A Remark on Martin's Conjecture

We prove that thc strong Martin conjecture is Ihlse. The counterexample is the first-ordcr thcory of inlinitc atomic Boolea11 algcbras. We show that for this class ol'Roolean algchras. the classilication of their (OJ + 01)-clcmcntary theorics can hc rcduccd t:, thc classification of thc elementary thcories of theilquotient algebras ~llodulo thc Frcchi-t ideals. Martin's conjecture is a strengthening of Vaught's conjecture. Let T be a complete consistent theory in L,,,,,. Define L 1 ( T )to be the smallest fragment of L,,,,., , , , , containing L TI('U) be the complete theory of 'U in the language L I( T ) .Martin's conjecture says that, if T has fewer than 2Nncountable models, then T I('21) is No-categorical for all countable models 'U of T. The statement implies that, if T has fewer than 2Nu countable models, then the Scott ranks of all countable models of T are 5 w + w : thus in particular, T has at most countably many models up to isomorphism. In [W] C. M. Wagner considered a strengthening of Martin's conjecture which she called strong Martin conjecture. This is by adding to the conclusion of Martin's conjecture the statement that, if T has 2 N ~ countable models, then T has 2Nodistinct U {/\ ,, cp(x)/ p E S , , ( T ) , n < a}. For a model 'U of T, let , L,,,+,,J is a countable fragment of ( T ) LI Since ( T ) . LI completions in this implies that T has 2 N ~ ,,,. models with distinct complete theories in LC,)+, Wagner verified the strong Martin conjecture for theories of linear orders and of one unary function. In this paper we make the remark that the strong Martin conjecture is false. The counterexample is in the theory of Boolean algebras. Throughout this paper let us fix the signature L = (U, C1, C,0 , l )of Boolean algebras. Our main result is the following theorem. THEOREM1. There is a complete consistent theory T in L , ,,, such that (i) T has 2N0countable mo~/els, and (i i) There are at most countably many models of T with distinct complete theories in L, , ,!I 0,. Before further explaining the theorem let us recall Tarski's analysis of the elementary theories of Boolean algebras (c.f. [K], $18). For any Boolean algebra 'U, the Ershov-Tarski ideal I (%) is the ideal of 'U defined by I ( % ) = {x uy I x is atomic and J] is atomless). Rece~ved November 8. 1999; Rev~sed June 26 1999 By induction we can define for n E w the n-th itemtedErshov-Turslci ideal I " ('U) of 'U as follows. Let 10('U) = (0) and I[(%) = I(%). For n > 0, let 71, : 'U -+ %/I1'(%) be the canonical homomorphism. Then let If'+' (a)= ~;l[I('U/In('U))]. The Tarski invariants are triples taken from the countable set { ( l , O , O ) , ( w , O , O ) } ~ { ( k , l , m ) / l c ~ w , l { O , l } , m ~ c o U { c o } , l + m # 0). To be specific, for a Boolean algebra 'U, the elementarjl invariant inv('U) of 'U is defined as follows. inv('U) = ( 1,0 ,0) if 'U = (0) is the trivial Boolean algebra. inv(2) = (w, 0,O) if In (%) # '21 for all n E w. Otherwise, let lc be the least n E w such that I"(%) # 'U but = 'U: let 1 = 0 iff there is no atomless element in 'U/Z~(a);and let m be the number of atoms in 'U/Ik('U). Then inv('U) = (k, I, m). Tarski's analysis culminates in his theorem that the elementary invariants are complete for elementary theories of Boolean algebras. Thus in particular, there are only countably many distinct complete elementary theories of Boolean algebras. Now coming back to our theorem. the theory T in question is just the theory of infinite atomic Boolean algebras. By Tarski's analysis, the models of T are exactly those which have elementary invariant ( O , O , w), and T is in fact a complete consistent theory. It is also well-known that T has 2N"countable models (c.f., e.g., [I]). Thus to establish Theorem 1, it only remains to verify clause (ii) of the conclusion. For this we prove the following theorem. For any Boolean algebra 'U, we denote by F('U) the ideal generated by the atoms of 'U, and call it the Frechtt ideal of 'U. THEOREM2. Let '21 and T3 be infinite atomic Boolean algebms. Then 'U -,I ,23 zff %IF(%) = B/F(%) . Now Theorem 1 follows from Theorem 2 immediately by Tarski's analysis. Note also that to get Theorem 1 we only use one direction of the equivalence in the conclusion of Theorem 2, so we are obtaining additional information from Theorem 2. The rest of the paper is organized as follows. Section 1 contains the proof of the inessential direction of Theorem 2. In section 2 we prove the other direction. Then we conclude with some corollaries of the proofs and a discussion of some related problems, which constitute section 3. $1. From algebras to quotients. In this section we show that if two arbitrary Boolean algebras are (w +w)-elementarily equivalent, then their quotients modulo the FrechCt ideals are elementarily equivalent. This is done by observing that any first order sentence describing a property of the quotient can be translated to an infinitary sentence describing a property of the original algebra. LEMMA3. For any sentence cp E L,,,, there is a sentence cp* E L,, ,,such that, for anjl Boolean algebra 'U, PROOF. For a ,h E 2,let a \ h be an abbreviation of a n C(h). Let ty(x1, x2) be an L,,,,,, formula expressing that (xl \ x2)U (x2\ X I ) is in the FrechCt ideal, e.g., V, B, 1 1 1 ( ~ 1 , x2) where B, ,,, states that there are distinct atoms a1,. . . ,a, and h l , . . . ,h,,, such that X I U a1 U . . . U a, = x2 U hl U . . . U h,,. For all formulas cp(x) we define cp*(x) by induction on the form of cp. A REMARK ON MARTIN'S CONJECTLJRE 403 If cp is the atomic formula X I = x2, then cp* is just t y (x l , x2 ) . If cp is xl Ux2 = x3, then cp* is y ( x l U x2, x3 ) . If cp is xl C1 x2 = x3, then cp* is ty(xl n x2, x 3 ) If cp is C ( X I ) = xz, then cp* is ty ( C ( X I ) , x2) .For nonatomic formulas the induction is the trivial one, i.e., cp* is built up from the atomic cases in exactly the same fashion as cp is. By induction it is easy to see that for any formula c p ( $ E L,.,, Boolean algebra 'U, tuples 7 = ( c l , .. . ,c,) E %IF(%)and 2 = ( d l , . .,d,) E 'U with dl E c, for every i = 1 , . . . , n , we have 'U/F('U)+ cp(7)iff 'U b cp*(z). In particular, if cp is a sentence, then %IF(a)b cp iff 'U b cp* . 4 Now using Lemma 1 we can show one direction of Theorem 2 in its full generality. LEMMA 4. Let '21 and % he Boolean algehvas. If' 'U =, % then %IF ('21) = BIF(23) . PROOF. Suppose 'U =o %. Then for any sentence cp E L,,,, %IF(%)b cp @ a b cp* @ % b cp* H% I F ( % )b cp. i In fact a lot more is true in this direction. defining an ideal in an arbitrary Boolean algebra, we have that 'U =, +, % implies 'U/J('U)= !B/J(%) . §2. From quotients to algebras. In this section we prove the backward direction of Theorem 2, i.e., if two models of T have elementarily equivalent quotients modulo Frechet ideals, then they are (cu + co)-elementarily equivalent. We use games to deal with these notions of elementary equivalence. For the first-order elementary equivalence we use the standard Ehrenfeucht-Fraissk game (c.f., e.g., [HI).For the (co + a)-elementary equivalence we use the game notion defined below. Let n E w and %R, TI be models of T. The game G, (%R, T I ) is played in the same manner as an n + 1 step Ehrenfeucht-Fraisse game: where for each i 5 n the elements x, and y, are from different structures. Suppose ao, . . . ,a , E %R, ho,. . . ,h, E TI are all the elements played by the players. Then player I1 wins if (9X, ao, .. . ,a,) = (TI, ho, . . . ,b,). By a standard argument we have that %R =, ,X iff player I1 has a winning strategy in G,(%R,T I ) for any n E w. Let us denote by E,(9X, T I ) the n + 1 step Ehrenfeucht-Fraisse game played on the structures %R and TI. For notational convenience we also denote by H, (9X,T I ) the n + 1 step Ehrenfeucht-Fraisse game played on the structures %R/F(%R) and TI/F ( X ) . Then %R/F(%R) = TI/F ( T I ) iff player 11 has winning strategies in the games H, (s337,TI) for all n E w . For any Boolean algebra 'U and a E 'U, we denote by 6 the restricted algebra with domain { x E 'U 1 x 5 a } . LEMMA5. Let n E w , ao, . . . ,a , E 9X and ho, . . . ,h, E TI. Then (%R, ao, . . . , a,) = (TI,bo, . . . , h,) z j j