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.

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