the structure of pseudocomplemented distributive

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 169, July 1972

THE STRUCTURE OF PSEUDOCOMPLEMENTED DISTRIBUTIVE LATTICES. III: INJECTIVE AND ABSOLUTE SUBRETRACTS BY

G. GRATZER AND H. LAKSER( 1) ABSTRACT. Absolute subretracts are characterized in the classes !B n , n ~ w. This is applied to describe the injectives in !B 1 (due to R. Balbes and G. Gratzer) and !B2.

1. Introduction. In Parts I and II ([41 and [61) we have acquired a rather thorough knowledge of the structure of pseudocomplemented distributive lattices. In this paper we use this knowledge to extend the results of R.Balbes and G. Gratzer

[11 on injective Stone algebras to any ~n. (Recall that ~ 1 is the class of Stone algebras; for the nota tion, see

§ 2.)

It turns out, however, that there are rather

few injectives in ~n. It appears that weak injectives and absolute subretracts are more appropriate to investigate in general. In

93 1

they coincide with injectives.

Accordingly, our main result is a description of weak injectives as presented in Theorem 1. An explic it construction (unique up to isomorphism) is given in (c) of Theorem 1 (corresponding to Theorem 2 of [1]) and an internal description in (d) of Theorem 1 (corresponding to Theorem 1 of [11). It turns out that weak injectives are injectives in ~ 1 and ~2; thus our result implies the results of [11, in fact, it yields a somewhat sharper form of Theorem 1 of [11. A description of injectives in ~2 is given in Theorem 3. In

§ 2 we

introduce weak injectives, absolute subretracts, and investigate

their interrelationships with injectives. These observations are applied in

§ 3,

where a series of lemmas are given leading up to Theorem 1. The applications are given in §4 including a proof of a result of R. A. Day [21. In the last section we describe a first order property n shared by the weak injectives in ~ n such that any first order property of weak injectives follows from n. 2. General algebraic preliminaries. We first recall some notations and results of Part II [4]. The equational classes of pseudocomplemented distributive lattices

Received by the editors July 29, 1971. AMS 1970 subject classifications. Primary 06A3S; Secondary 08A2S, 18A20, 18A30, 18COS. Key words and phrases. Distributive lattice, pseudocomplemented, injective, absolute subretract, weak injective. ( 1) Work of both authors was supported by the National Research Council of Canada.

475

Copyright © 1972, Amencan Mathematical Society

476

are the trivial class ~ -1' the classes ~n' n n

[July

G. GRATZER AND Ho LAKSER

2: 0, and the class ~w· Each ~n'

2: 0, is generated by the subdirectly irreduc ible pseudocompleme nted distributive

Bn ,where

B n is the n-atom Boolean lattice 2 n and

Bn

is B n with a new unit element. We recall that all these equational classes satisfy the Congruence lattice

Extension Property and that ~ n satisfies the Amalgamation Property if and only if n

1, 0, 1, 2, or w.

= -

If

K is

any class of algebras an algebra C

given any algebras A, B

¢: A

-+

K,

E

An algebra C

E

K

K is

said to be injective if,

A a subalgebra of B, and given a homomorphism

C, there is a homomorphism

enough injectives if each algebra in

K.

E

¢: B --+ C extending ¢. The class K has K is a subalgebra of an algebra injective

is a weak injective if, given algebras A, B

E

algebra of B, and given a surjective (i.e. onto) homomorphism ¢: A

¢:

is a homomorphism

B

--+

C extending

¢. An algebra C

subretract if C is a retract of each of its extensions in

E

K is

K,

in

A a sub-

--+

C, there

an absolute

K.

K will

We present the various relations among these concepts. The class

be

assumed to be an equational class. We first have a well-known lemma. Lemma 1. If C is an injective in

K,

weak injective in

K,

then C is a weak injective. If C is a

then C is an absolute suhre tract.

As converses to Lemma 1, we have the following results.

K satisfy subretract in K is a

Lemma 2. Let

absolute

the Congruence Extension Property. Then each weak injective.

Proof. Let C be an absolute subretract in of B, and let ¢: A

--+

K,

sion Property there is an extension C' of C in such that t/JiA

p: C'

=

E

K,

A a subalgebra

K

and a homomorphism

t/J:

B --+ C'

¢. Since C is an absolute subretract there is a retraction t/Jp is the required extension of ¢, proving the lemma.

--+ C. Thus

Lemma 3. If

K

satisfies the Congruence Extension Property and the Amalga-

mation Property, then any absolute subretract in

Proof. Let C be an absolute subretract in B be an embedding. Let ¢: A embedding. Let

let A, B

C be a homomorphism onto C. By the Congruence Exten-

¢ 1: A

--+

--+

K K.

is injective. E K and let a: A --+ /3: C 1 --+ C be the ¢ 1 f3 = ¢. By the Congru-

Let A, B

C, let C 1 = 1m ¢, and let

C 1 be defined by

¢; that is,

ence Extens ion Property there is an algebra B 1

E

K,

an embedding y: C 1 - 4 B

l'

and a homomorphism tjJ: B --+ B 1 such that at/J = ¢ 1 Y (see Figure 1). Since f3 and yare embeddings and K has the Amalgamation Property there is an algebra D in

.K

and embeddings A: B

1 --+

D, p: C

--+

D such that yA =

f3f1.

Since C is an absolute subretract there is a homomorphism p: D -+ C such that

llP is the identity mapping on C. Let

¢ = t/JAp;

injective, concluding the proof of the lemma.

then

a¢ = ¢ 1 f3 = ¢.

Thus C is

19721

477

PSEUDOCOMPLEMENTED DISTRIBUTIVE LATTICES. III

B

cP1

:r

A ... D

~

1 f3

Cl~C

A ----.. ...

Figure We now relate the concepts of absolute subretract and subdirect irreducibility. If A, B are algebras, A a subalgebra of B, t!"len B is said to be an es sential extension of A if, given any congruence ® on B such that W B'

8A

=

(8 A denotes the restriction of ® to A. If A is any algebra,

congruence: x == y(wA) if and only if x

=

wA then W

e=

A denotes the

y.) A maximal subdirectly irreducible

algebra is a subdirectly irreduc ible algebra every proper extension of which is

not subdirectly irreducible. Lemma 4. Any maximal subdirectly irreducible algebra A zn

lute subretr act in

K

is an abso-

K.

Proof. Let A be a subalgebra of B in K. By Zorn's Lemma (see Lemma 3(b) of [41) there is an essential extension B 1 of A and a surjective homomorphism ¢: B

--+

B 1 such that x¢

=

x for all x EA. Since A is subdirectly irreduc-

ible and B 1 is an essential extension of A it follows that B 1 is subdirectly irreducible (see [21). Thus, by the maximality of .A., B 1

=

A and so ¢ is a retraction

onto A. Thus we have shown that A is an absolute subretract. It is well known that a retract of an injective algebra is itself injective. A weaker result holds for weak injectives. Let (A il i E I) be a family of algebras and let ¢: A

--+

Il(A il i E I) be an embedding of A as a subdirect product. If

¢

also embeds A as a retract of II(A il i E I) we say that A is a subdirect retract of the family (A il i E I). Lemma 5. A subdi~ect retract of a family of we-a-k injective algebras is itself

a weak injective algebra.

Proof. Let (C iii E I) be a family of weak injectives and let y: C -+ I1(Cili E I) be a representation of C as a sudirect retr act; let p: II(C il i E I) --+ C be the retraction, that is, let yp

=

1 C'

Let A, B be algebras, let a: A

--+

a surjective homomorphism. If i Eland C z., then Y17.z is surJoective. Thus 'P r/....: A Z

B be an embedding, and let ¢: A ---. C be 17 i

--+

it

is the projection of I1(C i E I) onto C 1° defined by ¢.Z = ¢yrr.Z for each

i E 1 is a surjection. Since each C i is a weak injective, there is a homomorphism

t/J i:

B --+ C i such that atjJ i

=

1; t Consequently there is a homomorphism tj.J: B

--+

478

[July

G. GRATZER AND H. LAKSER

11(C i l i

¢ = t/Jp

E

t/J7T i = t/J i for each i E I; it follows that at/J = ¢y. a¢ = ¢, showing that C is a weak injective.

I) such that

we find that

Thus if

3. Absolute subretracts in ~n. Let B be a Boolean algebra and let B[n+1]

= {(X(\I x I'···' x n ) 'j.l

E

B n + 1\x Q -< x /\···/\ x n I. 1

B[ 21 was introduced in [11, and B In + d is the obvious gener alization of B[ 21. It follows quite easily that B[ n+ 11 is a pseudocomplemented distributive lattice and that

(X 0' xl'· · · , Xn) * = (x; /\ · · · /\ X~, X'},. • • , X~).

Bn

2[n+11, that if B' is a subalgebra of the Boolean algebra B then (B')[n+ll is a subalgebra of .B[n+ll, and that if B = mByly E r) (B, By" Y E r, are Boolean algebras), then B[n+ll '" mB~n+I]IY E r); thus B[n + 11 E ~n for any Boolean algebra B. In this section we show that the absolute It is also clear that

"J

subretracts in ~n are precisely the pseudocomplemented distributive lattices of the form B' x B[n+11, where Band B' are complete Boolean algebras.

Lemma 6. Any complete Boolean algebra is an injective pseudocomplemented distributive lattice.

Proof. Let B be a complete Boolean algebra, let C be a pseudocomplemented distributive lattice, and let A be a subalgebra (*-sublattice) of C. Let ¢: A --+ B be a *-homomorphism. Let f be the restriction of ¢ toe 2) S(A). Since S(A) is a subalgebra of the Boolean algebra S(C) and since B is an injective Boolean algebra,

f

mapping

¢:

lifts to a homomorphism of the Boolean algebras

T: S(C) --+ B.

The

C --+ B given by x¢ = (x**)T is a homomorphism that is the required

lifting of ¢. Thus B is an injective pseudocomplemented distributive lattice.

Lemma 7. Let B be a complete Boolean algebra. Then B[n +1] is a weak injective in ~ . n

Proof. Since B is an injective Boolean algebra and since the equational class of Boolean algebras is generated by the two-e lement Boolean algebra 2, it follows that B is a subdirect retract of a family (B 'Y lyE r) of Boolean algebras all isomorphic to 2. We claim that consequently B[n + 11 is a subdirect retract of the family (B;n+l]ly

B' is a homomorphism of Boolean algebras, then ¢[n+ll: B[n+l1-+(B')[n+l], given by (x o'··.' x ) ¢[n+l1= (x ¢'···' x ¢), o n n is a *-hom om 0 rphi sm. If 1T: B -+ B y is 0 n tot he n so is 1T[ n + 11: B[ n +I] -+ (B y )[n+ll; thus B[n+ll is a sudirect product of (B~+l1IY E r). If p: II(B 'Y Iy

E

E

r). If ¢: B

-+

r) --+ 3 is a retraction of Boolean algebras, then

(2)Recall that S(A)

= {x**lx

E

AI, the skeleton of A, is a Boolean algebra.

1972]

479

PSEUDOCOMPLEMENTED DISTRIBUTIVE LATTIC.c:So III

[12+1] p[n+d: ( n(By!Y

E

r) )

"'"'n(B.)[n+Il!y

E

r) _ B[n+l]

is a retraction of pseudocomplemented distributive lattices; thus B[n + r] is a sub-

direct retract of the family

(B~n+I1IY E

r). Since

B~n+I1 "'"' 2[n+I] "'"' En we con~n' proving the

clude, by Lemmas 2,4, 5, that B[n+ll is a weak injective in lemma.

It follows from Lemmas 6 and 7 that if Band B' are complete Boolean algebras, then B' x B[n + r] is a weak injective (equivalently, by Lemmas 1 and 2, an absolute subretract) pseudocomplemented distributive lattice in ~ . We now

characterize pseudocomplemented distributive lattice s of the form B' x

~[n + 11,

B,

B' Boolean. Let A be a pseudocomplemented distributive lattice. Recall that the set of dense elements of A, D(A) =

Ix

E A Ix* =

oJ, is a dual ideal of A. First

we present a result characterizing ~n'

Lemma 8. Let L be a pseudocomplemented distributive lattice. Then L if and only if L has the following property: let x 0' • • • ,

a whenever

i 1= j; then x~ V··· V x~

=

X

E

~

n

n E L satisfy x i A x j =

1.

Proof. K. B. Lee [71 gave an identity characterizing ~n' to wit (a 1 /\ ••• A a )* V (a* A a n

Let x.Z A x J. =

1

a if

i /:. j, i, j

2

1\ ... A a )* V·· · V (a

=

n

1

1\ ..• 1\ a

n-l

1\ a*)* n

= 1.

0, • • • ,n. Then x.Z -< x J~ whenever i 1= j. Hence x 0

< X *1 -

A··· A x *, x 1 < x * * A. x * A·.· A x * ••• x < x * A x * A... 1\ x * *. Applyn - 1 2 n" n- 1 2 n ing Lee's identity with a.Z = x~, i = 1,· • • , n, we conclude that Z

x~ V··· V x: ~ (x~ 1\ ... /\ x:)* V (x~* 1\ .. . I\x:)* V··· V (x~ A···/\ x:*)*=I. Thus the required property holds in ~ . n

Now let L be a pseudocomplemented distributive lattice such that x. A x . = Z

J

0, i/:.j, i, j= 0, ••• , n, implies that x~V ••• V x:= 1. Let a l , · · · , an ELand let x 0 = a 1 A... A a x = a * A a A ••• A a · . . x = a A. • • A a A a n*. n' 1 1 2 n" n 1 n-l Then x.Z A x.] = if i /:. j. Thus x*o V··· V x*n = 1, that is, Lee's identity holds

°

for a l~

-01

•

•

,

a n . Thus L E ~ n .

If L is a lattice with 0, 1 and x E L, a dual pseudocomplement of x, denoted

x +, is the obvious dual of a pseudocomplement; x V y

=

1 if and only if y ~ x +.

Lemma 9. Let L be a pseudocomplemented distributive lattice and let n The following two conditions are equivalent: (a) There are Boolean algebras Band B' such that L

(b) L has the following five properties: (i) L

E

33 n ;

I"'ttJ

B' x B[n + 1].

> 1.

480

[July

Go GRATZER AND Ho LAKSER

(ii) D(L) has a smallest element d; (iii) d has a dual pseudocomplement which is central (complemented) zn L; (iv) there are elements e l' • • • ,e n E L such that e"Z A e J" == 0 il i 1= J", e . V e ~ == d lor a II i, and e *1 A... A e * == 0; Z Z n (v) given u 1'·· ., u n E D(L) there is an x E L such that (x A e Z.)* V d == U i lor i == 1, · · · , n.

Proof. Let L == B' x B[n+1]. Then L is d == (1, (0, 1,.·· 1»

E

~n •

The smallest dense element in L

and d+ == (0, (1, 1,···, 1»

whose complement in L is

(1, (0, 0, .... , 0». For each i == 1,··· , n, let e i == (aIi,(O, a Ii'···' ani» where if i 1= j and a"" == 1. Then (iv) of (b) holds. To establish (v) note that zz

ail" ==

°

D (L)

==

l( 1, (x,I, .. . , 1»

E

B' x B[n+1]J.

Let u i == (1, (x i' 1, ... , 1» , i == 1,.. · , n. Th en x == ( 0 , (0, xl' • • • , x~ » . Now let (b) hold. We show that L I"'ttJ B' x B[n+Il where B' I"'ttJ (d+i (d++ exists, and equals d+*, because d+ is central) and B

I"'ttJ

D(L).

We first show that (d++] is a Boolean lattice; we claim, for each x

E

L, that

x* A d++ is the complement in (d++] of x A d++. Clearly (x A d++) A (x* A d++) == x A x* A d++ == O. Now (x A d++) V (x* A d++) == (x V x*) A d++; since x V x* E

D(L), x V x*

2

d

2

d++ and consequently (x Ad++) V (x* A d+l == d++, estab-

lishing our claim. That D(L) is Boolean will emerge during our proof of the representation. For each i == 1, · • • , n, define Ii: L ~ D(L) by requiring that xli ==

(x* A e i)* V d. The Ii will correspond to the projections B' x B[n+1] ~ B. We claim that each Ii is a homomorphism. First note that, for all x, y E L, (x A y)* == (x* * A y) * == (x* * A y* * )* •

Indeed (x A y)* == (x A y)* * * == (x* * A y* *)* ,

xA Y$x** AYSx**Ay**; thus (x A y)*

2 (x* * A y)*

~ (x* *

A y ** )* ,

establishing the triple equality. We can now show t!"tat Ii preserves A. Recall that U denotes the join operation in S(L), that is, a U b == (a* A b*)*. Now if x, y

E

L then

481

PSEUDOCOMPLEMENTED DISTRIBUTIVE LATTICES. III

1972]

(x A y) I.z = «x A y)* A e z.)* V d

=

«x A y)* A e~z *)* V d

=

«x 1\ y) * * U e ~) Vd z

=

«x * * A y * * ) U e ~) Vd z

=

«x * * U e ~z ) A (y * * U e ~z )) V d = «x * * U e ~z') V d) A «y * * U e ~z ) V d)

= «x*

A e.)* V d)A «y* A e.)*V d) = xl.Ayl .. z z z z

I.z preserves a

and 1 (the nO" of D(L) is d), since e z~ -< d. Before showing that Ii preserves V we observe that x Ii and x *Ii are complementary for Clearly

each x

E

L. Indeed

x I·1 A x* I·Z = 01.Z = d and

x I z. V x* I·z = (x* A e z.)* V (x* * A e.)* V d z ( * )* (* * 1\ e. )* Ve,V···Ve. * * IVe·+ * * > - x Ae.z V x zz 1 V···Ve n Z

1

=

1,

by (iv) of (b) and (the last equality) by Lemma 8. Since, by (v) of (b), Ii is onto (recall (x A e z.)* = (x** A e z.)*) W'e conclude that D(L) is Boolean; denote the complement of u in D(L) by u'. We now show that I.z preserves V. Note that xl·z = x** I·z since x* I·z = (xl·)' . z Thus (x V y)/.Z = (x V y)**I.Z = (x* A y*)*I.1 = «x* A y*)/.)' = (x*I.Ay*I.)' = Z Z Z

«xl·)' A (YI·)')' =xl·Z Vyl··Z Z Z Now let B' = (d+i and let 13 = D(L). We show that L ~ B' x B[n+l]. Define ¢: L --+ B' x B[n + 1] by setting rI... X'P

Note that (x V d, xl l'

••"

= (x A d++ , (x V d, xI 1 '

•••,

xl n) E B[n+ I] since x V d

Ii are lattice homomorphisms so is ¢. Also, O¢

=

x In) ) .

~ x**

(0, (d,

Vd

~ xli"

d~ • •• , d»

Since the and I¢

=

(d++, (1, 1,···, 1»; thus ¢ preserves 0, 1. To show that ¢ preserves pseudocomplements observe that

x*/ 1 A··· A x* In = «x** A e t )* A··· A (x** A en)*)V d =

«x*u e;)A ... A(x*U e:))v d

= (x* u

=

(x*u (e; A··· A e~)) V d

0) V d = x* V d.

Thus

x*¢ = (x* A d++, (x* V d, x*1 1'···' x* In»

= (x*A d++, (x*/ 1 A···A x*ln, x*/ 1 , · · · , x*l n » =

(x* A d++, «xiI)' A·· · A (xl n)' , (xiI)" · · · , (xln)'»

G. GRATZER AND Ho LAKSER

482

and, since we have shown above that x* /\ d++ is the complement in (d++] of x A d++, we conclude that x*¢ = (x¢)*.

We now show that -+

¢ is an isomorphism. It suffices to show that ¢ D(L): D(L)

D(B' x B[n+rl) and ¢S(L): S(L)

-+

S(B' x B[n+r]) are isomorphisms. Observe

that

and that u

E

D(L) implies u¢ = (d++, (u, 1,···, 1». Thus ¢D(L) is an isomor-

phism. We now show that ¢S(L) is an isomorphism. Let x E S(L) and let x¢ = 1, that is, x¢ = (d++, (1, •.. , 1». Thus x /\ d++ = d++, that is, x 2: d++, and

= 1 for i = 1,.··, n. Consequently (x* /\ e I.)* -> d+ for each i; = (x* 1\ e~)* = (x* /\ e.)* > d+ and, since e*1 /\ ••• /\ e n* = 0, we I I I conclude that x = (x U e~) /\ ••• /\ (x u e:) 2: d+. Thus x 2:. d+ V d++ = 1, showing that ¢s (L) is one-to-one. To complete the proof of the lemma we need only show that ¢s (L) is onto. First observe that d+ < (d++ /\ e .)* for i = 1, · · · , n, since d+ /\ d++ /\ e. = 0; I I thus d+l. = (d++ /\ e .)* V d> d+ V d 1, and so d++I. = d. Observe also that if I I I Y E (d++] then y E S(L); indeed, y* * ~ (d++) * * = d++ ~ d ~ y V y * and, since y =

(x* /\ e .)* V d I

thus x U e~

y** /\ (y V y*), thus y = y**. Each element of S(B' x B[n+ 11) is of the form (y, (u 1 /\ ••• /\ Un' U 1'···' Un»

condition (b) there is an x

E

where y E (d+i and u 1' ••• ' un E D(L). By (v) of L such that u.I (x 1\ e I.)* V d (x** A e I.)* V d,

= u.I for i = 1, · · · , n. Since d+l.I = 1 we conclude that = u.I for i = 1,· .. , n. Clearly x* A d+ E S(L); thus, since (x* /\ d1/.1 =

that is, such that x* I.I (x* /\ dl/.I

u i and x* A d+ /\ d++ y ~ d++ and d++l

=

0, (x* /\ d+)¢ = (0, (u 1 /\ ..• 1\ un' u 1'···' Un». Since

= d it follows that y¢ = (y, (d, d, ••• , d». Thus (y u (x* /\ d1)¢ i

(y, (u 1

/\ •• ·Au n, u 1'··· , Un»' showing that ¢S(L) is onto, and thus an isomorphism. Consequently ¢: L --+ B' x B[n+l] is an isomorphism, concluding the proof of the

lemma. We are now in a position to prove the main theorem of this paper.

Theorem 1. Let n 2: 1 be an integer and let L be a pseudocomplemented distributive lattice. The lollowing lour conditions are equivalent. (a) L is an absolute subretract in (b) L is a weak injective in

:Bn ;

93 n ;

(c) there are complete Boolean algebras Band B' such that L ~ B' x

B[n+ 11;

(d) L is complete and has the lollowing live properties:

(i) L E

'B n ;

(ii) D(L) has a smallest element d;

483

PSEUDOCOMPLEMENTED DISTRIBUTIVE LATTICES. III

1972]

(iii) d has a dual pseudocomplement which is central in L; (iv) there are elements e 1'··· ,e n E L such that e.l 1\ e.]

e.l V e~l

0 if i -/:- j,

=

= d for all i, and e*1 1\ ••• 1\ e*n = 0;

(v) given u 1'···' u n E D(L) there is an x E L such that (x 1\ e l.)* V d = U i for i = 1, · · · , n.

Proof. Since ~ n satisfies the Congruence Extension Property (a) and (b) are equivalent by Lemmas 1 and 2. Note that if L is complete so are a 11 principal ideals and principal dual ideals in L, and that if B' and B are complete so

is B' x B[n+l]. Thus, by Lemma 8, conditions (c) and (d) are equivalent. By Lemmas 6 and 7, condition (c) implies (b). To complete the proof of the theorem we need only show that condition (a) implies (d). Let L E ~

n

be an absolute sub-

retract. Since the subdirectly irreducible members of gj n are subalgebras of Bn "" 2[n + 1] it follows that L is a subalgebra, and thus a retract (preserving *) of a power of 2[n + 1]. Each power of 2[n + 1] is of the form B[n + 11, where B is a complete atomic Boolean lattice. Since B[n+l] is complete and satisfies (i) to (v), and since completeness and properties (i) to (v) are preserved under retraction of pseudocomplemented distributive lattices, it follows that L satisfies condition (d). We have thus concluded the proof of the theorem.

4. Injective pseudocomplemented distributive lattices. As was shown in Part

II [4], ~ l' the equational class of Stone algebras, and ~ 2 both satisfy the Congruence Extension Property and the Amalgamation Property. By Lemma 3, absolute subretracts and injectives agree in these classes; thus Theorem 1 specializes to the following two theorems.

Theorem 2. Let L E ~2. The following three conditions are equivalent: (a) L is injective in ~ 2; (b) there are complete Boolean algebras Band B' such that L

"J

B' x B[ 3];

(c) L is a complete lattice and L has the following four properties: (i) D(L) has a smallest element d;

(ii) d has a dual pseudocomplement which is central in L;

(iii) there is an element eEL such that e* V e** (iv) given u 1' u 2 and (x 1\ e *)* V d = u 2.

E

D(L) there is an x

E

=

d;

L such that (x 1\ e)* V d = u

1

Proof. We need.only show that (iii) and (iv) are equivalent to (iv) and (v) of Theorem 1 for n = 2. Clearly (iii) and (iv) imply (iv) and (v) of Theorem 1

Ve;

Ve;

with n = 2. Set e 1 = e, e 2 = e*. We need only establish that e 1 = e 2 = e V e* ~ e** V e* = d. On the other hand, let el' e satisfy 2

= d; d ~ e 1 V e~

(iv) and (v) of Theorem 1 with n = 2. Then e 1 1\ e 2 = 0, that is e 2 ~ e;, imply-

. e *2 2: e ** 1ng 1· S·1nce e *1 1\ e *2

= O·1mp1·1es e *2 S e *1 *

we conc 1u d e t h at e *2

= e ** 1 .

484

G. GRATZER AND H. LAKSER

= e; and note that (x 1\ e)* = (x 1\ e;)* = (x 1\ e;*)* = (x 1\ e 2)* and that

Set e

(x 1\ e *) *

(x 1\ e ~ *) *

=

=

(x 1\ e 1) * •

Theorem 3. Let L be a Stone algebra. The following three conditions are equivalent: (a) L is an injective Stone algebra;

(b) there are complete Boolean algebras Band B' such that L ~ B' x B[ 21; (c) L is a complete lattice and has the following three proper ties: (i) D(L) has a smallest element d;

(ii) d has a dual pseudocomplement which is central in L; (iii) given u

E

D(L) there is an x

E

L such that x* V d

=

u.

Proof. We need only show that condition (c) is equivalent to condition (c) = 1. From (iv) of Theorem 1, e 1 V e; = d and e~ = 0, that

of Theorem 1 where n

is, e =

1=

d, and (v) of Theorem 1 implies x* V d = (x 1\ d**)* V d = (x 1\ d)* V d

u, since d**

theorem with e

= 1

1. Thus (c) of Theorem 1 is equivalent to (c) of the present

= d,

thereby completing the proof.

Injective Stone algebras were characterized in [1]. Condition (b) of Theorem 3 appears there, but rather than condition (c) the following characterization

is given: a Stone algebra L is injective if and only if the following conditions hold:

(a) L is complete;

(f3) L has a smallest dense element; (y) L is also a dual Stone algebra; (o) a * = b * and a + = b + imply a = b.

We wish to remark that these conditions of [11 easily imply those of Theorem 3, indeed, (y) implies that d has a dual pseudocomplement which is central, and (y), (8) imply (iii) of Theorem 3. For u E D(L), let x = u +; we claim that x* V d =

u. Since u + is central (u +)* = U ++; thus (x* V d) + = (u ++ V d) + = (u V d) + = U +

since u

2: d. Also, (x * V d)* =

a = u*.

Consequently, by (8), x* V d

= u.

To complete our discussion we present a result of R. A. Day [21, giving a proof in the spirit of this paper.

Theorem 4 (H. A. Day [21). The only nontrivial equational classes of pseudocomplemented distributive lattices that have enough injectives are ~ 0'

:B l'

and

~2. If j3 is any other equational class of pseudocomplemented distributive lat-

tices, then L is injective in gj if and only if L is a com_plete Boolean algebra.

Proof. That

:B 0'

~ l' and gj 2 have enough injectives follows from the fact

that their respective subdirectly irreducible generators are injective in the respective class. The injectivity of complete Boolean algebras is Lemma 6. We now show that j3n' n

> 2,

has no other injectives-" Let L E ~n be

485

PSEUDOCOMPLEMENTED DISTRIBUTIVE LATTICES. III

1972]

injective. If L is a Boolean algebra, then L is a retract of any complete extension and so L is complete. Otherwise there is a dense element u E D(L) such that u 1= 1. Map e to u. Since

81

8n is

< e < 1) of 8 1, we

(= 10, e, 11,0

an extension

separating e (the smallest dense element in

to L by sending 0 to 0,1 to 1, and

¢: Bn -... L Bn) from 1. Since the subdirectly of Bn , there is a *-homomorphism get a *-homomorphism

irreducible members of ~ n are subalgebras p: L -... 8 n such that up 1= 1, hence e¢p = e. As was noted in Part I [61, the congruence collapsing e and 1 is the smallest nontrivial congruence in quently "/"'p: ~

Bn -... Bn is

an isomorphism and thus

Bn is

Bn.

Conse-

a retract of the injective

algebra L and is therefore injective. This conclusion would show that ~n has enough injectives and consequently satisfies the Amalgamation Property. Thus

n = 1 or 2. Thus

:B n ,

n

> 2, can have no injectives other than the complete

Boolean algebras. For ~w' an injective Boolean algebra must be complete, as above. If L E ~w and L is injective and non-Boolean, then-as above-there is a *-homomorphism

¢: 8 -...

L separating the two dense elements of

B for any nontrivial Boolean

algebra B. Thus ¢ is one-to-one, and to obtain the desired contradiction we need only choose B so that its cardinality is greater than that of L. 5. First order properties of weak injectives. The internal characterization of weak injectives in ~n' Theorem l(d), is in terms of first order properties (conditions (i)-(v) of Theorem l(d)) in addition to a second order property, namely that L be complete. In this section we show that all first order properties of weak injectives follow from a single first order property. Let B be a Boolean lattice. We say that B splits if

IB I =

1 or B "" Box

B l' where Bois atomic and there are no atoms in B 1 (therefore, either IB 11

=

1

or B 1 is infinite).

Lemma 10. E~ery complete Boolean lattice splits. Proof. Let B be a complete Boolean lattice and let a be. the join of all atoms in B. Then B "" (a] x (a'J and (a 1 is atomic while (a'J has no atoms. To prove that (a 1 is atomic one has to use the Inf inite Distributive Identity whic h is known to hold since B is complete. It is easy to see that not all Boolean lattices split.

Lemma 11. There is a first order sentence such that, for any Boolean lattice B, holds for B if and only if B splits.

Proof. The sentence should state that there is a smallest element a such that all atoms are contained in a. Let n be the first order sentence (in the language of pseudocomplemented

486

G. GRATZER AND H. LAKSER

[July

distributive lattices) that states (i)-(v) of Theorem led) and requires that (d+i and D(L) split. This makes sense since by the proof of Lemma 9, conditions (i)-(v) imply that (d+l and D(L) are Boolean.

Theorem 5. The first order sentence n holds for any weak injective in

93 n .

Conversely, if 'P is any first order sentence that holds in any weak injective in ~ n , then 'II follows from n •

The first part of Theorem 5 is already known. An equivalent form of the second part of Theorem 5 is the following:

Theorem 5'. A pseudocomplemented distributive lattice L is elementarily equivalent to a weak injective in ~ n if and only if n holds in L. Theorem 5' implies Theorem 5. Indeed, let Theorem 5' hold, let 'P be a first order sentence that holds in any weak injective in ~ n , and assume that n does not imply 'P. Then there exists a pseudocomplemented distributive lattice L satisfying n but not 'P. By Theorem 5', there exists a weak injective L 1 in ~ n that is elementarily equivalent to L. Hence 'P does nor hold in L l' contradicting the definition of 'P.

Proof of Theorem 5'. The ceonly if" part is trivial. Now let L be a pseudocomplemented distributive lattice satisfying n. By Lemma 9, we can assume that L

=

B' x B [n + 1], where B' and B are Boolean lattices that split.

By A. Tarski [8], any two infinite nonatomic Boolean lattices are elementarily equivalent and (see also Theorem 38.5 of G. Gratzer [3]) every atomic Boolean lattice is elementarily equivalent to a complete and atomic Boolean lattice. Since by S. Kochen [5] finite direct products preserve elementary equiv~lence, there are complete Boolean lattices C and C' such that B == C and B' ==

c'

(== denotes elementary equivalence).

It is easy to see that B -+ B[n + 1] commutes with prime limits. Therefore by S. Kochen [5] (or by direct computation) B[n + 11 == c[n + 11. Thus

L

=

B' x B[n+l] == C' x c[n+l],

and the right side is a weak injective in ~n by Theorem 1, completing the proof of Theorem 5'. REFERENCES L R. Balhes and G. Gratzer, Injective and projective Stone algebras, Duke Math. J. 38 (1971), 339-347. 2. R. A. Day, Injectivity in congruence distributive equational classes, Ph. D. Thesis, McMaster University, 1970. 3. G. Gratzer, Universal algebra, Van Nostrand, Princeton, N. J. 1%8. MR 40 #1320.

1972]

PSEUDOCOMPLEMENTED DISTRIBUTIVE LATTICES. III

487

4. G. Gratzer and H. Lakser, The structure of pseudocomplemented distributive lattices. II. Congruence extension and amalgamation, Trans. Amer. Math. Soc. 156 (1971), 343-358. MR 43 #124. 5. S. Kochen, Ultraproducts in the theory of models, Ann. of Math. (2) 74 (1961), 221-261. MR 2S #1992. 6. H. Lakser, The structure of pseudocomplemented distributive lattices. I. Subdirect decomposition, Trans. Amer. Math. Soc. 156 (1971), 335-342. MR 43 #123. 7. K. Bo Lee, Equational classes of distributive pseudocomplemented lattices, Canad. J. Math. 22 (1970),881-891. MR 42 #151. 8. Ao Tarski, Arithmetical classes and types of Boolean algebras, Bull. Amer. Math. Soc. SS (1949), 64. DEPARTMENT OF MATHEMATICS, UNIVERSITY OF MANITOBA, WINNIPEG 19, MANITOBA, CANADA