A uniform refinement property for congruence lattices

0.15 MB·English

by  Friedrich Wehrung

A UNIFORM REFINEMENT PROPERTY
5 FOR CONGRUENCE LATTICES
0
0
2 FRIEDRICHWEHRUNG

n
a
Abstract. The Congruence Lattice Problem asks whether every algebraic
J
distributivelattice isisomorphictothe congruence lattice ofalattice. Itwas
5 hoped that a positive solution would follow from E. T. Schmidt’s construc-
2 tion or from the approach of P. Pudla´k, M. Tischendorf, and J. T˚uma. In
a previous paper, we constructed a distributive algebraic lattice A with ℵ2
] compactelementsthatcannotbeobtainedbySchmidt’sconstruction. Inthis
M
paper,weshowthatthesamelatticeAcannotbeobtainedusingthePudla´k,
G Tischendorf,T˚umaapproach.
Thebasicideaisthateverycongruence latticearisingfromeithermethod
h. satisfiestheUniformRefinementProperty,thatisnotsatisfiedbyourexample.
t This yields,inturn,corresponding negative results about congruence lattices
a of sectionally complemented lattices and two-sided ideals of von Neumann
m
regularrings.
[

1
v
Introduction
8
5
E. T. Schmidt introduces in [11] the notion of a weakly distributive (resp., dis-
4
tributive)homomorphismofsemilattices,andprovesthefollowingimportantresult
1
0 (see, for example, [11, Satz 8.1] and [12, Theorem 3.6.9]):
5
Schmidt’s Lemma. Let A be an algebraic distributive lattice. If the semilattice
0
of compact elements of A is the image under a distributive homomorphism of a
/
h generalized Boolean semilattice, then there exists a lattice L such that ConL is
t
a isomorphic to A.
m
This result yields important partial positive answers to the Congruence Lattice
:
v Problem (see, for example, [9, 14] for a survey). On the other hand, we prove [15,
i Theorem 2.15], which implies the following result:
X
r Theorem. For any cardinal number κ≥ℵ2, there exists a distributive semilattice
a
S of size κ that is not a weakly distributive image of any distributive lattice.
κ
In this paper, we introduce a monoid-theoretical property, the Uniform Refine-
ment Property that is not satisfied by the semilattices S of the Theorem above,
κ
but it is satisfied by the congruence semilattice of any lattice satisfying an addi-
tional condition, the Congruence Splitting Property (Theorem 3.3). The latter is
satisfiedbyany latticewhichis eithersectionallycomplemented, orrelativelycom-
plemented, or a direct limit of atomistic lattices (Proposition 3.2). In particular,
for all κ ≥ ℵ , the semilattice S of the above Theorem is not isomorphic to the
2 κ
Date:February8,2008.
1991 Mathematics Subject Classification. Primary06A12,06B10;Secondary 16E50.
Key words and phrases. Semilattices; weakly distributive homomorphisms; congruence split-
tinglattices;uniformrefinementproperty;vonNeumannregularrings.
1
2 F.WEHRUNG
congruencesemilattice of any sectionally complemented lattice; this gives a partial
negative answer to [8, Problem II.8]. It follows that no semilattice S can be re-
κ
alized as the semilattice of finitely generated two-sided ideals of a von Neumann
regular ring.
Notation and terminology
The refinement property is the monoid-theoretical axiom stating that for every
equationof the forma +a =b +b , there exist elements c (i, j <2) such that
0 1 0 1 ij
foralli<2,a =c +c andb =c +c . Allsemilatticeswillbejoin-semilattices
i i0 i1 i 0i 1i
(not necessarily bounded). A semilattice is, as usual, distributive, if it satisfies the
refinement property.
If u and v are elements of a lattice L, then Θ (u,v) (or Θ(u,v) if there is no
L
ambiguity) denotes the least congruence of L identifying u and v. Furthermore,
ConL (resp., Con L) denotes the lattice (resp., semilattice) of all congruences
c
(resp., compact congruences) of L. If L has a least element (always denoted by 0),
an atom of L is a minimal element of L\{0}. A lattice L with zero is atomistic, if
every element of L is a finite join of atoms.
Our lattice-theoreticalresults willbe applied to ringsin Section 4. All ourrings
are associative and unital (but not necessarily commutative). Recall that a ring
R is (von Neumann) regular, if it satisfies the axiom (∀x)(∃y)(xyx = x). If R is
a regular ring, then every principal right ideal of R is of the form eR, where e is
idempotent, andthe setL(R)ofallprincipalrightidealsofR,partiallyorderedby
inclusion, is a complemented modular lattice.
1. Weak-distributive homomorphisms
Weshallfirstmodifyslightlytheoriginaldefinition,duetoE.T.Schmidt[12,13]
of a weakly distributive homomorphism. A homomorphism of semilattices f: S →
T is weakly distributive at an element u of S, if for all y , y ∈ T such that
0 1
f(u)= y +y , there are x , x ∈ S such that x +x = u and f(x )≤ y , for all
0 1 0 1 0 1 i i
i<2. Say that f is weakly distributive, if it is weaklydistributive at everyelement
of S.
In particular, if f is surjective, one recovers the usual definition of a weakly
distributive homomorphism. Furthermore, note that with this new definition, any
composition of two weakly distributive homomorphisms remains weakly distribu-
tive.
Lemma1.1. Letf: S →T beahomomorphismofsemilattices,withT distributive.
Then the set of all elements of S at which f is weakly distributive is closed under
join.
Proof. Let X be the set of all elements of S at which f is weakly distributive. It
suffices to prove that if u′ and u′′ are any two elements of X, then u = u′ +u′′
belongs to X. Thus let y (i <2) be elements of T such that f(u)=y +y , that
i 0 1
is, since f is a homomorphismof semilattices, f(u′)+f(u′′)=y +y holds. Since
0 1
T isdistributive,itsatisfiestherefinementproperty;thustherearedecompositions
y =y′+y′′ (i<2) such that f(u′)=y′ +y′ and f(u′′)=y′′+y′′. Since both u′
i i i 0 1 0 1
and u′′ belong to X, there are decompositions u′ =x′ +x′ and u′′ =x′′+x′′ such
0 1 0 1
that f(x′)≤y′ and f(x′′)≤y′′ (i<2). Put x =x′+x′′ (i<2). Then f(x )≤y
i i i i i i i i i
and x +x =u. Therefore, u∈X. (cid:3)
0 1
CONGRUENCE LATTICES 3
The following result yields a large class of weakly distributive homomorphisms:
Proposition 1.2. Let e: K → L be a lattice homomorphism with convex range.
Then the induced semilattice homomorphism f: Con K → Con L is weakly dis-
c c
tributive.
Proof. It is well-known that Con L is a distributive semilattice (see, for example,
c
[8, Theorem II.3.11]). Hence, by Lemma 1.1, it suffices to prove that f is weakly
distributive at every α of the form Θ (u,v) where u≤v in K. Thus let β (i<2)
K i
such that f(Θ (u,v)) = β ∨ β , that is, Θ (f(u),f(v)) = β ∨ β . Thus, by
K 0 1 L 0 1
[8, Lemma III.1.3], there exist a positive integer n and elements w′ (i ≤ 2n) of
i
L such that f(u) = w′ ≤ w′ ≤ ··· ≤ w′ = f(v) and for all i < n, w′ ≡
0 1 2n 2i
w′ (mod β ) and w′ ≡ w′ (mod β ). But the range of e is convex in L,
2i+1 0 2i+1 2i+2 1
thus the w′ belong to the range of e, thus there are elements w ∈ K such that
i i
e(w )=w′. One can of course take w =u and w =v, and, after replacing each
i i 0 2n
w by W (w ∨u)∧v, one can suppose that u = w ≤ w ≤ ··· ≤ w = v.
i j≤i j 0 1 2n
Then put α = W Θ (w ,w ) and α = W Θ (w ,w ). We have
0 i<n K 2i 2i+1 1 i<n K 2i+1 2i+2
α ∨α = Θ (u,v) = α, and f(α ) = W Θ (f(w ),f(w )) ⊆ β ; similarly,
0 1 K 0 i<n L 2i 2i+1 0
f(α )⊆β . (cid:3)
1 1
2. The Uniform Refinement Property
Inordertoillustratethe terminologyofthesectiontitle,letusfirstconsiderany
equation system (in a given semilattice) of the form
Σ: a +b =constant (for all i∈I).
i i
WhenI ={i,j},asatisfactorynotionofarefinement ofΣconsistsoffourelements
cuv (u,v <2) satisfying the equations
ij
a =c00+c01 and b =c10+c11,
i ij ij i ij ij
(2.1)
a =c00+c10 and b =c01+c11,
j ij ij j ij ij
see Figure 1.
a a b b
i j i j
c00 c10 c11 c01
ij ij ij ij
Figure 1
Note that (2.1) implies immediately the following consequence:
(2.2) a ≤a +c01.
i j ij
When I is an arbitrary finite set, one can extend this in a natural way and thus
define a refinement of Σ to be a P(I)-indexed family of elements of S satisfying
suitablegeneralizationsof(2.1). Nevertheless,thiscannotbeextendedimmediately
to the infinite case, so that we shall focus instead on the consequence (2.2) of
refinement, together with an additional “coherence condition” c01 ≤ c01 + c01.
ik ij jk
More precisely, we state the following definition:
4 F.WEHRUNG
Definition 2.1. Let S be a semilattice, let e be an element of S. Say that the
Uniform Refinement Property holds at e, if for all families (a ) and (b ) of
i i∈I i i∈I
elementsofS suchthat(∀i∈I)(a +b =e),there arefamilies (a∗) ,(b∗) and
i i i i∈I i i∈I
(c ) of elements of S satisfying the following properties:
ij (i,j)∈I×I
(i) For all i∈I, a∗ ≤a and b∗ ≤b and a∗+b∗ =e.
i i i i i i
(ii) For all i, j ∈I, c ≤a∗,b∗ and a∗ ≤a∗+c .
ij i j i j ij
(iii) For all i, j, k ∈I, c ≤c +c .
ik ij jk
Say that S satisfies the Uniform Refinement Property, if the Uniform Refinement
Property holds at every element of S.
It is to be noted that this is far from being the only possible “reasonable” defi-
nitionfora“UniformRefinementProperty”,see[15,Theorem2.8,Claim1],which
suggestsaquite differentUniform RefinementProperty(in the contextofpartially
orderedvector spaces). The following resultallows us to focus the investigationon
a generating set in the case where our semilattice is distributive:
Proposition2.2. LetS beadistributivesemilattice. ThenthesetX ofallelements
of S at which the Uniform Refinement Property holds is closed under join.
Proof. Let e and e be two elements of X, and put e = e +e . Let (a ) and
0 1 0 1 i i∈I
(b ) be two families of elements of S such that for all i∈I, a +b =e. Since S
i i∈I i i
is distributive, there are decompositions a =a +a , b =b +b (for all i∈I)
i i0 i1 i i0 i1
such that e = a +b (for all i ∈ I and ν < 2). Since both e and e belong to
ν iν iν 0 1
X,to the latter decompositionscorrespondelements a∗ , b∗ andc (i, j ∈I and
iν iν ijν
ν <2) witnessing the Uniform Refinement Property at e and e .
0 1
Now put a∗ =a∗ +a∗ , b∗ =b∗ +b∗ andc =c +c . It is obvious that (i)
i i0 i1 i i0 i1 ij ij0 ij1
to (iii) of Definition 2.1 aboveare satisfiedwith respectto (a ) and(b ) , thus
i i∈I i i∈I
proving that X+X ⊆X. (cid:3)
Note also the following result, whose easy proof we shall omit:
Proposition 2.3. Let f: S →T be a weakly distributive homomorphism of semi-
lattices and let u∈S. If the Uniform Refinement Property holds at u in S, then it
also holds at f(u) in T. (cid:3)
3. Congruence splitting lattices; property (C)
We will say throughout this paper that a lattice L is congruence splitting, if for
all a ≤ b in L and all congruences α and α of L such that Θ(a,b) ⊆ α ∨α ,
0 1 0 1
there are elements x and x of [a, b] such that x ∨x = b and for all i < 2,
0 1 0 1
Θ(a,x )⊆α . Of course, it suffices to consider the case where both α and α are
i i 0 1
compact congruences and Θ(a,b)=α ∨α .
0 1
For all elements a, b and c of a given lattice, write a⋖ b, if there exists x such
c
that a∨x=b and a∧x≤c.
Definition 3.1. A lattice L has property (C), if for all a≤b and all c in L, there
exist a positive integer n and elements x (i ≤ n) of L such that a = x ⋖ x ⋖
i 0 c 1 c
···⋖ x =b.
c n
The letter ‘C’ stands here for ‘complement’. In the following proposition, we
recordafewelementarypropertiesoflatticeseitherwithproperty(C)orcongruence
splitting.
CONGRUENCE LATTICES 5
Proposition 3.2. The following properties hold:
(a) Every relatively complemented lattice, or every sectionally complemented
lattice, has property (C).
(b) Every atomistic lattice has property (C).
(c) The class of all lattices satisfying property (C) is closed under direct limits.
(d) Every lattice satisfying property (C) is congruence splitting.
(e) The class of all congruence splitting lattices is closed under direct limits.
Proof. (a) is obvious (take n=1).
(b)LetLbeanatomisticlattice. Notethenthatforeverya∈Landeveryatom
p, a⋖ a∨p: indeed, if p ≤ a, then in fact a = a∨p, and otherwise, a∧p = 0.
0
Now let a < b in L. There exist a positive integer n and atoms p (i ≤ n) such
i
that b = W p . For all i ≤ n, put c = a∨W p . Then, by previous remark,
i≤n i i j≤i j
a=c ⋖ c ⋖ ···⋖ c =b.
0 0 1 0 0 n
(c) is straightforward.
(d)LetLbealatticesatisfyingproperty(C),letα (j <2)becongruencesofL,
j
leta≤binLsuchthatΘ(a,b)⊆α ∨α . Toreachthedesiredconclusion,weargue
0 1
byinductionontheminimallengthn=n(a,b)ofachaina=c ⋖ c ⋖ ···⋖ c =b
0 a 1 a a n
suchthatforalli<n,thereexistsj <2suchthatc ≡c (mod α )(suchachain
i i+1 j
exists by [8, Lemma III.1.3] and property (C)). If a < b, then there exists c such
thata≤c,n(a,c)<n(a,b),c⋖ b,and,withoutlossofgenerality,c≡b (mod α ).
a 0
By induction hypothesis, there are y ∈ [a, c] (j < 2) such that y ∨y = c and
j 0 1
y ≡ a (mod α ). Furthermore, there exists z such that z∧c ≤ a and z∨c = b.
j j
Put x =y ∨z and x =y ; then x ∈[a, b], x ≡a (mod α ), and x ∨x =b.
0 0 1 1 j j j 0 1
(e) Let L be a direct limit of a direct system (L ,e ) of lattices and
i ij i≤j in I
lattice homomorphisms (I is a directed poset), with limiting maps e : L →L (for
i i
all i ∈ I). It is well-known that Con L is then the direct limit of the Con L
c c i
with the corresponding transition maps and limiting maps. Once this observation
is made, it is routine to verify that if all the L are congruence splitting, then so
i
is L. (cid:3)
Theorem 3.3. Let L be a congruence splitting lattice. Then Con L satisfies the
c
Uniform Refinement Property.
Proof. PutS =Con L. By Proposition2.2, it suffices to provethat S satisfies the
c
Uniform Refinement Property at every element of S of the form ε=Θ(u,v) where
u≤v in L. Thus let (α ) and (β ) be two families of elements of S such that
i i∈I i i∈I
for all i∈I, α +β =ε. Since L is congruence splitting, there are elements s and
i i i
t of [u, v] such that s ∨t = v and Θ(u,s ) ⊆ α and Θ(u,t ) ⊆ β . Now, for all
i i i i i i i
i,j ∈I, put
α∗ =Θ(u,s ), β∗ =Θ(u,t ) and γ =Θ(s ,s ∨s ).
i i i i ij j i j
It is immediate that α∗ ⊆ α , β∗ ⊆ β and α∗ +β∗ = ε. Furthermore, for all
i i i i i i
i,j ∈ I, we have γ ⊆ Θ(u,s ) = α∗ and γ ⊆ Θ(u,t ) = β∗. Finally, γ is
ij i i ij j j ij
the least congruence θ of L such that θ(s ) ≤ θ(s ); it follows immediately that
i j
γ ≤γ +γ , thus completing the proof. (cid:3)
ik ij jk
Ontheotherhand,aninspectionoftheproofoftheTheoremstatedintheIntro-
duction[15,Theorem2.15]showsinfactthefollowingpropertyofthecorresponding
semilattices S :
κ
6 F.WEHRUNG
Lemma3.4. For everycardinal numberκ≥ℵ , the semilatticeS does notsatisfy
2 κ
the Uniform Refinement Property at its largest element. (cid:3)
By putting together Proposition2.3 and Lemma 3.4, one deduces the following:
Corollary 3.5. For every cardinal number κ≥ℵ , there are no congruence split-
2
ting lattice L and no weakly distributive semilattice homomorphism µ: Con L →
c
S with range containing the largest element of S . (cid:3)
κ κ
Corollary 3.6. Let κ ≥ ℵ be a cardinal number. Consider both following state-
2
ments:
(i) ThereexistalatticeLandaweaklydistributivehomomorphismµ: Con L→
c
S with range containing (as an element) the largest element of S .
κ κ
(ii) For every bounded lattice L, there exists a congruence splitting lattice L′
such that Con L∼=Con L′.
c c
Then (i) and (ii) cannot be simultaneously true.
Proof. Suppose that both (i) and (ii) are simultaneously true, and let L, µ be
as in (i). Since L is the direct union of its closed intervals, we obtain, by using
Proposition 1.2, the existence of a closed interval K of L such that the restriction
of µ to K satisfies (i). Then, applying (ii) to K contradicts Corollary 3.5. (cid:3)
In particular, the Congruence Lattice Problem and the problem whether every
lattice has a congruence-preserving embedding into a sectionally complemented
lattice cannot both have positive answers.
4. Applications to von Neumann regular rings
For any ring R, we denote by FP(R) the class of all finitely generated projective
right R-modules, and by V(R) the (commutative) monoid of isomorphism classes
of elements of FP(R), the addition of V(R) being defined by [A]+[B] = [A⊕B]
([A] denotes the isomorphismclassof A). ThenV(R) is conical, that is, it satisfies
the axiom (∀x,y)(x+y = 0 ⇒ x = y = 0). Moreover, if R is regular, then V(R)
satisfies the refinement property. Furthermore, we equip V(R) with its algebraic
preordering ≤, defined by x ≤ y if and only if there exists z such that x+z = y.
References about this can be found in [1, 5, 6, 10].
Lemma 4.1. Let R be a ring, let J ∈IdR, let a and b be idempotent elements of
R such that aR∼=bR (as right R-modules). Then a∈J if and only if b∈J.
Proof. Since aR∼=bR anda andb areidempotent, there areelements x∈aRband
y ∈ bRa such that a = xy and b = yx. Suppose, for example, that a ∈ J. Then
x∈aRb⊆J (because J is a right ideal of R), thus b=yx∈J (because J is a left
ideal of R). (cid:3)
IfLisanylatticewith0,anidealaofLisneutral,ifx∈aandx∼yimpliesthat
y ∈a (∼ is the relationofperspectivity). We will denote by NIdL the lattice ofall
neutral ideals of L. Recall [3, 8] that if L is a sectionally complemented modular
lattice, then ConL and NIdL are (canonically) isomorphic.
Lemma 4.2. Let R be a regular ring. Then an ideal a of L(R) is neutral if and
only if it is closed under isomorphism (that is, J ∈a and I ∼=J implies I ∈a).
CONGRUENCE LATTICES 7
Proof. It is trivial that if a is closed under isomorphism, then it is neutral. Con-
versely, suppose that a is neutral. Let I,J ∈ L(R) such that I ∼= J and J ∈ a.
There exists I′ ∈ L(R) such that (I ∩J)⊕I′ = I. Since I ∩J ≤ J and J ∈ a,
we have I ∩J ∈ a. Furthermore, there exists J′ ≤ J such that I′ ∼= J′. Since
I′∩J = {0}, we also have I′ ∩J′ = {0}, thus, by [5, Proposition 4.22], I′ ∼ J′.
Since J′ ≤ J and J ∈ a, we have J′ ∈ a, thus, since a is neutral, I′ ∈ a. Hence,
I =(I ∩J)⊕I′ ∈a. (cid:3)
We denote by IdR the (algebraic) lattice of two-sided ideals of any ring R, and
byId Rthesemilatticeofallcompact(thatis,finitelygenerated)elementsofIdR.
c
Theorem 4.3. Let R be a regular ring. Then one can define two mutually inverse
isomorphisms by the following rules:
ϕ: NIdL(R)→IdR, a7→{x∈R: xR∈a},
and
ψ: IdR→NIdL(R), I 7→{J ∈L(R): J ⊆I}.
Proof. First of all, we must verify that for all a∈NIdL(R), ϕ(a) as defined above
is a two-sided ideal of R. It is obvious that ϕ(a) is an additive subgroup of R. Let
x∈ϕ(a) andλ∈R. Then [xλR]≤[xR] and, since R is regular,[λxR]≤[xR] (the
natural surjective homomorphism xR → λxR splits). Therefore, it is sufficient to
prove that if J ∈ a and [I] ≤ [J], then I ∈ a. However, this results immediately
from Lemma 4.2. Conversely, the fact that ψ takes its values in NIdL(R) results
immediately from Lemma 4.1. The verification of the fact that both ϕ and ψ are
order-preservingand mutually inverse is straightforward. (cid:3)
Corollary 4.4. For any regular ring R, Con L(R) is isomorphic to Id R. (cid:3)
c c
In [2], G. M. Bergman proves among other things that any countable bounded
distributive semilattice is isomorphic to Id R for some regular (and even ultrama-
c
tricial)ringR. ByCorollaries3.5and4.4,onecannotgeneralizethistosemilattices
of size ℵ (the size ℵ case is still open):
2 1
Corollary 4.5. For any cardinal number κ ≥ ℵ , S is not isomorphic to the
2 κ
semilattice of finitely generated two-sided ideals of any regular ring. (cid:3)
Note that this also provides a negative answer to [15, Problem 2.16], via the
following easy result:
Proposition4.6. LetRbearegularring. ThenId Risisomorphictothemaximal
c
semilattice quotient of V(R).
Proof. Let π :V(R)→P(R) be the mapping defined by the rule
π(α)={x∈R: [xR]≤nα for some positive integer n}.
Using the refinement property of V(R), it is easy to verify that π is a monoid
homomorphism taking its values in IdR. Moreover, for every I ∈ L(R), we have
π([I]) = RI and those elements generate Id R, thus π maps V(R) onto Id R.
c c
Finally, by [5, Corollary 2.23], for all I, J ∈ L(R), π([I]) ≤ π([J]) if and only if
there exists a positive integer n such that [I]≤n[J]; again by using refinement, it
follows that for allelements α, β ∈V(R), π(α)≤π(β) if and only if there exists a
positive integer n such that α≤nβ. The conclusion follows. (cid:3)
8 F.WEHRUNG
Noteadded. Recently,theauthor,inajointpaperwithM.PloˇsˇcicaandJ.T˚uma,
hasprovedthatthere exists a bounded latticeL suchthatConLis not isomorphic
to ConL′, for any congruencesplitting lattice L′. Comparethis with Corollary3.6
of this paper.
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D´epartementde Math´ematiques,Universit´ede Caen,14032Caen Cedex,France
E-mail address: [email protected]