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Author: William A. Stein
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Free boolean algebras and generating functions
Next: Another property of parity Up: Parity structures and generating Previous: Introduction

# Free boolean algebras and generating functions

Call a finite set even or odd according to the parity of its size. Let S be a finite set, and let T be any subset of  , the power set of S. We say that T is a parity structure for S if, for each odd subset b of S, the intersection  is even; that is, if each odd subset b has an even number of subsets that lie in T. A logician might imagine that T consists of the subsets of S that are in some model, so that  is the power set of b within the model. We are simply requiring then that, in this model, every odd set have an even number of subsets. (This analogy is not perfect, of course, since b need not be in T, and  T is not necessarily closed under unions, but it is an interesting point of view to take.)

In order to study parity structures for S, we introduce a ring whose elements correspond naturally to subsets of  . Let RS denote the free (commutative) boolean algebra over  generated by idempotents corresponding to the elements of S; that is, define

(Note that this is similar to, but different from, the definition of a Stanley-Reisner ring, in which we take  xs2 = 0, rather than  xs2 = xs.) Since RS is spanned by idempotents and has characteristic 2, it follows that all of its elements are idempotents. Define a map  from  to RS by

Under this map, each subset of S maps to the monomial that is the product of the indeterminants corresponding to the elements of the subset, and each subset of  maps to the sum of the images of its elements. As RS is spanned by its monomials,  is a bijection.

We distinguish the element  of RS; it corresponds under  to the collection of subsets of S of size at most 1. Using RS we obtain a ring-theoretic characterization of the parity structures for S.

Theorem 1   Let S be a finite set.
(a)
A subset T of  is a parity structure for S if and only if  is in the ideal  .
(b)
The elements , with t a monomial in an even number of the generators xs, form an  -basis for the ideal  .

PROOF:Let S have size n, write R for RS, and let P be the set of all images under  of parity structures for S. If S is empty, then all subsets of  are parity structures, and  generates R as an ideal, so the theorem holds. Hence we may assume that n is positive.

We first translate the set-theoretic condition for the subset T of  to be a parity structure into an algebraic condition on . For any element t of R and any subset b of S, write t(b) for the value of the polynomial t with each indeterminant xs set equal to 1 if s is in b and set to 0 otherwise. This can be thought of as the evaluation of t on the characteristic vector of b. For each subset b of S and each element c of T, the term  of  will vanish on b if c is not a subset of b and will give the value 1 otherwise. Hence  gives the parity of the number of subsets of b that lie in T. Therefore, T is a parity structure if and only if, for every odd subset b of S, the value of  is 0.

Since polynomial specialization gives ring homomorphisms, for each odd subset b of S, the map  from R to  given by  is a homomorphism of rings. And, as the previous remarks show that P is the intersection of the kernels of these maps for all such subsets b, we see that P is an ideal of R. Now order the odd subsets of S by increasing size (with any order chosen for subsets of the same size) as  . Then the  matrix with (i,j)-entry equal to  is upper triangular with 1's on the main diagonal, so the homomorphisms  are linearly independent over  . Hence P has dimension  2n - 2n-1 = 2n-1 as an  -vector space.

Next we verify that P contains the ideal I generated by  . If b is any odd subset of S, then b has |b| + 1 subsets of size 0 or 1, and |b| + 1 is even. Hence the collection of subsets of S of size at most 1 is a parity structure, and as the image of this collection under  is , we have  . Since P is an ideal, it contains I as a subset.

Now we show that I and P have the same size, hence must be equal. Let J be the ideal  . Notice that  and  are orthogonal idempotents with sum 1, so R is the direct sum of I and J as rings. Also, the linear map from R to itself sending xs0 to  1 + xs0 (for some element s0 of S) and fixing xs for all other values of s is an automorphism of R (since R can be viewed as the free boolean algebra generated by  1 + xs0 and the other indeterminants xs). This automorphism interchanges  and  and, hence, also I and J, so I and J have the same dimension as vector spaces over  . Since their sum R has dimension 2n, the ideals I and J have dimension 2n-1 over  . Thus I and P have the same  -dimension, and since I is a subset of P, they must be equal. This completes the proof of part (a).

Finally, we prove part (b). For every even-degree monomial t, the product  is the sum of t and some odd-degree monomials, so these products, with t ranging over all even-degree monomials, are linearly independent over  . But there are 2n-1 such products, and by the above calculations, the dimension of P is 2n-1, so the products form a basis for P over  . This completes the proof of the theorem.

This result allows us to count immediately the parity structures for a finite set.

Corollary 1   For , a set of size n has  22n-1 parity structures.

PROOF:Theorem 1 shows that the parity structures are in bijection with an  -vector space of dimension 2n-1.

Next: Another property of parity Up: Parity structures and generating Previous: Introduction
William Arthur Stein
1999-10-27