# Boolean algebra-Boolean ring. Stone Theorem?

I am interested in knowing which theorem is responsible for the following statement:

Every Boolean algebra can become a Boolean ring by taking the ring addition to be $A\oplus B = (A \land \lnot B) \lor (\lnot A \land B)$ and the ring multiplication to be $A\odot B = A \land B$.

In which way are sigma ideals a special case of ideals?

Is it Stone Theorem?

• This is pretty unreadable. Does A¯¯¯¯ mean $\overline{A}$? Does A⊓B mean $A\cap B$, as I suspect? – rschwieb Jun 9 '15 at 15:06
• @rschwieb. Sorry, I do not know how to edit it....You find the correct writing in the link provided..... – Javier Arias Jun 9 '15 at 15:16
• No, this is much easier than Stone's theorem. It's a straightforward exercise. You in fact get an equivalence of categories this way; try to write down the inverse functor. – Qiaochu Yuan Jun 10 '15 at 5:16
• @Qiaochu Yuan. Is it then fair to phrase it as follows: According to the equivalences set up by Category Theory.......?? – Javier Arias Jun 12 '15 at 13:22

Boolean rings are rings with unit all elements of which are idempotent:

$x^2 = x$

Boolean algebras can be regarded as boolean rings through the definitions you have given for product and sum. Also boolean rings can be regarded as boolean algebras through the following definitions of join, meet and complement:

$A \lor B = A \oplus B \oplus (A\otimes B)$

$A \land B = A \odot B$

$A' = A \oplus 1$

All this is basic theory and has nothing to do with some named theorem.