Download splitting in relation algebras - American Mathematical Society

proceedings of the
american mathematical society
Volume 111, Number 4, April 1991
SPLITTING IN RELATION ALGEBRAS
H. ANDRÉKA, R. D. MADDUX, AND I. NÉMETI
(Communicated by Andreas R. Blass)
Abstract.
We define a way, which we call splitting, of getting new relation
algebras from old ones. We characterize those algebras to which splitting can
be applied. We show how to split representable relation algebras in order to
obtain nonrepresentable ones, and we give many examples.
Introduction
The method of splitting atoms, originating with L. Henkin, is well known
in cylindric algebra theory and is used to obtain nonrepresentable cylindric
algebras from representable ones (see [HMT, 3.2.67 and 3.2.69]). In this paper
we adapt this method to relation algebras. The conditions for splittability in
relation algebras seem to be more complex than in cylindric algebras, where
every atom below d(a x a) can be split.
We use the terminology and notation of [Mai]. In particular, we will deal
with the varieties NA, WA, SA, and RA, of nonassociative, weakly associative,
semiassociative, and associative relation algebras. These varieties are defined
in [Ma 1] and are obtained from the variety RA of relation algebras by weakening the axiom postulating associativity of relative multiplication. If 21 is any
Boolean algebra with additional operations, then At 21 denotes the set of all
atoms of the Boolean reduct of 21. Finally, our convention is that ; binds more
strongly than •. For example, x ; x • 0' denotes the term (x ; x) • 0'.
1. Definition
and basic properties
of splitting
Definition 1. Let 21 and 03 be atomic NA's. We say that 21 is obtained from
03 by splitting if the following three conditions are met:
(1) 213»;
(2) every atom x of 21 is contained in an atom c(x) of 03, called the
cover of x ; and
Received by the editors October 30, 1988 and, in revised form, May 18, 1990.
1980 Mathematics Subject Classification (1985 Revision). Primary 03G15.
Research supported by the Hungarian National Foundation for Scientific Research grant No.
1810.
©1991 American Mathematical Society
0002-9939/91 $1.00+ $.25 per page
1085
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
H. ANDREKA, R. D. MADDUX, AND I. NÉMETI
1086
(3) for all x, y e At 21, if x, y < 0' then
= (c(x);c(y)-0'
if x ¿ y,
\c(x);c(y)
ifx = y.
Let an and as be functions mapping At 03 to cardinals. We say that 21 is
obtained from 03 by splitting along an and as if 21 is obtained from 03 by
splitting, and for all x G At 03,
an(x) = \{yeAt?í:y<x,y¿y}\
and
as(x) = \{yeAt<A:y<x,y
= y}\.
The following lemma says that the result of splitting a complete atomic NA
along two functions, if it exists, is unique up to isomorphism. Its proof is
straightforward and is therefore omitted.
Lemma 2. Let 21, 2l', and 03 be complete atomic NA's. Let an and as be
functions mapping At 03 to cardinals. If 21 and 2t' are obtained from 03 by
splitting along an and as, then 21 and 2l' are isomorphic by an isomorphism
which leaves 03 fixed.
The next theorem characterizes those functions along which an atomic NA,
WA, SA, or RA can be split so as to obtain an algebra in the same class. Intuitively, in NA we have only the obvious conditions, namely, (a) in Theorem 3;
in WA, the additional condition is that we cannot split "nondiscrete" identity
elements; in SA, the additional condition is that we cannot split nondiscrete
functional elements; and in RA we have a slightly stronger condition.
If 21G NA, then Fn2t is the set of functional elements of 21. Thus Fn2l =
{x G A :x ;x < 1'}.
Theorem 3. Let 03 be an atomic NA. Let an, as be functions mapping At 03
to cardinals, and let a(x) = an(x) + as(x). Consider conditions (a)-(d):
(a) For all x e At 03, a(x) > 1, an(x) — an(x), if x < 1 ', then an(x) = 0;
if x = x, then a„(x) = Ik for some k ; and if x ± x, then as(x) = 0.
(b) For all x G At 03, if a(x) > 1 and 0' ; x ^ 0, then x < 0 '.
(c) For all x e At 03, if a(x) > 1 and 0 ' ; x ^ 0, then x £ Fn2l.
(d) For all x,y € At®, if a(x) > 1, y ;x ¿ 0, and y < 0', then y <
y;(x;x-0r).
Then
(i) There is an atomic 2t G NA obtained from 03 by splitting along an and
as iff (a) holds.
Now suppose 21G NA, 21 is atomic, and 21 obtained from 03 by splitting along
an and as. Then
(ii) 21G WA iff 03G WA and (b) holds,
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
SPLITTING IN RELATION ALGEBRAS
1087
(iii) 21G SA iff'Be SA and (c) holds,
(iv) 21g RA //J 03g RA and (d) holds.
Proof. First we prove the "only if parts. Assume 21 and 03 are atomic NA's
and 21 is obtained from 03 by splitting along an and a. Then (a) holds by
the elementary laws governing " in NA's (namely, by [Mai, 1.13(16)(12)(8)]).
Clearly, if 21 G WA, then 03 G WA by Definition 1(1), and similarly for SA
and RA. Next we show that (b), (c), or (d) holds whenever 21 is in WA, SA,
or RA, respectively.
Assume 21 G WA, 1' > x G At03, and 0';x ^ 0. Then there is some
y e At 03 such that y < 0' and y ; x ^ 0. Then y ; x = y since x < 1' and
y G At03. Recall that yr = y ;y ■V and yd = y ;y ■V (see [Mai, 5.10]).
Then y = x by [Ma2, 5.12(3)]. Suppose y > y0 e At21. Then yr0= x by
Definition 1(3); hence, x G At 21 since y0 e At SI, by [Ma2, 5.12(1)]. Therefore
a(x) = 1 .
Assume 21G SA, x G At 03, a(x) > 1, and 0' ; x ¿ 0. Then x < 0' by (b),
and there are x0, xx e At21 such that x0 ^ xx and c(x0) = c(xx) = x. Then
Xg = xáx= xd by Definition 1(3), so xx = xx ; xx = xd ; xx = (x0 ; x0 • 1') ; xx <
(1 ; x0) ; xx = 1 ; (x0 ; x,) by [Mai, 5.11(3)], and the semiassociative law. Thus
x0 ; x¡ ^ 0. We have x0 ; xx = x ; x • 0' by Definition 1(3), so x ;x £ 1'.
To prove that (d) holds when 21G RA, we will need (*) below. We note that
(*) is true if 21G NA.
(*)
If x G At03, y e At21, and x, y < 0', then x ;y = x ; c(y).
To prove (*), assume x G At03, y e At21, and x, y < 0'. Clearly x ;y <
x ; c(y). For the opposite direction, first note that if x = c(y), then x ; c(y) =
c(y) ; c(y) = y;y < x;y.
Assume x ^ c(y). Choose z e At21 so that
x = c(z). Then z ^ y and x ^ c(yj, and hence x ; c(y) = c(z) ; c(y) • 0' =
^ ; y < x ; y . (*) has been proved.
Assume 21GRA, x,j>GAt03,
y;x^0,
y < 0', and a(x) > 1. Then
x < 0' by (b). Let x0, x, G At 21 be such that x0 ^ xx and c(x0) = c(Xn) = x .
Then x0 ;xx = x ; x-0' by Definition 1(3), and 0 5¿y ;x = y ; c(x0)-y ; c(xx) =
y ; x0 • y ; xx by (*), so 0 / y ■(y ; x0) ; xx = y ■y ; (x0 ; x,) = y ■y ; (x ; x • 0')
by the associative law, and therefore y < y ; (x ; x • 0').
We now turn to proving the "if parts. Let 03, an, as, and a be as in the
statement of the theorem, and assume (a) holds. We may assume that 03 is
complete. We will construct a complete atomic 21 G NA that is obtained from
03 by splitting along an and as. Then we prove that 21G WA if (b) holds and
03 G WA, that 21G SA if (c) holds and 03 G SA, and that 21G RA if (d) holds
and 03 G RA.
First we will construct a copy of 21, namely St'. It will be the complex algebra
of a certain relational structure (U, C, f, I), where C ç U x U x U, f is an
involution of U, and I ç U. (See [Ma2, Definition 2.1].) For every x G At 03,
let S (x) and S (x) be sets such that \Sn(x)\ = a (x), \S (x)\ - a (x), and,
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1088
H. ANDREKA, R. D. MADDUX, AND I. NÉMETI
for all i, j e {n, s} and all x, y e At03 , we have S¡(x) nS ¡(y) = 0 if i =¿j
or x / y . Set U = \J{Sn(x)uSs(x)
: x e At 03}. Let / be a fonction mapping
U to U such that for all y e U we have ff (y) = y , f(y) = y iff y e Ss(x)
for some x G At03, and for all x G At03, if y e Sn(x), then f(y) e Sn(x).
Such a fonction / exists by (a). Let c be the fonction mapping U to At03
defined by c(y) = x if y e Sn(x) U Ss(x). For all x, y, z e U ,let
[x,y,
z] = {(x,y,
z), (fx, z, y),
(y, fz, fx), (fy, fx, fz), (fz,x, fy), (z, fy, x)}.
Let
C=\J{[x,y,z]:
c(x) ; c(y) > c(z) and c(x), c(y), c(z) < 0'}
U (J {[x, y, y] : c(x) ; c(y) > c(y), c(x) < V, and
either c(y) < 0'or y = x) ,
and let I = {x:xeU
and c(x) < 1'} . Set 2i' = €m(U, C, f, I).
It follows directly from the definitions and [Mai, 2.2(2)] that Si' G NA. For
every b e B let h(b) = {y : y e U and c(y) < b}. It is easy to check that «
is an embedding of 03 into 2l', and that 2l' is obtained from the image of 03
by splitting along an and as. We may then obtain 21 by replacing the image
of 03 in Si' with 03 itself.
Now assume (b) and 03 G WA. To show Si' G WA it suffices to check
condition [Mai, 2.2(d)]. Assume w e I and (w, x, x) e C and (x,y, z) e
C. Then c(w) ; c(x) > c(x) and c(x) ; c(y) > c(z ) in 03 ; hence, c(w) ; c(z) >
c(z), since 03 G WA . If c(z) < 0', then (w, z, z) e C, and we are done.
Assume c(z) < V. Then c(w) = c(z). If c(x) < V, then x = y = z, since
(x, y, z) e C. Hence (w, z, z) e C, and we are done. If c(x) < 0', then
c(xj < 0' and 0 ^ c(xj < c(xj ; c(w), so a(c(w)) = 1 by (b). Hence z = w ,
since c(z) = c(w), so (w , z, z) e C, and we are done.
Now assume (c) and 03 G SA. We want to check [Mai, 2.2(e)]. Let
(v , w , x), (x, y, z) e C. We have to find some u e U for which (v , u, z) e
C. We get (b) from (c), since for every x G At 21, either x < 1' or x < 0', but
if x < V then x G Fn2l. From what has been proved so far, we have 2l' G WA,
and [Mai, 2.2(d)] holds. If v e I, then w = x by the definition of C, so we
have (v , x, x), (x, y, z) e C, and hence (v , z, z) e C by 2.2(d), so u = z
will do. From (v , w , x), (x, y, z) e C, we get (z, f(y), x), (x, f(w), v) e
C. If z e I, then f(y) = x, so (z, x, x), (x, f(w), v) e C; hence,
(z, v, v) e C by 2.2(d), and finally (v, f(v), z) e C, so u = f(v) will do.
Therefore we may assume v £ I and z £ I. Since (v, w, x), (x, y, z) e
C, we have c(v);c(w) > c(x) and c(x) ; c(y) > c(z). Since 03 G SA,
c(z) < (c(v) ; c(w)) ; c(y) < (c(v) ; 1) ; 1 = c(v) ; 1, so there is some c(u) G
At03 such that c(v);c(u) > c(z). If c(u) < 0', then (v,u, z) e C, and
we are done. Assume c(u) < V. Then c(v) = c(z). If v = z, then
(v ,u, z) = (v,u,v)
G [u, f(v), f(v)] Ç C. If v ¿ z, then a(c(v)) > 1 ,
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1089
SPLITTING IN RELATION ALGEBRAS
and c(u) < c(z)~ ; c(v) < 0' ; c(v), so c(v) £ Fn2l by (c). Hence there is some
c(u) G At 03 such that c(u') < 0' and c(v) ; c(u) > c(z). Now (v, u , z) e C,
and we are done.
Assume (d) and 03 G RA. To check [Mai, 2.2(f)], let (v, w, x), (x, y, z)
e C. We have to find some u e U for which (v, u, z), (w , y, u) e C. First
note that (b) holds, since clearly (d) implies (c), so 2l' G WA and 2.2(d) holds. If
v , w , y , or z is in / then let u be z , y , w or f(v ), respectively. In each of
these four cases it is easy to show, by arguments similar to those in the previous
paragraph, that (v , u, z), (w , y, u) e C, using the fact that 2.2(d) holds.
Therefore assume v , w, y, z £ I, and hence c(v), c(w), c(y), c(z) < 0'.
Because 03 G RA, there is some c(u) e At03 such that c(v);c(u) > c(z)
and c(w);c(y) > c(u). If c(u) < 0', then we are done. Assume c(u) <
V. Then c(v) = c(z) and c(w) = c(yj. If v = z and w = f(y), then
(v , u, z), (w , y, u) e C and we are done. Assume v ^ z . (The case w ^
f(y) is completely analogous, and therefore we omit it.) Then a(c(v)) > 1 .
By (v , w , x) G C and (d), we have c(y) < c(w)~ ; (c(vj ; c(z) • 0'), so there is
some c(u') G At 03 such that c(y) < c(wj ; c(u) and c(u') < c(vj ; c(z) • 0'.
Then c(u') < c(w) ; c(y) and c(z) < c(v) ; c(u), so (v , u , z), (w , y, u) e C
and we are done. D
We now turn to investigating the condition of splittability in RA's.
Definition 4. Suppose 21G RA, 21 is atomic, and x G At 21. We say that x ¿s
splittable in 21 if x < 0' and y < y ; (x ; x • 0') whenever 0' > y G At 21 and
y ;x ^ 0, and y < (x ; x • 0') ; y whenever 0' > y e At2l and x ; y ^ 0. We
say that 21 is splittable iff 21 has a splittable atom.
By Theorem 3, x is splittable in 21 iff there is some 03 G RA obtained from
21 by splitting such that x ^ At 03 . Note that if x is splittable then there is no
y e Fn2lDAt2l such that y < 0' and either y = xd or yö = xr. The following
lemma says that the converse holds if x = x ; 1 ; x .
Lemma 5. Let 21G NA, x G At SI. Consider statements (l)-(4):
( 1) for every yGAt2l, if y ; x ¿ 0 and y < 0 ', then y < y ; (x ; x ■0 ");
(1) for every y e Fn2ln At SI, (0'-y) ;x = 0;
(3) for every y e Fn2l n At21, if y < 0' then y # xd ; and
(4) for every y e Fn2l n At 21, if y = xd, then y = xd.
Then (1) implies (1), (3), and (4). If 21G SA and x = x ; 1 ; x, then (l)-(4)
are all equivalent.
Proof. It is easy to show that (1) implies (2) in every NA. Assume SI e SA and
x = x ; 1 ; x . Assume that (2) holds.
want to show that y < y ; (x ; x • 0').
since y e At 21. From y e At 21 and
Using these equations, 21G SA, and
y;y<y;lA;y<x;l;lA;(l;x)
Let y G At 21, y ; x ^ 0, and y < 0'. We
We have y £ Fn2l by (2), so y < y ; 0',
y ; x / 0 we get y < 1 ; x and y < x ; I.
x = x ; 1 ; x, we get
= x;lA;x<(x;l;x);x
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
= x;x,
1090
H. ANDRÉKA, R. D. MADDUX, AND I. NEMETI
so y = y ■y ; 0' < y ; (y ; y • 0') < y ; (x ; x • 0'), as desired.
It is easy to see that if 21G SA then (2)-(4) are equivalent by the elementary
laws governing x and xr [Mai, 5.11 and 5.12]. D
2. Examples
Examples 1, 2, and 3 show that the condition x = x ; 1 ; x cannot be omitted
in Lemma 5. Examples 4, 5, and 6 show how to split atoms in order to obtain
nonrepresentable RA's from representable ones.
Example 1. Let 21 be the following RA: AtSl = {1\ a, ¿}, a = a, b = b,
a; a = b + 1', b ; b = a + V, and a;b = b;a = a + b . It is fairly easy to
prove that 21 has only one representation, the one given in Figure 1. (More
precisely, 21 is embeddable in 9\tU, the algebra of all relations on U, just in
the case that |t/| = 5 and the relations correlated with a and b look like those
in Figure 1.)
Now 21 is not splittable, because a -¿ a ; (b ; b • 0') and b jÇb ; (a ; à • 0').
Example 2. Nonminimal RA's in which all the atoms are equivalence elements
are not splittable, because of the following: Suppose 21 G RA, 21 is atomic,
|At2l| > 3, and for every a G At 21, à = a and a;a = a + V. (It follows
that 1' G At 21.) Let 0' > a e At SI. Then there is some b e At 21 such that
b < 0' and a ^ b. Now a ; ä • 0' = a, but b £ b ;a, showing that a is not
splittable. Examples of such algebras are the Lyndon algebras (see [L] or [J]).
In a Lyndon algebra, if a and b are atoms distinct from each other and from
1', then a = a , a ; a = a + V, and a ; b = (a + b + V)~ .
Example 3. We now examine the three minimal RA's. We will see that two of
them are splittable and that all the algebras obtained from them by splitting are
representable.
Let 21 be a minimal RA. Then At2l ç {V, 0'} and 0'; 0' G {0, 1', 1}. If
0' = 0, then 1' = 1 . Hence 21 is splittable. The algebras obtained from 21 by
splitting are all Boolean relation algebras, and hence are representable. Assume
0';0' = 1'. Then 0' G FnSl; hence, 1' and 0' are not splittable in 21 since
a
/
•.-./■•
\
a.
6A.•;.•
W VI
a
! b
b \
a
\/AXA\/
•'-a-•
Figure
1. Representation of 21
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1091
SPLITTING IN RELATION ALGEBRAS
RA ¡ \
kL
R"
•¡A.■/■■■
S'X.•••
\\
?'
R'
/"'S'
jR'Í
•1y-.f»S""\.■:.•
k ..••••""/
S'"'\
/ \
v "•••s" k' .■■•""/
R'
R"
\zXy
¡"'S". S'r'\
\//\y
-R'-'■'•
R"
••■-R'
Figure 2
0' ; 0V 0 and 0' ; 1' ¿ 0. Thus SI is not splittable. Assume 0' ; 0' = 1. Then
0' is splittable in 21 and the algebras 03 obtained from 21 by splitting are the
following: At© = {V} U S U N, à = a for all a e S, à # a for all a e N,
and a ; a = 1, a ; b = 0' for all a, b G S U N with a ^ b . These algebras are
all representable by [Mai, 5.19]. In the case N = 0, these algebras are called
£„({1, 2, 3}) in [Mai, 2.4], where « = |At03|.
Now we use splitting to obtain nonrepresentable
ones.
RA's from representable
Example 4. Let 21 be the following proper relation algebra: Set U = U' U U" ,
where U' CIU" = 0 and \U'\ = \U"\ = 5. Let R', S', R", S" be the
relations shown in Figure 2, and let 21 be the proper relation algebra with
atoms {Idy,, R!, S', Id^» , R", S", U' x U", U" x U'} .
Now by Lemma 5, II' x U" is splittable in 21. Let 03 be the RA obtained
from 21 by splitting U' x U" into at least six atoms. Then 03 is not representable, because the only representations of Id^-, R', S' and Id^" , R", S"
are the ones shown in Figure 2, and clearly there cannot be six disjoint relations
below U' x ¡j" with domain U'.
It is also true that splitting U' x If" into two parts already yields a nonrepresentable RA (with ten atoms). When showing this in Example 4a below,
we modify Example 4 so that we obtain a symmetric integral nonrepresentable
03 G RA with five atoms.
Example 4a. Let everything be as in Example 4. Let R = R' + R" , S = S'+S",
and D = (U'xU")+(U"xU1).
Let 2l' be the proper relation algebra with atoms
Idy, R, S, and D. It can be checkedthat D ; Ù ■0' = Ù ; D ■0' = R + S, and
y <y;(R
+ S) for all y e{R,S,D},
so D is splittable in St'. Let 03' G RA
be obtained from St' by splitting D into two symmetric parts, say into D1 and
D" . Then 03' has five atoms, Id^ , R, S, D', D" . Assume 03' is representable.
It can be shown that Si' has only one representation (up to isomorphisms of
the ten-element set U ). Therefore we may assume that the representation of
03' is the same as that of Si' and D = tí + D" . Let u e U be arbitrary. Let
D'(u) = {v e U" : (u, v) e D'} and D"(u) = {v e U" : (u, v) e tí'}. Then
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
H. ANDREKA, R. D. MADDUX, AND I. NÉMETI
1092
tí(u) n tí'(u) = 0 and \tí(u) U tí'(u)\ = 5 . Now, since tí < D = tí ; R =
tí ; S and 1' • (R + S) = 0, we have \tí(u)\ > 3, and similarly |2?"(k)| > 3,
since D" < D" ;R = D" ;S. This contradicts \tí(u)UD"(u)\ = 5 . Thus 03' is
a nonrepresentable symmetric integral RA with five atoms.
The essence of Example 4 was that we had an atom U' x U" such that the
structure of the algebra forced U' to be small in any representation. We then
split U' x U" into too many parts. Now we use the same idea to obtain a series
of nonrepresentable RA's.
Example 5. Let p = In + 1 be any prime number. We will construct a nonrepresentable 03n e RA. We start with a proper relation algebra 21^ on p =
{0, 1, ... , p - 1} . In the following, + and - are understood modulo p . For
i e p , define
Rj = {(«, m) : |« - m\ = i (modp),
«, m e p}.
This gives us a proper relation algebra 21'^ with atoms {R0, Rx, ... , Rn} . In
Si' we have R0 = ldp = V, and if i < j < n, then R¡ = RI, R¡ • Rj = 0,
1 = p x p = R0 u ■• ■U Rn , and R¡ ; Rj = R¡+j U R,_j .
Let 2ln be the proper relation algebra obtained from Ql'n by putting two
disjoint copies of 21'^ together as we did in Example 4; set U = if U U"
where U' n U" = 0 and \U'\ = \U"\ =p. Let R\,...,R'n
and R"x,..., R"n
be the corresponding relations on U' and U", respectively. Then At 2lw =
{Idy. ,R\,...,R'n,
Idy» ,R"X,..., R"n, U' x U", U" x [/'} . Let 03„ be the
algebra obtained from 2ln by splitting U' x U" into two atoms. By a slight
generalization of the argument in Example 4a, we see that 03n is not representable. First, it is easy to show that 21'^ has just one representation on a
p -element set, and hence that 2ln has just one representation on a lp -element
set, namely the one used to describe 2ln . But U' x u" = D + tí in 2ln , so for
any u e U' the images D(u) and D'(u) are disjoint subsets of U", each of
which contains at least « + 1 elements (since D = D ;R¡ and tí = tí ; R¡ for
/ = 0, ... , « - 1 ), contradicting \U"\ = p .
Exampleó. The algebras <£„({2, 3}) (see [Mai, 2.4]) are splittable. €„({2, 3})
has « atoms, namely 1' and «-1 atoms below 0', such that a = a, a; a— a,
and a ; b = 0' whenever a and b are distinct atoms below 0'. It is not
known whether every £„({2, 3}) is representable, but if a given <£„({2, 3}) is
embeddable in iHec7, then there cannot be three elements u, v, w G U such
that (u,v),
(v, w), and (u,w) are all in the relation correlated with any
diversity atom a, since a- a; a = 0. Therefore, by Ramsey's Theorem, for
every « there is some integer r(n) such that if <£„({2, 3}) is embeddable in
ÍHet/, then \U\ < r(n). Thus r(n) is also a bound on the cardinality of the
underlying set used to represent any relation algebra obtained from £„({2, 3})
by splitting. If 21 is obtained from <£„({!, 3}) by splitting one of its atoms
into more than r(n) parts, then any representation of 21 must also contain
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1093
SPLITTING IN RELATION ALGEBRAS
more than r(n) elements. This contradiction shows that such an 21 is not
representable.
Algebras constructed in this way can be used to prove Monk's theorem, that
RA is not finitely axiomatizable [Mo]. For more details on how this can be
done, see [Ma2, Theorems 12 and 13]. (In [Ma2], algebras obtained from
£„({2, 3}) by splitting are further modified so that they are generated by a
single atom.) Monk's original proof used Lyndon algebras. The algebras obtained from £„({2, 3}) by splitting are similar to the nonrepresentable cylindric algebras used by Monk to show the representable CAQ's are not finitely
axiomatizable whenever 3 < a < to (see [HMT, 3.2.76 and 4.1]).
References
[HMT] L. Henkin, J. D. Monk, and A. Tarski, Cylindric algebras, Part II, North-Holland, Amsterdam, 1985.
[J]
B. Jónsson, Varieties of relation algebras, Algebra Universalis 15 (1982), 273-298.
[L]
R. C. Lyndon, Relation algebras and projective geometries, Michigan Math. J. 8 (1961),
21-28.
[Mai] R. D. Maddux, Some varieties containing relation algebras. Trans. Amer. Math. Soc. 272
(1982), 501-526.
[Ma2] _,
Nonfinite axiomatizability results for cylindric and relation algebras, J. Symbolic Logic
54(1989), 951-974.
[Mo] J. D. Monk, On representable relation algebras, Michigan Math. J. 11 (1964), 207-210.
(H. Andréka and I. Németi) Mathematical
(R. D. Maddux) Department
Institute,
of Mathematics,
Budapest Pf. 127, H-1364 Hungary
Iowa State University,
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
Ames, Iowa 50011