Talk:Boolean ring
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Sirs:
M.H.STONE defined a ring to be Boolean if every element is idempotent. This does not require the presence of an identity (neutral element for multiplication). When considering the equivalence of Boolean algebras with what you call Boolean rings, it is needed that the ring be with identity.
-- 193.205.5.2
I don't really understand the comment. If you square x + x (instead of x + 1 as on the page), don't you find 2x = 4x anyway?
Charles Matthews 11:32, 20 Aug 2003 (UTC)
The point is that a Boolean algebra must have an identity (a top element in lattice-theoretic terms). So a Boolean ring without identity would not yield a Boolean algebra, even though it would indeed satisfy all of the other requirements. (Instead, it would yield a relatively complemented lattice.) To some extent, 193.205.5.2 is still wrong, because every ring has an identity; otherwise it's just a rng. But this is a relatively modern version of the definition of "ring", which some people still haven't come around to; in particular, identities weren't required in Stone's day. You can still see the legacy of this in the term sigma-ring; the only difference between a sigma-ring and a sigma-algebra is that the former need not have an identity. In measure theory, this can have some rather important consequences, especially to the development of the theory of non-sigma-finite measure spaces. Similar (but less drastic) consequences arise in the study of Boolean rings, which are basically the same as sigma-algebras but without the countable unions that measure theory requires. Thus Stone's definition of a Boolean ring would (I suppose) today be called a Boolean rng; and you can certainly find people today that haven't changed their definitions. -- Toby Bartels 03:11, 24 Aug 2003 (UTC)