Is a Boolean ring always a commutative ring with unity?
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?
Answers:
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.
This is author dependent. It would not be unreasonable to just exclude the empty set, however.
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Answers:
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.
This is author dependent. It would not be unreasonable to just exclude the empty set, however.
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