Is a Boolean ring always a commutative ring with unity?

Question

My question relates to the power set of any set S, denoted P(S). + is defined by +: u+v-> (uUv)\(unv) (U=union, n=intersection, \=complement), x is defined by x: uv-> unv, for u, v members of P(S). The identity law states that there exist 1 a member of P(S) such that 1 is not equal to 0 and a1=1a=a for all a members of S. But when S is the empty set, 1=0=empty set. So, is it still a commutative ring with unity?

Answer 1

Strangely, it depends on who you ask. Some abstract algebra authors specifically require that 1 and 0 are distinct, thus specifically excluding trivial rings that contain only one element. Not all require this, though.

Answer 2

This is author dependent. It would not be unreasonable to just exclude the empty set, however.