Lattices n Boolean Algebra

8/6/2019 Lattices n Boolean Algebra

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Lattices and Boolean algebras

This includes topics of combinatorial interest, with applications in many areas including

computer science and logic. The areas are well represented in the UH Math Department.

Definition. A poset or partially ordered set is a set S with a binary relation which isreflexive, transitive and antisymmetric. is called a partial ordering.

Definition. A lattice is a poset in which any two elements a, b have a least upper bounda b, called the join of a and b, and a greatest lower bound a b, called the meet of a andb.

Examples. 1. The lattice of all subsets of a set: The meet is intersection and the join is

union.2. The lattice of ideals of a ring (or submodules of a module): The meet is intersection andthe join is the sum I+ J.

Algebraic definition of a lattice

Theorem 1. L is a lattice iff L is a set with binary operations , such that for alla,b,c L,

(1) a b = b a and a b = b a. (Commutative laws)

(2) (a b) c = a (b c) and (a b) c = a (b c). (Associative laws)(3) a a = a and a a = a. (Idempotent laws)(4) (a b) a = a and (a b) a = a (Absorption laws)

Notice that the connection with the original definition is that a b a = a b b = a b.

Theorem 2 (Principle of Duality). Any valid statement deduced from the axioms for alattice remains true if meet and join are interchanged and is replaced by .

Definition. A sublattice of a lattice L is a subset ofL closed with respect to meet and join.

Definition. A lattice homomorphism is a function f: L L which preserves meet andjoin (hence also preserves the partial ordering).

Definition. A lattice is complete if every subset of L has a lub and glb.

For example, the lattice P(S) of all subsets of a set S is complete. The lattice (Q, ) isnot.

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Definition. The identity element of a lattice is an element 1 L such that 1 x, x L.The zero element of a lattice is an element 0 L such that 0 x, x L.

Such an element is clearly unique if it exists. For example, the 1 in P(S) is S and the 0is . But (Q, ) has no 0 or 1.

Definition. A lattice is distributive if a,b,c L, a (b c) = (a b) (a c).

The lattice of subsets of a set is distributive, the lattice of ideals of a ring is not.

Definition. A lattice is modular if a,b,c L, a b = a (b c) = b (a c).

Remark: duality applies to both distributive and modular lattices. Distributive latticesare modular.

Theorem 3. The lattice of normal subgroups of a group is modular.

Theorem 4. A lattice is modular iff it has no sublattice isomorphic to N5.

Corollary 5. For any ring R and R-module M, the lattice of R-submodules of M is mod-ular.

Corollary 6. The ideal lattice of a ring is modular.

Definition. Let L be a lattice with 0 and 1. A complement of an element x L is anelement y L such that x y = 0 and x y = 1. We write y = x. L is a complementedlattice if all of its elements have complements.

For any set S, the lattice P(S) is complemented, using the usual set-theoretic complement.

Definition. A Boolean algebra is a complemented, distributive lattice. A Boolean algebra

homomorphism is a function preserving meet, join and complement.

Examples. 1. Any field of sets is a Boolean algebra. That is, a collection of subsets of afixed set S which is closed under union, intersection and complement and contains both and S. Stones Representation Theorem says that all Boolean algebras are of this form.

2. Propositional Calculus. This consists of the set of all propositions (statements withknown truth values T or F) and connectives , , which satisfy the Boolean algebraaxioms for meet, join and complement. Note that p q means p q. The identity 1corresponds to a tautology such as p p, and 0 corresponds to any proposition which is


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always false, such as p p. This is the origin of lattice theory and was first studied byGeorge Boole.

Proposition 7. Let B be a Boolean algebra. Each x B has a unique complement x.Furthermore, (x) = x, (x y) = x y, (x y) = x y.

Given any Boolean algebra A, we can construct an associated Boolean ring A as follows:The ring A has the same set of elements as A; multiplication is defined by x y = x y;addition is defined by x+y = (xy)(xy) (symmetric difference ofx and y). Conversely,given a Boolean ring A, one obtains a Boolean algebra A by setting x y = xy, x y =x + y + xy and x = x + 1. These processes are inverse to one another.

Proposition 8. Let A, B be Boolean algebras and f: A B a function. The following

are equivalent:(1) f is a Boolean algebra homomorphism.(2) f preserves meet and complement.(3) f preserves join and complement.(4) f preserves meet and join and f(0) = 0, f(1) = 1.(5) f is a homomorphism of the associated Boolean rings.

Definition. Let C1,C2 be categories. They are equivalent if there exists a functor F: C1 C2 such that for all C, D ob(C1), Fgives a bijection between M orC1(C, D) and MorC2(F(C),F(D[or M orC2(F(D),F(C)) ifF is contravariant] and for all B ob(C2), there exists C ob(C1)

such that F(C) = B. If also F: ob(C1) ob(C2) is bijective, we say that F is an isomorphismof categories.

Theorem 9. The category of Boolean algebras and the category of Boolean rings are iso-morphic.

Definition. A subalgebra of a Boolean algebra is a sublattice containing 0 and 1 whichis itself a Boolean algebra. Equivalently, B is a subalgebra of A if the canonical injectionB A is a Boolean algebra homomorphism by Proposition 8.

We define ideals, prime ideals and maximal ideals of a Boolean algebra to be the corre-sponding objects in the associated ring. By your homework, every prime ideal is maximal.Doing this translation between categories gives

Proposition 10. Let A be a Boolean algebra.

(1) A subset I is an ideal of A iff it is closed under join and for all a I, b a =b I.

(2) An ideal I is maximal iff I is proper and for all a A, either a I or a I.


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Theorem 11. Let X be a Boolean space and let R be the Boolean ring of all clopen subsetsof X. Then X is homeomorphic to Spec R.

Theorem 12. Let R be a Boolean ring and let B be the Boolean ring of all clopen subsetsof Spec R. Then R = B.

Corollary 13. The category of Boolean algebras is equivalent to the category of Booleanspaces.

Corollary 14 (Stones Representation Theorem). Every Boolean algebra is isomorphic toa field of sets.

Note that the field of sets is not generally a power set of some set. For example, forthe Boolean algebra associated with the Cantor set, every clopen set is infinite. (Finitesets cannot be open since the topology is inherited from the reals.) We can make somedefinitions to clarify this further.

Definition. Let B be a Boolean algebra. An element b B is an atom if b = 0 and forall x B, x b = x = b or x = 0. The Boolean algebra B is called atomic if for all0 = x B, there exists b x with b an atom. The Boolean algebra B is called atomless ifit has no atoms.

Any finite Boolean algebra is atomic. The Boolean algebra associated with the Cantorset is atomless.

Lemma 15. Let b be an atom with b < x1 xn. Then b xi for some i.

Theorem 16. Let B be a Boolean algebra. There exists a set A such that B is isomorphicto P(A) iff B is complete and atomic.

Rings which are not Boolean but have Boolean spectrum.

Definition. A (not necessarily commutative) ring R is von Neumann regular if for allx R, there exists y R such that xyx = x.

Proposition 17. If R/ Nil R is von Neumann regular, then Spec R is Boolean. (The con-verse also holds.)

Example. Let G be an abelian torsion group and F a field with characteristic not dividingthe order of any element ofG. Given x =

n1 figi F[G], let H be the subgroup generated

by {g1, g2, . . . , gn}. Then F[H] is semisimple by Proposition 3.8, hence is a finite product


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of fields by the Wedderburn theorem. Look at x = (x1, . . . , xn) in this product. Let

yi =

x1i , if xi = 0

0, if x = 0. Then y = (y1, . . . , yn) satisfies xyx = x and y F[H] F[G].

This shows that F[G] is von Neumann regular (known as Maschkes Theorem).

The Mobius function of a poset. See [J1, 8.6] for more details.

This is a generalization of work done by Mobius in number theory. We will do his exampleat the end.

Problem. Count the number of permutations of a finite set V without fixedpoints. For T V, let f(T) be the number of permutations of V which fix all the elementsof T and no element ofT = V \ T. Let g(T) equal the number of permutations ofV which

fix all elements of T, so g(T) = |T

|!. Then

g(T) =UT

f(U) (U P(V))

We want a formula for f(U) in terms of g(T). Our original question asks for f().

In general, let S be a finite poset and and let A be an abelian group. Let f, g : S Abe such that

() g(y) =xSxy

f(x).

We want to write f in terms of g.

Lemma 18. The poset S can be listed as x1, . . . , xn such that if xi < xj in the partialordering, then i < j.

Proof. Let x1 be a minimal element of S. Choose x2 to be a minimal element in S\ {x1},and in general, xi to be a minimal element in S\ {x1, . . . , xi1}.

Define : S S Z by

(x, y) =

1, if x y

0, otherwise

Then () is equivalent to

() g(xi) =n

j=1

(xi, xj)f(xj) (i = 1, . . . , n)


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where we think of A as a Z-module. Since j < i = (xi, xj) = 0, the matrix Z = (i,j) =(xi, xj) is upper triangular with ones on the diagonal. Its inverse is easily computed. LetN = I Z, so that Nn = 0. Then M = I+ N + N2 + + Nn1 has inverse I N = Z.

We can write () in matrix form as G = ZF, where G =

g(x1)...

g(xn)

and F =

f(x1)...

f(xn)

.Then solve and obtain F = M G.

Definition. Write M = (ij). The Mobius function of the poset S into Z is the function : S S Z defined by (xi, xj) = ij for all xi, xj S.

Then F = M G gives us

f(y) = xS

(y, x)g(x).

Theorem 19. Let S be a finite poset. There exists a unique function : S S Zsuch that if A is any abelian group and f, g : S A are functions satisfying (), then

f(y) =xS

(y, x)g(x) for all y S.

Replacing S by the same set with the reverse partial ordering, we obtain a dual statementto Theorem 19.

Corollary 20. Let S be a finite poset. There exists a unique function : S S Z suchthat if A is any abelian group and f, g : S A are functions satisfying g(y) =

xSxy

f(x),

then f(y) =xS

(x, y)g(x) for all y S.

Rewriting the equation MZ = I or ZM = I gives

Corollary 21. The Mobius function is the unique function SS Z such that(x, y) = 0unless x y and

(1)yS

xyz

(x, y) = (x, z),

where (x, z) = 1 if x = z and (x, z) = 0 if x = z. Or, (1) can be replaced by

(2)yS

xyz

(y, z) = (x, z),


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Definition. If x z in a poset S, we define the interval I[x, z] in S to be { y S | x y z } with the induced partial ordering. The length of the interval is |I[x, z]| 1.

Corollary 22. If the intervals I[x, z] andI[w, t] are isomorphic inS, then(x, z) = (w, t).

Computation for Chains: Consider the totally ordered chain C = {0, . . . , n}.

(i, i) = 1

(i 1, i) = 0 (i 1, i 1) = 1

i > j = (i, j) = 0 since M is upper triangular

(i, i + 2) = 0 (i, i) (i, i + 1) = 1 + 1 = 0

(i, i + 3) = 0 1 + 1 0 = 0

(i, j) = 0 if i = j, j 1

Therefore for C we have

M =

1 1 0 . . . 00 1 1 . . . 0

0 . . . 0 0 1

Definition. The product of posets S1, S2 is S1 S2 with the partial ordering (x1, x2) (y1, y2) x1 y1 and x2 y2.

Proposition 23. Let S = S1 S2 be posets with Mobius functions , 1, 2, respectively.Then for all xi, yi Si,

((x1, x2), (y1, y2)) = 1(x1, y1)2(x2, y2).

Corollary 24. LetP(S) be the Boolean algebra of the power set of S = {1, 2, . . . , n}. TheMobius function onP(S) is given by

(U, V) =

(1)|V\U|, if U V

0, if U V

Solution of original problem. The number of permutations of S = {1, 2, . . . , n} withoutfixed points equals

f() =

UP(S)

(, U)g(U)

=

UP(S)

(1)|U||U|! =ni=0

(1)i

n

i

(n i)!

= n!ni=0

(1)i

i!

n!

efor large n.


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Application to number theory. Let Dn be the lattice of positive integer divisors ofn ordered by divisibility. If n =

r1p

eii , then Dn

= C1 Cr, where Ci is the chain

{0, 1, . . . , ei}, via the mapping d =

pdii (d1, . . . , dr). Ifc =

pcii | d, then

(c, d) =

(ci, di) =

(1)l if dc

is a product of l distinct primes

0, if a2 | dc

for some a > 1

1, if dc

= 1

In number theory, one writes (d/c) = (c, d) = (1,d/c). Let (n) be the Euler totientfunction, counting the number of positive integers less than n and relatively prime to n,

(1) = 1. Let S = {1, 2, . . . , n} =d|n

Sd, where i Sd gcd(n, i) = d i = jd, 1

j nd

, gcd(j, nd

) = 1. Thus |Sd| = (n/d).

Taking f = and g equal to the identity in Corollary 20 gives n =

d|n (nd

), hence

(n) =

d|n (d)nd

. Thus we recover the facts we proved algebraically earlier: For a primep,

(pe) =d|pe

(d)pe

d=

ei=0

(pi)pei

= (1)pe + (p)pe1 = pe pe1 = pe(1 1/p)

We know that is multiplicative by the computation above, so if gcd(m, n) = 1, then

(mn) =d|mn

(d)mn

d

=d1|m

d2|n

(d1)(d2)m

d1

n

d2

= (m)(n).

Therefore, if n =

peii , (n) =

(peii ) = n

p|n

1 1

p

.

References

[J1] N. Jacobson, Basic Algebra I, W. H. Freeman and Co., 1974.

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Lattices n Boolean Algebra