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ON POSET BOOLEAN ALGEBRAS

URI ABRAHAM, MATATYAHU RUBINDEPARTMENT OF MATHEMATICS AND CS, BEN GURION

UNIVERSITYBEER-SHEVA ISRAEL

ROBERT BONNETLABORATOIRE DE MATHEMATIQUES, UNIVERSITE DE SAVOIE

LE BOURGET-DU-LAC FRANCE

HAMZA SI KADDOUR

LAPCS, UNIVERSITE CLAUDE-BERNARDVILLEURBANNE FRANCE

Abstract. Let 〈P, ≤〉 be a partially ordered set. We definethe poset Boolean algebra of P , and denote it by F (P ). Theset of generators of F (P ) is {xp : p ∈ P}, and the set of re-lations is {xp · xq = xp : p ≤ q}. We say that a Booleanalgebra B is well-generated, if B has a sublattice G suchthat G generates B and 〈G, ≤G〉 is well-founded. A well-generated algebra is superatomic.Theorem 1 Let 〈P, ≤〉 be a partially ordered set. The fol-lowing are equivalent. (i) P does not contain an infinite setof pairwise incomparable elements, and P does not containa subset isomorphic to the chain of rational numbers. (ii)F (P ) is superatomic. (iii) F (P ) is well-generated.The equivalence (i) ⇔ (ii) is due to M. Pouzet [P]. A par-tially ordered set P is well-ordered, if P does not contain astrictly decreasing infinite sequence, and P does not containan infinite set of pairwise incomparable elements.Theorem 2 Let F (P ) be a superatomic poset algebra.Then there are a well-ordered set W and a subalgebra Bof F (W ), such that F (P ) is a homomorphic image of B.This is partly analogous to the fact (proved in [AB1]), thata superatomic interval algebra is isomorphic to a subalgebraof an ordinal algebra.

Mathematics Subject Classification 2000 (MSC2000) Primary: 03G05,06A06.Secondary: 06E05, 08A05, 54G12.Keywords: stucture theory, superatomic Boolean algebras, posets, well-quasi-orderings.E-mail: [email protected] (U. Abraham), [email protected] (R. Bonnet),[email protected] (M. Rubin), [email protected] (H. Si Kaddour).

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1. Introduction

In this paper we investigate superatomic poset Boolean algebras.

The construction of a Boolean algebra from a partially ordered set

(poset) is very natural. However, not much is known about the rela-

tionship between the properties of the poset and the properties of the

Boolean algebra constructed from it. The two theorems proved in this

work are of this kind.

We start with the definition of a poset algebra. Let 〈P,≤〉 be a par-

tially ordered set. The poset algebra of 〈P,≤〉 denoted by F (〈P,≤〉),

is defined by specifying a set of generators for F (〈P,≤〉) together with

a set of relations between the generators. We assume the familiarity

with this type of construction.

The set of generators for F (〈P,≤〉) is {xp : p ∈ P}. The set of

relations is {xp · xq = xp : p ≤ q}.

We use the following abbreviations. P stands for 〈P,≤〉. Thus

F (〈P,≤〉) is abbreviated by F (P ). F (P ) is called the poset algebra of

P . If the context involves more than one poset, then we use ≤P to

denote the partial ordering of P .

We next define the notions used in the statement of the three main

theorems of this work.

A poset P is well-founded, if P has no strictly decreasing infinite

sequences.

For Boolean algebras we use the notations of [Ko]. Thus +, · , −

and ≤ denote the join, meet, complementation and partial ordering of

a Boolean algebra B. The zero and one of B are denoted by 0 and 1.

ON POSET BOOLEAN ALGEBRAS 3

Let B be a Boolean algebra. A member a ∈ B is called an atom of B,

if a is a minimal element of B − {0}.

A Boolean algebra B is superatomic, if every homomorphic image

of B has an atom. A Boolean algebra B is called a well-generated

algebra, if B has a sublattice G such that:

(1) G generates B;

(2) 〈G,≤B�G〉 is well-founded.

Every well-generated algebra is superatomic. This is proved in [BR1]

Proposition 2.7(b), and [Sk]. However, the proof is easy, and can be

found by the reader.

We now turn to partial orderings. A subset of a poset P consisting

of pairwise incomparable elements is called an antichain of P .

A poset P is narrow, if P does not contain infinite antichains.

A poset P is scattered, if P does not contain a subset isomorphic to

the chain of rational numbers.

We can now state the first main theorem. This theorem will be

proved in Section 3.

Theorem 1.1. Let P be a partially ordered set. Then the following

properties are equivalent.

(1) P is narrow and scattered.

(2) F (P ) is superatomic.

(3) F (P ) is well-generated.

The implication (1) ⇒ (3) is the difficult part of Theorem 1.1.

M. Pouzet [P] proved the implication (1) ⇒ (2). The direct proof of

(1) ⇒ (2) due to Pouzet is tricky: see R. Fraısse [F] Section 4.13. But

now this fact follows easily from the implication (1) ⇒ (3).

4 ABRAHAM ET ALL

The interval algebra of a linearly ordered set 〈L,<〉 is the subalgebra

of ℘(L) generated by the set {(−∞, a) : a ∈ L} of closed left rays

of L. We denote the interval algebra of L by B(L). It is easy to see

that the interval algebra of L is isomorphic to the poset algebra of L,

and that the function xa �→ (−∞, a) extends to such an isomorphism.

In [AB1] Theorem 3.4, it is proved that an interval algebra of a

scattered linear ordering is embeddable in the interval algebra of a

well-ordering. The second main theorem in this work is an analogue

of the above fact for partially ordered sets.

A partially ordered set is well-ordered, if it is narrow and well-

founded.

Theorem 1.2. Let P be a narrow scattered partially ordered set. Then

there is a well-ordered poset W and a subalgebra B of F (W ) such that

F (P ) is a homomorphic image of B.

Theorems 1.1 and 1.2 use a structure theorem for narrow scattered

partially ordered sets, which is similar to the theorem of Hausdorff

on the structure of scattered linear orderings. This theorem is due to

Abraham and Bonnet. It is proved in [AB2] Theorem 3.4. We next

quote this theorem.

Suppose that P is a poset and for every a ∈ P , Qa is a poset. We

define the lexicographic sum of the indexed family {Qa : a ∈ P}, and

denote it by∑

{Qa : a ∈ P}. The universe of∑

{Qa : a ∈ P} is⋃{{a} × Qa : a ∈ P}. The partial ordering on

∑{Qa : a ∈ P} is

defined as follows: 〈a, q〉 ≤ 〈b, r〉, if either (i) a <P b or (ii) a = b and

q ≤Qa r.


We say that a poset 〈P,≤〉 is anti well-ordered, if 〈P,≥〉 is well-

ordered. Let W denote the class of posets which are either well-ordered

or anti well-ordered.

Let 〈P,≤1〉 and 〈P,≤2〉 be posets. We say that 〈P,≤2〉 is an exten-

sion of 〈P,≤1〉, if ≤1 ⊆ ≤2 .

Theorem 1.3. ([AB2] Theorem 3.4)

Let H 0 be the class of all posets whose universe is a singleton. Let H

be the closure of H 0 under lexicographic sums indexed by members of

W, and under isomorphism.

Then for every poset P the following are equivalent.

(1) P is narrow and scattered.

(2) P is an extension of a member of H.

Theorem 1.2 follows easily from Theorem 1.3 and the following fact.

Theorem 1.4. Let P ∈ H. Then there is a well-ordered poset W

such that F (P ) is embeddable in F (W ).

Theorem 1.4 will be proved in Section 4.

We assume Theorems 1.3 and 1.4, and prove Theorem 1.2.

Proof of Theorem 1.2. Let 〈P,≤〉 be narrow and scattered. By

Theorem 1.3 we may assume that there is ≤′ ⊆ ≤ such that

〈P,≤′〉 ∈ H. By Theorem 1.4, F (〈P,≤′〉) is embeddable in F (W ),

where W is a well-ordered poset, and obviously F (〈P,≤′〉) is a homo-

morphic image of F (〈P,≤〉) .

We conjecture that Theorem 1.2 cannot be strengthened by requir-

ing that for some well-ordered posetW , F (P ) is embeddable in F (W ).

We thus ask the following question.

6 ABRAHAM ET ALL

Question 1.5. (a) Is there a narrow scattered poset P such that

F (P ) cannot be embedded in F (W ), whereW is a well-ordered poset ?

(b) Is there a narrow scattered poset P such that F (P ) cannot be

embedded in F (W ), where W ∈ H ?

In [AB1] Theorem 1 it is proved that a superatomic subalgebra of

an interval algebra is embeddable in an interval algebra of an ordinal.

We ask whether the analogous fact for partially ordered sets is also

true.

Question 1.6. Let B be a superatomic subalgebra of a poset algebra.

Does there exist a narrow scattered poset such that B is embeddable

in F (P ) ?

Section 2 is devoted to the proof of a theorem that says that super-

atomic subalgebras of interval algebras have very well-behaved sets of

generators.

We define the notions used in the statement of this theorem. We

first describe the Boolean algebraic dual of the Cantor Bendixon’s

derivation. Let B be a Boolean algebra. At(B) denotes the set of

atoms of B, and IAt(B) denotes the ideal of B generated by At(B).

Note that IAt(B) may be equal to B; that is, we regard B as an ideal

of B. We define by induction on ordinals the sequence of canonical

ideals of B. Let I0(B) = {0B}. Suppose that the ideal Iα(B) has been

defined. Let ϕα : B → B/Iα(B) denote the canonical homomorphism

from B onto B/Iα(B). We define

Iα+1(B) = ϕ−1α (IAt(B/Iα(B))).


If δ is a limit ordinal, then we define

Iδ(B) =⋃γ<δ

Iγ(B).

It is well-known and easy to show that a Boolean algebra B is

superatomic, iff for some ordinal α, B/Iα(B) is finite. We call the first

such α the rank of B, and denote it by rk(B). However, if B = {0B},

then rk(B) is defined to be −1.

For α ≤ rk(B) let Atα(B) = {a ∈ B : a/Iα(B) ∈ At(B/Iα(B))}.

Let At(B) =⋃

{Atα(B) : α ≤ rk(B)}.

We define the following equivalence relation on B. If a, b ∈ B,

then a ∼B b means that either (i) a = b = 0 or (ii) there is α such

that a/Iα(B) = b/Iα(B) �= 0.

Definition 1.7. Let B be a superatomic Boolean algebra.

(a) A subset H ⊆ B is a complete set of representatives (CSR) for

B, if H ⊆ At(B), and for every a ∈ At(B) there is a unique b ∈ H

such that b ∼B a.

(b) A subset H ⊆ B is tree-like, if for every a, b ∈ H : either a ·b = 0,

or a ≤ b, or b ≤ a.

We need to know the following easy facts. Their proof is left to the

reader.

Proposition 1.8. Let B be a superatomic Boolean algebra.

(a) If H is a CSR for B, then H generates B.

(b) Suppose that H is a tree-like CSR for B, then the sublattice of

B generated by H is well-founded.

8 ABRAHAM ET ALL

Proposition 1.8 implies that a Boolean algebra that has a tree-like

CSR is well-generated.

We now state the main result of Section 2. Slightly strengthened,

it is restated as Theorem 2.1.

Theorem 1.9. Let B be a subalgebra of an interval algebra of an

ordinal. Then B has a set of generators T such that: (1) T is tree-

like; (2) T is a CSR for B.

The above theorem was proved by Bonnet and Si-Kaddour in 1987

(see [Sk]). It has since appeared in several preprints of Bonnet, Rubin

and Si-Kaddour.

We combine the results of this work and those of [AB1], and obtain

the following summarizing statement.

Theorem 1.10. (a) Let B be a Boolean algebra. The following are

equivalent.

(1) B is a superatomic algebra embeddable in an interval algebra.

(2) B is embeddable in an interval algebra of an ordinal.

(3) B is superatomic, and has a tree-like CSR.

(b) Let B be a subalgebra of an interval algebra of an ordinal. Then

there are an ordinal γ, a subalgebra B∗ of B(γ) and a subset T of B∗

such that B∗ ∼= B and

(1) T is a tree-like CSR for B∗;

(2) At(B∗) = At(B(γ));

(3) every member of T is an interval of B(γ).

Part (b) of the above theorem uses a standard construction. This

will be described at the end of Section 2.


A Boolean algebra B is canonically well-generated, if it has a CSR

H such that the sublattice of B generated by H is well-founded.

This strengthening of well-generatedness was considered extensively

in [BR1] and [BR3]. Combining Proposition 1.8 and Theorem 1.10(a)

we obtain the following corollary.

Corollary 1.11. Every superatomic subalgebra of an interval algebra

is canonically well-generated.

We conjecture that this is not the case for poset algebras of narrow

scattered posets.

Question 1.12. (a) Is every poset algebra of a narrow scattered poset

canonically well-generated ?

We even do not know the answer to the following special case.

(b) Is every poset algebra of a well-ordered poset canonically well-

generated ?

One last comment. By Theorem 1.10, every subalgebra of a super-

atomic interval algebra is well-generated. This is not true for poset

algebras. Let P be the disjoint sum of two copies of ω1. It will be

shown in [BR2] that F (P ) has a non well-generated subalgebra.

At some points we find it easier to use the topological dual of a

poset algebra. In section 5 we define the dual notion, and prove the

duality.

The results of Section 2 are due to Bonnet and Si-Kaddour, in 1987.

Section 3 contains results of Abraham and Bonnet, obtained in 1993.

The results of section 4 are due to Bonnet and Rubin in 1992.

10 ABRAHAM ET ALL

2. The well-generatedness of subalgebras of interval

algebras

In this section we prove Theorem 1.9. In fact, we prove a stronger

statement. An application of Theorem 2.1, can be found in [BR4].

Theorem 2.1. Let B be a subalgebra of an interval algebra of an or-

dinal. Suppose that H is a CSR for B. Then B has a set of generators

T such that: (1) T is tree-like; (2) T is a CSR for B; (3) For every

a ∈ H there is b ∈ T such that b ∼B a and b ≤ a.

Let 〈L,<〉 be a chain with a first element 0L. Every element

of the interval algebra B(L) can be represented in the form a =⋃{[a2i, a2i+1) : i < n}, where n ∈ ω, 0L ≤ a0 < a1 < · · · < a2n−1 ≤ ∞.

Such a representation of a is unique. It is called the canonical decom-

position of a. We denote �(a)def= n and ep(a)

def= { a0 , . . . , a2n−1}. n is

called the length of a, and ep(a) is called the set of endpoints of a.

Note that �(∅) def= 0.

We first prove two lemmas, 2.2 and 2.3.

Lemma 2.2. Let B(L) be an interval algebra and a, b ∈ B(L).

(a) �(a ∩ b) + �(a ∪ b) ≤ �(a) + �(b).

(b) �(a− b) + �(b− a) ≤ �(a) + �(b).

Proof. (a) See Pouzet and Rival [PR].

(b) Let a =⋃

{[a2i, a2i+1) : i < �(a)} and b =⋃

{[b2i, b2i+1) : i <

�(b)}. Let c0< · · ·<cq−1 be such that

Cdef= {c0, . . . ,cq−1} = {a0, . . . , a2�(a)−1} ∪ {b0 , . . . , b2�(b)−1}.

Let E = {a∪b, a∩b, a−b, b−a, a, b}. It is obvious that for every e ∈ E


and an interval I in the canonical decomposition of e, the endpoints

of I belong to C. Suppose that we take a chain L+ ⊇ L, and we define

a+ =⋃

{[a2i, a2i+1)L+

: i < �(a)} and b+ =⋃

{[b2i, b2i+1)L+

: i < �(b)}.

Then a+ and a have the same canonical decomposition, and this is also

true for a − b, b − a and all other members of E. So we are allowed

to replace L by L+. We can thus assume that c0 �= 0L, cq−1 �= ∞,

and that for every i < q−1, ci+1 is not the successor of ci in L. That is

(1) 0L,∞ �∈ ep(a) ∪ ep(b).

Hence, for e ∈ E, �(−e) = �(e) + 1. Let b′ = −b. So a − b = a ∩ b′ ;

and b− a = −b′ − a = −(a ∪ b′), that is a ∪ b′ = −(b− a). Hence

(2) �(a ∩ b′) = �(a− b) ,

and

(3) �(b′) = �(−b) = �(b) + 1 and �(a ∪ b′) = �(−(b− a)) = �(b− a) + 1 .

By Part (a) applied to a and b′:

(4) �(a ∩ b′) + �(a ∪ b′) ≤ �(a) + �(b′).

By (2) – (4),

(5) �(a− b) + �(b− a) + 1 ≤ �(a) + �(b) + 1 .

That is �(a− b) + �(b− a) ≤ �(a) + �(b) .

We define the relation ≤lex on B(L). Let a =⋃

{[a2i, a2i+1) : i <

�(a)} and b =⋃

{[b2i, b2i+1) : i < �(b)}. a ≤lex b, if

(i) �(a)= �(b), (ii) a= b or for the first i such that ai �= bi, ai<bi.

Obviously for every m, 〈{c ∈ B(L) : �(c) = m},≤lex〉 is a linear

ordering.

Lemma 2.3. Let B(L) be an interval algebra and a, b ∈ B(L).

(a) Assume that �(a ∩ b) + �(a ∪ b) = �(a) + �(b). If a <lex a ∩ b,

then a ∪ b <lex b.

12 ABRAHAM ET ALL

(b) Assume that �(a − b) + �(b − a) = �(a) + �(b). If a <lex a − b,

then b− a <lex b.

Proof. (a) We use the notations of the proof of Lemma 2.2. In

particular for a, b ∈ B(L), C def= {c0, . . . ,cq−1} = ep(a) ∪ ep(b),

E = {a ∪ b, a ∩ b, a − b, b − a, a, b}, and we assume that c0 �= 0L,

cq−1 �= ∞, and that for every i < q − 1, ci+1 is not the successor of ci

in L.

Claim 1 If I and J are intervals in the canonical decompositions of

a and −b respectively, then they do not have the same left endpoint.

Proof. Suppose by contradiction that cp is the left endpoint of I and

J . Let c ∈ L be such that c < cp, and for every 0 ≤ i < p, ci < c.

Let a′ = a ∪ [c, cp). Then �(a′) = �(a), �(a′ ∩ b) = �(a ∩ b) + 1 and

�(b ∪ a′) = �(b ∪ a). So

�(a′ ∩ b) + �(a′ ∪ b) = �(a ∩ b) + 1 + �(b ∪ a)

> �(a ∩ b) + �(b ∪ a) = �(a) + �(b) = �(a′) + �(b) .

This contradicts Lemma 2.2(a), so Claim 1 is proved.

We return to the proof of Part (a). Since a �= a ∩ b, a− b �= ∅. Let

cp be the minimal element of C ∩ (a− b). So [cp, cp+1) ⊆ a− b.

Claim 2 p > 0.

If p = 0, then cp is the left endpoint of intervals in the canonical

decompositions of a and of −b. This contradicts Claim 1.

Claim 3 [cp−1, cp) �⊆ a− b.

This follows from the fact that cp = min(C ∩ (a− b)).

Claim 4 [cp−1, cp) �⊆ a ∩ b.

Suppose by contradiction that [cp−1, cp) ⊆ a∩b. Since (a−b)∩[0L, cp) =


∅, ep(a) ∩ [0L, cp) = ep(a ∩ b) ∩ [0L, cp). Since [cp−1, cp), [cp, cp+1) ⊆ a,

cp �∈ ep(a). Since [cp, cp+1) ⊆ a − b, cp is the right endpoint of an

interval in the canonical decomposition of a ∩ b. So cp ∈ ep(a ∩ b).

This means that a �<lex a ∩ b. A contradiction.

Claim 5 [cp−1, cp) �⊆ b− a.

If [cp−1, cp) ⊆ b − a, then cp is a left endpoint of intervals in the

canonical decompositions of a and of −b. This contradicts Claim 1.

It follows from Claims 2 - 5 that [cp−1, cp) ⊆ −(a ∪ b). Since

[cp, cp+1) ⊆ a, cp ∈ ep(−(a ∪ b)). Since [cp−1, cp), [cp, cp+1) ⊆ −b, cp �∈

ep(b). Since (a− b)∩ [0L, cp) = ∅, ep(b)∩ [0L, cp) = ep(a∪ b)∩ [0L, cp).

It follows that cp is the first point on which ep(b) and ep(a∪ b) differ.

So a ∪ b <lex b.

(b) If L0, L1 are chains with a first element and L+ = L0 + L+ L1,

then the lengths of a, b, a ∪ b, a ∩ b are the same in L and L+. Also,

a <lex a − b holds in L iff it holds in L+, and b − a <lex b holds in L

iff it holds in L+. So we may assume that

(1) 0L,∞ �∈ ep(a) ∪ ep(b).

Hence, for e ∈ {a ∪ b, a ∩ b, a− b, b − a, a, b}, �(−e) = �(e) + 1. Let

b′ = −b. So a− b = a∩ b′. Since a <lex a− b, �(a) = �(a− b). That is

(2) �(a) = �(a ∩ b′).

Since �(b) = �(b − a), �(−b) = �(b) + 1 = �(b − a) + 1 = �(−(b − a)).

That is

(3) �(b′) = �(a ∪ b′).

(4) a <lex a ∩ b′.

By (2) - (4) and Part (a) applied to a and b′, a ∪ b′ <lex b′. That is,

(5) a ∪ (−b) <lex −b.

14 ABRAHAM ET ALL

From (1) it is obvious that

(6) ep(a ∪ (−b)) = ep(−(a ∪ (−b))) ∪ {0L,∞},

and that

(7) ep(−b) = ep(b) ∪ {0L,∞}.

By (5) - (7), b− a <lex b.

For a ∈ B let B � a = {b ∈ B : b ≤ a}. So B � a is a Boolean

algebra. We define rkB(a) = rk(B � a).

Proof of Theorem 2.1. Let L be a well-ordering, B be a subalgebra

of B(L), and H be a CSR for B. For a ∈ H let

m(a) = min({�(b) : b ∼B a and b ≤ a}) and

Ua = {b ∈ B : �(b) = m(a), a ∼B b and b ≤ a}

Since L is a well-ordering, 〈Ua,≤lex〉 is a well-ordering. Since Ua �= ∅,

it has a minimum, which we denote by µ(a). Let

T = {µ(a) : a ∈ H} .

Clearly, T is a CSR for B, and for every a ∈ H , µ(a) ∼B a and

µ(a) ≤ a.

We shall prove that T is tree-like.

Let B be a superatomic Boolean algebra. We shall use the following

easy to prove facts.

(i) If b ∈ At(B) and a ≤ b, then either a ∼B b and rkB(b−a) < rkB(b);

or b− a ∼B b and rkB(a) < rkB(b).

(ii) If a, b ∈ At(B), a �∼B b and rkB(a) ≤ rkB(b), then b− a ∼B b.

(iii) If b ∈ At(B) and rkB(a) < rkB(b), then b ∪ a ∼B b.

Let a, b ∈ T be distinct, and suppose that a∩b �= 0. We may assume

that rkB(a) ≤ rkB(b).


Since T is a CSR and a �= b, a �∼B b. By (ii), b − a ∼B b. By (i),

a− b ∼B a or a ∩ b ∼B a.

Case 1 Assume that a− b ∼B a.

By Lemma 2.2(b),

�(a− b) + �(b− a) ≤ �(a) + �(b).

Suppose that a′, b′ ∈ H are such that a = µ(a′) and b = µ(b′). So

a − b ≤ a ≤ a′ and b − a ≤ b ≤ b′. Hence by the minimality of �(a)

and �(b) in the sets {�(c) : c ∼B a′ and c ≤ a′} and

{�(c) : c ∼B b′ and c ≤ b′},

�(a) ≤ �(a− b) and �(b) ≤ �(b− a).

So �(a) = �(a−b) and �(b) = �(b−a). Since a∩b �= 0, a �= a−b. So by

the minimality of a in 〈Ua′ , <lex〉, a <lex a − b. So by Lemma 2.3(b),

b − a <lex b. This contradicts the minimality of b in 〈Ub′ , <lex〉. So

Case 1 cannot occur.

Case 2 Assume that a ∩ b ∼B a.

By (i), rkB(a − b) < rkB(a). So rkB(a − b) < rkB(b). By (iii),

a ∪ b ∼B b. By Lemma 2.2(a),

�(a ∩ b) + �(a ∪ b) ≤ �(a) + �(b).

Repeating the argument of Case 1 we conclude that

�(a) ≤ �(a ∩ b) and �(b) ≤ �(a ∪ b).

So �(a) = �(a ∩ b) and �(b) = �(a ∪ b). Suppose by contradiction that

a �= a ∩ b. So by the minimality of a in 〈Ua′ , <lex〉, a <lex a ∩ b. So

by Lemma 2.3(a), a∪ b <lex b. This contradicts the minimality of b in

〈Ub′, <lex〉. It follows that a = a ∩ b, that is, a ≤ b.

We have proved that T is tree-like.

16 ABRAHAM ET ALL

We now turn to the proof of Theorem 1.10. We have not presented

a proof of (3) ⇒ (1) in Part (a) of 1.10, and we have not proved

Part (b). The following lemma implies both.

Lemma 2.4. Let B be a superatomic Boolean algebra and T be a tree-

like CSR for B. Then there is a well-ordering L, and an embedding

ψ : B → B(L) such that At(B) = At(B(L)) and for every a ∈ T ,

ψ(a) is an interval.

Remark It is well-known that (∗) a Boolean algebra which is gener-

ated by a tree-like set is embeddable in an interval algebra. The proof

of the present lemma is similar to the proof of (∗), and the argument

is routine. So we just sketch the proof.

Proof. Let τ = {a ∈ T : rkB(a) = rk(B)}. Obviously, τ is finite.

We may assume that∑τ = 1. It suffices to prove the claim of the

lemma for each algebra of the form B � a, a ∈ τ . So we may assume

that τ = {1B}. Note that At(B) ⊆ T . Let T+ = T − At(B). For

every a ∈ T+ let

T<a = {b ∈ T : b < a}, and Ea be the following relation on T<a :

bEa c if there is d ∈ T such that b, c ≤ d < a.

Ea is an equivalence relation. For every a ∈ T+ let <a be a well-

ordering of T<a/Ea. We define a relation ≺ on At(B). For every

distinct b, c ∈ At(B) there is a unique a ∈ T+ such that b, c < a,

and ¬(bEa c). We define b ≺ c, if b/Ea <a c/Ea. Then ≺ is a well-

ordering of At(B). Let L = 〈At(B),≺〉. For a ∈ B let

ψ(a) = {b ∈ At(B) : b ≤ a}. ψ is an embedding of B in ℘(At(B)),


and it is easy to check that for every a ∈ T , ψ(a) is an interval in B(L).

Obviously, ψ(At(B)) = At(B(L)). It follows that ψ is as required.

Proof of Theorem 1.10. (a) (1) ⇒ (2) is the result of [AB1], (2) ⇒

(3) is Theorem 2.1 and (3) ⇒ (1) is Lemma 2.4.

(b) This follows from Lemma 2.4 and the equivalence proved in

Part (a).

Remark Theorem 2.1 suggests the following strengthening of canon-

ical well-generatedness.

A superatomic Boolean algebra is decreasingly canonically well-generated

(DCWG), if for every CSR H of B there is a CSR E of B such that

the lattice generated by E is well-founded, and for every a ∈ H there

is b ∈ E such that b ∼B a and b ≤ a.

That this property is not implied by canonical well-generatedness

is shown in [BR3] Proposition 3.1 and Theorem 3.6.

A superatomic Boolean algebra is hereditarily canonically well-generated

(HCWG), if for every subalgebra of B is canonically well-generated.

Obviously, a superatomic subalgebra of an interval algebra is both

HCWG and DCWG.

Question 2.5. Is every HCWG Boolean algebra DCWG?

The converse is not true. It is easily checked that the free product of

two DCWG Boolean algebras is DCWG. So if P is the poset which is

the disjoint sum of two copies of ω1, then F (P ) is DCWG. However, it

will be shown in [BR2] that F (P ) has a non well-generated subalgebra.

18 ABRAHAM ET ALL

3. The well-generatedness of poset algebras

In this section we prove Theorem 1.1. It says that the poset algebra

of a narrow scattered poset is well-generated. The key is Theorem 3.4.

Recall that W is the class of posets which are either well-ordered or

anti well-ordered.

Definition 3.1. (a) We define by induction on the ordinal α the class

of posets H α.

H 0 is the class of all posets whose universe is a singleton.

If α is a limit ordinal, then Hα =⋃

{H β : β < α}.

If α = β + 1, then H α is the class of all posets P such that there is

W ∈ W and an indexed family of posets {Pw : w ∈ W} ⊆ H β such

that P is isomorphic to∑

{Pw : w ∈ W}.

We define H =⋃

{Hα : α ∈ Ord}.

(b) Let P be a poset. An element b ∈ F (P ) is said to be bounded if

there is a nonempty finite subset σ of P such that b ≤∑

{xp : p ∈ σ}.

Let I bnd(P ) be the set of bounded elements of F (P ).

The easy proof of the following proposition is left to the reader.

Proposition 3.2. Let F (P ) be a poset algebra. Then Ibnd(P ) is a

maximal ideal of F (P ).

The following proposition is a restatement of [BR1] Lemma 2.9(c),

and of [Sk] p. 21.

Proposition 3.3. Let B be a well-generated algebra and I be a maxi-

mal ideal of B. Then there is a sublattice G ⊆ I such that G generates

B, and G is well-founded.


Proof. We prove the following claim.

Claim 1 Let G′ be a well-founded sublattice of B and b ∈ B. Let H

be the sublattice of B generated by G′ ∪ {b}. Then H is well-founded.

Proof. We may assume that 0B, 1B ∈ G′. Then every member of H

has the form g + h · b, where g, h ∈ G′.

Let {gn + hn · b : n ∈ ω} be a decreasing sequence in H . Let

dn = gn + hn · b and en = gn + hn. Suppose by contradiction that

{dn · b : n ∈ ω} is not eventually constant. dn · b = en · b . So

{en ·b : n ∈ ω} is not eventually constant. Then pndef=

∏i≤n ei, n ∈ ω,

is a decreasing sequence of members of G′ which is not eventually

constant. So

(i) {dn · b : n ∈ ω} is eventually constant.

Suppose by contradiction that {dn − b : n ∈ ω} is not eventually

constant. dn−b = gn−b. So {gn−b : n ∈ ω} is a decreasing sequence,

which is not eventually constant. Then qndef=

∏i≤n gi, n ∈ ω, is a

decreasing sequence of members of G′ which is not eventually constant.

So

(ii) {dn − b : n ∈ ω} is eventually constant.

It follows from (i) and (ii) that {dn : n ∈ ω} is eventually constant.

So H is well-founded. We have proved Claim 1.

We prove the proposition. If G′ ⊆ I, then there is nothing to prove.

Assume that G′ �⊆ I. Let G1def= G′ − I. By the maximality of I, G1 is

a lattice, and since in addition, G1 is well-founded and not empty, it

has a minimum. Denote the minimum by a. Let G be the sublattice

of B generated by (G′∩I)∪{b−a : b ∈ G1}∪{−a}. G is contained in

the lattice generated by G′ ∪ {−a}. So by Claim 1, G is well-founded.

20 ABRAHAM ET ALL

By the maximality of I, −a ∈ I, so G ⊆ I. Let b ∈ G′. If b ∈ I, then

b ∈ G. If b �∈ I, then a ≤ b. So b = (b − a) + a. This means that b

belongs to the subalgebra generated by G. So G generates B.

Let 〈T,∨,∧〉 be a lattice. x ∨ y and x ∧ y denote the join and the

meet of x and y in T . The partial ordering of T is denoted by ≤. So

x ≤ y if x ∧ y = x.

Theorem 3.4. Let 〈T,∨,∧〉 be a distributive lattice. We assume that

T has a minimum 0 and a maximum 1. Let 〈I,≤〉 be a poset, and

〈Ti : i ∈ I〉 be a family of sublattices of T . Suppose that:

(W1) 〈I,≤〉 is a well-ordered poset.

(W2) For every i ∈ I, Ti is well-founded.

(W3)⋃

{Ti : i ∈ I} generates the lattice T .

(W4) For every i, j ∈ I, s ∈ Ti and t ∈ Tj: if i < j and s �= 1, then

either s < t or s ∧ t = 0.

(W5) Let σ ⊆ I, � ∈ I, and for every i ∈ σ ∪ {�}, ui ∈ Ti. Suppose

that the following properties hold:

(1) σ is finite and � �∈ σ;

(2) for every i ∈ σ, i �< �;

(3)∧i∈σ ui �= 0 and u� �= 1.

Then∧i∈σ∪{�} ui <

∧i∈σ ui .

Then T is a well-founded lattice.

We shall need the following lemma. It appears in [BR1] Lemma 2.8,

and [Sk].

Lemma 3.5. Let T be a distributive lattice.


(a) If T is well-founded, then every nonempty subset L of T closed

under meet has a minimum m.

(b) Let S be a subset of T such that:

(1) S generates T .

(2) The meet of two elements of S is a finite join of elements of S.

(3) S is well-founded with respect to the partial ordering of T .

Then T is well-founded.

In particular, if S ⊆ T is closed under meet, generates T , and is

well-founded, then T is well-founded.

Proof. (a) Let m1 and m2 be minimal elements of L. Then

mdef= m1 ∧m2 ∈ L, and thus m = m1 = m2.

(b) Clearly, every element of T is a finite join of elements of S. It is

easy to check that the following holds.

(∗) If w0 > w1 > · · · is a strictly decreasing sequence in T , and

w0 =∨i<n vi, then there is � < n such that {wj ∧ v� : j < ω} contains

a strictly decreasing infinite subsequence.

The proof uses the distributivity of T .

Next, suppose by contradiction that u0 > u1 > · · · is a strictly

decreasing sequence in T . We define by induction a strictly decreasing

sequence 〈vn : n < ω〉 in S. Assume by induction that vn has the

following property.

(∗∗) There is a strictly decreasing sequence w0 > w1 > · · · in T such

that w0 < vn.

Let U be a finite subset of S such that u0 =∨U . By (∗), there is

v0 ∈ U such that {uj ∧ v0 : j < ω} contains a strictly decreasing

22 ABRAHAM ET ALL

subsequence. Hence v0 satisfies the induction hypothesis. Suppose

that vn has been defined, and let vn > w0 > w1 > · · · be as in

the induction hypothesis. Let W be a finite subset of S such that∨W = w0. By (∗), there is vn+1 ∈ W such that {wj ∧ vn+1 : j <

ω} contains a strictly decreasing subsequence. So vn+1 satisfies the

induction hypothesis and vn > w0 ≥ vn+1. The sequence 〈vn : n <

ω〉 ⊆ S is strictly decreasing. This contradicts the well-foundedness

of S, so T is well-founded.

Proof of Theorem 3.4. We may assume that for every i ∈ I, 1 ∈ Ti.

Let T be the set of all functions 6t such that:

(1) Dom(6t) is a finite subset of I,

(2) for every i ∈ Dom(6t), 6t(i) ∈ Ti.

Let∧6t

def=

∧{6t(i) : i ∈ Dom(6t)}, and

Ldef= {

∧6t : 6t ∈ T }.

The proof of the theorem follows easily from the following claim.

Main Claim L is well-founded.

Let 6t ∈ T and i ∈ Dom(6t). We denote 6t(i) by ti,

Claim 1 Let u ∈ L − {0, 1}. Let 6t ∈ T be such that u =∧

{ti : i ∈

Dom(6t)}. Let σ be the set of minimal elements of Dom(6t). Then σ

is a finite antichain in I and u =∧

{ti : i ∈ σ}.

Proof. For each j ∈ Dom(6t) − σ, let m(j) ∈ σ be such that m(j) < j.

Since u �= 0, tm(j) ∧ tj �= 0. By (W4), tm(j) < tj. Hence

u =∧6t ≤

∧{tj : j ∈ σ} ≤ u .

Claim 2 For u ∈ L− {0, 1}, there is 6tu ∈ T satisfying the following

properties.


(1) Dom(6tu) is an antichain.

(2) u =∧6tu.

(3) If i ∈ Dom(6tu), then tui �= 0, 1.

(4) For every 6t ∈ T such that u =∧6t : Dom(6tu) ⊆ Dom(6t), and

if i ∈ Dom(6tu) then tui ≤ ti.

Proof. Let i ∈ I. The set

T uidef= { t′ ∈ Ti : there is 6t ∈ T such that i ∈ Dom(6t ), ti = t′ and u =

∧6t }

is nonempty and closed under ∧ . By (W2), Ti is well-founded, and

by Lemma 3.5(a), t�idef= min(T ui ) exists.

Let 6s ∈ T be such that u =∧6s. Since u �= 1, we may assume that

for every k ∈ Dom(6s), sk �= 1.

By Claim 1 and since u �= 0, we may assume that Dom(6s) is an

antichain. For every k ∈ Dom(6s), t∗k ≤ sk, so∧k∈Dom(�s) t

∗k ≤

∧k∈Dom(�s) sk. That is,

(i)∧

{t∗k : k ∈ Dom(6s)} ≤ u.

For every k ∈ Dom(6s), let 6t k ∈ T be such that u =∧6t k and tkk = t∗k .

Let ρ = {〈k, j〉 : k ∈ Dom(6s), j ∈ Dom(6t k)}. Obviously,

u =∧

{∧6t k : k ∈ Dom(6s)} =

∧{tkj : 〈k, j〉 ∈ ρ}.

Since for every k ∈ Dom(6s), t∗k ∈ {tkj : 〈k, j〉 ∈ ρ},∧{tkj : 〈k, j〉 ∈ ρ} ≤

∧{t∗k : k ∈ Dom(6s)}. That is,

(ii) u ≤∧

{t∗k : k ∈ Dom(6s)}.

It follows from (i) and (ii) that u =∧

{t∗k : k ∈ Dom(6s)}.

Let 6tu be defined as follows: Dom(6tu) = Dom(6s) and for every

i ∈ Dom(6tu), 6tu(i) = t∗i , that is 6tui = t∗i .

24 ABRAHAM ET ALL

We show that 6tu is as required.

(1) Dom(6tu) is an antichain, since Dom(6s) was chosen to be an an-

tichain.

(2) We have proved that u =∧6tu.

(3) Since for every i ∈ Dom(6s), t∗i ≤ si < 1, (3) holds.

(4) Let 6t ∈ T be such that u =∧6t, and suppose by contradiction that

� ∈ Dom(6tu)−Dom(6t). By Claim 1, we may assume that Dom(6t) is an

antichain. Suppose by contradiction that for some k ∈ Dom(6t), k < �.

Since Dom(6tu) is an antichain k �∈ Dom(6tu) : for every i ∈ Dom(6tu),

i �< k. Since u �= 1, we may assume that for every k ∈ Dom(6t), tk �= 1.

We apply (W5) to Dom(6tu), k, {tui : i ∈ Dom(6tu)}, and tk, and

conclude that

u ∧ tk = (∧

{tui : i ∈ Dom(6tu)} ) ∧ tk <∧

{tui : i ∈ Dom(6tu)} = u.

Hence u ∧ tk < u, and so u �≤ tk. A contradiction. So there is no

k ∈ Dom(6t) such that k < �.

We now apply (W5) to Dom(6t), �, {ti : i ∈ Dom(6t)}, and tu� , and

conclude that

u ∧ tu� = (∧

{ti : i ∈ Dom(6t)}) ∧ tu� <∧

{ti : i ∈ Dom(6t)} = u.

Hence u ∧ tu� < u, and so u �≤ tu� . A contradiction. So Dom(6tu) ⊆

Dom(6t).

Let i ∈ Dom(6tu), and we show that tui ≤ ti. Since∧6tu ∧

∧6t = u,

tui ∧ ti ∈ T ui . Also, tui = min(T ui ). So tui ≤ tui ∧ ti. That is, tui ≤ ti.

We have proved Clause (4) of Claim 2. So Claim 2 is proved.

It is obvious that the element 6tu which satisfies the conclusion of

Claim 2 is unique. For every u ∈ L− {0, 1} we denote

σ(u) = Dom(6tu).


Claim 3 Let u, v ∈ L− {0, 1} be such that u ≤ v.

(a) For every i ∈ σ(u) ∩ σ(v), tui ≤ tvi .

(b) For every i ∈ σ(v) there is j ∈ σ(u) such that j ≤ i.

Proof. (a) For w ∈ L− {0, 1} and i ∈ I − σ(w) we set twi = 1. Hence∧{tui : i ∈ σ(u)} = u = u∧v =

∧{tui : i ∈ σ(u)} ∧

∧{tvi : i ∈ σ(v)}

=∧

{tui ∧ tvi : i ∈ I}.

So by Claim 2, for every i ∈ σ(u) ∩ σ(v), tui ≤ tui ∧ tvi , and thus

tui ≤ tvi .

(b) Let i ∈ σ(v), and suppose that there is no j ∈ σ(u) such that

j ≤ i. We apply (W5) to σ(u), i, {tuj : j ∈ σ(u)} and tvi . Then

u ∧ tvi = (∧

{tuj : j ∈ σ(u)}) ∧ tvi <∧

{tuj : j ∈ σ(u)} = u.

That contradicts the fact that u = u ∧ v ≤ u ∧ tvi . So the required j

exists. We have proved Claim 3.

Let Q be a poset and J be a subset of Q. J is an initial segment of

Q if for every q ∈ Q and p ∈ J : if q ≤ p, then q ∈ J .

Claim 4 Let Q be a well-ordered poset. Then there is no strictly

decreasing sequence of initial segments of Q with respect to set inclu-

sion.

Proof. A proof appears in [H], [Kr] and [F]. For completeness, we

sketch a proof. Assume by contradiction that 〈Jn : n ∈ ω〉 is a strictly

decreasing sequence of initial segments of Q. For every n ∈ ω, let

qn ∈ Jn − Jn+1. Then 〈qn : n ∈ ω〉 has no increasing pair. Hence Q is

not well-ordered. A contradiction.

We define a partial ordering on the set Ant(Q) of finite antichains

of a poset Q. σ ≤m τ , if for every i ∈ τ there is j ∈ σ such that j ≤ i.

26 ABRAHAM ET ALL

Claim 5 If Q is a well-ordered poset, then 〈Ant(Q),≤m〉 is well-

founded.

Proof. For σ ∈ Ant(Q) let Q≥σ = {q ∈ Q : (∃p ∈ σ)(p ≤ q)} and

I(σ) = Q−Q≥σ. Q≥σ is a final segment of Q, and so its complement

I(σ) is an inital segment. The function σ �→ Q≥σ is one-to-one, and

if σ ≤m τ , then Q≥σ ⊇ Q≥τ , and so I(σ) ⊆ I(τ). If {σn : n ∈ ω} is

a strictly decreasing sequence in Ant(Q), then {I(σn) : n ∈ ω} is a

strictly decreasing sequence in the set of initial segments of Q, which

by Claim 4 is impossible. So 〈Ant(Q),≤m〉 is well-founded.

Proof of the Main Claim. We prove that L is well-founded.

Note that by Claim 3(b), for every u, v ∈ L − {0, 1}: if u ≤ v, then

σ(u) ≤m σ(v).

Let 〈un : n < ω〉 be a decreasing sequence in L. Then for every

m > n, σ(um) ≤m σ(un). So by Claim 5, we may assume that {σ(un) :

n ∈ ω} is a constant sequence. Let σ = σ(u0). By Claim 3(a), ifm > n

then tmi ≤ tni . Recall that tni ∈ Ti and that Ti is well-founded. Since

σ is finite, there is n0 ∈ ω such that for every i ∈ σ and m > n0,

tmi = tn0i . So um = un0. We have proved the Main Claim.

We prove Theorem 3.4. By the Main Claim, L is well-founded. It

is obvious that L is closed under ∧. Since T is a distributive lattice,

and T is generated by⋃

{Ti : i ∈ I}, every member of T is a finite

sum of members of L. So L generates T . By Lemma 3.5(b), T is

well-founded.

Proof of Theorem 1.1. We prove by induction on α, that for every

P ∈ Hα, F (P ) is well-generated. There is nothing to prove for α = 0


and for limit ordinals. Suppose that the claim is true for α, and let

P ∈ Hα+1 − Hα .

Claim 1 For a poset P let P ∗ denote the inverse ordering of P . Then

F (P ) ∼= F (P ∗).

Proof. (A sketch.) Suppose that {xp : p ∈ P} is the set of generators

of F (P ) and {yp : p ∈ P} is the set of generators of F (P ∗). It

is easy to check that the function xp �→ −yp can be extended to an

isomorphism between F (P ) and F (P ∗). (A complete proof appears in

the Appendix, Proposition 5.4(a).)

By Claim 1 and the definition of Hα, we may assume that

P =∑

{Pv : v ∈ V }, where V is a well-ordered poset, and for every

v ∈ V , Pv ∈ Hα. By the induction hypothesis, for every v ∈ V ,

F (Pv) is well-generated. By Lemma 3.2, I bnd(Pv) is a maximal ideal

in F (Pv). So by Proposition 3.3, there is a well-founded sublattice Gv

of F (Pv) such that Gv generates F (Pv) and Gv ⊆ I bnd(Pv). Let G be

the sublattice of F (P ) generated by⋃

{Gv : v ∈ V }. We verify that

G and {Gv : v ∈ V } satisfy the hypotheses of Theorem 3.4.

By the definition, V is well-ordered, and thus (W1) holds.

Since each Gv is a well-founded lattice, (W2) holds.

(W3) follows from the definition of G.

(W4) Let v < w in V , g ∈ Gv − {1F (P )} and h ∈ Gw. We show

that either g < h or g · h = 0. Let γ be a finite subset of Pv such that

g ≤∑

{xp : p ∈ γ}. Let b =∑

{xp : p ∈ γ}. It suffices to show that

either b < h or b ·h = 0. h has a representation of the following form.

There is � ∈ ω and for every i < � there are finite disjoint subsets η(i)

and τ(i) of Pw such that η(i) �= ∅ and

28 ABRAHAM ET ALL

h =∑i<�

( ∏{xp : p ∈ η(i)} ·

∏{−xq : q ∈ τ(i)}

).

Case 1 There is i0 < � such that τ(i0) = ∅.

Hence b <∏

{xp : p ∈ η(i0)} ≤ h, and thus b < h.

Case 2 For every i < �, τ(i) �= ∅.

For every i < �, b ·∏

{−xq : q ∈ τ(i)} = ∅, and thus

b ·∏

{xp : p ∈ η(i)} ·∏

{−xq : q ∈ τ(i)} = 0. Hence b · h = 0.

We have proved that (W4) holds.

(W5) Let ρ ⊆ V , w ∈ V , and {gv : v ∈ ρ ∪ {w}} ⊆ G. We assume

that: (1) ρ is finite and w �∈ ρ; (2) for every v ∈ ρ, v �< w; (3) for

every v ∈ ρ ∪ {w}, gv ∈ Gv; (4)∏

{gv : v ∈ ρ} �= 0.

We show that gw ·∏

{gv : v ∈ ρ} <∏

{gv : v ∈ ρ}.

By (W4) we may assume that ρ is an antichain.

Claim 2 Let P be a poset and Q be a subset of P . Then F (Q) is a

subalgebra of F (P ).

Proof. The claim is proved in the Appendix Proposition 5.4(c).

Let η = {v ∈ ρ : w < v} and τ = ρ− η. Let g1 =∏

{gv : v ∈ η}.

Claim 3 gw · g1 < g1.

Proof. We may assume that:

(∗) V = {w} ∪ {u} ∪ η; for every v ∈ η: w < u < v; and Pu is a

singleton {p}.

Here is the explanation. Let u �∈ V and V + = V ∪{u}. Let ≤V +be the

order relation on V + defined as follows : ≤V +� V = ≤V ; w <V +u; and

for every v ∈ V : if w <V v then u <V +v and if v <V w then v <V +

u.

So V + is a well-ordered poset. Let Pu = {p} and P+ =∑

{Pv : v ∈

V +}. By Claim 2, F (P ) is a subalgebra of F (P+). Hence, we may


assume that V = V +, and thus P = P+. Let V − = {w} ∪ {u} ∪ η and

P− =∑

{Pv : v ∈ V −}. By Claim 2 again, F (P−) is a subalgebra of

F (P ). Hence, we may assume that V = V −, and thus P = P−. That

is 〈V, P 〉 satisfies (∗).

By (∗), gw < xp. Suppose first that for some v ∈ η , xp · gv = 0. So

gw · gv = 0, and thus gw · g1 = 0 < g1. Suppose next that for every

v ∈ η , xp · gv �= 0. Since Pu = {p}, xp ∈ Gu , and by (W4), xp ≤ gv.

So gw < xp ≤∏

{gv : v ∈ η} = g1, and thus gw · g1 < g1. Claim 3 is

proved.

Let B1 ∗B2 denote the free product of B1 and B2. Let η1 = η∪{w}.

Since for every u ∈ η1 and u′ ∈ τ , u and u′ are incomparable,

(i) F (η1 ∪ τ) ∼= F (η1) ∗ F (τ).

Let g2 =∏

{gv : v ∈ τ}. We have thus the following situation.

(ii) g2 ∈ F (τ) and g2 �= 0.

(iii) gw · g1 , g1 ∈ F (η1) and gw · g1 < g1.

By (i) - (iii), (gw · g1) · g2 < g1 · g2 .

That is, gw ·∏

{gv : v ∈ ρ} <∏

{gv : v ∈ ρ}. So (W5) holds.

We have shown that G and {Gv : v ∈ V } satisfy the hypotheses

of Theorem 3.4. So G is well-founded. For every v ∈ V , Gv generates

F (Pv) and⋃

{F (Pv) : v ∈ V } generates F (P ). So G =⋃

{Gv : v ∈ V }

generates F (P ).

30 ABRAHAM ET ALL

4. The embeddability of poset algebras in well-ordered

poset algebras

In this section we prove Theorem 1.4. In fact, we prove the following

slight strengthening of 1.4.

Theorem 4.1. Let P ∈ H. Then there is a well-ordered posetW and

an embedding ϕ of F (P ) in F (W ) such that ϕ(Ibnd(P )) ⊆ I bnd(W ).

We shall prove by induction on α, that the claim of the theorem is

true for every P ∈ Hα.

The next lemma contains the main claim in the proof of Theo-

rem 4.1. It will be used in the inductive step.

Lemma 4.2. Let V be a poset.

For every v∈V let Pv,Qv be posets and ϕv :F (Pv)→F (Qv) be an em-

bedding such that ϕv(Ibnd(Pv))⊆ I bnd(Qv). Let P =

∑{Pv : v∈V }.

For every v∈V let Q+v = {m(v)}+Qv. That is, Q

+v is the lexicographic

sum of a singleton and Qv over a chain with two elements.

Let Q+ =∑

{Q+v : v ∈V }. For every v∈V let Xv= {xp : p∈Pv}, and

let ϕ+v :Xv →F (Q+

v ) be defined as follows: ϕ+v (xp)= xm(v) +ϕv(xp).

Let ϕ+ =⋃v∈V ϕ

+v .

Then ϕ+ can be extended to an embedding ψ of F (P ) in F (Q+),

and ψ(I bnd(P )) ⊆ I bnd(Q+).

We first introduce some notations and prove some preliminary facts.

Let {xi : i ∈ I} be a set of generators for a Boolean algebra B. Let

σ, τ ⊆ I be finite. We denote xσ,τ =∏

{xp : p ∈ σ} ·∏

{−xq : q ∈ τ}.


If ϕ is a function from {xi : i ∈ I} to a Boolean algebra C, then we

denote x ϕσ,τ =

∏{ϕ(xp) : p ∈ σ} ·

∏{−ϕ(xq) : q ∈ τ}.

To prove that ϕ+ is extendible to an embedding, we shall use the

following lemma due to Sikorski. See [Ko] Theorem 5.5.

Proposition 4.3. Let {xi : i∈ I} be a set of generators for a Boolean

algebra B. Let ϕ be a function from {xi : i∈ I} to a Boolean algebra

C. Then

(a) ϕ can be extended to a homomorphism from B to C iff for every

finite σ, τ ⊆ I: if xσ,τ = 0, then x ϕσ,τ = 0.

(b) Suppose that in addition, for every finite σ, τ ⊆ I: if x ϕσ,τ = 0,

then xσ,τ = 0.

Then ϕ can be extended to an embedding of B into C.

We assume that V , {Pv : v ∈ V }, {Qv : v ∈ V } etc. are as in

Lemma 4.2.

Lemma 4.4. (a) Let R =∑

{Ri : i ∈ I} be a lexicographic sum of

posets. Suppose that σ, τ ⊆ I are finite, and for every i ∈ σ and j ∈ τ ,

i < j. For every i ∈ σ let ai ∈ I bnd(Ri). For every j ∈ τ let rj ∈ Rj.

Then∑

i∈σ ai <∏

j∈τ xrj .

(b) For every v ∈ V , ϕ+v can be extended to an embedding of F (Pv)

in F (Q+v ).

(c) Let p ∈ Pv and q ∈ Pw. Suppose that v < w. Then

ϕ+(xp) < xm(w) < ϕ+(xq).

(d) Let v ∈ V and σ be a finite subset of Pv. Then:

(1)∑

{ϕ+v (xp) : p ∈ σ} = xm(v) +

∑{ϕv(xp) : p ∈ σ}.

(2)∏

{ϕ+v (xp) : p ∈ σ} = xm(v) +

∏{ϕv(xp) : p ∈ σ}.

32 ABRAHAM ET ALL

Proof. We prove only (a) and (b). The other parts are easy, and

are left to the reader.

(a) By Propostion 5.4(c), we may assumme that (1) I = σ∪τ ∪{s};

(2) for every i ∈ σ and j ∈ τ : i < s < j; (3) Rs is the linear ordering

{0, 1}. Then∑

i∈σ ai ≤ x0 < x1 ≤∏

j∈τ xrj .

(b) Let us denote F (Pv) and F (Q+v ) by B and C respectively, and

let cdef= xm(v). We define a homomorphism ψ1 : B → C � c and a

homomorphism ψ2 : B → C � −c. Let b ∈ B. Then

ψ1(b) = c, if for some finite σ ⊆ P ,∏

p∈σ xp ≤ b; and ψ1(b) = 0

otherwise.

ψ2(b) = ϕv(b) · (−c).

Since ψ1, ψ2 are homomorphisms, ψ1+ψ2 : B → C is a homomorphism.

Obviously, ψ1 + ψ2 extends ϕ+v . Since ψ2 is an embedding of B into

C � −c, ψ1 + ψ2 is also an embedding.

Proof of Lemma 4.2. We show that ϕ+ can be extended to a homo-

morphism from F (P ) to F (Q+).

We prove that it satisfies Part (a) of Sikorski’s criterion. Let σ, τ ⊆

P be finite. Suppose that xσ,τ = 0. By Proposition 5.4(b), there

are p ∈ σ and q ∈ τ such that p ≤ q. So xp · −xq = 0. Clearly,

x ϕ+

σ,τ ≤ ϕ+(xp) · −ϕ+(xq). Let v, w be such that p ∈ Pv and q ∈ Pw.

Suppose first that v = w. By Proposition 4.4(b), ϕ+v can be extended

to an embedding ψ : F (Pv) → F (Q+v ). So

ϕ+(xp) · −ϕ+(xq) = ϕ+v (xp) · −ϕ+

v (xq) = ψ(xp · −xq) = 0. Hence

x ϕ+

σ,τ = 0.

Suppose that v �= w. Then since p ≤ q, v < w. By Proposi-

tion 4.4(c), ϕ+(xp) < xm(w) < ϕ+(xq). So ϕ+(xp) · −ϕ+(xq) = 0.


Hence x ϕ+

σ,τ = 0. We have shown that ϕ+ can be extended to a homo-

morphism from F (P ) to F (Q+).

We prove that the homomorphism extending ϕ+ is an embedding.

By Part (b) of Sikorski’s criterion. It suffices to show that:

(=) if x ϕ+

σ,τ = 0, then xσ,τ = 0.

For a finite subset η ⊆ P let ρ(η) = {v ∈ V : η ∩ Pv �= ∅}. Let

ρ+ def= ρ+(σ, τ) = ρ(σ) and ρ− def

= ρ−(σ, τ) = ρ(τ).

Claim 1 If ρ+(σ, τ) and ρ−(σ, τ) are antichains and x ϕ+

σ,τ = 0, then

xσ,τ = 0.

Proof. By induction on |ρ+(σ, τ) ∪ ρ−(σ, τ)|. Let

n = n(σ, τ) = |ρ+(σ, τ) ∪ ρ−(σ, τ)|. We show that the induction holds

for n = 1. Suppose that x ϕ+

σ,τ = 0 and that |ρ+ ∪ ρ−| = 1. Then

there is v ∈ V such that σ ∪ τ ⊆ Pv. By Proposition 4.4(b), let

ψ be the embedding of F (Pv) in F (Q+v ) which extends ϕ+

v . Then

0 = x ϕ+

σ,τ = x ϕ+v

σ,τ = ψ(xσ,τ ). Since ψ is an embedding, xσ,τ = 0.

Suppose that the induction hypothesis holds for n, and let σ, τ be

such that x ϕ+

σ,τ = 0, ρ+ and ρ− are antichains and |ρ+ ∪ ρ−| = n + 1.

Case 1 There is v ∈ ρ+ and w ∈ ρ− such that v < w.

Let p ∈ Pv ∩ σ and q ∈ Pw ∩ τ . Then p < q. Hence xp · −xq = 0. So

xσ,τ = 0.

For the other cases we need some additional notations. For v ∈ ρ+

let

b+vdef= b+v (σ, τ) =

∏{ϕ+(xp) : p ∈ Pv ∩ σ} and

c+vdef= c+v (σ, τ) =

∏{ϕ+(xp) : p ∈ σ − Pv} . For v ∈ ρ− let

b−vdef= b−v (σ, τ) =

∏{−ϕ+(xq) : q ∈ Pv ∩ τ} and

c−vdef= b−v (σ, τ) =

∏{−ϕ+(xq) : q ∈ τ − Pv} .

34 ABRAHAM ET ALL

Let B and C be Boolean algebras. Then B ∗ C denotes their free

product.

Case 2 ρ+ ∩ ρ− �= ∅.

Let u ∈ ρ+ ∩ ρ−. Let ρ′ = (ρ− ∪ ρ+) − {u}. Since ρ+ and ρ− are

antichains, u is incomparable with every member of ρ′. So

(i) F (∑

{Q+v : v ∈ ρ− ∪ ρ+}) ∼= F (Pu) ∗ F (

∑{Q+

v : v ∈ ρ′}) .

We have 0 = x ϕ+

σ,τ = (b+u · b−u ) · (c+u · c−u ). Note that b+u · b−u ∈ F (Pu) and

that c+u · c−u ∈ F (∑

{Q+v : v ∈ ρ′}). So by (i), either b+u · b−u = 0, or

c+u · c−u = 0.

Suppose that b+u · b−u = 0. If we denote σ′ = σ ∩ Pu and τ ′ = τ ∩Pu,

then b+u · b−u = x ϕ+

σ′,τ ′ and ρ+(σ′, τ ′) = ρ−(σ′, τ ′) = {u}. So by the

induction claim for n = 1, xσ′,τ ′ = 0. Obviously, xσ,τ ≤ xσ′,τ ′ = 0, and

thus xσ,τ = 0. Suppose next that c+u · c−u = 0. If we denote σ′ = σ−Puand τ ′ = τ − Pu, then c

+u · c−u = x ϕ+

σ′,τ ′ , ρ+(σ′, τ ′) = ρ+ − {u}, and

ρ−(σ′, τ ′) = ρ− − {u}. Clearly, |ρ+(σ′, τ ′)∪ ρ−(σ′, τ ′)| = |ρ+ ∪ ρ−| − 1.

So by the induction hypothesis, xσ′,τ ′ = 0. Obviously,

xσ,τ ≤ xσ′,τ ′ = 0, and thus xσ,τ = 0.

Case 3 Cases 1 and 2 do not occur.

So

(1) For every v ∈ ρ+ and w ∈ ρ−, either w < v or w and v are

incomparable.

We shall show that in this case x ϕ+

σ,τ �= 0. This will contradict our

assumption. We shall thus conclude that Case 3 cannot occur. We

compute x ϕ+

σ,τ . For v ∈ ρ+ let σv = σ ∩Pv. For v ∈ ρ− let τv = τ ∩Pv.

Let v ∈ ρ+. Then x ϕ+

σv,∅ =∏

{ϕ+v (xp) : p ∈ σv}. So by Lemma

4.4(d)(1), x ϕ+

σv ,∅ has the form xm(v) + dv, where dv ∈ I bnd(Q+v ). So


(2) xϕ+

σ,∅ =∏

{xϕ+

σv ,∅ : v∈ ρ+} =∏

{xm(v)+dv : v ∈ ρ+} ≥∏

{xm(v) : v∈ ρ+}.

Let v ∈ ρ−. then

x ϕ+

∅,τv =∏

{−ϕ+v (xq) : q ∈ τv} = −

∑{ϕ+

v (xq) : q ∈ τv}.

Then by Lemma 4.4(d)(2), x ϕ+

∅,τv has the form −(xm(v) + ev), where

ev ∈ I bnd(Q+v ). So

(3) x ϕ+

∅,τ =∏

{x ϕ+

∅,τv : v ∈ ρ−} =∏

{−(xm(v) + ev) : v ∈ ρ−}

= −(∑

{xm(v) + ev : v ∈ ρ−}) .

By (2) and (3),

(4) xϕ+

σ,τ = (xϕ+

σ,∅ )·(xϕ+

∅,τ ) ≥∏

{xm(v) : v ∈ ρ+}−∑

{xm(v)+ev : v∈ ρ−}.

Let ρ = ρ− ∪ ρ+, ≤1 = ≤V� ρ and

≤2 = {〈w, v〉 : w = v or (w ∈ ρ− and v ∈ ρ+) }.

The relations ≤1 and ≤2 are partial orderings of ρ. Since ρ+ and ρ− are

antichains in V , and by (1), for every v, w ∈ ρ: if w ≤1 v, then w ≤2 v.

Let Q1 =∑1{Q+

v : v ∈ ρ}, where the lexicographic sum is taken with

respect to ≤1. Q2 is defined similarly. Let Q′ =⋃

{Q+v : v ∈ ρ}.

So IdQ′ is a homomorphism from Q1 to Q2. We denote the gener-

ators of F (Q2) by {yp : p ∈ Q′}. By Proposition 5.4(c), there is

a homomorphism θ : F (Q1) → F (Q2) such that for every p ∈ Q′,

θ(xp) = yp.

We apply Lemma 4.4(a) to

I = 〈ρ,≤2〉 ; Ri = Q+i , i ∈ ρ ; σ = ρ− ; τ = ρ+ ; ai = θ(xm(i) + ei),

i ∈ ρ− ; rj = m(j), j ∈ ρ+ .

Obviously, for every w ∈ ρ− and v ∈ ρ+, w <2 v. Also, aw ∈

Ibnd(Q+w), since xm(w) + ew ∈ Ibnd(Q+

w). We conclude that in F (Q2),∑w∈ρ− θ(xm(w) + ew) <

∏v∈ρ+ θ(xm(v)). That is,∏

v∈ρ+ θ(xm(v)) −∑

w∈ρ− θ(xm(w) + ew) �= 0. Hence

36 ABRAHAM ET ALL

θ(∏

v∈ρ+ xm(v) −∑

w∈ρ−(xm(w) + ew)) �= 0. So∏v∈ρ+ xm(v) −

∑w∈ρ−(xm(w) + ew) �= 0. By (4), x ϕ+

σ,τ �= 0. A contra-

diction, so Case (3) does not occur.

Hence xσ,τ = 0. We have proved Claim 1.

Claim 2 Let σ, τ be such that x ϕ+

σ,τ = 0. Then there are σ′ ⊆ σ and

τ ′ ⊆ τ such that x ϕ+

σ′,τ ′ = 0 and ρ+(σ′, τ ′), ρ−(σ′, τ ′) are antichains.

Proof. Let ρ+ = ρ+(σ, τ) and ρ− = ρ−(σ, τ). Let η+ be the set of

minimal elements of ρ+, and η− be the set of maximal elements of ρ−.

Let σ′ = σ ∩⋃

{Pv : v ∈ η+} and τ ′ = τ ∩⋃

{Pv : v ∈ η−}.

Clearly, ρ+(σ′, τ ′) = η+ and ρ−(σ′, τ ′) = η−. So ρ+(σ′, τ ′) and

ρ−(σ′, τ ′) are antichains.

Let p ∈ σ − σ′. Let v ∈ ρ+ be such that p ∈ Pv. There is w ∈ η+

such that w < v. Let q ∈ Pw ∩ σ. By the definition of σ′, q ∈ σ′. By

Lemma 4.4(c), ϕ+(xq) < ϕ+(xp). It follows that∏

{ϕ+(xp) : p ∈ σ′} ≤∏

{ϕ+(xp) : p ∈ σ}. Since σ′ ⊆ σ,∏{ϕ+(xp) : p ∈ σ′} ≥

∏{ϕ+(xp) : p ∈ σ}. So x ϕ+

σ′,∅ = x ϕ+

σ,∅ .

An identical argument shows that x ϕ+

∅,η′ = x ϕ+

∅,η . So xϕ+

σ′,η′ = x ϕ+

σ,η = 0.

We have proved Claim 2.

We now prove (=). Suppose that x ϕ+

σ,τ = 0. Let σ′, τ ′ be as assured

by Claim 2. By Claim 1, xσ′,τ ′ = 0. Since σ′ ⊆ σ and τ ′ ⊆ τ , xσ,τ = 0.

We have proved (=).

We have shown that ϕ+ fulfills the two parts of Sikorski’s criterion.

So there is an embedding ψ of F (P ) into F (Q+) which extends ϕ+.

It remains to show that ψ(Ibnd(P )) ⊆ Ibnd(Q+). Since {xp : p ∈ P}

generates Ibnd(P ), it suffices to show that for every p ∈ P , ψ(xp) ∈

Ibnd(P ). But this follows from the definition of ϕ+.


Proof of Theorem 4.1. We prove by induction on α, that for every

P ∈ Hα, F (P ) is well-generated. There is nothing to prove for α = 0

and for limit ordinals. Suppose that the claim is true for α, and let

P ∈ Hα+1 − Hα .

By Proposition 5.4(a), F (P ∗) ∼= F (P ). It follows from the definition

of Hα that if P ∈ H α+1 − H α, then P∗ ∈ Hα+1 − Hα.

Hence we may assume that P =∑

{Pv : v ∈ V }, where V is a

well-ordered poset, and for every v ∈ V , Pv ∈ H α. By the induction

hypothesis, for every v ∈ V there is a well-ordered poset Qv and an

embedding ϕv of F (Pv) in F (Qv) such that ϕv(Ibnd(Pv)) ⊆ I bnd(Qv).

We may apply Lemma 4.2 to V , {Pv : v ∈ V }, {Qv : v ∈ V } and

{ϕv : v ∈ V }.

Let Q+v , v ∈ V , Q+ and ψ be as assured by the lemma. Since Q+

v

is obtained from Qv by adding only one element, Q+v is well-ordered.

V is well-ordered.

It is easy to check that if in a lexicographic sum the index poset

is well-ordered and each summand is well-ordered, then the sum is

well-ordered. So Q is well-ordered.

Finally, ψ is an embedding of F (P ) in F (Q+), and ψ(I bnd(P )) ⊆

I bnd(Q+). So Q+ and ψ are as required.

38 ABRAHAM ET ALL

5. Appendix

Definition 5.1. A subset A of a poset Q is a final segment of Q, if

for every a ∈ A and b ∈ Q: if a ≤ b, then b ∈ A.

For a poset Q let M(Q) be the set of order preserving functions

from Q to {0, 1}. So the function f �→ f−1({1}), f ∈ M(Q), is a

bijection between M(Q) and the set of final segments of Q.

For every p ∈ Q let Vp = {f ∈ M(Q) : f(p) = 1}. Let B(Q) be the

subalgebra of ℘(M(Q)) generated by the set {Vp : p ∈ Q}.

Note that M(Q) is a subset of the set X = {0, 1}Q of all functions

from Q to {0, 1}, and if we equip X with the product topology, then

B(Q) is the set of clopen sets in the topology induced by X onM(Q).

Proposition 5.2. Let P be a poset. There is an isomorphism ϕ be-

tween F (P ) and B(P ) such that for every p ∈ P , ϕ(xp) = Vp. We

denote ϕ by ϕP .

Proof. By the definition of B(P ) for every p, q ∈ P : if p ≤ q, then

Vp ⊆ Vq. So the relation Vp ∩ Vq = Vp holds between the generators

Vp and Vq of B(P ). Recall that F (P ) was defined as the free Boolean

algebra with generators {xp : p ∈ P} and relations {xp · xq = xp :

p ≤ q}. Since the set of generators {Vp : p ∈ P} of B(P ) satisfy the

same relations, the relation {〈xp, Vp〉 : p ∈ P} is a function, and it

extends to a homomorphism from F (P ) to B(P ).

If p �= q, then Vp �= Vq. From this it follows that in F (P ), xp �= xq.

Since {Vp : p ∈ P} generates B(P ), ϕ is surjective.


Let σ, τ be finite subsets of P . It is easy to see that

(i) if⋂

{Vp : p ∈ σ} ∩⋂

{−Vq : q ∈ τ} = ∅, then for some p ∈ σ and

q ∈ τ , p ≤ q.

By the definition of the relations of F (P ),

(ii) if p ≤ q, then in F (P ), xp · −xq = 0.

Now suppose that a ∈ F (P ) and ϕ(a) = ∅. The element a is the

sum of elements of the form∏

{xp : p ∈ σ} ·∏

{−xq : q ∈ τ}. Let

us take one of these summands. Then

∅=ϕ(∏

{xp : p∈σ} ·∏

{−xq : q∈ τ})=⋂

{Vp : p∈σ} ∩⋂

{−Vq : q∈ τ}.

By (i), there are p ∈ σ and q ∈ τ such that p ≤ q. By (ii),∏{xp : p ∈ σ} ·

∏{−xq : q ∈ τ} = 0. It follows that a = 0. So ϕ is

injective.

Definition 5.3. Let 〈P,≤〉 be a poset.

(a) We denote by 〈P,≤〉∗ the inverse ordering of 〈P,≤〉. That is,

〈P,≤〉∗ = 〈P,≥〉. We abbreviate 〈P,≤〉∗, and denote it by P ∗.

(b) Let a ∈ P . Then P≥a denotes the set {b ∈ P : a ≤ b}. The set

P≤a is defined analogously.

(c) Let R ⊆ P . Then χR denotes the characteristic function of R.

That is, χR(p) = 1 for p ∈ R, and χR(p) = 0 for p �∈ R. So if R is a

final segment of P , then χR ∈ M(P ).

(d) Let P and Q be posets and α : P → Q. We say that α is a

homomorphism from P toQ, if for every s, t ∈ P : if s ≤ t, α(s) ≤ α(t).

We say that α is an embedding of P in Q, if for every s, t ∈ P : s ≤ t,

iff α(s) ≤ α(t).

40 ABRAHAM ET ALL

Proposition 5.4. (a) Let P be a poset. We denote by {xp : p ∈ P}

the set of generators of F (P ), and by {yp : p ∈ P} the set of generators

of P ∗. Then the function xp �→ −yp, p ∈ P extends to an isomorphism

between F (P ) and F (P ∗). We denote this isomorphism by ψP .

(b) Let σ and τ be finite subsets of P . The following are equivalent.

(i)∏

{xp : p ∈ σ} ·∏

{−xq : q ∈ τ} = 0.

(ii) There are p ∈ σ and q ∈ τ such that p ≤ q.

(c) Let P and Q be posets and α : P → Q be a homomorphism. Let

Xdef= {xp : p ∈ P} and Y

def= {yq : q ∈ Q} be the sets of generators

of F (P ) and F (Q) respectively. We define ρ : X → Y . ρ(xp) = yα(p).

Then

(1) ρ extends to a homomorphism ρα from F (P ) to F (Q).

(2) If α is an embedding of P in Q, then ρα is an embedding of F (P )

in F (Q).

Proof. (a) The function ψ′P defined by xp �→ −yp , p ∈ P , is an or-

der preserving function from {xp : p ∈ P} to F (P ∗). That is, for every

relation xp ·xq = xp in the defining set of relations of F (P ), the relation

(−yp) · (−yq) = −yp holds for the images of the x ’s under ψ′P . So ψ

′P

extends to a homomorphism ψP between F (P ) and F (P ∗). Since

Rng(ψ′P ) generates F (P

∗), ψP is surjective.

The homomorphism ψP ∗ : F (P ∗) → F (P ) is defined similarly.

ψP ∗(yp)=−xp. So ψP ∗(−yp)=xp. Hence

(ψP ∗ ◦ ψP ) �{xp : p∈P}=Id. It follows that ψP ∗ ◦ ψP = Id. So ψP is

injective.


(b) Suppose that there are p ∈ σ and q ∈ τ such that p ≤ q. Then

in F (P ), xp · xq = xp. So in F (P ), xp − xq = 0. So in F (P ),∏{xp : p ∈ σ} ·

∏{−xq : q ∈ τ} = 0.

For the second direction we use the isomorphism ϕP between F (P )

and B(P ). Suppose that for every p ∈ σ and q ∈ τ , p �≤ q. The set

R =⋃

{P≥p : p ∈ σ} is a final segment of P . For every p ∈ σ, p ∈ R;

and for every q ∈ τ , q �∈ R. So χR ∈⋂

{Vp : p ∈ σ}∩⋂

{−Vq : q ∈ τ}.

It follows that

ϕP (∏

{xp : p∈ σ} ·∏

{−xq : q ∈ τ}) =⋂

{Vp : p∈ σ} ∩⋂

{−Vq : q ∈ τ} �= ∅ .

So∏

{xp : p ∈ σ} ·∏

{−xq : q ∈ τ} �= 0.

(c) (1) A relation between generators of F (P ), has the form xs ·xt =

xs, where s ≤ t in P . Since α is a homomorphism, α(s) ≤ α(t). So

yα(s) · yα(t) = yα(s) holds in F (Q). That is, ρ(xs) · ρ(xt) = ρ(xs). This

implies (1).

(2) If σ and τ be finite subsets of P we denote

xσ,τ =∏

{xp : p ∈ σ} ·∏

{−xq : q ∈ τ}. Let ρ = ρα. It suffices to

show that for every finite σ, τ ⊆ P , if ρ(xσ,τ ) = 0, then xσ,τ = 0.

Now, ρ(xσ,τ ) =∏

{yα(p) : p ∈ σ} ·∏

{−yα(q) : q ∈ τ}. So if

ρ(xσ,τ ) = 0, then by Part (b), there are p ∈ σ and q ∈ τ such that

α(p) ≤ α(q). Since α is an embedding, p ≤ q. So by Part (b), xσ,τ = 0.

42 ABRAHAM ET ALL

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Soc., 2, 1952, 326-336.

[Ko] S. Koppelberg: Handbook on Boolean Algebras, Vol. 1, Ed. J.D. Monk, North

Holland 1989.

[Kr] J. B. Kruskal: Well-quasi-ordering, the tree theorem and the Vazsony’s con-

jecture, Trans. Amer. Math. Soc., 95, 1960, 210-255.

[P] M. Pouzet: On the set of initial interval of a scattered poset satisfying FAC,

1981, Private communication. (Annonced in Ordered sets, Edit. I.Rival . Nato

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[Sk] H. Si Kaddour: Ensembles de generateurs d’une algebre de Boole, Diplome

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Permanent addresses

Uri Abraham, Matatyahu Rubin

Departement of Mathematics

Ben-Gurion University of the Negev

P.O.B. 653 Beer-Sheva 84 105(Israel)

Robert Bonnet

Laboratoire de Mathematiques (Le Chablais)

Universite de Savoie – Campus Scientifique

73 376 Le Bourget-du-Lac Cedex (France)

Hamza Si Kaddour

Laboratoire de Probalite, Combinatoire et Statistique (Batiment 101)

Universite Claude-Bernard

69 622 Villeurbanne Cedex (France)

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