On Fri, 25 Apr 2003, Peter Freyd wrote:
Varieties of algebras when viewed as categories can be unexpectedly equivalent. For a reason explained at the end, I was looking at varieties of unital rings satisfying the equations p = 0 and x^p = x, one such variety for each prime integer p.
The equivalence type of these categories is independent of p. The easiest way of establishing that is to show that each is equivalent to the category of Boolean algebras (a well-known fact when p = 2) and all the equivalences can by established by just one functor. Given a unital ring, R, define B(R) to be the boolean algebra of its central idempotents where the meet of a and b is ab and the join is a + b - ab. Then the restriction of B to the p'th variety described above is always an equivalence of categories.
The fastidious will note (one would certainly hope) that B is not a functor in general (homomorphisms don't preserve centrality). But in a ring "without nilpotents" (that is, in which x^2 = 0 implies x = 0) all idempotents are central. The equations x^p = x, of course, imply the absence of nilpotents.
(Given p the inverse functor to B can be described as follows: for a Boolean algebra C consider the set of "p-labeled partitions of unity", that is, the set of functions f:Z_p -> C whose values are pairwise disjoint and have unity as their join. Given two such, f and g, define their sum by setting (f+g)i to be the join of the set { fj ^ gk | j+k = i } and their product by setting (fg)i to be the join of { fj ^ gk | jk = i }.)
The equivalence of these varieties for all p is well known. It's best understood by seeing that they are all dual to the category of Stone spaces: given a Stone space, the ring of continuous Z_p-valued functions on it (where Z_p is given the discrete topology) is a ring satisfying p1=0 and x^p=x; conversely, given such a ring, its prime (=maximal) ideal spectrum is a Stone space. Not having my copy of "Stone Spaces" to hand as I write this, I can't remember whether this fact was in the book. But it certainly should have been. Peter Johnstone