T. S. Blyth and H. J. Silva 2
Abstract. We consider, in the context of an Ockham algebraL= (L;f), the ideals
IofLthat are kernels of congruences onL. We describe the smallest and the biggest congruences having a given kernel ideal, and show that every congruence kernelI6=L is the intersection of the prime idealsPsuch thatI⊆P,P∩f(I) =∅, andf2(I)⊆P. The congruence kernels form a complete lattice which in general is not modular. For every non-empty subsetX ofLwe also describe the smallest congruence kernel to containX, in terms of which we obtain necessary and sufficient conditions for modularity and distributivity. The case whereLis of finite length is highlighted.
An algebra L= (L;∧,∨, f,0,1) of type (2,2,1,0,0) in which (L;∧,∨,0,1) is 3
a bounded distributive lattice and the unary operation x 7→ f(x) is a dual 4
lattice endomorphism is known as an Ockham algebra [2]. It is well-known 5
that the set I(L) of ideals of L, ordered by set inclusion, forms a complete 6
lattice. Given a subsetAofI(L) we shall denote byVAthe infimum ofAin 7
I(L) and byWAthe supremum ofAin I(L). 8
An ideal I of L is said to be a congruence kernel of L if there exists a 9
congruenceϑonLsuch that I= Kerϑ= [0]ϑ. The set IK(L) of congruence
10
kernels ofL, ordered by set inclusion, clearly has a smallest element, namely 11
{0L}. Now it is easy to verify that the intersection of any family of congruence
12
kernels ofLis also a congruence kernel ofL. It then follows that the setIK(L)
13
of congruence kernels ofL, ordered by set inclusion, forms a complete lattice, 14
the biggest element of which isLitself. Given a subsetAofIK(L), we shall
15
denote by VKA the infimum of Ain IK(L) and byWKAthe supremum of
16
AinIK(L). It is clear that
17
V
K
A=TA and W
K
A=T{J ∈ IK(L)| SA ⊆J}.
We shall also have occasion to consider the subset 18
I2(L) ={I∈ I(L)|f2(I)⊆I},
which is clearly a complete sublattice ofI(L). 19
We recall from [2, Theorem 2.12] that ifL= (L;f) is an Ockham algebra 20
then an idealI ofLis a congruence kernel ofLif and only if 21
(α) I∈ I2(L); 22
(β) (∀x∈L)(∀j∈f(I)) x∧j ∈I =⇒ x∈I. 23
Presented by . . .
Received . . . ; accepted in final form . . .
2010Mathematics Subject Classification: Primary: 06D30; Secondary 06D05.
We note here that in general IK(L) is not a sublattice ofI2(L). This is
24
suitably illustrated by the following example in which the principal congruence 25
identifyingxand ywithx6y is denoted as usual byϑ(x, y). 26
We recall that, as established by Berman [1], these principal congruences 27
are determined in terms of principal lattice congruences as follows: 28
ϑ(x, y) = W
n>0
ϑlat fn(x), fn(y).
Example 1. Consider the finite Ockham algebraLdescribed as follows: 29
r ⊙1
r
a9
ra6 r
a8
❅ ❅ ❅
ra3 r
a5⊙ r
a7
❅ ❅ ❅
r ⊙a2
r
a4 ❅ ❅ ❅
⊙
r ⊙a1
r ⊙0 ❅
❅❅
❅ ❅
❅❅
❅ ❅❅❅❅ x 0a1a2a3a4a5a6a7a8a91
f(x) 1a5a4a4a2a1a1a2a1a10
30
Here the ideals I1 = (a7] = {0, a1, a4, a7} and I2 = (a3] = {0, a1, a2, a3} are 31
such thatI1= Kerϑ(0, a7) andI2= Kerϑ(0, a3). LetJ=I1∨KI2∈ IK(L).
32
Then there existsγ∈ConLsuch that J = Kerγ. Since a7∈I1 anda3∈I2 33
we see thata9=a7∨a3∈J = [0]γ. Then we have 1 γ
≡f(a9) =a1≡γ 0 whence 34
Kerγ=Land thereforeI1∨KI2=L. On the other hand,I1∨I2= (a9]6=L.
35
ThusIK(L) is not a sublattice ofI2(L).
36
In order to obtain a description of the structure of congruence kernels, we 37
first characterise the ideals that satisfy property (β) above. 38
For this purpose, given an idealI of the latticeL, consider the set 39
e
I={x∈L| (∃i∈I) x∧f(i)∈I}.
Clearly,Ieis a down-set ofL. It is in fact an ideal of L since ifx, y∈Iethen 40
there existi1, i2∈Isuch thatx∧f(i1)∈Iandy∧f(i2)∈I whence, sinceI 41
is an ideal, 42
(x∨y)∧f(i1∨i2) = [x∧f(i1)∧f(i2)]∨[y∧f(i1)∧f(i2)]∈I
and therefore x∨y ∈I. Moreover, ife i ∈I then i∧f(0) =i∈I gives i∈Ie 43
whenceI⊆I.e 44
Theorem 2. LetL= (L;f)be an Ockham algebra and letI be an ideal ofL.
45
Then each of the following statements is equivalent to(β) : 46
(1) (∀x∈L)(∀i∈I) x∧f(i)∈I =⇒ x∈I; 47
(2) (∀x∈L) {x} ∪f(I)∩I6=∅ =⇒ x∈I; 48
(3) I=Ie.
49
Proof. (β)⇒ (1): Suppose that (β) holds and let x∈ L and i ∈ I be such 50
(1)⇒(2): Letx∈Lbe such that {x} ∪f(I)∩I6=∅. Then there exists 52
t∈I such thatt∈{x} ∪f(I), whence there exist y1, . . . , ym∈ {x} ∪f(I)
53
such thatt>y1∧ · · · ∧ym. Consider the sets
54
A1={r|yr=x}, A2={r|yr6=x}.
Letz= V
r∈A2
yr. Clearly, there existsi∈Isuch thatz=f(i). IfA16=∅ then
55
m
V
r=1
yr=x∧z=x∧f(i)6t∈I
and sox∧f(i)∈I whence, by the hypothesis, we havex∈I. On the other 56
hand, ifA1 =∅ then f(i) = Vm
r=1
yr 6t∈I gives 1∧f(i) =f(i)∈I whence
57
the hypothesis gives 1∈I. ThenI=Land consequentlyx∈I. 58
(2)⇒(3): Suppose that (2) holds and let x∈I. Then there existse i ∈I 59
withx∧f(i)∈I. Clearly,x∧f(i)∈{x} ∪f(I)∩I and so, by (2),x∈I. 60
ThusIe⊆I and equality follows. 61
(3) ⇒ (β): Suppose that Ie = I. Let x ∈ L and let j ∈ f(I) such 62
that x∧j ∈ I. Then there exists t ∈ I such that j > f(t). Consequently,
63
x∧f(t)6x∧j∈I whencex∧f(t)∈I and thereforex∈Ie=I. ⊡
64
Corollary 3. An ideal I of an Ockham algebra L is a congruence kernel if
65
and only if I=IeandI∈ I2(L).
66
In the case where (L;f) belongs to the subvarietyS1 of the Berman class 67
K1,1, this subvariety being characterised by the axiomf2(x)∧f(x) = 0 [2], an
68
idealI ofLis a kernel ideal if and only ifI= ˜I. Indeed, in this case, ifI= ˜I 69
then, for each i∈ I, f2(i)∧f(i) = 0∈I gives, by Theorem 2(1), f2(i)∈ I 70
whenceI∈ I2(L). The conclusion then follows by Corollary 3. 71
Theorem 4. If I ∈ IK(L) then the smallest congruence on L with kernel I
72
is the relationΘI given by
73
(x, y)∈ΘI ⇐⇒ (∃i∈I) (x∨i)∧f(i) = (y∨i)∧f(i).
Proof. By [2, Theorem 2.11 and Lemma 2.1] we know that ΘI is given by
74
(x, y)∈ΘI ⇐⇒ (∃i∈I)(∃j ∈
f(I)) (x∨i)∧j = (y∨i)∧j.
Consider now the relationϕ(I) defined onLby 75
(x, y)∈ϕ(I) ⇐⇒ (∃i∈I) (x∨i)∧f(i) = (y∨i)∧f(i).
We show as follows thatϕ(I) = ΘI. It is clear that ϕ(I) ⊆ ΘI. To obtain
76
the reverse inclusion, suppose that (x, y) ∈ ΘI. Then there exist i ∈ I and
77
j∈[f(I)) such that (x∨i)∧j= (y∨i)∧j. Letk∈I be such thatj>f(k). 78
Then 79
It follows from this that 80
(x∨i∨k)∧f(i∨k) = (x∨i∨k)∧f(i)∧f(k) = (y∨i∨k)∧f(i)∧f(k) = (y∨i∨k)∧f(i∨k)
withi∨k∈I, and therefore (x, y)∈ϕ(I) whence ΘI ⊆ϕ(I) as required. ⊡
81
Having thus described the smallest congruence with a given kernel, we now 82
proceed to determine the biggest such congruence. 83
Given an idealI ofL, for eacha∈Land for eachn∈N, consider the set
84
WI
a,n={x∈L|fn(a)∧x∈I}
and the binary relation ΘI
n onLdefined by
85
(a, b)∈ΘI
n ⇐⇒Wa,nI =Wb,nI .
Clearly, ΘI
n is a lattice congruence onL. Consider now the relation ΘI onL
86
defined by 87
ΘI = T
n∈N
ΘIn.
It is clear that ΘI is a congruence on L.
88
Theorem 5. If I∈ IK(L) then the biggest congruence onL with kernelI is
89
the relationΘI.
90
Proof. We show first that Ker ΘI =I. Ifa∈Ker ΘI then
91
{x∈L|a∧x∈I}=WI
a,0=W0I,0={x∈L|0∧x∈I}=L
whence a =a∧1 ∈ I and so Ker ΘI ⊆ I. Conversely, suppose that a ∈ I.
92
Clearly, we have 93
Wa,I0={x∈L|a∧x∈I}=L=W0I,0
and therefore (a,0)∈ΘI
0. 94
Now letn∈Nbe such that n>1. There are two cases to consider. 95
(a) neven. In this caseWI
0,n={x∈L|fn(0)∧x∈I}=L. Sincea∈I,
96
nis even andf2(I)⊆Iwe have fn(a)∈Iand consequently
97
WI
a,n={x∈L|fn(a)∧x∈I}=L=W0I,n.
(b) nodd. Withn= 2k+ 1 we haveWI
0,n={x∈L|fn(0)∧x∈I}=I.
98
On the other hand, 99
I⊆Wa,nI ={x∈L|f(f2k(a))∧x∈I}.
Since f2(I) ⊆ I and therefore f2k(a) ∈ I, it follows by Theorem 2(1) that
100 WI
a,n⊆I. Thus we see that in this case
101
WI
a,n=I=W0I,n.
It follows from the above that for everyn∈Nwe have (a,0)∈ΘI
n whence
102
(a,0)∈ΘI and consequentlyI⊆Ker ΘI. ThusI= Ker ΘI.
Suppose now that γ ∈ ConL is such that I = Kerγ. Let (a, b) ∈ γ 104
and suppose that x ∈ WI
a,n. Then fn(a)∧x ∈ I. Since I = Kerγ, we
105
see that (0, fn(a)∧x) ∈ γ. On the other hand, (fn(a), fn(b)) ∈ γ gives
106
(fn(a)∧x, fn(b)∧x)∈γ. Then (0, fn(b)∧x)∈γ, whencefn(b)∧x∈Kerγ=
107
I, and sox∈WI
b,n. It follows from this that Wa,nI ⊆Wb,nI . Interchanging the
108
roles ofaandb, we obtainWI
b,n⊆Wa,nI . Hence Wa,nI =Wb,nI for everyn∈N
109
and therefore (a, b)∈ΘI. Henceγ⊆ΘI and the conclusion follows. ⊡
110
Theorem 6. LetL= (L;f)be an Ockham algebra and letI be an ideal ofL.
111
Then
112
(1) if I∈ I2(L) thenIe∈ I2(L); 113
(2) the mapping ξ : I2(L) → I2(L) defined by ξ(I) = Ieis a closure on
114
I2(L)and Imξ=IK(L).
115
Proof. (1) Suppose thatI∈ I2(L) and letx∈I. Then there existse i∈Isuch 116
thatx∧f(i)∈I. Since by hypothesisf2(I)⊆I, we havef2(x)∧f(f2(i))∈I 117
withf2(i)∈I, and thereforef2(x)∈I. Thuse f2(I)e ⊆IewhenceIe∈ I2(L). 118
(2) It is clear thatξ is isotone with id 6ξ. Given I ∈ I2(L), let J =Ie 119
and letx∈Je. Then there existsu∈Iesuch thatx∧f(u)∈I. Consequentlye 120
there exist t, s ∈I such thatx∧f(u)∧f(t)∈I and u∧f(s)∈ I. Clearly, 121
[u∧f(s)]∨t∈I andf2(s)∈I. Now 122
x∧f [u∧f(s)]∨t= [x∧f(t)∧f(u)]∨[x∧f(t)∧f2(s)]∈I
and therefore x ∈ Ie= J. Hence Je ⊆ J and the resulting equality shows 123
that ξ is idempotent. Thus ξ is a closure onI2(L) and, by the Corollary 3, 124
IK(L) = Imξ. ⊡
125
We recall that an idealI of Lis said to be proper ifI 6=L, and thatI is 126
primeifI is proper andx∧y∈Iimplies thatx∈I ory∈I. In what follows 127
we shall denote byIp(L) the set of prime ideals ofL.
128
Theorem 7. LetL= (L;f)be an Ockham algebra and letIbe a proper ideal
129
ofL. Then the following statements are equivalent: 130
(1) I=I;e 131
(2) I=T P∈ Ip(L)|I⊆PandP∩f(I) =∅ .
132
Proof. (1)⇒(2): Suppose thatIe=I and consider the set 133
V={P ∈ Ip(L)|I⊆P andP∩f(I) =∅}.
Observe first thatV 6=∅. In fact, sinceIis a proper ideal we have that 16∈I. 134
It therefore follows by Theorem 2(2) thatf(I)∩I=∅. We may now call on 135
Stone’s celebrated Prime Ideal Theorem [3] to assert the existence of a prime 136
idealQsuch thatI⊆Qandf(I)∩Q=∅. ThenQ∈ V. 137
Clearly, we have I ⊆ TV. To establish the converse inclusion suppose, 138
by way of obtaining a contradiction, that there exists x ∈ TV with x 6∈ I. 139
Then, again by the Prime Ideal Theorem, there exists a prime idealP0 such 141
that I⊆P0 andF ∩P0 =∅. Then P0∩f(I) =∅ and consequently P0 ∈ V. 142
Sincex∈F, it follows fromF∩P0=∅that x6∈P0, and this contradicts the 143
hypothesis thatx∈TV. 144
(2)⇒(1): Suppose that (2) holds and let x∈I. Then there existse i ∈I 145
withx∧f(i)∈I. LetP be a prime ideal ofLsuch thatI⊆PandP∩f(I) =∅. 146
Then we havex∧f(i)∈P andf(i)6∈P. SinceP is a prime ideal, we see that 147
x∈P. It follows that 148
x∈T P ∈ Ip(L)|I⊆P andP∩f(I) =∅ =I.
ThusIe⊆I and henceIe=I. ⊡
149
Corollary 8. A proper idealIof Lis a congruence kernel ofLif and only if
150
I=T P ∈ Ip(L)|I⊆P, P ∩f(I) =∅, f2(I)⊆P ⊡
In what follows, for eachx∈Rwe let⌊x⌋= max{i∈Z|i6x}. Also, for 151
eachn∈Nwithn>2 we letndenote the chain 0<1<· · ·< n−1. 152
Example 9. Letn∈Nwithn>2. Consider the mappingf:n→ngiven by 153
f(i) =n−1−i. LetCn denote the Ockham algebra (n;f). It is clear thatCn
154
is a de Morgan algebra and thereforef2(I) =I for every idealI. It is readily 155
seen by Theorem 2(1) that 156
IK(Cn) ={(k]|2k < n−1} ∪ {n}
and consequently IK(Cn) consists of a chain of ⌊n+22 ⌋elements. In this case
157
the lattice of congruence kernels ofCn is therefore distributive.
158
Example 10. LetE be an infinite set and leta,bbe distinct elements of E. 159
LetL be the boolean lattice consisting of the finite and cofinite subsets ofE. 160
We shall denote byPf(E) the set of finite subsets ofEand byPcf(E) the set
161
of cofinite subsets ofE, so thatL=Pf(E)∪ Pcf(E).
162
Consider the mappingf :L→Lgiven by 163
f(X) = (
E\(X∪ {a}) ifb∈X; (E\X)∪ {a} ifb /∈X.
Clearly,L= (L;f) is an Ockham algebra. Moreover, 164
f2(X) = (
X∪ {a} ifb∈X; X∩(E\{a}) ifb /∈X,
so that f3 = f and f(X)∩f2(X) = ∅ for every X ∈ L. Consequently, L 165
belongs to the subvarietyS1 of the Berman classK1,1[2]. 166
SinceI∈ I2(L) if and only iff2(I)⊆I, it follows that 167
I2(L) =I∈ I(L)| {b}∈/I ∪ I∈ I(L)| {a, b} ∈I
We now observe that IK(L) = I2(L). To see this, let J ∈ I2(L). Let
168
X ∈ L and W ∈ J be such that X ∩f(W) ∈ J. SinceJ ∈ I2(L) we also 169
have X∩f2(W) ∈J. But from L ∈S
1 it follows that f(W)∪f2(W) =E. 170
Consequently 171
X=X∩E= X∩f(W)∪ X∩f2(W)∈J
whence, by Theorem 2, J =Jeand then, by Corollary 3,J ∈ IK(L). Hence
172
I2(L) =IK(L). Consequently,IK(L) is a distributive lattice.
173
Now let{∞}be a singleton that is disjoint fromE, and for everyX∈ P(E) 174
letX∞=X∪ {∞}. Consider the set
175
P⊛(E) =P(E)∪ {X∞
|X ∈ Pcf(E)}
which is a sublattice of the boolean latticeP(E∞
) with bottom element∅and 176
top elementE∞. Consider also the mappingǫ:P⊛(E)→ I(L), defined by
177
ǫ(X) = (
Pf(X) ifX ∈ P(E);
(Y]L ifX =Y∞ whereY ∈ Pcf(E).
It is readily seen that ǫ is a lattice isomorphism with ǫ−1 : I(L) → P⊛(E)
178
given by 179
ǫ−1(I) = (
{x∈E| {x} ∈I} ifI⊆ Pf(E);
{x∈E| {x} ∈I}∞
ifI6⊆ Pf(E).
Consequently,I(L)≃ P⊛(E).
180
We can conclude from the above that 181
IK(L) =I2(L)≃ǫ−1 I2(L)
=P E\{b} ∪ X∞|X ∈ P
cf E\{b}
∪X |X ∈ P(E) and {a, b} ⊆X
∪X∞|
X ∈ Pcf(E) and {a, b} ⊆X .
As we shall see in the next example,for an Ockham algebra L the lattice
182
IK(L) of congruence kernels is not in general modular.
183
For this purpose, we observe that ifI is an ideal ofL then 184
e
I=L ⇐⇒ 1∈Ie ⇐⇒ (∃i∈I) f(i) = 1∧f(i)∈I
⇐⇒ I∩f(I)6=∅.
Example 11. Given n∈Nwithn >3, letHn =n×nand endow Hn with
185
the cartesian order. Consider the mappingf:Hn →Hn given by
186
f(i, j) = (n−1−j, n−1−i).
Clearly,Hn = (Hn;f) is a de Morgan algebra. Consider the following ideals
187
ofHn:
188
R= (0, n−2), S= (2,0), T = (0, n−3).
It follows from Theorem 2(1) and Corollary 3 thatR, S, T ∈ IK(Hn). Clearly,
189
(2, n−3) = (2,0)∨(0, n−3)∈S∨KT. Now sincef(2, n−3) = (2, n−3) we
see thatS∨KT contains a fixed point and therefore, by the above observation,
191
S∨KT =L. Consequently,
192
R∧K(S∨KT) =R∩(S∨KT) =R= (0, n−2)
.
On the other hand, we haveR∧KS=R∩S={(0,0)}and therefore
193
(R∧KS)∨KT =T = (0, n−3)
.
It follows from this that 194
R∧K(S∨KT)6= (R∧KS)∨KT
withR⊇T. HenceIK(Hn) is not modular.
195
SinceIK(L) is a complete lattice, for every non-empty subsetX ofLthere
196
is a smallest congruence kernel of L that contains X. We now proceed to 197
describe this. For this purpose, consider the subsetX⋆ defined by declaring
198
y∈X⋆ ⇐⇒ (∃x1, . . . , xr∈X)(∃i1, . . . , ir∈N) y= r
W
j=1 f2ij(x
j).
In what follows we shall be particularly interested in the ideal ofLgiven by 199
κ(X) =(Xg⋆].
In particular, whenX={x}we shall write κ(X) asκ(x). 200
Theorem 12. LetL= (L;f)be an Ockham algebra and letXbe a non-empty
201
subset ofL. Then
202
(1) κ(X)is a congruence kernel and is the smallest congruence kernel that
203
containsX; 204
(2) (∀y∈L) y∈κ(X) ⇐⇒ (∃x∈X⋆) y∧f(x)6x.
205
Proof. (1) If y ∈ (X⋆] then by definition there exist x1, . . . , xr ∈ X and
206
i1, . . . ir ∈N such thaty 6
Wr
j=1f2ij(xj), whence f2(y)6
Wr
j=1f2(ij+1)(xj)
207
and thereforef2(y)∈(X
⋆]. Hence (X⋆]∈ I2(L). It now follows by Theorem
208
6(2) thatκ(X)∈ IK(L). Moreover, it is clear thatX ⊆X⋆⊆(Xg⋆] =κ(X).
209
Suppose now thatIis a congruence kernel ofLwithX ⊆I. Sincef2(I)⊆I, 210
we haveX⋆⊆I. It then follows by Theorem 6(2) thatκ(X) =(Xg⋆]⊆Ie=I.
211
Thusκ(X) is the smallest congruence kernel that contains X. 212
(2) Ify ∈ κ(X) = (Xg⋆] then there exists x ∈ (X⋆] such that y∧f(x) ∈
213
(X⋆], and then there exist c1, . . . , cp ∈ X and m1, . . . , mp ∈ N such that
214
y∧f(x) 6 Wpj=1f2mj(c
j). Now since x ∈ (X⋆] there exist d1, . . . , dq ∈ X
215
andn1, . . . , nq ∈Nsuch that x6Wqj=1f2nj(dj). Letα=Wpj=1f2mj(cj) and
216
β=Wqj=1f2nj(dj). Sincex6α∨β, we see that
217
y∧f(α∨β)6y∧f(x)6α6α∨β
withα∨β∈X⋆. The reverse implication is trivial. ⊡
Corollary 13. For everya∈L,
219
κ(a) =y∈L|(∃k∈N) y∧fWk
j=0
f2j(a)6 Wk j=0
f2j(a) .
In particular, ifLbelongs to the Berman classK1,1 then 220
κ(a) ={y∈L|y∧f(a)6a∨f2(a)}.
Proof. Ify∈κ(a) then by (2) above there existi1, . . . , ir∈Nsuch that
221
y∧f
r
W
j=1 f2ij(a)
6
r
W
j=1
f2ij(a).
If now we letk= max{ij|j∈ {1, . . . , r}}then clearly{i1, . . . , ir} ⊆ {0, . . . , k}
222
and consequentlyWrj=1f2ij(a)6Wk
j=0f2j(a). It follows from this that 223
y∧fWk
j=0
f2j(a)6 Wk j=0
f2j(a).
The reverse implication is immediate from (2) above. 224
The final statement is an immediate consequence of the fact that ifL ∈K1,1 225
thenf3=f. ⊡
226
Example 14. For theK1,1-algebra Lof Example 10 we have that
227
X∪f2(X) = (
X∪ {a} ifb∈X; X ifb /∈X.
It follows from this and Corollary 13 that 228
κ(X) = (
(X∪ {a}] ifb∈X; (X] ifb /∈X.
Corollary 15. IfI, J ∈ IK(L)then
229
I∨KJ =κ(I∪J) ={x∈L|(∃y∈I)(∃z∈J) x∧f(y∨z)6y∨z}.
Proof. Let I, J ∈ IK(L). It is clear that I ⊆ κ(I∪J) and J ⊆ κ(I∪J).
230
Since, by (1),κ(I∪J)∈ IK(L) we see that I∨KJ ⊆κ(I∪J). On the other
231
hand,I∪J ⊆I∨K J whence, again by (1), we obtain the reverse inclusion
232
κ(I∪J)⊆I∨KJ whence the first equality.
233
Now let 234
M ={x∈L|(∃y∈I)(∃z∈J) x∧f(y∨z)6y∨z}.
Suppose thatx∈M. Then there existy∈Iandz∈J such thatx∧f(y∨z)6
235
y∨z. Sincey∨z∈I∨KJ we see thatx∈I^∨KJ =I∨KJ.
236
Conversely, letx∈κ(I∪J). Then by (2) there exist c1, . . . , cr∈I∪J and
237
i1, . . . , ir ∈N such thatx∧fWrj=1f2ij(cj)
6Wrj=1f2ij(c
j). Consider the
238
sets B1 =j ∈ {1. . . , r} | cj ∈ I and B2 =
j ∈ {1. . . , r} |cj ∈J . Let
239
b1=Wj∈B1f 2ij(c
j) andb2=
W
j∈B2f 2ij(c
j). Clearly, we havex∧f(b1∨b2)6
240
b1∨b2with b1∈I andb2∈J. Thenx∈M. ⊡
We can now settle the question of when, for a given Ockham algebraL, the 242
latticeIK(L) is modular. For this purpose, we note the following property.
243
Theorem 16. LetL= (L;f)be an Ockham algebra. For alla, b∈L, we have
κ(a∨b) =κ(a)∨Kκ(b).
Proof. Let a, b∈L. By Theorem 12(1), it follows from a ≤a∨b∈ κ(a∨b) 244
thata∈κ(a∨b) and consequentlyκ(a)⊆κ(a∨b). Likewise,κ(b)⊆κ(a∨b). 245
Sinceκ(a∨b) is a congruence kernel it follows that κ(a)∨K κ(b)⊆κ(a∨b).
246
On the other hand,a∈κ(a)⊆κ(a)∨Kκ(b) andb∈κ(b)⊆κ(a)∨Kκ(b) give
247
a∨b∈κ(a)∨Kκ(b). Henceκ(a∨b)⊆κ(a)∨Kκ(b). ⊡
248
Theorem 17. If L = (L;f) is an Ockham algebra then the following
state-249
ments are equivalent: 250
(1) IK(L)is a modular lattice;
251
(2) for all a, b, c∈L,
κ(a∨c)∩κ(b∨c)⊆ κ(a∨c)∩κ(b)∨Kκ(c).
Proof. (1) ⇒(2): Let a, b, c∈L. It follows from κ(c)⊆κ(a∨c) and modu-252
larity that 253
κ(a∨c)∩ κ(b)∨Kκ(c)
⊆ κ(a∨c)∩κ(b)∨Kκ(c).
whence (2) follows by Theorem 16. 254
(2)⇒(1): Suppose that (2) holds and letI, J, T ∈ IK(L) withT ⊆I. Let
255
a ∈ I∩(J ∨KT). By Corollary 15 there exist b ∈ J and c ∈ T such that
256
a∧f(b∨c)6b∨c. By Corollary 15 again, we see thata∈κ(b)∨Kκ(c) and
257
therefore, by Theorem 16,a∈κ(a∨c)∩κ(b∨c). It now follows from (2) that 258
a∈ κ(a∨c)∩κ(b)∨Kκ(c).
259
On the other handc∈T ⊆Iand thereforea∨c∈I. Thenκ(a∨c)∩κ(b)⊆
260
I∩J and κ(c)⊆T, whence κ(a∨c)∩κ(b)∨Kκ(c)⊆(I∩J)∨KT. Since
261
a∈ κ(a∨c)∩κ(b)∨Kκ(c), we conclude thata∈(I∩J)∨KT.
262
We deduce from this that 263
T ⊆I ⇒ I∩(J∨KT)⊆(I∩J)∨KT.
HenceIK(L) is a modular. ⊡
264
Finally, we can settle the question of when, for a given Ockham algebraL, 265
the latticeIK(L) is distributive.
266
Theorem 18. If L = (L;f) is an Ockham algebra then the following
state-267
ments are equivalent: 268
(1) IK(L)is a distributive lattice;
269
(2) for all a, b, c∈L,
270
κ(a)∩κ(b∨c)⊆ κ(a)∩κ(b)∨K κ(a)∩κ(c)
Proof. (1)⇒(2): Leta, b, c∈L. Since, by hypothesis,IK(L) is distributive,
271
we have 272
κ(a)∩ κ(b)∨Kκ(c)
⊆ κ(a)∩κ(b)∨K κ(a)∩κ(c)
,
from which (2) follows by Theorem 16. 273
(2)⇒(1): Suppose that (2) holds. LetI, J, T ∈ IK(L) anda∈I∩(J∨KT).
274
As in the proof of Theorem 17, there exist b ∈ J and c ∈ T such that a ∈
275
κ(a)∩κ(b∨c). It then follows by (2) thata∈ κ(a)∩κ(b)∨K κ(a)∩κ(c)
. 276
On the other handκ(a)∩κ(b)⊆I∩J andκ(a)∩κ(c)⊆I∩T. Thus 277
a∈ κ(a)∩κ(b)∨K κ(a)∩κ(c)
⊆(I∩J)∨K(I∩T).
We deduce from this that 278
I∩(J∨KT)⊆(I∩J)∨K(I∩T).
HenceIK(L) is distributive. ⊡
279
We now consider an Ockham algebra L = (L;f) in which L is of finite 280
length. In this case every ideal ofLis principal. Consequently, for everya∈L, 281
κ(a) has a maximum element. Consider therefore the mapping µ : L → L 282
defined by 283
(∀a∈L) µ(a) = maxκ(a). Clearly,κ(x) = (µ(x)] for every x∈L.
284
Theorem 19. IfL= (L;f)is an Ockham algebra of finite length thenµis a
285
closure mapping. Moreover,
286
IK(L) =
(x]|x∈Imµ .
Proof. Let a, b∈Lbe such that a6b and lety ∈κ(a). Then, by Corollary 287
13, there existsk∈Nsuch that
288
y∧fWk
j=0
f2j(b)6y∧fWk j=0
f2j(a)6 Wk j=0
f2j(a)6 Wk j=0
f2j(b)
whence it follows thaty∈κ(b). Thus κ(a)⊆κ(b) and thereforeµ(a)6µ(b). 289
Sincea∈κ(a) we also have thata6µ(a).
290
We now observe that 291
(⋆) (x]∈ IK(L) ⇐⇒ κ(x) = (x] ⇐⇒ µ(x) =x.
Indeed, it suffices to note that if (x]∈ IK(L) then since x∈(x] it follows by
292
Theorem 12(1) thatκ(x)⊆(x]. On the other hand,x∈κ(x) gives (x]⊆κ(x). 293
Since (µ(a)] =κ(a)∈ IK(L) it follows from this thatµ(µ(a)) =µ(a).
294
Thus we see thatµis an isotone idempotent which is bigger than the identity 295
onL and is therefore a closure mapping. 296
The final statement follows immediately from (⋆). ⊡
297
Corollary 20. The lattices IK(L)and Imµare isomorphic.
298
Proof. Clealy,ϑ: Im µ→ IK(L) given byϑ(a) = (a] is an isomorphism. ⊡
Example 21. In view of the above, we may revisit Example 11. There,Hn
300
is a de Morgan algebra and therefore, by Corollary 13, 301
(∀a∈Hn) µ(a) = max{y∈Hn|y∧f(a)6a}.
For eacha = (i, j) ∈ Hn we have that aand f(a) are comparable. Indeed,
302
either a>f(a) [i.e., i+j >n−1] ora < f(a) [i.e., i+j < n−1]. Simple 303
calculations reveal that 304
µ(a) = (
a ifa < f(a); 1 ifa>f(a).
Then the lattice IK(L) ≃ Imµ fails to be modular. In fact, the maximal
305
elements of Imµ are of the form (i, j) where i+j =n−2 and, for such an 306
element (i, j) with 0 < i < n and 0 < j < n, it is readily seen that Imµ 307
contains the sublattice 308
r
r r
r r r
r
❅ ❅ ❅ ❅❅ ✁✁
✁✁ ✁✁
❆ ❆ ❆ ❆ ❆ ❆
(i−1,j−1) (i−1,j) (i−1,j+1) (i,j)
(i,j−1) (i+1,j−1) (n−1,n−1)
which is non-modular. 309
Acknowledgment. This work was supported by the Portuguese Foundation 310
for Science and Technology, project UID/MAT/00297/2013 (CMA). 311
References 312
[1] J. Berman, Distributive lattices with an additional unary operation,Aequationes Math.
313
16, 1977, 165–171.
314
[2] T. S. Blyth and J. C. Varlet,Ockham Algebras, Oxford University Press, 1994.
315
[3] G. Gr¨atzer,General Lattice Theory, Birkh¨auser, 1998.
316
T. S. Blyth 317
Mathematical Institute, University of St Andrews, Scotland.
318
e-mail:tsb@st-and.ac.uk
319
H. J. Silva 320
Centro de Matem´atica e Aplica¸c˜oes (CMA), Departamento de Matem´atica, F.C.T.,
321
Universidade Nova de Lisboa, Portugal.
322
e-mail:hdjs@fct.unl.pt
