# How 'commutative' can a non-commutative ring be?

Let $R$ be a finite non-commutative ring. Let $P(R)$ be the probability that two elements chosen uniformly at random commute with each other. Consider the value

$$S=\sup_RP(R)$$

where the supremum is taken over all finite, non-commutative rings, with unity. Is anything known about $S$? Do we know its value, or do we know any bounds? Does there exist a ring that achieves the supremum? What if we consider rings without unity?

This question is motivated by the notion of commutativity degree of finite groups. In that case, it is known that $$\sup_GP(G)=\frac{5}{8}$$

and in fact, there exist groups $G$ such that $P(G)=5/8$. Here, the supremum is taken over finite, nonabelian groups.

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Here is the reference for the group case: What is the probability that two group elements commute? by W. H. Gustafson, ime.usp.br/~rbrito/docs/2318778.pdf One may ask the same question for monoids, semigroups, rings and non-unital rings. I'm pretty sure that the bound will be larger than $5/8$. –  Martin Brandenburg Jun 3 '13 at 19:35
For semigroups or monoids the answer is "any rational probability" (except 0 of course). See this recent paper. –  vadim123 Jun 3 '13 at 19:51
Is there any chance to compute $P(R)$ for $R=M_2(\mathbb{Z}/n)$? I've written a program which computes some values, but no pattern emerges. But of course matrix rings are very non-commutative, i.e. $P$ is very small. –  Martin Brandenburg Jun 3 '13 at 20:03
@MartinBrandenburg, At least when $n$ is prime this should be doable. –  Stephen Jun 3 '13 at 20:06
@MartinBrandenburg: Another class of rings to check would be the upper triangular subring of $M_2(\mathbb{Z}/n)$. –  Jared Jun 3 '13 at 20:15

## 1 Answer

This question has been studied:

• Desmond MacHale, Community in Finite Rings, The American Mathematical Monthly, Vol. 83, No. 1 (Jan., 1976), pp. 30-32, online

Theorem 1 states that if $R$ is a non-commutative ring, then $P(R) \leq 5/8$, with equality if and only if $(R:Z(R))=4$. The proof is quite similar (but not equal) to the case of groups. Besides, the bound is achieved for the ring of upper-triangular $2 \times 2$-matrices over $\mathbb{F}_2$. Hence $S=5/8$ as in the case of groups. It doesn't matter if we consider rings or non-unital rings.

Theorem 4 states that $P(R) \leq P(R')$ if $R'$ is a subring of $R$. Intuitively, this says that a subring of a finite ring is at least as commutative as the ring itsself.

More about the distribution of the probabilities $P(-)$ can be found in the following preprints (and these seem to be the only ones):

• S. M. Buckley, D. MacHale, A. Ní Shé, Finite rings with many commuting pairs of elements, online
• S. M. Buckley, D. MacHale, Contrasting the commuting probability of groups and rings, online
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