ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.284962 J. Algebra Comb. Discrete Appl. 4(2) • 189–196 Received: 12 June 2015 Accepted: 25 February 2016 Journal of Algebra Combinatorics Discrete Structures and Applications Commuting probability for subrings and quotient rings Research Article Stephen M. Buckley, Desmond MacHale Abstract: We prove that the commuting probability of a finite ring is no larger than the commuting probabilities of its subrings and quotients, and characterize when equality occurs in such a comparison. 2010 MSC: 05E15 Keywords: Commuting probability, Subring, Quotient ring 1. Introduction Suppose R is a finite (possibly nonunital) ring. The commuting probability of R is Pr(R) := |{(x,y) ∈ R×R : xy = yx}| |R|2 , where | · | denotes cardinality. There has been much written on the commuting probability of a finite group: see for instance [5], [7], [8], [10], [4], and [6]. The commuting probability of a ring has been discussed in [9], [3], [2], and [1]. Work on the commuting probability of rings R has so far mainly concentrated on the possible values of Pr(R). However, it was shown in [9] that Pr(R) is no larger than Pr(S) whenever S is a subring of R. Our first result gives a new proof of this result, one that allows us to characterize when equality occurs. Theorem 1.1. Suppose S is a subring of a finite ring R. Then Pr(R) ≤ Pr(S). Equality holds if and only if [x,S] = [x,R] for all x ∈ R. Our second result is similar, but involves a comparison with quotient rings. Stephen M. Buckley (Corresponding Author); Department of Mathematics and Statistics, Maynooth University, Maynooth, Co. Kildare, Ireland (email: stephen.buckley@maths.nuim.ie). Desmond MacHale; School of Mathematical Sciences, University College Cork, Cork, Ireland (email: d.machale@ucc.ie). 189 S. M. Buckley, D. MacHale / J. Algebra Comb. Discrete Appl. 4(2) (2017) 189–196 Proposition 1.2. Suppose I is an ideal in a finite ring R. Then Pr(R) ≤ Pr(R/I). Equality holds if and only if [x,R]∩ I = {0} for all x ∈ R. In the above results, [x,S] = {[x,s] : s ∈ S}, and [x,s] = xs−sx is the commutator of x and s. After some preliminaries in Section 2, we prove generalizations of the above results in Section 3. In Section 4, we give various counterexamples which rule out seemingly plausible variants of the above conditions for equality. 2. Preliminaries Given a set S of finite cardinality, and a function f : S → R, we write |S| for the cardinality of S, and define the arithmetic mean ——— ∑ x∈S f(x) = 1 |S| ∑ x∈S f(x) . In this paper, a ring is not necessarily unital. Our results do not use associativity either and, to emphasize this, we sometimes talk of PN rings (where “PN” stands for “possibly nonassociative”). In the absence of the “PN” qualifier, rings and algebras are assumed to be associative. However, an ideal in a PN ring is not assumed to be associative. Suppose R is a PN ring and x ∈ R. The annihilator Ann(R), center Z(R), and centralizer CR(x) are defined by Ann(R) := {u ∈ R : uv = vu = 0 for all v ∈ R} , Z(R) := {u ∈ R : [u,v] = 0 for all v ∈ R} , CR(x) := {u ∈ R : [u,x] = 0} . If A and B are finite subsets of a PN ring R, then we define the commuting probability for the triple (A,B;R) to be PrR(A,B) := |{(x,y) ∈ A×B : xy = yx}| |A| · |B| , where juxtaposition indicates multiplication in R. We also write PrR(A) := PrR(A,A) and Pr(R) := PrR(R). If x,y are elements of a PN ring R, and S is an additive subgroup of R, then we define the commutator [x,y] := xy −yx, and we write [x,S] := {[x,s] : s ∈ S}. Note that [x,S] is always an additive subgroup of R. If T is another additive subgroup of R, we define [S,T] to be the additive subgroup of R given by the set of finite sums of commutators [s,t], s ∈ S, t ∈ T . A + B denotes the additive subgroup {a+b : a ∈ A, b ∈ B} whenever A,B are additive subgroups of a PN ring R, and spanS is the subspace of finite linear combinations of elements of a subset S of an algebra R. If a PN ring R is the direct sum of PN rings R1 and R2, it follows easily that Pr(R) = Pr(R1) Pr(R2). 3. Proofs Theorem 1.1 follows immediately from the following more general result. Theorem 3.1. Suppose a PN ring R has finite additive subgroups A1,A2,B1,B2 satisfying A1 ⊆ A2 and B1 ⊆ B2. Then PrR(A2,B2) ≤ PrR(A1,B1). Furthermore, the following conditions are equivalent: 190 S. M. Buckley, D. MacHale / J. Algebra Comb. Discrete Appl. 4(2) (2017) 189–196 (AB1) PrR(A1,B1) = PrR(A2,B2). (AB2) [x,A1] = [x,A2] and [y,B1] = [y,B2], for all x ∈ B1, y ∈ A2. (AB3) [x,A1] = [x,A2] and [y,B1] = [y,B2], for all x ∈ B2, y ∈ A2. Proof. Note that for any finite subsets A,B of R, we have PrR(A,B) = ——— ∑ x∈B ——— ∑ y∈A f(y,x) , where f : R×R →{0,1} is the function defined by f(y,x) = 1 if xy = yx, and f(x,y) = 0 otherwise. We first prove the result in the special case B1 = B2. For each x ∈ B2, define a surjective homo- morphism of additive groups, φx : A2 → [x,A2], by φx(y) = [x,y] , y ∈ A2 . For x ∈ B2, y ∈ A2, and f as in the previous paragraph, we have f(x,y) = 1 if and only if y ∈ kerφx. By the first isomorphism theorem, it follows that ——— ∑ y∈A2 f(y,x) = |kerφx| |A2| = 1 |[x,A2]| . and so PrR(A2,B2) = ——— ∑ x∈B2 1 |[x,A2]| . By the same argument, we have PrR(A1,B2) = ——— ∑ x∈B2 1 |[x,A1]| . It follows readily that PrR(A2,B2) ≤ PrR(A1,B2), with equality if and only if [x,A1] = [x,A2] for all x ∈ B2. This proves the equivalence of (AB1)–(AB3) in the special case B1 = B2. We wish to employ symmetry between the A- and B-subgroups. For this, we note that (AB2) can be written in a simpler form in our special case B1 = B2: (AB2′) [x,A1] = [x,A2], for all x ∈ B2. Moreover, let us say that (AB2′) has data (A1,A2;B2). By symmetry, we can now handle the special case A1 = A2. In fact, we have PrR(A2,B2) ≤ PrR(A2,B1), with equality if and only if [y,B1] = [y,B2] for all y ∈ A2, and we deduce the equivalence of (AB1)–(AB3) as before. For the special case A1 = A2, (AB2) can be written in the simpler form (AB2′′) [y,B1] = [y,B2], for all y ∈ A2. Moreover, let us say that (AB2′′) has data (B1,B2;A2). We now consider the general case. By the two special cases considered above, we have PrR(A2,B2) ≤ PrR(A2,B1) ≤ PrR(A1,B1) , (1) as required. Moreover, PrR(A1,B1) = PrR(A2,B2) if and only if both of inequalities in (1) are equal- ities, which is equivalent to the conjunction of (AB2′) with data (A1,A2;B1), and (AB2′′) with data (B1,B2;A2). This conjunction is just the required general form of (AB2). Thus, (AB1) is equivalent to (AB2). Because of the symmetry between the A- and B-subgroups in (AB1) that is lacking in (AB2), we get a version of (AB2) where the equations are instead true for all y ∈ A1 and all x ∈ B2. Putting this together with the original form of (AB2), we derive the formally stronger (AB3). Thus, all three conditions (AB1)–(AB3) are mutually equivalent. 191 S. M. Buckley, D. MacHale / J. Algebra Comb. Discrete Appl. 4(2) (2017) 189–196 Next, we tackle Proposition 1.2. In fact we prove a slight generalization of it in the context of PN rings. Proposition 3.2. Suppose I is an ideal in a finite PN ring R. Then Pr(R) ≤ Pr(R/I), with equality if and only if [x,R]∩ I = {0} for all x ∈ R. Proof. Take x + I,y + I ∈ R/I, where x,y ∈ R. Since [x + I,y + I] = [x,y] + I is independent of the representatives x,y of these elements in R/I, it follows that Pr(R/I) = |{(x,y) ∈ R×R : xy −yx ∈ I}| |R|2 . It is now clear that Pr(R) ≤ Pr(R/I). Furthermore, we have equality if and only if the condition xy −yx ∈ I is equivalent to xy = yx for all x,y ∈ R. This is equivalent to the desired condition. 4. Counterexamples Here we pose three questions and give a negative answer in each case. Together, these answers show that our results cannot be simplified or improved in any obvious way. Question 4.1. Can we improve Theorem 3.1 by dropping one equation from (AB2) or (AB3), or by significantly restricting the set of elements for which the equations hold, and still obtaining a condition equivalent to (AB1)? Question 4.2. Can we strengthen the first conclusion of Theorem 3.1 by dropping one of the assumptions that Ai,Bi are additive subgroups of R? Question 4.3. Can we strengthen the statement of either Theorem 3.1 or Proposition 1.2 by replacing the necessary and sufficient condition for equality of commuting probabilities by a simpler quantifier-free commutator subgroup property? We will see that the first two questions are easily answered, but that the third one is rather more interesting (although we will need to make clearer what we have in mind by this question separately for each of the two results to which it refers). Our first proposition gives a negative answer to Question 4.1. Proposition 4.4. Neither of the following conditions are equivalent to conditions (AB1)–(AB3) in The- orem 3.1. (AB4) [x,A1] = [x,A2] and [y,B1] = [y,B2], for all x ∈ B1, y ∈ A1. (AB5) [x,A1] = [x,A2] for all x ∈ B2. Proof. We get counterexamples in an arbitrary finite noncommutative ring R. In (AB4), let A1 = B1 = Z(R) and A2 = B2 = R. Then [x,A1] = [x,A2] = [y,B1] = [y,B2] = {0} for all x ∈ B1, y ∈ A1. However, PrR(A1,B1) = 1 > PrR(A2,B2). For (AB5), let B1 = Z(R) and A1 = A2 = B2 = R. Trivially, [x,A1] = [x,A2] for all x ∈ B2. However, PrR(A1,B1) = 1 > PrR(A2,B2). The following proposition gives a negative answer to Question 4.2. Proposition 4.5. If we drop any one of the assumptions that Ai,Bi are additive subgroups of R in Theorem 3.1, then the main inequality PrR(A2,B2) ≤ PrR(A1,B1) may fail. 192 S. M. Buckley, D. MacHale / J. Algebra Comb. Discrete Appl. 4(2) (2017) 189–196 Proof. By symmetry, it suffices to show that the inequality may fail if either B1 or B2 is not an additive subgroup. Below, p is any prime number. Let R be the Zp-algebra with basis {u,v}, where u2 = vu = u and v2 = uv = v. Let B2 = A1 = A2 = R, and let B1 = {u}. It is clear that PrR(A1,B1) = 1/p, whereas it is well known and straightforward to verify (see [3, Theorem 5.1]) that PrR(A2,B2) = Pr(R) = p2 + p−1 p3 > 1 p . Next, let S be the ring of order p3 given by an internal direct sum of Zp and the ring R of the previous paragraph. Let A1 = A2 = B1 = R, and let B2 be any subset of S such that B2 = R ∪{z} where z ∈ Z(S) \ {0}; note that |Z(S) \ {0}| = p − 1 and Z(S) \ {0} does not intersect R. Then PrR(A1,B1) = Pr(R) = (p 2 + p−1)/p3 as before, but PrR(A2,B2) = (p3 + p2 −p) + p2 p2(p2 + 1) = p2 + 2p−1 p3 + p > p2 + p−1 p3 . since (p2 + 2p−1)/(p3 + p) is a weighted mean of Pr(R) and 1. Remark 4.6. The counterexamples in Proposition 4.5 do not immediately imply that the assumption that S is a subring in Theorem 1.1 is essential. However, this is easily shown. For instance, if R is any finite non-commutative ring and S = {a,b}, where a,b ∈ R do not commute, then Pr(S) = 0 < Pr(R). We next address Question 4.3 in relation to Theorem 3.1. Assume that the hypotheses of Theorem 3.1 are in effect and that (AB1)–(AB3) hold. Suppose u ∈ [A2,B2]. By definition, u is a finite sum of terms of the form [y,x], y ∈ A2, x ∈ B2. By (AB2), we may assume that x ∈ B1 for each such term, and so [A2,B2] = [A2,B1]. By symmetry, it is also true that [A2,B2] = [A1,B2]. If we restrict y to A1 then, by essentially the same argument, it follows that [A1,B1] = [A1,B2]. Thus, (AB1)–(AB3) imply the following quantifier-free condition: (AB6) [A1,B1] = [A1,B2] = [A2,B1] = [A2,B2]. If (AB6) were equivalent to (AB1)–(AB3), then we could weaken the condition for equality in The- orem 1.1 to [S,S] = [R,R]. However, we will see that this is false. In fact, we can say more. Let us consider the following four conditions for a subring S of a ring R. (S1) S + Z(R) = R. (S2) Pr(R) = Pr(S). (S3) [S,S] = [R,R]. (S4) [R,S] = [R,R]. If (S1) holds, then R is a disjoint union of cosets of the form z+S, z ∈ Z(R). Since [z1+s1,z2+s2] = [s1,s2] for s1,s2 ∈ S, z1,z2 ∈ Z(R), it follows readily that (S2) holds. Since (AB1) implies (AB6), it follows in particular that (S2) implies (S3), and trivially (S3) implies (S4). The above implications cannot be reversed. First, it is easy to see that (S4) does not imply (S3): just take R to be a two-dimensional non-commutative Zp-algebra (as in the proof of Proposition 4.5), where p is a prime, and let S be a one-dimensional subalgebra. Then [R,S] = [R,R] has order p, but [S,S] = {0}. The following pair of results show that the other two reverse implications also fail. Proposition 4.7. For each prime p, there exists a 5-dimensional Zp-algebra R with a subalgebra S of codimension 1 such that [S,S] = [R,R] and Pr(R) < Pr(S). Theorem 4.8. There exists a 7-dimensional Z2-algebra R with a subalgebra S of codimension 1 such that Pr(S) = Pr(R) and S + Z(R) 6= R. 193 S. M. Buckley, D. MacHale / J. Algebra Comb. Discrete Appl. 4(2) (2017) 189–196 In the proofs of the above pair of results and the proof of one subsequent result, R will in each case be a finite nilpotent Zp-algebra. In each proof, we list a basis B of R and define xy for all x,y ∈ B. By distributivity, this defines R uniquely as a PN Zp-algebra. However the products of basis elements will be of a special type that in each case allows us to drop the “PN” qualifier: for x,y ∈B, we will either have xy = 0 or xy = z for some z ∈ B ∩ Ann(R). By distributivity, it follows that (uv)w = u(vw) = 0 for all u,v,w ∈ R, and so in particular R is associative. We will in each proof denote elements of B∩ Ann(R) as zi (or simply z if there is only one such element). Proof of Proposition 4.7. Let R be the Zp-algebra with basis{x1,x2,y1,y2,z}, where the only nonzero products of basis elements are x1y1 = x2y2 = z. Letting S be the 4-dimensional subalgebra of R with basis {x1,x2,y1,z}, it is clear that [S,S] = [R,R] = span{z}. Next, let T be the 3-dimensional subalgebra of S with basis {x1,y1,z}. Thus, S = T ⊕ span{x2} is isomorphic to a direct sum of T and Zp, and so it is readily verified that Pr(S) = Pr(T) Pr(Zp) = p2 + p−1 p3 ·1 = p2 + p−1 p3 . The ring R is what we call an augmentation of T in [3, Section 4], so it follows from that paper, or by direct calculation, that Pr(R) = p4 + p−1 p5 < Pr(S) . Proof of Theorem 4.8. Let R be the algebra with basis B = {u1,u2,v1,v2,w,z1,z2} , where the only nonzero products of basis elements are v1v2 = z1 and uivi = uiw = z2 for i = 1,2. Let S be the codimension 1 subalgebra with basis B′ := B\{w}. Let Ai := spanBi for i = 1,2, where B1 := {u1,u2,v1,v2,w} and B2 := {z1,z2}. It is clear that (R,+) is a direct sum of A1 and A2. We claim that Z(R) = A2. Clearly A2 ⊆ Ann(R) ⊆ Z(R), so we need only show that Z(R) ⊆ A2. First, note that dim[v1,R] = dimA2 = 2, so CR(v1) has codimension 2, and we easily deduce that CR(v1) = span{u2,v1,w,z1,z2} . By symmetry, CR(v2) = span{u1,v2,w,z1,z2}. Now A2 ⊆ Z(R) ⊆ CR(v1)∩CR(v2) = span{w,z1,z2} . Moreover, w is not central, so we deduce that Z(R) = A2, as claimed. Since A2 = Z(R) ⊂ S ⊂ R, we have also proved that S + Z(R) 6= R. The fact that Pr(S) = Pr(R) is a routine exercise, but we indicate how to carry out the required work efficiently. We need to show that [x,S] = [x,R] for all x ∈ R. Since A2 = Z(R), it suffices to examine x ∈ A1. Let us write Z2 := span{z2}. Since R = span(S ∪{w}) and [w,R] = Z2, we have [x,S] ⊆ [x,R] ⊆ [x,S] + Z2 , x ∈ R. Consequently, [x,S] = [x,R] if either z2 ∈ [x,S] or x ∈ CR(w). We claim that one of these two conditions holds for all x ∈ A1. Let x = a1u1 + a2u2 + b1v1 + b2v2 + cw, where a1,a2,b2,b2,c ∈ Z2 . Now, [x,u1] = (b1 +c)z2, so z2 ∈ [x,S] if b1 +c = 1. Thus, without loss of generality, it suffices to consider the case b1 = c and, by symmetry, we may also assume that b2 = c. It follows that [x,v1+v2] = (a1+a2)z2 so, again without loss of generality, it suffices to consider the case a1 = a2. Since u1 + u2, v1, v2, and w all lie in CR(w), our claim is proved, and we have shown that [x,S] = [x,R]. 194 S. M. Buckley, D. MacHale / J. Algebra Comb. Discrete Appl. 4(2) (2017) 189–196 Finally, we address Question 4.3 in relation to Proposition 1.2. The quantifier-free condition [R,R]∩ I = {0} certainly implies that [x,R] ∩ I = {0} for x ∈ R. However, the next result shows that this quantifier-free condition is not necessary for Pr(R) = Pr(R/I). Theorem 4.9. There exists a 15-dimensional Z2-algebra R containing a nontrivial ideal I such that Pr(R) = Pr(R/I) and I ⊂ [R,R]. Proof. Let R be the Z2-algebra with basis B = {xi,yi | i = 1,2,3}∪{zi,j : (i,j) ∈ S} , where S = {(i,j) | 1 ≤ i,j ≤ 3} , and the only nonzero products of basis elements are xiyj = zi,j , (i,j) ∈ S . It is readily verified that Ann(R) = R2 = span{zi,j : (i,j) ∈ S} . Let I = span{s}, where s := z1,1 + z2,2 + z3,3. Since I ⊂ Ann(R), I is an ideal. We claim that s is not a commutator in R. Suppose that c := [u,u′] is a sum of the form ∑3 i,j=1 ci,jzi,j for some u,u′ ∈ R, where ci,j ∈ Z2 and ci,i = 1 for 1 ≤ i ≤ 3. We claim at least one of the coefficients ci,j, i 6= j, equals 1, regardless of the choice of u,u′; note that it follows from this claim that s is not a commutator. It suffices to assume that both u and u′ are linear combinations of the six basis elements that lie outside Ann(R): u := 3∑ i=1 (aixi + biyi) u′ := 3∑ i=1 (a′ixi + b ′ iyi)   , ai,bi,a ′ i,b ′ i ∈ Z2 , i = 1,2,3 . Note that ci,j = aib′j + a ′ ibj. Since zi,i occurs as a term in c, it follows that exactly one of aib′i and a ′ ibi is nonzero for each i ∈ {1,2,3}. By swapping u and u′ if necessary, we may assume that aib′i = 1 and a ′ ibi = 0 for at least two indices i. In fact, by symmetry of the indices, we may assume that these two equations hold for i ∈{1,2} and, in particular, a1 = a2 = b′1 = b′2 = 1. Because 0 = (a′1b1)(a ′ 2b2) = (a ′ 2b1)(a ′ 1b2) , it also follows that either a′1b2 = 0 or a ′ 2b1 = 0. Thus, either a1b ′ 2 + a ′ 1b2 = 1 or a2b ′ 1 + a ′ 2b1 = 1, and so either c1,2 = 1 or c2,1 = 1, as claimed. We have shown that [x,R] ∩ I = {0} for all x ∈ R and so, by Proposition 1.2, Pr(R) = Pr(R/I). However, s ∈ [R,R] because s = [x1,y1] + [x2,y2] + [x3,y3], and so I ⊂ [R,R]. References [1] S. M. Buckley, Distributive algebras, isoclinism, and invariant probabilities, Contemp. Math. 634 (2015) 31–52. 195 http://dx.doi.org/10.1090/conm/634/12689 http://dx.doi.org/10.1090/conm/634/12689 S. M. Buckley, D. MacHale / J. Algebra Comb. Discrete Appl. 4(2) (2017) 189–196 [2] S. M. Buckley, D. MacHale, Commuting probabilities of groups and rings, preprint. [3] S. M. Buckley, D. MacHale, Á. Ní Shé, Finite rings with many commuting pairs of elements, preprint. [4] J. D. Dixon, Probabilistic group theory, C. R. Math. Acad. Sci. Soc. R. Can. 24(1) (2002) 1–15. [5] P. Erdös, P. Turán, On some problems of a statistical group–theory, IV, Acta Math. Acad. Sci. Hung. 19(3) (1968) 413–435. [6] R. M. 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Math. 82(1) (1979) 237–247. 196 http://www.maths.nuim.ie/staff/sbuckley/Papers/bm_g-vs-r.pdf http://www.maths.nuim.ie/staff/sbuckley/Papers/bms.pdf http://www.ams.org/mathscinet-getitem?mr=1882359 http://dx.doi.org/10.1007/BF01894517 http://dx.doi.org/10.1007/BF01894517 http://dx.doi.org/10.1016/j.jalgebra.2005.09.044 http://dx.doi.org/10.1016/j.jalgebra.2005.09.044 http://dx.doi.org/10.2307/3615961 http://dx.doi.org/10.2307/3615961 http://dx.doi.org/10.2307/2318829 http://dx.doi.org/10.2140/pjm.1979.82.237 http://dx.doi.org/10.2140/pjm.1979.82.237 Introduction Preliminaries Proofs Counterexamples References