BOOLEAN ALGEBRAS AND BOOLEAN RINGS

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1 BOOLEAN ALGEBRAS AND BOOLEAN RINGS AMBRUS PÁL Definition 0.1. A ring R is Boolean if x 2 = x for every x R. Clearly the direct product of Boolean rings is Boolean, and the field with two elements F 2 is a Boolean ring, so for every set X the direct product ring: is a Boolean ring. F X 2 = i X F 2 Proposition 0.2. In a Boolean ring R the following hold: (i) we have 2x = 0 for every x R, (ii) every prime ideal p is maximal, and R/p is the field with two elements, (iii) we have (x, y) = (x + y xy) for every x, y R, (iv) every finitely generated ideal is principal. Proof. Since 2x = (2x) 2 = 4x 2 = 4x, we get that 2x = 0 by subtracting 2x from both sides. Now let p be a prime ideal in R. Then the quotient R/p is a Boolean ring. For every x R/p we have x(1 x) = 0 which implies that x = 0 or x = 1 since R/p is an integral domain. Claim (ii) follows. Note that x(x + y xy) = x 2 + xy x 2 y = x + xy xy = x. Hence x, y (x + y xy). Since x + y xy (x, y), clam (iii) is clear. Let I = (x 1, x 2,..., x n ) be an finitely generated ideal of R. Since I = ((x 1, x 2,..., x n 1 ), x n ), we may assume by induction on n that I = (x, y) for some x, y R. The claim now follows from part (iii). Proposition 0.3. Every Boolean ring R can be imbedded as a subring in F X 2 some set X. Proof. First note that N (R) = 0. So by Krull s theorem there is an injective map: R R/p. p pr By the above every ring appearing in the product on the right hand side is F 2. The claim is now clear. Since a finite Boolean ring is Artinian, we get the following immediate corollary to the structure theorem for Artinian rings: for Date: October 25,

2 2 AMBRUS PÁL Corollary 0.4. Let R be a finite Boolean ring and let X denote the set of proper maximal ideals of R. Then the map is an isomorphism. R m X F 2 = F X 2 Definition 0.5. A Boolean algebra is a six-tuple consisting (B,,,, 0, 1) of a set B, equipped with two binary operations (called intersection or and ), (called union or or ), a unary operation (called complement or not ) and two elements 0 and 1, such that for all elements a, b and c of B the following axioms hold: a (b c) = (a b) c a (b c) = (a b) c (associativity) a b = b a a b = b a (commutativity) a (a b) = a a (a b) = a (absorption) a 0 = a a 1 = a (identity) a (b c) = (a b) (a c) a (b c) = (a b) (a c) (distributivity) a a = a a a = 0 (complements) Boolean algebras are just another class of algebraic structures which are defined with the means of operations and axioms for these operations. We can define the analogue of all basic notations of algebra for them, for example homomorphisms, isomorphisms and sub-structures. Remark 0.6. It is traditional to include the associativity and the absorption laws in these axioms, but note that they can be excluded from the set of axioms as they can be derived from the other four axioms (called Huntington s postulates). There are a lot of other identities which Boolean algebras satisfy, for example: a = a 1 = a (a a) = (a a) (a a) = (a a) 0 = a a. Note that all axioms of Boolean algebras come in pairs, where we can get one from the other by swiching between and, and between 0 and 1. This is called the duality of Boolean algebras. For example, by duality we also have the identity a a = a, too. Definition 0.7. Let B be a Boolean algebra. Let be the following binary relation on B: a b if and only if a b = a. Lemma 0.8. Let B be a Boolean algebra. Then the following holds: (i) the binary relation is a partial ordering, (ii) we have a b if and only if a b = b. Proof. By the above a a = a, so a a. If a b and b a, then a = a b = b a = b, so the second axiom of partial orderings holds for, too. If a b and b c, then a c = (a b) c = a (b c) = a b = a, so a c. Claim (i) is now proved. If a b = a, then a b = (a b) b = b

3 BOOLEAN ALGEBRAS AND BOOLEAN RINGS 3 using the absorption law. The converse can be proved dually. Claim (ii) is now clear. Remark 0.9. By the above 0 is the smallest and 1 is the largest element with respect to the ordering. In particular using (ii) we get that for every a B. 0 a = 0 and 1 a = 1 Lemma Let B be a Boolean algebra. Then B satisfies the following: (i) the equalities a b = 0 and a b = 1 imply a = b, (ii) ( x) = x, (iii) (x y) = x y and (x y) = x y (De Morgan s laws). Proof. If a b = 0 then a = a 0 = a (a b) = ( a a) ( a b) = 1 ( a b) = a b, and hence b a. If a b = 1 then a = a 1 = a (a b) = ( a a) ( a b) = 0 ( a b) = a b, so a b. Therefore from the equalities a b = 0 and a b = 1 we get b = a, and claim (i) is now clear. Now a a = 0 and a a = 1, so ( a) = a by part (i). Claim (ii) follows. Finally and (x y) ( x y) = x (y ( x y)) = x ((y x) (y y)) = x (y x) = (x x) y = 1 y = 1, (x y) ( x y) = (x ( x y)) (y ( x y)) = ((x x) y)) ((y y) x)) = (0 y) (0 x) = 0 0 = 0, so from part (ii) we get the first De Morgan law. The second follows by duality. By part (i) above 0 = 1 and 1 = 0. Example Let X be a set; let P(X) denote its power set. The Boolean algebra of subsets of X is by definition the six-tuple (P(X),,, ( ) c,, X), where ( ) c denotes the complement in X. It is a pleasant exercise in elementary set theory to check that the axioms of Boolean algebras hold. Theorem 0.12 (Stone). (a) Let (B,,,, 0, 1) be a Boolean algebra. Define B to be the algebraic structure (B, +,, 0, 1), where a + b def = (a b) ( a b), a b def = a b. Then B is a Boolean ring. (b) Let (R, +,, 0, 1) be a Boolean ring. (R,,,, 0, 1), where Define R to be the algebraic structure a b def = a b, a b def = a + b + a b, a def = 1 + a.

4 4 AMBRUS PÁL Then R is a Boolean algebra. (c) Given a B and an R as above we have (B ) = B and (R ) = R. Proof. (a) Let a, b, c B. Then a + 0 =(a 0) ( a 0) = (a 1) 0 = a, a + b =(a b) ( a b) = (b a) ( b a) = b + a, a + a =(a a) ( a a) = 0 0 = 0, a + (b + c) =(a (b + c)) ( a (b + c)) =(a ((b c) ( b c))) ( a ((b c) ( b c))) =(a (b c) ( b c)) ( a b c) ( a b c) =(a ( b c) (b c)) ( a b c) ( a b c) =(a b b) (a b c) (a c b) (a c c) ( a b c) ( a b c) =(a b c) (a b c) ( a b c) ( a b c). The value of this last expression does not change if we permute a, b and c in any manner, so c + (a + b) = a + (b + c), and hence associativity for + follows from the already established commutativity. Moreover: while a 1 =a 1 = 1 a = 1 a = a, a b =a b = b a = b a, a (b c) =a (b c) = (a b) c = (a b) c, a (b + c) =a ((b c) ( b c)) = (a b c) (a b c), a b + a c =a b + a c = (a b (a c)) ( (a b) a c) =(a b ( a c)) (( a b) a c) =(a b a) (a b c) ( a a c) ( b a c) =(a b c) ( b a c), so a (b + c) = a b + a c, Therefore B is a Boolean ring. (b) Let a, b, c R. Then a a = a a = a. a b =a + b + a b = b + a + b a = b a, a b =a b = b a = b a, a (b c) =a + (b c) + a (b c) =a + (b + c + b c) + a (b + c + b c) =a + b + c + b c + a b + a c + a b c. The value of this last expression does not change if we permute a, b and c in any manner, so c (a b) = a (b c), and hence associativity for follows from the

5 BOOLEAN ALGEBRAS AND BOOLEAN RINGS 5 already established commutativity. Moreover: a 0 =a a 0 = a + 0 = a, a 1 =a 1 = a, a a =a + (1 + a) + a (1 + a) = a + a a + a 2 = 4a + 1 = 1, a a =a (1 + a) = a + a 2 = 2a = 0, a (b c) =a (b c) = (a b) c = (a b) c, a (a b) =a + (a b) + a (a b) = a + a b + a (a b) =a + 2a b = a, a (a b) =a (a + b + a b) =a 2 + a b + a 2 b = a + 2a b = a. Therefore R is a Boolean algebra. (c) Let B be a Boolean algebra and let a, b B. Then the new Boolean algebra operations in the Boolean ring B are: a b =a b, 1 + a =(1 a) ( 1 a) = a (0 a) = a 0 = a, a + b + a b =a + b (1 + a) = a + b a =(a (b a)) ( a b a) =(a ( b a)) ( a b) =a ( a b) = (a a) (a b) = 1 (a b) = a b. Therefore (B ) is B. Now let R be a Boolean ring and let a, b R. Then the new ring operations in the Boolean algebra R are: (a b) ( a b) =(a (1 + b)) ((1 + a) b) Therefore (R ) is R. =(a (1 + b)) + ((1 + a) b) + a b (1 + a) (1 + b) =a + a b + b + a b + a b + a 2 b + a b 2 + a 2 b 2 =a + b + 6a b = a + b, a b =a b. Theorem Every Boolean algebra B can be imbedded as a subalgebra in P(X) for some set X. If B is finite, that it is isomorphic to P(X) for a finite set X. Proof. Note that (F X 2 ) = P(X) for every set X. So the claim follows from Propositions 0.3 and 0.4.