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146 (1995)

M -rank and meager types

by

Ludomir N e w e l s k i (Wrocław)

Abstract. Assume T is superstable and small. Using the multiplicity rank M we find locally modular types in the same manner as U -rank considerations yield regular types.

We define local versions of M-rank, which also yield meager types.

0. Introduction. Throughout, T is superstable, small, and we work in C = C eq . In [Ne1] we defined the multiplicity rank M and proved that M has additivity properties similar to those of U -rank. In [Ne2] we defined the notion of meager regular type and proved that every such type is locally modular. It turns out that using M-rank we can produce locally modular regular types of prescribed M-rank. These types are either trivial or meager (the second case always holds when T has < 2

0

countable models).

We use the following notation. If s(x) is a (partial) type over C, then [s]

denotes the class of (partial) types over C containing s. For p ∈ S(A), St(p) is the set of stationarizations of p over C, St A (p) = {r|acl(A) : r ∈ St(p)}.

For B ⊇ A, S p (B) = S(B) ∩ [p] and S p,nf (B) = {q ∈ S(B) ∩ [p] : q does not fork over A}. We regard strong types over A as types over acl(A). We define Tr A (s) (the trace of s on A) as the set of types r(x) ∈ S(acl(A)) consistent with s(x). Thus if p ∈ S(A) then Tr A (p) = St A (p). In general, Tr A (s) is closed. Tr A (a/B) abbreviates Tr A (tp(a/B)). If a ∪ B (A) then sometimes | we use St A (a/B) to denote Tr A (a/B). Also, x A denotes the tuple of variables x indexed by elements of A. We will often tacitly use the following easy fact.

Fact 0.1. Assume A ⊆ B ⊆ C. Then either Tr A (a/C) is open in Tr A (a/B) or Tr A (a/C) is nowhere dense in Tr A (a/B).

P r o o f. Suppose Tr A (a/C) is not nowhere dense in Tr A (a/B). Since Tr A (a/C) is closed, this means that for some ϕ = ϕ(x, c 0 ) with c 0 ∈ acl(A),

∅ 6= [ϕ] ∩ Tr A (a/B) ⊆ Tr A (a/C). Thus we can choose a 0 realizing ϕ(x, c 0 )

1991 Mathematics Subject Classification: Primary 03C45.

Research supported by KBN grant 2-1103-91-01.

[121]

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with a 0 ≡ a (C). Now let r ∈ Tr A (a/C). So r = stp(a 00 /A) for some a 00 a (C). Choose c 00 with a 00 c 00 ≡ a 0 c 0 (C). Clearly, r ∈ [ϕ(x, c 00 )] ∩ Tr A (a/B) ⊆ Tr A (a/C). This shows that Tr A (a/C) is open in Tr A (a/B).

We define the multiplicity rank M on complete types p over finite sets A, with values in Ord ∪ {∞}, by the following conditions.

(1) M(p) ≥ 0.

(2) M(p) ≥ α + 1 iff for some finite B ⊇ A and q ∈ S p,nf (B), M(q) ≥ α and St(q) is nowhere dense in St(p).

(3) For limit δ, M(p) ≥ δ iff M(p) ≥ α for every α < δ.

M(a/A) abbreviates M(tp(a/A)).

Notice that we have proved in [Ne2] that if I(T, ℵ 0 ) < 2

0

then M(p) <

∞ for every p.

Suppose P is a closed subset of S(acl(A)) for some finite set A. We say that forking is meager on P if for every formula ψ forking over A, the set of types r ∈ P consistent with ψ is nowhere dense in P , that is, P ∩ Tr A (ψ) is nowhere dense in P .

Suppose p is a stationary non-trivial regular type. Let ϕ be a p-simple formula over a finite set A such that the p-weight of ϕ is 1 and for each a ∈ ϕ(C), if w p (a/A) > 0 then stp(a/A) is regular non-orthogonal to p. Let P ϕ = {r ∈ S(acl(A)) ∩ [ϕ] : w p (r) > 0}. Assume P ϕ is closed in S(acl(A)), and for each a ∈ ϕ(C) and b ⊆ C, if w p (a/b) = 0 then for some formula ψ(x, y) over acl(A) true of a, b, whenever ψ(a 0 , b 0 ) holds then w p (a 0 /b 0 ) = 0.

We call any ϕ with the above properties a p-formula (over A). Given p, a p-formula ϕ over some finite set A exists by [Hr-Sh]. We say that p is meager if forking is meager on P ϕ . In [Ne2] we show that this definition does not depend on the choice of ϕ, and also that if p is meager then p is locally modular. A complete regular type p is meager iff every stationarization of p is meager.

In the next lemma we collect the properties of M we shall use.

Lemma 0.2. (1) M(ab/A) ≤ M(a/Ab) ⊕ M(b/A).

(2) If p ∈ S(A) and A ⊆ B are finite, then M(p 0 ) = M(p) for some p 0 ∈ S p,nf (B).

(3) If a ∪ b (A) then M(a/Ab) + M(b/A) ≤ M (ab/A). |

(4) Assume A ⊆ B and a ∪ B (A). If St(a/B) is open in St(a/A) then | M(a/B) = M(a/A). If M(a/B) = M(a/A) < ∞ then St(a/B) is open in St(a/A).

(5) (Symmetry) If a ∪ b (A) and St(a/Ab) is open in St(a/A) then | St(b/Aa) is open in St(b/A) and M(ab/A) = M(a/A) ⊕ M(b/A).

P r o o f. (1)–(3) are proved in [Ne1]. (4) and (5) are easy.

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The next theorem is a kind of open mapping theorem. It is an easy conse- quence of Shelah’s finite equivalence relation theorem ([Sh]). However, it has non-trivial applications. In fact, the open mapping theorem of Lascar–Poizat ([Ba]) follows from the finite equivalence relation theorem in a similar way.

Theorem 0.3. Assume A ⊆ B ⊆ C and f : Tr B (ab/C) → Tr A (a/C) is restriction to acl(A) and to formulas with free variable x a . Then f is an open surjection.

P r o o f. Clearly f is a surjection. To show that f is open, let E(x ab , x 0 ab ) ∈ FE(B). Let X = {[E(x ab , a 0 b 0 )] ∩ Tr B (ab/C) : a 0 b 0 ≡ ab (C)}. We see that X is finite, elements of X are closed, C-conjugate, and cover Tr B (ab/C).

Hence Tr A (a/C) = S

{f (S) : S ∈ X} and f (S) is closed for each S ∈ X.

It follows that for some (hence every) S ∈ X, f (S) has non-empty interior in Tr A (a/C). It is easy to see that in fact f (S) is open in Tr A (a/C). This shows that f is open.

In [Ne1] I pointed out that M-rank may be defined in a way similar to the definition of Morley rank. It was puzzling whether there are local versions of M-rank, just as there are local versions of Morley rank. Local Morley ranks measure (type-)definable sets by means of instances of formulas from some (finite) set ∆. A local M-rank should measure St(p) with respect to some fixed (finite) set C ⊆ Dom(p). We define some local versions of M-rank in

§2. This must be done carefully, so that the resulting rank has the additivity properties of M. We show that these local versions of M-rank can also be used to produce meager types.

1. M-rank and meager types. In this section we show how to produce meager types with the help of M-rank. The next lemma improves Lemma 0.2(2).

Lemma 1.1. Assume A ⊆ B are finite.

(1) If Tr A (a/B) is open in St A (a/A) then M(a/B) ≥ M(a/A).

(2) If p(x) ∈ S(A) and s(x) is a (partial) type over B and Tr A (s) ∩ St A (p) is open in St A (p) then there is a p 0 ∈ S p (B)∩[s] with M(p 0 ) ≥ M(p).

P r o o f. (1) First we prove the following.

(a) For every finite B 0 ⊇ A with B 0 ∪ a (A) there is a B | 00 with B 00 B 0 (Aa), B 00 ∪ B (A), a | ∪ B | 00 (B) and with Tr B

00

(a/BB 00 ) open in St B

00

(a/B 00 ).

Given B 0 as in (a) choose B ≡ B 0 (Aa) with B ∗ | ∪ B (A). Let p(x) =

tp(a/A), p 0 (x) = tp(a/B) and q = tp(a/B ). Since Tr A (p 0 ) is open in St A (p)

and stp(a/A) ∈ Tr A (q) ∩ Tr A (p 0 ) ⊆ St A (p), we see that Tr A (q) ∩ Tr A (p 0 )

is non-empty and open in Tr A (q). Let s(x) be the type over BB saying

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that x ∪ B | (B). Let f : St B

(q) → Tr A (q) be restriction and let U = f −1 (Tr A (q) ∩ Tr A (p 0 )). Then U is open in St B

(q). Also we have

(b) U ⊆ Tr B

(p 0 (x) ∪ s(x)).

Indeed, let r ∈ U . Then r|acl(A) ∈ Tr A (p 0 ), hence we can choose c realizing p 0 ∪ r|acl(A), and without loss of generality, c ∪ B | (B). Now, B ∗ | ∪ B (A) implies c ∪ B | (A), hence c realizes r ∪ p 0 (x) ∪ s(x).

Since T is small, there is a p 00 ∈ S q (BB )∩[p 0 ∪s] such that Tr B

(p 00 )∩U is open in St B

(q). By Fact 0.1 we see that Tr B

(p 00 ) is open in St B

(q). Let a 00 realize p 00 . We see that a 00 ≡ a (B), a 00 | ∪ B (B), Tr B

(a 00 /BB ) is open in St B

(a 00 /B ) and a 00 B ≡ aB 0 . Choose B 00 with a 00 B ≡ aB 00 (B). Clearly, B 00 satisfies our demands in (a).

Now we prove by induction on α that M(a/A) ≥ α implies M(a/B) ≥ α.

We check only the successor step. Suppose M(a/A) ≥ α + 1. So there is a finite B 0 ⊇ A with a ∪ B | 0 (A), M(a/B 0 ) ≥ α and St(a/B 0 ) nowhere dense in St(a/A). By (a), without loss of generality, a ∪ B | 0 (B) and Tr B

0

(a/BB 0 ) is open in St B

0

(a/B 0 ). Thus by the inductive hypothesis, M(a/BB 0 ) ≥ α.

To finish, we must show

(c) St(a/BB 0 ) is nowhere dense in St(a/B).

In the following diagram all the maps α i are restrictions.

St BB

0

(a/BB 0 ) St B (a/B)

St B

0

(a/B 0 ) St A (a/A)

α

0

//

α

1

²²

α

2

²² α

3

//

This diagram commutes. Also, Tr A (a/B) = Rng α 2 , hence by Theo- rem 0.3 and our assumptions, α 2 is open. By (a) and Theorem 0.3 also α 1

is open. St BB

0

(a/BB 0 ) and St B (a/B) are naturally homeomorphic to St(a/BB 0 ) and St(a/B) respectively. So to prove (c) it suffices to show that α 0 (St BB

0

(a/BB 0 )) is nowhere dense in St B (a/B).

If not, then α 0 (St BB

0

(a/BB 0 )) is open in St B (a/B), hence also α 2 α 0 (St BB

0

(a/BB 0 )) is open in St A (a/A). We have

α 2 α 0 (St BB

0

(a/BB 0 )) = α 3 α 1 (St BB

0

(a/BB 0 )) ⊆ α 3 (St B

0

(a/B 0 )).

Since St(a/B 0 ) is nowhere dense in St(a/A), we see that α 3 (St B

0

(a/B 0 )) is nowhere dense in St A (a/A). Hence the more so α 2 α 0 (St BB

0

(a/BB 0 )) is no- where dense in St A (a/A), a contradiction.

(2) By the smallness of T , S p (B) ∩ [s] is countable. The sets Tr A (p 0 ),

p 0 ∈ S p (B) ∩ [s], cover Tr A (s) ∩ St A (p). Hence for some p 0 ∈ S p (B) ∩ [s],

Tr A (p 0 ) is not nowhere dense in St A (p). By Fact 0.1, Tr A (p 0 ) is open in

St A (p). Hence (2) follows from (1).

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The range of M is of the form I ∪ {∞} or I, for some proper initial segment I of ω 1 . Let α = ω β for some β ∈ Ord, or α = β = ∞. Assume α is in the range of M. Let γ β be the minimal γ such that M(p) ≥ α for some type p of ∞-rank γ. Notice that if β 0 < β and α 0 = ω β

0

then γ β

0

≤ γ β .

Theorem 1.2. Assume A is finite, q ∈ S(A), R (q) = γ β and M(q) ≥ α. Then q is regular and locally modular. If q is non-trivial then q is meager and for some isolated q 0 ∈ S(A), R (q 0 ) = γ β , M(q 0 ) ≥ α and q 0 is non- orthogonal to q.

We begin the proof of Theorem 1.2 with two lemmas.

Lemma 1.3. q is regular and orthogonal to every type with ∞-rank < γ β . P r o o f. Suppose not. Then for some finite B ⊇ A and r ∈ S(B) with R (r) < γ β , r is not almost orthogonal to some nf extension q 0 of q over B. Choose ψ ∈ r with R (ψ) < γ β . Since r a 6⊥ q 0 , for some a, b realizing r, q 0 respectively, a ∪ b (B). This is witnessed by a formula δ(x, y) over B, - true of a, b. Since forking means decreasing some local rank, without loss of generality we have

(a) if δ(a 0 , b 0 ) holds and b 0 satisfies a nf extension of q over B, then ψ(a 0 ) and a 0 - ∪ b 0 (B).

Let δ 0 (y) = ∃x δ(x, y). So δ 0 (y) does not fork over A. Let s(y) be the type over B saying:

“δ 0 (y) and q(y) and tp(y/B) does not fork over A.”

We see that Tr A (s) is open in St A (q). By Lemma 1.1(2) for some q 00 S q,nf (B) ∩ [δ 0 ], we have M(q 00 ) ≥ α. Let b 00 realize q 00 , and choose a 00 with δ(a 00 , b 00 ). By (a), R (b 00 /Ba 00 ) < γ β and R (a 00 /B) < γ β (as R (ψ) <

γ β ). By the choice of γ β , we have M(a 00 /B), M(b 00 /Ba 00 ) < α. By Lemma 0.2 we have M(b 00 /B) ≤ M(a 00 b 00 /B) ≤ M(a 00 /B) ⊕ M(b 00 /Ba 00 ) < α, a contradiction.

Choose p 0 ∈ S(A) non-orthogonal to q, with R (p 0 ) = γ β , M(p 0 ) ≥ α and Cantor–Bendixson rank CB(p 0 ) minimal possible. Choose ϕ ∈ p 0 with R (ϕ) = γ β , CB(p 0 ) = CB(ϕ) and CB-multiplicity of ϕ equal to 1. By Lemma 1.3, p 0 is regular.

Lemma 1.4. p 0 is orthogonal to every type in S(A)∩[ϕ]\{p 0 } and forking is meager on St A (p 0 ).

P r o o f. Suppose r ∈ S(A) ∩ [ϕ] \ {p 0 } is non-orthogonal to p 0 . Then

for some finite B ⊇ A and a, b realizing over B nf extensions of r, p 0

respectively, we have a ∪ b (B). As in Lemma 1.3, for some b - 0 realizing p 0 ,

we have b 0 | ∪ B (A), M(b 0 /B) ≥ α and for some a 0 ∈ ϕ(C)\p 0 (C), a 0 - ∪ b 0 (B).

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If M(a 0 /A) ≥ α and a 0 | ∪ B (A), then by the choice of p 0 and ϕ, tp(a 0 /A) is orthogonal to q, hence to p 0 . Thus also tp(a 0 /B) is orthogonal to p 0 , a contradiction. If a 0 - ∪ B (A) then R (a 0 /B) < γ β , so again (by Lemma 1.3 applied to q := p 0 ), tp(a 0 /B) is orthogonal to p 0 .

Hence we get M(a 0 /A) < α and a 0 | ∪ B (A). Thus M(a 0 /B) < α and M(b 0 /Ba 0 ) < α, because R (b 0 /Ba 0 ) < γ β . Hence M(b 0 /B) ≤ M(a 0 b 0 /B)

≤ M(a 0 /B) ⊕ M(b 0 /Ba 0 ) < α, a contradiction.

To prove the second clause, suppose forking is not meager on St A (p 0 ).

Then for some forking formula δ(x) over a finite set B ⊇ A, Tr A (δ) ∩ St A (p 0 ) is open in St A (p 0 ). By Lemma 1.1, for some p 00 ∈ S p

0

(B) ∩ [δ], we have M(p 00 ) ≥ M(p 0 ) ≥ α. Since p 00 forks over B, R (p 00 ) < γ β . Hence M(p 00 ) <

α, a contradiction.

Now we can conclude the proof of Theorem 1.2. Suppose q is non-trivial.

So p 0 is non-trivial. Let p be any stationarization of p 0 . We shall prove that ϕ satisfies the conditions from the definition of a meager type, and forking is meager on P ϕ , hence that p is meager. By Lemmas 1.3 and 1.4, every type in S(A) ∩ [ϕ] \ {p 0 } is hereditarily orthogonal to p. Also, p is orthogonal to any forking extension of p 0 , and every type in St A (p 0 ) is regular. Hence ϕ is p-simple of p-weight 1. By the claim in the proof of Lemma 1.6(2) in [Ne2], the set of r ∈ St A (p 0 ) such that r is non-orthogonal to p is clopen in St A (p 0 ).

Thus P ϕ is clopen in St A (p 0 ). Since all stationarizations of p 0 have the same local ranks, as in Lemma 1.3 we see that the p-weight 0 is definable on ϕ(C).

Thus ϕ is a p-formula. By Lemma 1.4, forking is meager on P ϕ , hence p is meager.

Next notice that if some type in P ϕ is modular, then every type in P ϕ is, since they are all stationarizations of the complete type p 0 . But then if c realizes one of them, then they all have forking extensions over Ac, contradicting meagerness. It follows that every type in P ϕ is not modular.

By [Ne2, Corollary 1.8], P ϕ is open in S(acl(A)). Hence also St A (p 0 ) is open in S(acl(A)). This means that p 0 is isolated, proving the theorem.

Notice that by Lemma 1.4 and [Ne2, Lemma 1.6(1)], if I(T, ℵ 0 ) < 2

0

then in the proof of Theorem 1.2 the type p 0 is non-trivial. It follows that if I(T, ℵ 0 ) < 2

0

then the type q in Theorem 1.2 is non-trivial and meager.

Lemma 1.5. Let q ∈ S(A) be as in Theorem 1.2. Assume A 0 is finite, q 0 ∈ S(A 0 ) is isolated, R (q 0 ) = γ β and q 0 is non-orthogonal to q. Then M(q 0 ) ≥ α and M(q) ≤ M(q 0 ) ⊕ α 0 for some α 0 < α.

P r o o f. First we prove that M(q) ≤ M(q 0 ) ⊕ α 0 for some α 0 < α. For

some finite B ⊇ A ∪ A 0 there are a, b realizing over B nf extensions of q, q 0

respectively, with a ∪ b (B). As in Lemma 1.3, by Lemma 0.2, changing a -

somewhat we can assume that M(a/B) ≥ M(q). This change affects b, but

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since q 0 is isolated, b still realizes q 0 , and since q is orthogonal to any type with ∞-rank < γ β , we have b ∪ B (A | 0 ). Hence M(b/B) ≤ M(q 0 ). Also, a ∪ b (B) yields M(a/Bb) < α. Let α - 0 = M(a/Ba). Hence we get

M(q) ≤ M(a/B) ≤ M(ab/B) ≤ M(q 0 ) ⊕ M(a/Bb) = M(q 0 ) ⊕ α 0 . If M(q 0 ) < α, then M(q) ≤ M(q 0 ) ⊕ α 0 < α, a contradiction.

Suppose M(p) = n 1 ω β

1

⊕ . . . ⊕ n k ω β

k

, where n i are finite, n 1 6= 0, and β 1 > . . . > β t ≥ β > β t+1 > . . . > β k = 0. We define M β (p) as n 1 ω β

1

⊕ . . . ⊕ n t ω β

t

. If M(p) = ∞ then M β (p) = ∞. If β = ∞ then M β (p) = 0 if M(p) < ∞ and M β (p) = ∞ otherwise.

Given β and α = ω β , for isolated q over a finite set such that R (q) = γ β

and M(q) ≥ α, M β (q) is determined by the non-orthogonality class of q.

The next corollary clarifies the orthogonality relations between the locally modular types q obtained in Theorem 1.2.

Corollary 1.6. Assume α = ω β , α 0 = ω β

0

are in the range of M, A, A 0 are finite and q ∈ S(A), q 0 ∈ S(A 0 ) are isolated with R (q) = γ β , R (q 0 ) = γ β

0

, M(q) ≥ α and M(q 0 ) ≥ α 0 . If q, q 0 are non-orthogonal then γ β = γ β

0

and M β (q) = M β

0

(q) = M β (q 0 ) = M β

0

(q 0 ).

In the case when T has < 2

0

countable models, we can strengthen Theorem 1.2.

Theorem 1.7. Assume I(T, ℵ 0 ) < 2

0

. For q and A as in Theorem 1.2, we have M β (q) = α.

P r o o f. By Theorem 1.2 there is an isolated p over A non-orthogonal to q, with R (p) = γ β and M(p) ≥ α. By the claim in [Ne2, Lemma 1.6], extending A by an element of acl(A), we can assume that all types in St A (p) are non-orthogonal. Let ϕ isolate p over A. By the proof of Theorem 1.2, ϕ and P ϕ = St A (p) witness that p is meager. By Lemma 1.5 it suffices to prove that M(p) < α ⊕ α.

We shall use Corollary 2.15 from [Ne2], which says that in our case, for every finite B extending A, for every a realizing p with a ∪ B (A), exactly | one of the following holds.

(a) For some b realizing p with b ∪ B (A), tp(b/B) is isolated and | a ∪ b (A). -

(b) There are finitely many a 0 , . . . , a n realizing r = tp(a/B) such that for every b realizing r, for some i ≤ n and b 0

s b (A), we have b 0 - ∪ a i (A).

Suppose for a contradiction that M(p) ≥ α ⊕ α. Let a realize p. We can

find a finite set B ⊇ A with a ∪ B (A) and M(a/B) = α. The proof splits |

into two cases, depending on which of conditions (a), (b) holds.

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C a s e 1: (a) holds. So choose b realizing p with tp(b/B) isolated, b | B (A) and a ∪ b (A). It is easy to see that M(b/B) = M(p), hence M(b/B) -

≥ α ⊕ α. Since a ∪ b (B), by the choice of γ - β we have M(b/Ba) < α.

Hence α ⊕ α ≤ M(b/B) ≤ M(ab/B) ≤ M(a/B) ⊕ M(b/Ba) < α ⊕ α, a contradiction.

C a s e 2: (b) holds. Extending B by an element from acl(B), we can assume that n = 0, that is, for every b realizing r = tp(a/B), for some b 0

s b (A), we have b 0 - ∪ a (A). In other words, every s ∈ St B (r) has a forking extension over Ba. Since T is countable, there is a single formula ψ(x) over Ba, forking over B, such that the set of s ∈ St B (r) consistent with ψ has non-empty interior in St B (r). Without loss of generality, R (ψ) < γ β . Since M(r) = α, by Lemma 1.1 there is a type r 0 ∈ S r (Ba) ∩ [ψ] with M(r 0 ) ≥ α.

Since R (r 0 ) < γ β , this contradicts the choice of γ β . Corollary 1.8. Assume I(T, ℵ 0 ) < 2

0

.

(1) Assume β < β 0 , α = ω β , α 0 = ω β

0

and α 0 is in the range of M.

Then γ β

0

> γ β + 2. Also, γ 0 ≥ 1 and γ 1 ≥ ω.

(2) Assume A is finite, p ∈ S(A) and M(p) ≥ ω β . Then R (p) ≥ 1+2β.

P r o o f. (1) Without loss of generality, β 0 = β + 1. Choose a finite set A and an isolated p ∈ S(A) with R (p) = γ β

0

and M(p) ≥ α 0 . So p is non- trivial, regular and meager. Without loss of generality, all stationarizations of p are non-orthogonal, and the formula ϕ isolating p witnesses that p is meager. Again we apply Corollary 2.15 from [Ne2] (see the proof of Theorem 1.7 above). Let a realize p. Since β < β 0 , for some finite B ⊇ A with a ∪ B (A) we have M(a/B) = α ⊕ α. Since α ⊕ α < α | 0 , as in Theorem 1.7 we see that in Corollary 2.15, case (b) holds, and (without loss of generality) n = 0 there. As in Theorem 1.7, we find a forking extension r 0 of tp(a/B) over Ba with M(r 0 ) ≥ α ⊕ α. In particular, R (r 0 ) ≥ γ β . In fact, R (r 0 ) ≥ γ β + 1. If not, then R (r 0 ) = γ β and M(r 0 ) ≥ α ⊕ α, which contradicts Theorem 1.7. Since r 0 forks over B, we get γ β

0

≥ R (r 0 ) + 1. Altogether we get γ β

0

≥ γ β + 2.

We have γ 0 ≥ 1 trivially, and γ 1 ≥ ω because by [Ne1], every type with finite U -rank has finite M-rank.

(2) Notice that R (p) ≥ γ β , and by (1), γ β ≥ 1 + 2β.

Corollary 1.8 shows that there is a bound on M-rank, depending on the

∞-rank. Can we improve the bound obtained there? Assuming I(T, ℵ 0 ) <

2

0

, is it true that M(p) ≤ U (p)? This is proved in [Ne1] for types with finite U -rank.

Corollary 1.9. Assume I(T, ℵ 0 ) < 2

0

, β < β 0 , α = ω β , α 0 = ω β

0

and

α 0 is in the range of M. Suppose A and A 0 are finite, p ∈ S(A), p 0 ∈ S(A 0 ),

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R (p) = γ β , R (p 0 ) = γ β

0

, M(p) ≥ α and M(p 0 ) ≥ α 0 . Then p and p 0 are meager , regular and orthogonal.

To illustrate our ideas we shall give an example of a superstable theory with a meager type p of M-rank ω.

Example 1.10. Let V = ω×ω 2 and let + be pointwise addition on V , modulo 2. We think of V as the product Q

n {n}×ω 2. We define certain subgroups of V . For n > 0 let

P n = {f ∈ V : f |n × n ≡ 0} and G n = {f ∈ P n : f |(ω \ n) × ω ≡ 0}.

Clearly, G n and P n are subgroups of (V, +) and G n ∩ P n+1 ⊆ G n+1 . More- over, for each n > 1, G n /G n−1 is naturally isomorphic to ( {n−1}×(ω\n) 2, +), and G 1 ∩P n is isomorphic to {0}×(ω\n) 2. Thus the mapping g : {n−1}×(ω\n) 2

{0}×(ω\n) 2 given by g(f )((0, k)) = f ((n − 1, k)) induces a group isomor- phism f n : G n /G n−1 → G 1 ∩ P n .

Let M = (V ; +, P n , G n , f n+1 ) 0<n<ω and T = Th(M ). Then T is a su- perstable 2-dimensional 1-based theory. Up to non-orthogonality, there are two regular types in T : p 0 , the principal generic type of M , and p 1 , the principal generic type of G 1 . Both types are meager, p 1 is weakly minimal, R (p 0 ) = U (p 0 ) = ω. The functions f n , n > 1, ensure that the weakly minimal groups G n /G n−1 , n > 1, are non-orthogonal to G 1 .

Let p ∈ S(∅) be the type generated by ¬P 1 (x). Thus p is regular non- orthogonal to p 0 , R (p) = U (p) = ω and St (p) = Str(∅) ∩ [¬P 1 (x)]. Let a realize p and let ϕ n (x, a) be G n (x − a), n > 0. We see that Tr n (x, a)) ⊆ St (p) and Tr n (x, a)) ∩ [P n+1 (x − a)] is nowhere dense in Tr n+1 (x, a)) (the formula P n+1 (x − a) is almost over ∅). This is essentially because in V with the product topology, G n ∩ P n+1 is nowhere dense in G n+1 .

Let p n ∈ S(a) be one of the countably many non-forking extensions of p such that St (p n ) is an open subset of Tr n (x, a)). Of course every stationarization of p n is modular. We see that M(p n ) = n and M(p) = ω.

Also γ 0 = 1 and γ 1 = ω here.

Notice that γ 0 is the minimal γ such that some type without Morley

rank has ∞-rank γ. By Theorem 1.7, if I(T, ℵ 0 ) < 2

0

then for every type

p of ∞-rank ≤ γ 0 , M(p) ≤ 1, and if M(p) = 1 then p is meager (hence

locally modular). This is related to Proposition 1.11 in [Pi]. Pillay proves

there (under the few models assumption) that if G is a locally modular

superstable group of rank γ 0 , without Morley rank, then for every finite

A and a ∈ G, if tp(a/A) is non-isolated then it has finite multiplicity. In

general we cannot claim that much. The following example shows that it can

happen that a type p ∈ S(∅) has ∞-rank γ 0 , M(p) = 1 and p is non-isolated

(in this case necessarily γ 0 ≥ ω).

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Example 1.11. Let V = ω×ω 2 and + be as in Example 1.10. We define some subgroups of V . For n > 0 let

P n = {f ∈ V : f |{0} × n ≡ 0},

G n = {f ∈ V : f |{0} × ω ≡ 0 and f |(ω \ (n + 1)) × ω ≡ 0}, H n = {f ∈ V : f |(ω \ n) × 1 ≡ 0}.

Thus G n ⊆ G n+1 T

n P n , H n ⊆ H n+1 , [G n+1 : G n ] is infinite and [H n+1 : H n ] = 2. Consider the structures M 0 = (V ; +, P n , G n ) n>0 and M 1 = (V ; +, H n ) n>0 and their theories T 0 , T 1 respectively. They are 1- based. T 0 is superstable, with ∞-rank ω and γ 0 = ω, while T 1 has Morley rank 2 and ∞-rank 2. T 0 and T 1 have countably many countable models.

Now we do not have to include into M 0 the functions f n (as in Example 1.10) since G n /G n−1 are strongly minimal and ω-categorical.

We shall define a structure M with universe V 0 = V × V . Let E be the equivalence relation on V 0 defined by (a, b)E(a 0 , b 0 ) iff b = b 0 . We can naturally identify V 0 /E with V . Let Q n = {(a, b) ∈ V 0 : a ∈ H n }. For each a ∈ V 0 , a/E can be endowed with a structure isomorphic to M 1 , uniformly in a. However, we do not want to have in M a full group structure on a/E. Let R be the 4-ary relation on V 0 defined by: R((a 0 , b 0 ), (a 1 , b 1 ), (a 2 , b 2 ), (a 3 , b 3 )) iff b 0 = b 1 , b 2 = b 3 and a 1 − a 0 = a 3 − a 2 .

We see that R gives an affine group structure on each a/E, and addi- tionally a group isomorphism between any two E-classes, after fixing single points in them.

Finally, let M = (V 0 ; E, Q n , R; +, P n , G n ) n>0 , where +, P n , G n are de- fined on V 0 /E via the identification of V 0 /E with V . Let T = Th(M ). Then T is a superstable 1-based theory with countably many countable models.

Here γ 0 = ω. Let p(x) ∈ S(∅) be the type generated by {¬Q n (x), n <

ω} ∪ {¬P 1 (x/E)}. Clearly p is non-isolated, R (p) = ω and M(p) = 1.

Also, p is regular and meager. Notice, however, that since M/E is bi- interpretable with M 0 , M/E is a regular group of ∞-rank ω, non-orthogonal to p, with meager generics. If a realizes p, then a/E is generic in M/E and tp((a/E)/∅) is isolated, as promised by Proposition 1.11 in [Pi]. More- over, ϕ(x) = (x = x) is a p-formula in M , and isolated types are dense in P ϕ = Str(∅)∩ T

n>0 [¬G n (x/E)]. This is a good illustration of the phenomena encountered in [Ne2].

2. Localization. There are several motivations for this section. One is that unfortunately possibly not all meager types (even up to non-ortho- gonality) are obtained by the minimization process from §1. Another is that if, for example, S(acl(∅)) is uncountable (and hence M(q) ≥ 1 for some q ∈ S(∅); remember that T is small), then the minimization procedure from

§1 yields some meager (or trivial) p ∈ S(A) with M(p) ≥ 1. However, p may

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have nothing in common with the rich topological structure of S(acl(∅)), that is, possibly Tr (p) is finite. My feeling was that we should be able to find a type p as above such that the topological structure of St(p) is related to S(acl(∅)). In other words, that the complicated structure of S(acl(∅)) really forces the existence of a locally modular type. In this section we will show that this is the case. To do this we need some local versions of M-rank, which we define below. These local versions have many additivity properties of M. Another reason for trying to define local M-ranks is an attempt to find the most basic impact that the structure of acl(C) (for a given finite C ⊆ C) has on the structure of the whole theory. So local M-ranks are in- tended to play the same role in the “theory of multiplicities” as local Morley ranks in the theory of forking. I leave it to the reader to judge how much I succeeded in this attempt.

In this section let C be a fixed finite subset of C. We define local rank M C on types p ∈ S(A) for finite sets A ⊇ C, as follows. Suppose p = tp(a/A).

We define M C (p) = M C (a/A) by the following conditions:

(1) M C (a/A) ≥ 0.

(2) M C (a/A) ≥ α + 1 iff for some finite B ⊇ A with a ∪ B (A), | M C (a/B) ≥ α and Tr C (a/B) is nowhere dense in Tr C (a/A).

(3) M C (a/A) ≥ δ for limit δ iff M C (a/A) ≥ α for each α < δ.

From now on we usually assume that C is contained in the sets of pa- rameters A, B we consider. M C has the following properties.

Lemma 2.1. Assume C ⊆ A ⊆ B, A, B are finite.

(1) M C (a/A) ≤ M(a/A).

(2) If a ∪ A (C) then M | C (a/A) = M(a/A).

(3) If Tr C (b/A) = Tr C (a/A) then M C (b/A) = M C (a/A).

(4) For each a there is a b with b ∪ A (C) such that Tr | C (b/A) = Tr C (a/A), hence M(b/A) = M C (b/A) = M C (a/A).

(5) If a ∪ B (A) and M | C (a/A) < ∞ then M C (a/B) = M C (a/A) iff Tr C (a/B) is open in Tr C (a/A).

(6) For every a there is a b with b ≡ a (A), b ∪ B (A) and M | C (b/B) = M C (a/A).

P r o o f. (1) We prove by induction on α that M C (a/A) ≥ α implies M(a/A) ≥ α. Let us check the successor step. Suppose M C (a/A) ≥ α + 1.

Choose a finite B ⊇ A with a ∪ B (A), M | C (a/B) ≥ α and Tr C (a/B) nowhere dense in Tr C (a/A). Let f B : St B (a/B) → Tr C (a/B), f A : St A (a/A)

→ Tr C (a/A) and g : St B (a/B) → Tr A (a/B) ⊆ St A (a/A) be restrictions. By

Theorem 0.3, f A , f B and g are open. Also, f B = g ◦ f A . It follows that

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Tr A (a/B) is nowhere dense in St A (a/A). Otherwise Tr A (a/B) is open in St A (a/A), hence Tr C (a/B) = g(Tr A (a/B)) is open in Tr C (a/A), a contra- diction.

As Tr A (a/B) is nowhere dense in St A (a/A), and St A (a/A) and St B (a/B) are canonically homeomorphic to St(a/A) and St(a/B) respectively, we see that St(a/B) is nowhere dense in St(a/A). By the inductive hypothesis, M(a/B) ≥ α, hence by the definition of M, M(a/A) ≥ α + 1. The proofs of the other parts of the lemma are equally easy, so we omit them.

Unfortunately, M C does not have the additivity properties of M from Lemma 0.2. One of the reasons for that is as follows. Suppose b ⊆ A. Then St(ab/A) is homeomorphic to St(a/A), hence M(ab/A) = M(a/A). Let p(x ab ) = tp(ab/A). Consider Tr C (a/A) = Tr C (p|x a ) and Tr C (ab/A) = Tr C (p). Let f : Tr C (ab/A) → Tr C (a/A) be restriction to formulas with free variable x a . Then f is continuous and open; however, f need not to be injective. In particular, it could happen that Tr C (a/A) is finite and Tr C (ab/A) is infinite. Then M C (a/A) = 0 and M C (ab/A) ≥ 1, showing that M C (ab/A) ≤ M C (a/Ab) ⊕ M C (b/A) fails.

It turns out that the situation described above is the only obstacle pre- venting M C from having the additivity properties of M. We correct this by defining an eventual version of M C , denoted by M e C . It is defined just like M C , but with (2) replaced by the following condition:

(2 0 ) M e C (a/A) ≥ α + 1 iff for some finite B, D extending A with B ⊆ D and a ∪ D (A), we have M | e C (a/D) ≥ α and Tr C (aB/D) is nowhere dense in Tr C (aB/B).

Notice that M e C is an eventual version of M C just as the Morley rank is an “eventual” version of the Cantor–Bendixson rank. Also, in (2 0 ) we can additionally assume that Tr C (aA/B) is open in Tr C (aA/A) (by Lemma 2.2(3) below and Theorem 0.3). We can give another, equivalent definition of M e C . For finite B ⊆ C define an equivalence relation B on S(C) by r B r 0 iff r|acl(C)B = r 0 |acl(C)B. For X ⊆ S(C) let X/ B be {r/ B : r ∈ X}. We can equivalently replace (2 0 ) by the following condition:

(2 00 ) M e C (a/A) ≥ α + 1 iff for some finite B, D extending A, with B ⊆ D and a ∪ D (A), we have M | e C (a/D) ≥ α and S

St(a/D)/ B is nowhere dense in St(a/B).

We shall not use (2 00 ), however. If in (2 00 ) we required only that

S St(a/D)/ B is nowhere dense in St(a/A), we would end up with M e C =

M. The role of B in (2 0 ) and (2 00 ) is puzzling. It would be natural to re-

quire in (2 0 ), (2 00 ) additionally that St(a/B) is open in St(a/A). If we did

so, however, we might lose the property of M e C that M e C (p) ≥ M e C (p 0 ) for

any nf extension p 0 of p (see Lemma 2.2(3) below).

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Lemma 2.2. (1) M C (a/A) ≤ M e C (a/A) ≤ M(a/A).

(2) If a ∪ A (C) then M | C (a/A) = M e C (a/A) = M(a/A).

(3) If A ⊆ B and a ∪ B (A) then M | e C (a/B) ≤ M e C (a/A).

(4) If A ⊆ B, a ∪ B (A) and St(a/B) is open in St(a/A) (which implies | M(a/B) = M(a/A)), then M e C (a/B) = M e C (a/A).

(5) If A ⊆ B, then for every a there is an a 0 ≡ a (A) with a 0 | ∪ B (A) and M e C (a 0 /B) = M e C (a 0 /A). Also, if additionally B ⊆ acl(A) then M e C (a/A) = M e C (a/B).

(6) In (2 0 ) we can additionally assume that D \ A ⊆ C = (C = is the “real”

sort of C eq ).

P r o o f. (1) To show M C (a/A) ≤ M e C (a/A), we prove by induction on α that M C (a/A) ≥ α implies M e C (a/A) ≥ α. Let us check the successor step.

Assume M C (a/A) ≥ α + 1. Choose a finite B ⊇ A with a ∪ B (A) such that | M C (a/B) ≥ α and Tr C (a/B) is nowhere dense in Tr C (a/A). It follows that Tr C (aA/B) is nowhere dense in Tr C (aA/A). By the inductive hypothesis, M e C (a/B) ≥ α, and by the definition of M e C , M e C (a/A) ≥ α + 1. Similarly we prove M e C (a/A) ≤ M(a/A).

(2) follows from (1) and Lemma 2.1(2). (3) is easy.

(4) By (3), M e C (a/B) ≤ M e C (a/A). By Lemma 2.5(1) below, also M e C (a/A) ≤ M e C (a/B) (and no vicious circle arises).

(5) Since T is small, we can find an a 0 ≡ a (A) with a 0 ∪ B (A) and | St(a 0 /B) open in St(a/A). Hence (5) follows from (4).

(6) Suppose M e C (a/D) ≥ α and Tr C (aB/D) is nowhere dense in Tr C (aB/B) (for some B, D as in (2 0 )). Choose finite B 0 , D 0 with B 0 D 0 ⊆ C = , B \ A ⊆ acl(B 0 ), D \ A ⊆ acl(D 0 ) and a ∪ D | 0 (A). As in (4) we can choose B 0 , D 0 so that M e C (a/DD 0 ) ≥ α and Tr C (aBB 0 /DD 0 ) is nowhere dense in Tr C (aBB 0 /BB 0 ). As in (4) and (5), we see that M e C (a/AD 0 ) ≥ α and Tr C (aAB 0 /AD 0 ) is nowhere dense in Tr C (aAB 0 /AB 0 ), so we are done.

Notice that Lemma 2.2(4) naturally corresponds to the following prop- erty of forking: If A ⊆ B and U (a/B) = U (a/A) then R(a/B) = R(a/A) for any (local) rank R. This shows again a similarity between our treatment of multiplicities and the theory of forking.

To evaluate M e C (ab/A) in terms of M e C (a/Ab) and M e C (b/A) we need the following technical lemma.

Lemma 2.3. Assume C ⊆ A ⊆ B and b ∪ B (A). |

(1) If Tr C (a/Bb) is open in Tr C (a/Ab) and Tr C (bA/B) is open in Tr C (bA/A) then Tr C (a/B) is open in Tr C (a/A).

(2) If Tr C (ab/B) is open in Tr C (ab/A) then Tr C (a/B) is open in

Tr C (a/A).

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P r o o f. (1) Let ϕ(x a ; B, y b ) be a formula over B such that for some E(x a , x 0 a ) ∈ FE(C), for each b 0 , ϕ(C; B, b 0 ) is a union of some E-classes, and (a) Tr C (a/Ab) ∩ [ϕ(x a ; B, b)] = Tr C (a/Bb).

Let ϕ 0 (x a , x b ) be a formula over some c ∈ acl(A), saying: for B 0

s B (A) with B 0 ∪ x | a x b (A), ϕ(x a ; B 0 , y b ) holds. We can also assume that c ∈ dcl(B).

Now, ϕ 0 (x a , b) is equivalent to ϕ(x a ; B, b).

Indeed, ϕ 0 (a 0 , b) implies that ϕ(a 0 ; B 0 , b) holds for some B 0 ∪ a | 0 b (A) with B 0

s B (A). This gives B 0 b ≡

s Bb (A), hence ϕ(C; B, b) = ϕ(C; B 0 , b), and ϕ(a 0 ; B, b) holds. The other direction is similar.

So we can assume ϕ = ϕ 0 , that is, ϕ(x a ; B, y b ) is both over B and over c ∈ dcl(B) ∩ acl(A). We need the following claim.

Claim. Suppose C ⊆ A ⊆ B and c ∈ acl(b) ∩ A. If Tr C (b/B) is open in Tr C (b/A) then Tr C (bc/B) is open in Tr C (bc/A).

P r o o f. For b 0 ≡ b (A) let X b

0

= {stp(b 00 c/C) : b 00

s b 0 (C) and b 00 b (A)}. Since c ∈ acl(b) ∩ A, X b

0

is finite. Indeed, let C b

0

= {c 0 : c 0 ≡ c (b 0 ) and for some b 00 ≡ b (A), b 00 c ≡

s b 0 c 0 (C)}. Since c ∈ acl(b)∩A, C b

0

is finite and Ab 0 -definable. Moreover, if stp(b 00 c/C) ∈ X b

0

then for some c 0 , stp(b 00 c/C) = stp(b 0 c 0 /C), and c 0 ∈ C b

0

, hence X b

0

= {stp(b 0 c 0 /C) : c 0 ∈ C b

0

}. Also, if b 00 ≡ b 0 ≡ b (A) then X b

00

is the b 00 -copy of X b

0

over A, hence |X b

00

| = |X b

0

|.

Similarly, for b 0 ≡ b (B) let Y b

0

= {stp(b 00 c/C) : b 00

s b 0 (C) and b 00 ≡ b (B)}

= {stp(b 0 c 0 /C) : c 0 ≡ c (b 0 ) and for some b 00 ≡ b (B), b 00 c ≡

s b 0 c 0 (C)}.

We see that stp(b 0 c/C) ∈ Y b

0

⊆ X b

0

. Notice that Tr C (bc/A) = [

{X b

0

: b 0 ≡ b (A)} and Tr C (bc/B) = [

{Y b

0

: b 0 ≡ b (B)}.

Choose a formula χ(y bc ) over acl(C) with X b ∩ [χ(y bc )] = {stp(bc/C)}.

Clearly,

if b 0

s b (C) and b 0 ≡ b (A) then X b

0

= X b , and if b 0

s b (C) and b 0 ≡ b (B) then Y b

0

= Y b .

So by compactness there is a formula α(y b ) over acl(C), true of b, such that:

if b 0 ≡ b (A) and α(b 0 ) then X b

0

∩ [χ(y bc )] has size 1, and if b 0 ≡ b (B) and α(b 0 ) then Y b

0

∩ [χ(y bc )] has size 1.

Thus if b 0 ≡ b (B) and α(b 0 ) then moreover X b

0

∩ [χ(y bc )] = Y b

0

∩ [χ(y bc )].

Since Tr C (b/B) is open in Tr C (b/A), for some β(y b ) over acl(C),

Tr C (b/B) = Tr C (b/A) ∩ [β(y b )].

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Let γ(y bc ) be α(y b )&β(y b )&χ(y bc ). So γ(bc) holds. To finish, it suffices to show that

Tr C (bc/A) ∩ [γ(y bc )] ⊆ Tr C (bc/B).

So suppose b 0 c 0 ≡ bc (A) and γ(b 0 c 0 ) holds (and necessarily c 0 = c).

We want to find b 00 with b 00 c ≡

s b 0 c (C) and b 00 ≡ b (B). Now, β (b 0 ) yields a b 00

s b 0 (C) with b 00 ≡ b (B). So α(b 00 ) holds, which gives that X b

00

∩ [χ(y bc )]

equals Y b

00

∩ [χ(y bc )] and has size 1. By the definition of Y b

00

, there is a b

s b 00 (C) with b ≡ b (B) and {stp(b c/C)} = Y b

00

∩ [χ(y bc )]. Without loss of generality, b = b 00 . Since b 00

s b 0 (C) and b 00 ≡ b 0 (A), we get X b

0

= X b

00

. So stp(b 0 c/C) ∈ X b

0

∩ [χ(y bc )] = X b

00

∩ [χ(y bc )] = Y b

00

∩ [χ(y bc )] = {stp(b 00 c/C)}.

It follows that b 00 c ≡

s b 0 c (C) and b 00 ≡ b (B), hence stp(b 0 c/C) ∈ Tr C (bc/B).

This proves the claim.

Returning to the proof of the lemma, since Tr C (bA/B) is open in Tr C (bA/A), and c ∈ dcl(B) gives Tr C (bA/B) ⊆ Tr C (bA/Ac), we deduce that Tr C (bA/B) is open in Tr C (bA/Ac). Applying the claim to b := bA, A := Ac and B := Bc, we see that Tr C (bAc/B) is open in Tr C (bAc/Ac).

Moreover, c ∈ acl(A) gives that Tr C (bAc/Ac) is open in Tr C (bAc/A). Alto- gether we conclude that Tr C (bAc/B) is open in Tr C (bAc/A). Hence there is a formula δ(y bAc ; B) almost over C such that

(b) Tr C (bAc/A) ∩ [δ(y bAc ; B)] = Tr C (bAc/B).

Let S = S B = {stp(a 0 /C) : for some b 0 , a 0 b 0 ≡ ab (Ac) and δ(b 0 Ac; B) and ϕ(a 0 ; B, b 0 ) hold}.

Then S B is a closed subset of Tr C (a/A). Since c ∈ dcl(B), S B is definable over B (that is, Aut B (C)-invariant), and for B 0 ≡ B (A) we can define S B

0

as the B 0 -copy of S B over A. Since δ and ϕ are almost over A, the set X = {S B

0

: B 0 ≡ B (A)} is finite and Tr C (a/A) = S

X. It follows that S is open in Tr C (a/A) and stp(a/C) ∈ S. So to show that Tr C (a/B) is open in Tr C (a/A) it suffices to prove

(c) S ⊆ Tr C (a/B).

So suppose stp(a 0 /C) ∈ S. Thus for some b 0 we have a 0 b 0 ≡ ab (Ac) and δ(b 0 Ac; B) and ϕ(a 0 ; B, b 0 ) hold. We must find a 00 with a 00

s a 0 (C) and a 00 ≡ a (B). By (b), for some b 00 with b 00 Ac ≡

s b 0 Ac (C), we have b 00 ≡ b (B).

Hence we can find a 0 and a 1 with (d) a 0 b 00 Ac ≡

s a 0 b 0 Ac (C), ab ≡ a 1 b 00 (B).

In particular, we have

(e) a 0 b 00 ≡ a 0 b 0 ≡ ab ≡ a 1 b 00 (Ac).

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