Generalized Boolean lattices

J. Austral. Math. Soc. 19 (Series A) (1975), 225-237. GENERALIZED BOOLEAN LATTICES Dedicated to the memory of Hanna Neumann RICHARD D. BYRD, ROBERTO A. MENA, and LINDA A. TROY (Received 11 June 1973) Communicated by G. B. Preston 1. Introduction Hashimoto (1952; Theorems 8.3 and 8.5) proved the following theorems: THEOREM A. / / L is a distributive lattice, then there exists a generalized Boolean algebra Lr and an isomorphism from the lattice of all congruence relations of L onto the lattice of all congruence relations of Lr. THEOREM B. Any distributive lattice L is isomorphic with a sublattice of a relatively comolemented distributive lattice L* such that (1) the lattice of congruence relations on L* is isomorphic with that on L and (2) the length of the closed interval [a, b] in Lis equal to that of [a, b] regarded as an interval in L*. It has been noted that Hashimoto's proofs are somewhat difficult to follow and are proved with the apparatus of topology; hence these purely lattice theoretic theorems are not placed in their most natural setting. In 1958 Gratzer and Schmidt (1958; Theorem 1) asserted the following generalization of the Hashimoto theorems: To any distributive lattice L there exists a generalized Boolean algebra B having the properties: (1) L is a sublattice of B; (2) the lattice of all congruence relations of L is isomorphic to the lattice of all congruence relations of B; (3) if the interval [a,fc]of Lis of finite length, then [a, b] has the same length as an interval of B. In this note we give a counterexample to (2). In Section 2 we construct an ideal £ of Band prove that the lattice of all congruence relations of L is isomorphic to the lattice of all congruence relations of E (Corollary 2.11). We prove that E = B if and only if 0 e L (Theorem 2.4); otherwise, £ is a maximal ideal of B (Corollary 2.5). Our example shows that in general L cannot be embedded into E. The construction of E is algebraic and, moreover, we prove that E is unique up to isomorphism (Corollary 2.12). Thus we strengthen Theorem A. 225 Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 226 Richard D. Byrd, Roberto A. Mena and Linda A. Troy [2] Our example also shows that Gratzer (1971; Lemma 5, page 104)) is incorrect (Lemma 5 is in essence (2) above). We prove in Section 3 (also, see Corollary 4.3) that a necessary and sufficient condition that this lemma be true is that the lattice L has a smallest element. In Section 4 we investigate distributive lattices K and L, where Lis a sublattice of K and each congruence of Lhas a unique extension to K. In this case we prove that there is a generalized Boolean lattice B that is /^-generated by Land contains K as a sublattice (Theorem 4.5). From this result we obtain several interesting corollaries; one of which asserts that the lattice L*of Theorem B is unique up to isomorphism. In Section 5 we give the example. Throughout this note L will denote a distributive lattice, ^(L) will denote the lattice of all congruence relations on L, J{L) will denote the lattice of all ideals of L, N will denote the set of natural numbers, Z will denote the set integers, • will denote the empty set, and X\Y will denote the set of elements that belong to the set X but not to the set Y. Unless otherwise stated, isomorphism will mean a homomorphism that is one-to-one (not necessarily onto). For the standard results and definitions concerning lattices, the reader is referred to Gratzer (1971) in particular, to Sections 9 and 10 of Chapter 2. 2. Evenly generated ideals Throughout this section let B be a generalized Boolean lattice and let L be a sublattice of B that generates B, that is, the smallest subring of B that contains Lis B. LEMMA 2.1 (i) B = {ax + ••• + a n | n e N and au---,aneL). ( i i ) If T is a sublattice of Land xeB such that x = at + ••• + an (n eN), whereau---,ane T, thenx = bt + ••• + bn, where bu • • - , £ „ £ Tandbx ^ ••• ^ bn. The proof of this lemma is similar to the proof of Gratzer (1971; Lemma 3, page 102) and will be omitted. COROLLARY 2.2. IfTis a sublattice of L and xeBsuch ( n e N ) , where a1,---,a2n-l^T, that x = a1-] 1- a2n _ t then x ~2. a for some a in T. PROOF. By the lemma, x = bx -i— + b2n-i, where bu---,b2n-ieT and bt ^ ••• ^ &2B-1- Thus x ^ x A l » i = xbi = bx, since In - 1 is odd and b^bj = bt for each j . LEMMA 2.3. Let T be an ideal of L and ET = {x\xeB au---,a2,,e T(neN) such that x = at + ••• + a2n}. Then and there exists (i) ET is an ideal of B; (ii) ET = E x . n ( T ] B , where ( T ] B denotes the ideal of B generated by T; Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 [3] Generalised Boolean Lattices (iii) ET = {x\xeB and there exists and x — at-\— + a 2 n } . 227 such that a1 ^ ••• ^ a2n au---,a2neT The proof of this lemma is straightforward and will be omitted. We shall call the ideal ET of this lemma the ideal of B evenly generated by T. Also for the remainder of this note we shall denote EL simply by E. In Corollary 2.2, if T = Land x in B has a representation as a sum of an odd number of elements from L, then x exceeds an element of L. The next theorem shows that if L does not contain the zero of B and x in B has a representation as a sum of an even number of elements of L, then x does not exceed an element of L. Also, if x = at + ••• + an, where au---,aneL, then xe{al V ••• V fln]fl and so each element of B is exceeded by an element of L. We say that L R-generates B if Lgenerates B and if Lhas a least element, then it is the zero of B. (The definition of .R-generates given in Gratzer( 1971; page 102) also required that if L has a largest element, then it must be the largest element of B.) If L does not have a smallest element, then the definitions of .R-generates and generates coincide. THEOREM 2.4. / / L generates B, then the following assertions are equiv- alent: (i) (ii) (iii) (iv) OeL. LHE ¥= • • L c E. E = B. PROOF, (i) implies (ii) is immediate as 0 e L n £ . (ii) implies (iii). Let aeL and beLnE. a = a + b + beE. Then a + beE and hence (iii) implies (iv) is obvious as E is an ideal of B. (iv) implies (i). If beL, then b = at + ••• + a2n, where au •••,a2neL at ^ ••• ^ a2n. Then b A ateL and b /\at = bal = at + ••• + alt there are 2« summands. Therefore bat = 0 and so OeL. and where COROLLARY 2.5. IfO$L, then the index of E in B is two and hence E is a maximal ideal of B. Moreover, L is a sublattice of the relatively complemented lattice B\E and if M is an ideal of B such that M <~\L= • , then M s E. PROOF. By the theorem, £ ^ B and L s B\E. If xeB\E, then x = at + —|«2n-i > where au---,a2n_ If n = 1, x = a t a n d i f n > I,a1 + ••• + a2n_2eE. teL. Thus E + x = E + a for some aeL. If a,beL, then a + beE and so £ + a = £ + b . Therefore the index of £ in B is two. Since £ is an ideal of B and B is a generalized Boolean lattice, B\E is relatively complemented. If M e / ( B ) | / ( £ ) , then M H L # Q by Corollary 2.2. Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 228 Richard D. Byrd, Roberto A. Mena and Linda A. Troy [4] LEMMA 2.6. Let 0 e # ( L ) and Ig be the ideal of B generated by {a + b\a,beLand aOb}. Then Ig = { I ? = 1 a ; + fc,-|neN, a^b^L, at <L bh and OiObi} and so IoeJ(E). PROOF. For a,beL, a9b if and only if af\bOa\/b. Also, a + b = ab + a + b + ab = a A b + a\/ b. Thus, Ie is the ideal of B generated by {a + b\a,beL, a ^ b , and aOb}. Let I = { 'L?=1al +bt\neN, a^h^eL, at ^ bt, and afib-^. Then clearly {a + b\a,beL, a ^ b, and adb} £ / ^ Ie and I is closed with respect to addition. To prove that / is an ideal of B, it suffices to show that for xel and aeL, axel. Let x = X" = 1 a f + bt, wherea h b t eL, a ; ^ bi, and afib^ For each i, aat = a Afl;^ a Afrf= ab,- and a^b, implies a A «; 0 a A £>;• Thus a x e / and so I = Ie. Clearly 79e ./(£). 2.7. Let leJ{E) and 9t = {{a,b)\a,beL Qj is a congruence relation of L. LEMMA PROOF. and a + bel}. Then It is easily verified that 07 is an equivalence relation of L and if aOb and t e L , then a A tO b A t . N o w a \ / t + b \ / t = a + t + at + b + t + b t = a + b + (a + b)te/. Therefore a\J tOb\J t and so 0Te'g'(L). THEOREM 2.8. T/ie mapping g of J{E) into ^(L) given by (l)g = 6t is an isomorphism of •/(£) onto "^(L). T7ie inverse of g is given by {6)g~1 = Ig. PROOF. By Lemma 2.7, # is a mapping of J{E) into #(L). If 0 e <tf(L), then by Lemma 2.6, l9eJ{E). We will show that (78)^ = 0Ig = 6. If aOb, then a + fee/, and so a6Ieb. Conversely ifaGIob, then a + belg and, without loss of generality, we may assume that a ^ b. Now a + bele implies that a + b = E" =1 tfj + b^ where a,-, J^eL, a; ^ fo;, and a ^ . We induct on n. If n = 1, then a + b = ( a +fo)6= axb + &! b, axb ^ fcib, and a^bOb^. Since &tb ^ i, we may assume that a + b = ax + blt where ax 0 b t and a ^ i i j ^ f i . Therefore 0 = a t + at = (at + b^a! = aax + bat = aat + at. Hence ax ^ a. Now a + b = ax + bt g bx, as at ^ bt, and so n = a V (a + b) ^ a V i>j ^ a V & = b. Whence a\J bx = b and a V «i — a. Thus a ^ b ! implies a V axda\J bx or a 0 b . Next assume that n > 1. Again a + b = £ " = 1 a ^ + bjo and so we may assume that at^ bt ^ b for each i . Let d = V " = i bt. For 1 ^ k ^ n , aafc + a* = (a + b)ak = 2 "=i,i#kaiak + biak. Now a 0/efo implies that aak9Igak and hence by the inductive hypothesis aak0ak. Since ak0bk, we have aak9bk and so a 0 a V fyt. Also ad + d = (a + fo)d = Z r = i M + M = 2 " = i a , - + fc; = a + 6. Hence a V ^ = « + ^ + a d = b . Thus we have a0 V % i ( a V h;), which is equivalent to aQb. Therefore g maps S(E) onto ^"(L). Let / , JeJ(E) with / ^ J and let xej\l. Then there exists a x , - - - , a 2 n 6 L with at ^ ••• ^ a 2n and x = a x + ••• + a 2 n - Multiplying x respectively by a2> - "» f l 2n-2» w e obtain flj + a 2 , ••• ) a 2 n _ 1 + a 2 n e J . Hence for some fc,(l ^ fe^n) a 2 t - i + a2keJVTherefore ( a 2 t _ 1 , a 2 t ) G ^ j \ ^ r a n d so g is one-to-one. Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 [5] Generalised Boolean Lattices 229 If /, J e ./(£), then clearly / £ J if and only if 9j £ Qj. Therefore g is an isomorphism of J{E) onto ^(L). It is now clear that (6)g~l = Ie. An immediate consequence of this theorem is COROLLARY 2.9. (i) / / 0e#(L), then 9 = 0Ie. (ii) IfIeS(E), COROLLARY then I = Itl. 2.10. (Hashimoto, (1972; Theorem 7.2) </(£) (is isomorphic to «"(£). PROOF. Since £ is a generalized Boolean lattice and E generates itself, we have by the theorem that J(E) is isomorphic to COROLLARY 2.11. %(L) is isomorphic to COROLLARY 2.12. / / D is a generalized Boolean lattice such that ^(L) is isomorphic to ^(D), then D is isomorphic to E. PROOF. AS noted above, ^(D) is isomorphic to •/(!>)• Hence ./(D) and •?(E) are isomorphic. The compact elements of S(E) are the principal ideals of E, which are isomorphic to E. Since compact elements are preserved under isomorphism, it follows that E is isomorphic to D. If Te J{L), let 6T = {(a,b)|a,beL and a\/ t = b\/ t for some t eT}. It is easily verified that dT = {(a, b) I a, b e L and a\J b = (a /\b)\J t for some te T] and that the mapping that sends Tinto 9T is an isomorphism of «/(L) onto a sublattice of #(L) (Hashimoto (1973; Theorem 5.1)) . Moreover, for each t e T, [t]9T = T, where \t]9T denotes the congruence class of 9T containing t. COROLLARY 2.13. Let h be the mapping ofJ(L) into J(E) given by (T)h= ET. Then h is an isomorphism of -f(L) onto a sublattice of J(E). PROOF. It suffices to show that ET = IBT . If s, t e T, then s9Tt and so s + teI$T. It follows that ET = I9r. Conversely, let a,beLwith a9Tb and a + b eIgT. Then a \J b = (a A 6) V t for some (in Tandsoa + b + ab = ab + t + abt. Therefore a + b = t + abt and since /, abte T, we have a + beET. Again it follows that Igr £ ET. We have now obtained the Hashimoto theorems mentioned in the introduction. For let L be a distributive lattice and B be a generalized Boolean lattice R-generated by L. Then E = EL is a generalized Boolean lattice and ^(E) is isomorphic to ^(L) (Corollary 2.11). Moreover, we have proven that, up to isomorphism, E is unique (Corollary 2.12). If L has a smallest element, then E = B (Theorem 2.4). If L does not have a smallest element, then L £ B\E (Theorem 2.4). Let K = B\L- Then K is a prime dual ideal of B since £ is a prime ideal of B (Corollary 2.5). Thus K is a relatively complemented distributive lattice, Lis a sublattice of K, and K does not have a smallest element (Corollary 2.2). Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 230 Richard D. Byrd, Roberto A. Mena and Linda A. Troy [6] If we let EK = {ct + ••• + c2n\cl,---,c2neK}, then it is easily verified that EK = E. Since K generates B, <£(K) is isomorphic to V(EK) (Corollary 2.11) and so ^{K) is isomorphic to ^ ( L ) . We will prove in Section 4 that, up to isomorphism, K is unique (see Corollary 4.9). 2.14. Let F be a maximal ideal of B and K = B\F. Then F = {x|xe.B and there exists cly---,c2neK(ne N) such that x = c t + ••• + c2n}. Hence K generates B and ^(K) is isomorphic to COROLLARY PROOF. Let EK = {ct + ••• + c2n \ cu •••, cln e K]. Since F is maximal, K is a dual ideal of B and hence a sublattice of B. Let c, d e K. Since c + d = c f\d + c\J d, we may assume that c <j d. Thus c A (c + d) = 0 . Thus c + deF, as F is prime, and it follows that EK £ F. If xeF and c e X , x + c e F + c s x and so xeK + c s £ K . Therefore F = EK and -K generates B . Since EK is the ideal of B evenly generated by K we have by Corollary 2.11 that ^(K) is isomorphic to ^(F). Observe that K is a relatively complemented lattice, but in general does not have a smallest element. Hence in general J(K) and ^(K) are not isomorphic. To this end we prove 2.15. Let F be a maximal ideal of B and K = B\F. Then the are equivalent: K is a generalized Boolean lattice. K has a smallest element. K is isomorphic to F. There exists ceK such that cF - {0}. There exists an atom of B in K. F is a direct summand of B. THEOREM following (i) (ii) (iii) (iv) (v) (vi) PROOF, (i) implies (ii) is trivial. (ii) implies (iii). As noted above, K is always a relatively complemented lattice and so by (ii), K is a generalized Boolean lattice. Thus by Corollary 2.12, K is isomorphic to F. (iii) implies (iv). If K is ismorphic to F, K has a least element c0. Now F is generated by {c + d|c,deK and c ^ d}. For c,deK with d ^ c, c ^ c0 and so 0 g c 0 A (c + d) ^ c A (c + d) = 0 . It follows that c0F = {0}. (iv) implies (v). Let c e K such that cF = {0}. Let x e B with 0 ^ x <; c. If x e f , then 0 = ex = x. If xeK, then c + x e F and 0 = (c + x)c = c + x . Therefore x = c and hence c is an atom of B. (v) implies (vi). Let c be an atom of B in K. Then (c] B n F = {0}. Since F is a maximal ideal of B, F + (c] B = B and so F is a direct summand of B. (vi) implies (i). Suppose that B is the direct sum of F and M , where M is an ideal of B. Since the index of F in B is two, it follows that M = {0, c} for Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 [7] Generalised Boolean Lattices 231 some ceK. If x e K, then xf\c = c or x/\c = 0. Since F is prime, x A c ^ 0. Therefore c is the smallest element of K and hence K is a generalized Boolean lattice. In concluding this section, we note that Theorem 2.15 exemplifies a reason for requiring in the definition of L R-generates B that if L has a smallest element, it must be the zero of B. 3. Extending congruences Throughout this section let L denote a lattice without a smallest element, L o denote the lattice L with a smallest element 0 adjoined, and let «•„(!) = {6\0e%(L) and L/0 has a least element}. J. Hashimoto (1973; Theorem 5.1) proved that &0{L) is a dual ideal of C€{L). The hypothesis that L is distributive is not needed for the proofs of Lemma 3.1, Corollary 3.2, and Theorems 3.3 and 3.4. LEMMA 3.1. / / 9 is a congruence relation of L, then 9 can be extended to a congruence relation of Lo. Moreover, (i) if 9£t>0{L), then 9 has a unique extension to L o ; (ii) if 9e'£0(L), then 9 has exactly two extensions to L o . PROOF. If 90 = 6 U {(0,0)}, then 90 is a congruence relation of L o that extends 9. If 9 $ ^0(L), then it is easily verified that 90 is the only extension of 9. If 9e<£0(L) and T is the smallest element of LI9, then let 0 t = 9 u ( T x {0}) U({0} x T ) u { ( 0 , 0 ) } . Then 9l is a congruence relation of LO that extends 9 and 9i> 90. Again it is easily verified that these are the only extensions of 9 in COROLLARY 3.2. A congruence relation 9 of L has a unique extension to a congruence relation of Lo if and only if If T is an ideal of L, we observed in Section 2 that 9T e 'g'(L) and trivially T is the least element of ~L\9T. Thus by the theorem, 9T has two extensions to L o . It is evident that if B is a generalized Boolean lattice .R-generated by L o , then B is also R-generated by L. We note that Lemma 5 (Gratzer (1971; page 104)) is valid if the lattice has a smallest element and hence each congruence of Lo has a unique extension to B, but Lemma 3.1 shows that each element of ^0{L) has two extensions to B. Since ^(L) is isomorphic to ^(E) and co(L0) is isomorphic to 'tf(B), this raises the following question: which congruences of E have two extensions to B1 (By Gratzer (1971; Theorem 6, page 90) each congruence of E has at least one extension to B.) In this section we give an answer to this question. Let f0 and , : t be the mappings of r€(L) into ^(L o ) given by Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 232 Richard D. Byrd, Roberto A. Mena and Linda A. Troy 1 I 0/o [8] if otherwise, where 0 e ^(L) and 90 and 9l are given above. THEOREM 3.3. ( i ) / 0 is an isomorphism of #(L) into (ii)/! is a one-to-one inclusion preserving mapping of<£(L) into ^(Lo) and /j | ^ 0 (L) is an isomorphism. PROOF. The verification of (i) and that/ x is a one-to-one inclusion preserving mapping is straightforward and will be omitted. Let 9, vFe'£'0(L) and let S and T denote the smallest elements of 9 and ¥ respectively. Then SnTis the smallest element of 9 A ¥ (Hashimoto (1952; Lemma 5.1)). Then 91 A ^ j = (0U(Sx {0}) u({0} x S ) u { ( 0 , 0 ) } n C F U ( r x {0})u({0} x T) = (9 n «F) u ((S n T) x {0}) u ({0} x(Sn T)) u {(0,0)} = (0A ¥) ! • Since / x is inclusion preserving 6t V ^ s (0 V *P)i • Let (a,b)e(0 V ^ ) i and U be the least element of L/(0 V * ) • If (a,fc)e 0 V * , then clearly (a, b) e 0! V f i If (a,b)e(U x {0}) U({0} x C/), then, since (0 V *F)i is symmetric, we assume that aeU and b = 0. Now aeU implies that there exists bo,---,bmeL such that a = b0, bmeT, and (bi,bi+1)e9ux¥. Hence (a,b m )e9 t V ^ i • Since TzU (Hashimoto (1952; Lemma 5.1)). ( ^ , 0 ) 6 ^ . Therefore (a,0)e9i V ^ i • Trivially, if a = b = 0, (a,b)e91 V ^ i . We now describe the lattice ^(L o ) in terms THEOREM 3.4. #(L 0 ) is f//e disjoint union o/(^(L))/ 0 and {^0{L))ft. Moreover, (^(L))/! is a prime dual ideal ofV(L0) and for 9eV(L) and ¥ e ^ 0 ( L ) , and 0o A Vi = (0 A »P)o PROOF. Using Lemma 3.1, Theorem 3.3, and the fact that L is a sublattice of Lo, it is easily verified that ^(L o ) is the disjoint union of C#(L))f0 and (^o(£))/i and that (^ 0 (£))/i is a prime dual ideal of^(L0). Let 0e^(L), f e ^ L ) , and T be the least element of L/>P. Then 0o A ? \ = (0 u {(0,0)})nCF U(T x {0}) U({0} x T) U {(0,0)}) Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 [9] Generalised Boolean Lattices 233 Since ft is inclusion preserving 0O s (0 y ¥ ) ! and *Fj £ (0 V **0i • Therefore 0O V ^ i £ (0 V ¥ ) i . Let ( a , i ) £ ( 8 v *P)i and let U be the least element of L/(0 V *P) • If (a,fc)e 0 V ¥ , then clearly (a, b) e 0O V ? i . Suppose that (a,b)e[/ x {0}. Then there exists b0, •••,bmeL such that a — b0, bmeT, and ( ^ J j + 1 ) e 9 u 1 P . As above, (a, &J e 0O V ^ i and ( 6 . 0 ) 6 ? , . Hence (o,0) e 0O V ^ i • It follows that (0 V *)i £ 0O V ? i • For the remainder of this section, let B be a generalized Boolean lattice i?-generated by L and E be the ideal of B evenly generated by L. Then by Corollary 2.5, £ is a maximal ideal of B. Obviously B is also R-generated by Lo and by Theorem 2.4 the ideal of B evenly generated by Lo is B. Let p be the isomorphism of ^(L) onto J{E) given by (8)p = Ig (p is the inverse of the isomorphism g of Theorem 2.8), let p0 be the corresponding isomorphism of "^(LQ) onto let i be the inclusion mapping of •/(£) into^(B), and let/ 0 be as above. 3.5. / / 0e#(L), then (0)pi = (0)/ o p o The proof of this lemma is routine and hence will be omitted. LEMMA 3.6. Let F be a maximal ideal of B and I be an ideal of F. Then there are at most two ideals of B whose intersections with F is I. LEMMA PROOF. Let M,NeJ{B)\f(F) such that M\C\ F = AT nF = I. Since F is maximal, we have M = M A B = M A (F V N) = (M A F) V (M A N) = / V (M A N) S iV. Dually 7V,c M. Trivially / is the only ideal of F whose intersection with F is / . THEOREM 3.7. Lef / be an ideal of E. Then there exists such that M r\E = I if and only if (Z)p-1 e^ 0 (L). MeJ{B)\J{E) Let M e S{B)\J{E) such that M HE = I ,T = M r>L, and 9 = Ip~l. Then TeJ(L) and to prove that 06<g'o(L), it suffices to show that TeLjO. If a,beT, then a + beM (~\E and hence by Lemma 2.7, aOb. Conversely let a e T and be[a]0. Then a + foe/ £ M. Thus we have b =a + a + beM. Therefore beM nL= Tand so TeLjQ. Since TsJ{V), it is the smallest element of L/0. Hence 0 e # 0 (L). Next suppose that 0 = Ip-xe<tf0(L). Then 0O = 0ft n((Lx L)u{(0,0)}) = 0/t n(Lx L)/ o . By Theorem 3.4, 0/, n ( L x L)/ o = 0/o and by Lemma 3.5, Now 0/ l 3 o e^(B) and if I = 6/n = 6fogo = 0/lfiro n(Lx L)fogo = eftgonE. Tis the smallest element of L/0, then T £ 0/^0- Therefore Qfxg0§J(E). It follows from this theorem that a congruence of £ which has exactly two extensions to B is induced (see the discussion after Corollary 2.12) by an element from PROOF. 4. Extension Property Let K be a distributive lattice. We say that K has the extension property (EP) over L if L is a sublattice of K and each congruence of L can be uniquely Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 234 Richard D. Byrd, Roberto A. Mena and Linda A. Troy [10] extended to a congruence of K. If L is a sublattice of K, then K has (EP) over L if and only if the mapping of ^(K) into 'tf(L), which sends / onto x^(L x L), is one-to-one. (It is well known that this mapping is onto (Gratzer (1971; Theorem 6, page 90)). The next lemma is immediate from the definition. LEMMA 4.1. Let K be a distributive lattice and M and L be sublattices of K such that L ^ M. Then K has (EP) over L if and only if K has (EP) over M and M has (EP) over L. 4.2. Let B be a generalized Boolean lattice R-generated by L. (i) If OeL and K is a sublattice of B that contains L, then K has (EP) over L. (ii) If 0$Land K is a sublattice of B\E that contains L, then K has (EP) over L. THEOREM Let K be as either in (i) or (ii), 9 e %(L) and / e <£(K) such that x 1- c2n | cu---,c2neK), is an extension of 9. Trivially, K i?-generates B and E = {cl-{ where E is the ideal of B evenly generated by L. Let /„ and Ix be the ideals of E generated by {a + b | a, b e L and a9b} and {c + d | c, d e K and c%d} respectively. Clearly Ig £ Ix. Suppose (by way of contradiction) that Ig # Ix. Let f e ^ ( L ) such that OFJgT1 = Ix, where g'1 is the isomorphism given in Theorem 2.8. Since Ix=> Ie, we have by Lemma 2.6 that there exists a,beL such that PROOF. a + be Ix\Ie and (a, b) e *¥\9. Now a + belx implies (a, b)ex and hence (a, b)e 9, a contradiction. Therefore lx = Ie. Since K generates B and E is the ideal of B evenly generated by K, we have by Theorem 2.8 that x i s unique. COROLLARY 4.3. Let B be a generalized Boolean lattice R-generated by L. (i) If OeL, then B has (EP) over L. (ii) lfO$L, then B\E has (EP) over L. Note that in the preceding corollary, if 0 $ L, then BjE is a relatively complemented lattice without a smallest element. Also, it gives a corrected version of Gratzer (1971; Lemma 5, page 104). 4.4. Let K be a distributive lattice that has (EP) over L. (i) If a0 is the smallest element of L, then a0 is the smallest element of K. (ii) If c0 is the smallest element of K, then coeL. LEMMA PROOF, (i) Let eeK and x = {(c,d)\c,deK, c f\e f\a0 = d /\e A a0, an and c\/ a0 = d\/ a0}. It is readily verified that xe^(^0 d (e A a0,ao)exIf a,beLand axb, then a = b as a0 is zero the of L. Therefore / O(L x L) = {(a,a)| a eL} and since K has (EP) over L, it follows that x = {( c >c)\ceK}. Hence e A a0 = a0. (ii) If c 0 4 L, then by (i), Ldoes not have a smallest element. If D is the dual ideal of K generated by L, then D = {c I c 2: a for some aeL} and D n {c0} = • • Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 [11] Generalised Boolean Lattices 235 By Gratzer (1971; Theorem 15, page 75) there exists a prime ideal P of K such that c o eP and PIOD = • • Let / = {(c,d\c,deP or c,deKIP}. Then xe%{K) and K/x = {P,K\P}. Moreover, xr\{L x L) = L x L = (K xK) O(L x L). Thus K does not have (EP) over L. We now prove a converse of Theorem 4.2. THEOREM 4.5. Let K be a distributive lattice that has (EP) over L. Then there exists a generalized Boolean lattice B that is R-generated by Land such that K is a sublattice of B. Moreover, if L does not have a smallest element, then K £ B\E, where E is the ideal of B evenly generated by L. Let C be a generalized Boolean lattice .R-generated by K and let D = { x | x e C and there exists al,---,a2neL such that x = ay + ••• + a2n}• Since Lis a sublattice of C, D is a subring of C. Let c, d e K, Xi = {(*, y) \ x, y e K and x + y ^ c + d], and Xi — {(x» .V) | x> >' e K a n ^ x + y<Lef^c + d for some e e D } . Clearly X\ is reflexive and symmetric. If (x, y), (y,z)GXi> t n e n x + y <: c + d and ;; + z ^ c + rf. Hence (x + y) V (y + z) ^ c + cf. Now x + z = (x + y) + (y + z) is the relative complement of (x + y) A (>' + z) in [0, (x + y) V (y + z)] and hence x + z ^ c + d. Therefore Xi is a n equivalence relation on K. Let (x,y)ext and z E K . Then xz + yz = (x + y) A z g x + j> ^ c + d and so (x Az, v A z) e / t . Now xAz + y\Jz = x + z + xz + y + z + yz = (x + y) + (x + y)z ^ (x + y) V ((x + y)z) ^ x + y ^ c + d. Therefore, (x V z, y V z ) e x i and so X t G ^ X ) . A similar argument yields that and obviously Xi-X\- ^(a, b) e / x n (L x L), then a + b ^c + d and a Therefore (a, 6) e x 2 • Since X has (£P) over L, it follows that *i = Xz • s i n c e (c, d)exi, we have c + d ^ e g c + rffor some eeD so c + d e D . It follows that D is the ideal of C evenly generated by K. If /C has a smallest element c0, then by Lemma 4.4, coeL and c0 is the zero of C. Thus, since 0 e K, we have by Theorem 2.4 that D = C. Therefore C is a generalized Boolean lattice K-generated by Land D is the ideal evenly generated by L. If K does not have a smallest element, then by Lemma 4.4, L does not have a smallest element. Since K generates C and 0<£K, we have by Corollary 2.5 that D is a maximal ideal of index two in C and iC n D = Q . Thus if a e L, D + a = C\D. Again we have that C is a generalized Boolean lattice R-generated by L and D is the ideal evenly generated by L. Finally, if Ldoes not have a smallest element, then K does not have a smallest element and so K s C\D. PROOF. As a corollary to the proof of this theorem we have COROLLARY 4.6. Let K be a distributive lattice that has (EP) over L. If C is a generalized Boolean lattice R-generated by K, then C is R-generated by L. Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 236 Richard D. Byrd, Roberto A. Mena and Linda A. Troy [12] COROLLARY 4.7. / / K is a relatively complemented distributive lattice that has (EP) over L, then there exists a generalized Boolean lattice B that is R-generated by L and such that K = B or K = B\E. PROOF. By the theorem there is a generalized Boolean lattice .R-generated by Lsuch that K is a sublattice of B. If 0 e L, then OeK and so K is a generalized Boolean lattice. Let a, be L and let c be the relative complement of a A b in the interval [0, a\/ b~]Kof K. Then c = a + b and it follows that E £ K. By Theorem 2.5, E = B. If 0 $ L, then by the theorem K s B\E. Let x e B\E. Then x = a t + ••• + a 2 «-i where a,, ••• ) a 2n _ 1 e L with aj g ••• g a 2B _!. If n = 1, then ^ e K . Suppose that n > 1 and that a2 + ••• + a2n-2eK. Now a : g a 2 ^ a 2 + "• + a 2n-2 ^ a 2n-2 ^ a 2«-i • Let c be the relative complement of a2 + ••• + a2n^2 in the interval O i , a 2 n - i ] K of K. Then o^ = c A (a 2 + ••• + a2n.2) and a 2n _j = Hence, cy(a2 + •••+a2n_2) = c + a 2 + ••• + « 2 n _ 2 + c(a 2 + ••• +a2n-2). «i + ••• + a 2 n - i = c e X . COROLLARY 4.8. Let K be a relatively complemented distributive lattice that has (EP) over L. Then no proper sublattice of K contains L and is relatively complemented. PROOF. Let M be a relatively complemented sublattice of K that contains L. By the preceding corollary, there is a generalized Boolean lattice B that is R-gtnerated by Land such that K = B or K = B\E. Then the proof of Corollary 4.7 shows that M = K. COROLLARY 4.9. Let K^ and K2 be relatively complemented distributive lattices which have (EP) over L. Then there is an isomorphism of Kt onto K2 that is the identity on L. PROOF. Let B^ and B2 be generalized Boolean lattices that are i?-generated by Ki and K2 respectively. By Corollary 4.6, Bt and B2 are .R-generated by L. By Gratzer (1971; Theorem 6, page 104) there is an isomorphism q of B onto B2 that is the identity on L.) (We note that Theorem 6 of Gratzer (1971) is valid even though Lemma 5 which is invalid, is used in the proof.) If 0 e L, then B^ = iCt and B2 ~ K2. If 0 e L, then K t = B 1 /£ 1 and K2 = B2jE2, where E, is the ideal of Bj evenly generated by L. Since (K^q is a relatively complemented lattice that contains L, we ghave by Corollary 4.8 that (Kt)q = K2. 5. An example The motivation for many of the ideas in this note is the following example. Let P denote the power set of the naturally ordered set of integers and F denote the collection of all finite subsets of the integers. For n e Z , (n] will Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529 [13] Generalised Boolean Lattices 237 denote the ideal of Z generated by n. Let B = F u {(n] u S \ n e Z and S e F}. Then B is a sublattice of the complete Boolean lattice P and it is readily verified that B is a generalized Boolean lattice. Let r be the mapping of ^(Z) into P given by (0)r = l\{n \ n = l.u.b.[«]0}. Then r is an isomorphism of "^(Z) onto P and hence #(Z) is a complete Boolean lattice. If L = {(«] | n e Z}, then L is isomorphic to Z, B is /^-generated by L, and F is the ideal of B that is evenly generated by L. By Corollary 2.11, #(F) is isomorphic to #(L) and hence to ^(Z). Since F satisfies the descending chain condition, there does not exist an isomorphism of L into F . Suppose (by way of contradiction) that ^(Z) is isomorphic to ^(B). Since B is a generalized Boolean lattice, -f{B) is isomorphic to ^(B). Thus the ideals of B form a Boolean lattice. However, by the corollary to Theorem 4.3 (Hashimoto (1952; page 165)) this implies that B satisfies the descending chain condition. This is impossible as L is a sublattice of B. References G. Gratzer (1971), Lattice Theory, (W. H. Freeman and Company, San Francisco, 1971). G. Gratzer and E. T. Schmidt (1958), 'On the generalized Boolean algebra generated by a distributive lattice', Indag. Math. 20, 54-553. J. Hashimoto (1953), 'Ideal theory for lattices', Math. Japan. 2,149-186. University of Houston U.S.A. and University of Wyoming, U.S.A. Downloaded from https://www.cambridge.org/core. IP address: 18.206.13.133, on 04 Jun 2020 at 14:00:54, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700029529