 
 
 
 
 
   
 ,
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
,
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 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.)
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
.
Let RS 
denote the free (commutative) boolean algebra over 
 generated 
by idempotents corresponding to the elements of S; that is, define
generated 
by idempotents corresponding to the elements of S; that is, define
![\begin{displaymath}R_{S}
:= \frac{\mbox{\bf F}_2[x_{s}\, :\, s \in S]}{(x_{s}^{2}- x_{s}\, : \, s \in S)}.
\end{displaymath}](img6.gif) 
 from
from 
 to RS by
to RS by 
 
 maps to the sum of the images 
of its elements.  As RS is spanned by its monomials,
maps to the sum of the images 
of its elements.  As RS is spanned by its monomials,  is a 
bijection.
is a 
bijection.
We distinguish the element 
 of RS; it corresponds under
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.
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.  
 is a parity structure for S 
if and only if
is a parity structure for S 
if and only if  is in the ideal
is in the ideal 
 .
.
 ,
with t a monomial in an even number of the 
generators xs, form an
,
with t a monomial in an even number of the 
generators xs, form an 
 -basis for the ideal
-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
of parity structures for S.
If S is empty, then all subsets of 
 are parity structures, 
and
are parity structures, 
and 
 generates R as an ideal, so the theorem holds.  
Hence we may assume that n is positive.
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
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
.
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
of  will vanish on b 
if c is not a subset of b and will give the value 1 otherwise.  
Hence
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
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.
is 0.  
Since polynomial specialization gives ring homomorphisms, for each 
odd subset b of S, the map  from R to
from R to 
 given 
by
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
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
.
Then the 
 matrix with (i,j)-entry equal 
to
matrix with (i,j)-entry equal 
to 
 is upper triangular with 1's on the 
main diagonal, so the homomorphisms
is upper triangular with 1's on the 
main diagonal, so the homomorphisms  are linearly 
independent over
are linearly 
independent over 
 .
Hence P has 
dimension 
2n - 2n-1 = 2n-1 as an
.
Hence P has 
dimension 
2n - 2n-1 = 2n-1 as an 
 -vector space.
-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
.
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
is  ,
we have
,
we have 
 .
Since P is an ideal, it 
contains I as a subset.
.
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
.
Notice that  and
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
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 
 and, hence, also I and J, so I and J have the same dimension 
as vector spaces over
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
.
Since their sum R 
has dimension 2n, the ideals I and J have dimension 2n-1 
over 
 .
Thus I and P have the same
.
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).
-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
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
.
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 completes the proof of the theorem.   
This result allows us to count immediately the parity structures for a finite set.
 ,
a set of size n has 
22n-1 parity structures.
,
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.
-vector space of dimension 2n-1.   
 
 
 
 
