properties of Qn–rings and established sufficient conditions for Qn–rings to be commutative. In this paper, we extend our study to near-rings, and we localize our definition of Qn-ring to obtain center-like subsets. 1. Neumann Near-rings In this section N will denote a zero-symmetric left near-ring. Let U be the set of nilpotent elements of N and Z = Z(N) the multiplicative center of N . For any subset T of N , let A`(T ), Ar(T ) and A(T ) be the left, right, and two-sided annihilators of T ; and let |T | be the cardinal number of T . For a fixed integer n ≥ 1, we say that N is a Qn-near-ring if xS = Sx for all x ∈ N and all n− subsets S of N . (1) We call N a Neumann near-ring if it is a Qn-near-ring for some n. Clearly N is a Qn-near-ring if |N | < n; hence to deduce any real consequences of (1), we must assume |N | ≥ n. Our first lemma, the proof of which is trivial, lists some elementary properties of Qn-near-rings. Lemma 1.1. Let N be a Qn-near-ring with |N | ≥ n. Then N has the following properties: (a) xM =Mx for all x ∈ N and all subsets M of N with |M | ≥ n. (b) For each x, y ∈ N , there exist w, z ∈ N such that xy = wx = yz. (c) All idempotents of N are in Z. (d) If x is a left-invertible or right-invertible element of N , then x is invertible. (e) If |N | > n and x is a left or right zero divisor in N , then x is a two-sided zero divisor. Lemma 1.2. Let N be a Qn-near-ring with 1, such that |N | > n. Then every invertible element is in Z. In particular, if N is a near-field, then N is commutative. 1991 Mathematics Subject Classification Primary 16Y30. ∗Supported by the Natural Sciences and Engineering Research Council of Canada, Grant 3961. 32 HOWARD E. BELL AND ABRAHAM A. KLEIN Proof. Suppose x is invertible and x 6∈ Z. For y such that xy 6= yx, choose an n-subset S which contains y; and note that since xS = Sx, there exists a unique z 6= y such that xy = zx. Taking T to be an n-subset containing y but not z, we have xT 6= Tx – a contradiction. Our next theorem is particularly useful. Theorem 1.3. Every infinite Neumann near-ring is multiplicatively commutative. Proof. The proof employs in a crucial way the fact that in left near-rings, right annihilators (but not necessarily left annihilators) are additive subgroups. Suppose we have a counterexample N , which is a Qn-near-ring. Then n > 1; and we may assume N is not a Qn−1-near-ring, in which case there exist x ∈ N and an n− 1-subset H of N such that xH 6= Hx. We show first that Hx ⊆ xH. Suppose not, and take h ∈ H such that hx 6∈ xH. For any a ∈ N\H, x(H ∪ {a}) = (H ∪ {a})x, so hx = xa. For fixed b ∈ N\H, xa = xb, hence a− b ∈ Ar(x); consequently, N\H ⊆ Ar(x) + b. Also, if c ∈ Ar(x), hx = xb = x(c + b), so that c + b ∈ N\H; hence Ar(x) + b ⊆ N\H. Therefore N\H = Ar(x)+b, so H = N\(Ar(x)+b) and |H| = |N\Ar(x)|. Note that Ar(x) is a proper subgroup of (N,+), for otherwise xN = Nx = {0} and x ∈ Z. It follows that |N\Ar(x)| ≥ |Ar(x)|, so that Ar(x) and N\Ar(x) are both finite. But this contradicts the hypothesis that N is infinite. We now have xH 6⊆ Hx, so there exists ĥ ∈ H such that xĥ 6∈ Hx. Since x(H ∪ {a}) = (H ∪ {a})x for all a ∈ N\H, we get xĥ = ax for all a ∈ N\H ; (2) hence (N\H)x ⊆ xH. Having shown above thatHx ⊆ xH, we now have Nx ⊆ xH. But Nx = xN , so xN ⊆ xH and hence xN is finite. The map y 7→ xy is an additive epimorphism from N onto xN , so Ar(x) is infinite. Thus, {0} = xAr(x) = Ar(x)x, so that Ar(x) ⊆ A`(x) and A`(x) is infinite. It follows that there exists a ∈ (N\H) ∩ A`(x), and by (2) we get xĥ = 0. Therefore ĥ ∈ Ar(x) ⊆ A`(x), so that ĥx = xĥ, contradicting our choice of ĥ. Our proof is now complete. Further investigation of the finite case yields an extension of [3, Theorem 4.2]. Theorem 1.4. If N is a finite Qn-near-ring with |N | > 2n − 2, then N is commutative. Proof. Since Q1-near-rings are the commutative near-rings, we may assume n ≥ 2. We assume that N is a noncommutative Qn-near-ring and show that |N | ≤ 2n− 2. Since n− 1 < 2n− 2, we may assume that |N | ≥ n. Let x, y ∈ N with xy 6= yx. If H is any (n − 1)-subset of N which excludes y, the condition x(H ∪ {y}) = (H ∪ {y})x gives z ∈ H such that yx = xz. Thus, defining K to be {z ∈ N | yx = xz}, we have that every (n− 1)-subset of N either contains y or intersects K; and it follows that |N\K| ≤ n − 1. Arguing as in the previous proof, we see that for any z ∈ K, K = Ar(x) + z; hence |K| = |Ar(x)|. Since x 6∈ Z, we cannot have xN = Nx = {0}; hence Ar(x) is a proper subgroup of (N,+) by Lemma 1.1(a), and |Ar(x)| = |N | m for some integer m ≥ 2. It follows that |N | = |K|+ |N\K| ≤ |N | m + n− 1, and hence that |N | ≤ m m−1 (n− 1) ≤ 2(n− 1) as required. NEUMANN NEAR-RINGS AND NEUMANN CENTERS 33 We proceed now to reduced near-rings – i.e. those for which U = {0}. Lemma 1.5. Let N be a Qn-near-ring with no nonzero divisors of zero. If |N | > n, then N is a commutative ring. Proof. If N is infinite, it is commutative, and hence distributive, by Theorem 1.3. Thus N is additively commutative; and since A(N) = {0}, this forces (N,+) to be commutative, so that N is a ring. We now assume that N is finite, in which case each nonzero element has a nonzero idempotent power. Since idempotents are central by Lemma 1.1(c), we see that N has 1 and that 1 is the unique nonzero idempotent. It follows that N is a near-field, hence is commutative by Lemma 1.2; and since it is well known that near-fields have commutative addition, N is a ring. Theorem 1.6. Let N be a reduced near-ring with 1. If N is a Qn-near-ring for some n ≤ 8, then N is a commutative ring. Proof. We may assume that n = 8. By [1, Lemma 3], N is a subdirect product of a family {Nα|α ∈ Λ} of Qn-near-rings with 1, each of which has no nonzero divisors of zero. Let Ñ be a typical Nα. If |Ñ | > 8, then Ñ is a commutative ring by Lemma 1.5. Now suppose that |Ñ | ≤ 8. Since Ñ has 1 and has no zero divisors, idempotents are central by [1, Lemma 2(c)]; and we can argue as above that Ñ is a near-field. But it is known that any near-field with fewer than 9 elements is a field (cf. [5, p.257]); hence all Nα are commutative rings, and so is N . Our final results in this section deal with distributively-generated (d-g) nearrings. Lemma 1.7. If N is a d-g Qn-near-ring with |N | ≥ n, then U is an ideal of N . Proof. Theorem 1.3 gives U ⊆ Z if N is infinite; hence U is an ideal by Lemma 1 of [2]. Therefore, we assume that N is finite, in which case each element has a central idempotent power. For each positive integer m, let Um = {x ∈ U | x = 0}. It is immediate from Lemma 1.1(b) that NUm ⊆ Um and UmN ⊆ Um for each m. To show that U is an additive subgroup, we show by induction on m that U − Um = {u− v | u ∈ U, v ∈ Um} ⊆ U for each m. Accordingly, let u ∈ U and v ∈ Um, and let k > 1 be an integer for which (u− v) ∈ Z. Then (u− v) = (u− v)u− (u− v)v = u(u− v) − v(u− v) = u(u− v) − (vu− v)(u− v)k−1 = u(u− v) + (vu− v)(−(u− v)k−1) = u(u− v) + w(vu− v) for some w ∈ N ; and taking y ∈ N such that wvu = uy, we obtain (u− v) = u((u− v) + y)− wv . (3) Now if m = 2 – i.e. if v = 0, (3) yields (u − v) ∈ U and hence u − v ∈ U ; and since m > 2 and v ∈ Um implies v ∈ Um−1, (3) shows that the inductive assumption U −Um−1 ⊆ U implies U −Um ⊆ U . Thus U −Um ⊆ U for all m, and U is an additive subgroup of N . 34 HOWARD E. BELL AND ABRAHAM A. KLEIN We complete the proof by showing that U is a normal subgroup of (N,+). Let x ∈ N , u ∈ Um and k > 1 such that (x + u − x) ∈ Z. Letting y = x + u − x, we have y = yx+ yu− yx = yx+ uy − yx = yx+ (ux+ u − ux)yk−1 − yx = = yx+ w(ux+ u − ux)− yx for some w ∈ N ; hence y = (yx+ wux) + wu − (yx+ wux) . This equation is all we need to obtain our result by an induction on m. Theorem 1.8. Let N be a d-g Qn-near-ring with |N | ≥ n. Then the commutator ideal C(N) is a nil ideal. Proof. If N is infinite, then C(N) = {0} by Theorem 1.3; hence we assume N is finite. Then N = N/U is reduced, hence by Lemma 2 of [2], it has the x = x property. Therefore N is commutative by Theorem 2 of [1], so that C(N) ⊆ U . Theorem 1.9. Let N be a d-g Qn-near-ring with 1, such that |N | > n. Then N is a commutative ring. Proof. Since (−1) = 1, we have −1 ∈ Z by Lemma 1.2; and it follows that (N,+) is abelian. By a well-known theorem of Fröhlich [4], we get that N is a ring; and commutativity follows by Theorem 3.1 of [3]. Remark 1.10. Under the hypotheses of Theorem 1.8 we cannot prove commutativity, even if N is a ring. See Example 4.4 of [3]. 2. Neumann Centers for Rings In this section we deal with rings R, although some of the proofs would work for near-rings. We shall keep the notation from Section 1 for annihilators; and we shall denote the center of R by Z(R) or Z. For each x ∈ R, C(x) will be the centralizer of x in R. For each ring R and each positive integer n, we define the n-th Neumann center Tn(R) by Tn(R) = {x ∈ R | xS = Sx for all n− subsets S of R} ; and we define T∞(R) = {x ∈ R | xS = Sx for all infinite subsets S of R} . Clearly Z(R) = T1(R) ⊆ Tn(R) ⊆ Tn+1(R) ⊆ T∞(R) for all n and all R; hence the Tn(R) and T∞(R) are in some sense center-like. We are particularly interested in finding sufficient conditions for these sets to coincide with Z(R). Theorem 2.1. For any infinite ring R, T∞(R) = Z(R). Proof. Suppose x ∈ T∞(R)\Z(R), and take y ∈ R such that xy 6= yx. If Q = {z ∈ R | xy 6= zx}, then xy ∈ xQ\Qx; therefore Q must be finite and S = R\Q = {z ∈ R | xy = zx} contains all except finitely many elements of R. Therefore, xS = Sx = {xy}, so S ⊆ C(x) and hence R\C(x) ⊆ Q. It is now clear that C(x) NEUMANN NEAR-RINGS AND NEUMANN CENTERS 35 is an additive subgroup of R for which |R\C(x)| < |C(x)|; hence C(x) cannot be proper and x ∈ Z(R) – a contradiction. Corollary 2.2. If R is any infinite ring, then Tn(R) = Z(R) for all positive integers n. Exploring further the finite case, we have the following lemma, the proof of which is trivial. Lemma 2.3. Let R be a finite ring with |R| ≥ n, and let x ∈ Tn(R). Then (i) xR = Rx and |A`(x)| = |Ar(x)|; (ii) If x is idempotent, then x ∈ Z; (iii) If x 6∈ Z, then R\(A`(x) ∪ C(x)) and R\(Ar(x) ∪ C(x)) are non-empty. Using this lemma to make obvious modifications to the proofs of Theorems 4.2, 4.3, and 5.1 of [3], we obtain the following three results. Theorem 2.4. Let n ≥ 4, and let R be a ring. If |R| > 2n− 2, or if n is even and |R| > 2n− 4, then Tn(R) = Z(R). Theorem 2.5. Let n ≥ 4, and let R be a ring with |R| > 32 (n − 1). If |R| is odd or if (R,+) is not the union of three proper subgroups, then Tn(R) = Z(R). Theorem 2.6. If n ≤ 4, then Tn(R) = Z(R) for all rings R. Example 2.7. Let R be the ring of upper-triangular 2 × 2 matrices over GF (2), with basis {1, x, y}, where x = e12 and y = e22. Note that xy = x, yx = x = 0, and y = y. LetK = {1, y, x+y, 1+x} and L = {1, 1+x, 1+y, 1+x+y}. It is easily verified that xK = {x}, x(R\K) = {0}, Lx = {x} and (R\L)x = {0}. If S is any 5subset of R, S must intersect each of K,R\K,L, and R\L; hence xS = Sx = {0, x} and x ∈ T5(R)\Z(R). In fact, T5(R) = {0, 1, x}, for y, x + y, 1 + y, and 1 + x + y are all non-central idempotents (and hence not in T5(R) by Lemma 2.3 (ii)) and (1 + x){0, x, y, x+ y, 1 + y} 6 = {0, x, y, x+ y, 1 + y}(1 + x). This example shows that the integer 4 in Theorem 2.6 cannot be increased, and it also shows that Tn(R) need not be a subring of R. Of course, Tn(R) can be a subring different from the center, for there exist noncommutative Qn-rings – i.e. noncommutative R for which Tn(R) = R (cf. Example 4.4 of [3]). In fact, it is possible for Tn(R) to be a subring different from R and Z(R). Example 2.8. Let R be the algebra over GF (2) with basis {z, a, b} and multiplication defined by