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Proper congruence-preserving extensions of lattices
George Grätzer, Friedrich Wehrung
To cite this version:
George Grätzer, Friedrich Wehrung. Proper congruence-preserving extensions of lattices. Acta Math-
ematica Hungarica, Springer Verlag, 1999, 85 (1-2), pp.169-179. �10.1023/A:1006693517705�. �hal-
00004053�
ccsd-00004053, version 1 - 24 Jan 2005
LATTICES
G. GR ¨ATZER AND F. WEHRUNG
Abstract. We prove that every lattice with more than one element has a proper congruence-preserving extension.
1. Introduction
LetLbe a lattice. A latticeKis acongruence-preserving extensionofL, ifKis an extension and every congruence ofLhas exactly one extension toK. (Of course, then, the congruence lattice ofLis isomorphic to the congruence lattice ofK.)
In [4], the first author and E. T. Schmidt raised the following question:
Is it true that every latticeLwith more than one element has a proper congru- ence-preserving extensionK?
Hereproper means thatK properly containsL, that is,K−L6=∅.
The first author and E. T. Schmidt pointed out in [4] that in the finite case this is obviously true, and they proved the following general result:
Theorem 1. LetLbe a lattice. If there exist a distributive interval with more than one element in L, thenL has a proper congruence-preserving extensionK.
Generalizing this result, in this paper, we provide a positive answer to the above question:
Theorem 2. Every latticeLwith more than one element has a proper congruence- preserving extension K.
2. Background
LetK andLbe lattices. If Lis a sublattice ofK, then we callK anextension ofL. If Kis an extension of Land Θ is a congruence relation ofK, then ΘL, the restriction of Θ toLis a congruence ofL. If the map Θ7→ΘLis a bijection between the congruences of L and the congruences of K, then we call K a congruence- preserving extension of L. Observe that if K a congruence-preserving extension ofL, then the congruence lattice ofLis isomorphic to the congruence lattice ofK in a natural way.
The proof of Theorem 1 is based on the following construction of E. T. Schmidt [9], summarized below as Theorem 3. (A number of papers utilize this construction;
Date: February 20, 1998.
1991Mathematics Subject Classification. Primary 06B10; Secondary 08A30.
Key words and phrases. Lattice, congruence, congruence-preserving extension, proper extension.
The research of the first author was supported by the NSERC of Canada.
1
2 G. GR ¨ATZER AND F. WEHRUNG
see, for instance, E. T. Schmidt [10], [11] and the recent paper G. Gr¨atzer and E. T.
Schmidt [5].)
Let L be a bounded distributive lattice with bounds 0 and 1, and let M3 = {o, a, b, c, i}be the five-element nondistributive modular lattice. LetM3[L] denote the poset of tripleshx, y, zi ∈L3 satisfying the condition
(S) x∧y=y∧z=z∧x.
Theorem 3.
Let D be a bounded distributive lattice with bounds0 and1.
(i) M3[D] is a modular lattice.
(ii) The subset
M3={h0,0,0i,h1,0,0i,h0,1,0i,h0,0,1i,h1,1,1i}
of M3[D] is a sublattice ofM3[D]and it is isomorphic to M3.
(iii) The subposet D ={ hx,0,0i |x∈D} of M3[D] is a bounded distributive lattice and it is isomorphic toD; we identifyD with D.
(iv) M3 andD generateM3[D].
(v) Let Θ be a congruence relation ofD =D; then there is a uniquecongru- ence Θ of M3[D] such that Θ restricted to D is Θ; therefore, M3[D] is a congruence-preserving extension of D.
Unfortunately,M3[L] fails, in general, to produce a lattice, ifLis not distributive.
In this paper, we introduce a variant on theM3[L] construction, which we shall denote asM3hLi. This lattice M3hLiis a proper congruence-preserving extension ofL, for any latticeLwith more than one element, verifying Theorem 2.
3. The construction
For a latticeL, let us call the triplehx, y, zi ∈L3 Boolean, if x= (x∨y)∧(x∨z),
y= (y∨x)∧(y∨z), (B)
z= (z∨x)∧(z∨y).
We denote byM3hLi ⊆L3the poset of Boolean triples ofL.
Here are some of the basic properties of Boolean triples:
Lemma 1. Let Lbe a lattice.
(i) Every Boolean triple of Lsatisfies (S), soM3hLi ⊆M3[L].
(ii) hx, y, zi ∈L3 is Boolean iff there is a triplehu, v, wi ∈L3 satisfying x=u∧v,
y=u∧w, (R)
z=v∧w.
(iii) For every triplehx, y, zi ∈L3, there is a smallest Boolean triple hx, y, zi ∈ L3 such thathx, y, zi ≤ hx, y, zi; in fact,
hx, y, zi=h(x∨y)∧(x∨z),(y∨x)∧(y∨z),(z∨x)∧(z∨y)i.
(iv) M3hLiis a lattice with the meet operation defined as hx0, y0, z0i ∧ hx1, y1, z1i=hx0∧x1, y0∧y1, z0∧z1i and the join operation defined by
hx0, y0, z0i ∨ hx1, y1, z1i=hx0∨x1, y0∨y1, z0∨z1i.
(v) If L has 0, then the subposet { hx,0,0i | x∈ L} is a sublattice and it is isomorphic to L.
If L has 0 and 1, then M3hLi has a spanning M3, that is, a {0,1}- sublattice isomorphic toM3, namely,
{h0,0,0i,h1,0,0i,h0,1,0i,h0,0,1i,h1,1,1i}.
(vi) If hx, y, ziis Boolean, then one of the following holds:
(a) the components form a one-element set, so hx, y, zi = ha, a, ai, for somea∈L;
(b) the components form a two-element set and hx, y, zi is of the form hb, a, ai, orha, b, ai, or ha, a, bi, for some a,b∈L,a < b.
(c) the components form a three-element set and two components are comparable and L has two incomparable elements a andb such that hx, y, ziis of the form ha, b, a∧bi, orha, a∧b, bi, orha∧b, a, bi.
(d) the components form a three-element set and the components are pair- wise incomparable and L has an eight-element Boolean sublattice B so that the components are the atoms of B.
Proof.
(i) Ifhx, y, ziis Boolean, then
x∧y = ((x∨y)∧(x∨z))∧((y∨x)∧(y∨z))
= (x∨y)∧(y∨z)∧(z∨x), which is the upper median ofx,y, andz. So (S) holds.
(ii) Ifhx, y, ziis Boolean, then u=x∨y,v =x∨z, andw=y∨zsatisfy (R).
Conversely, if there is a triple hu, v, wi ∈ L3 satisfying (R), then by Lemma I.5.9 of [1], the sublattice generated by x, y, and z is isomorphic to a quotient of C32 (where C2 is the two element chain) andx, y, and z are the images of the three atoms ofC32. Thus (x∨y)∧(x∨z) =x, the first part of (B). The other two parts are proved similarly.
(iii) For hx, y, zi ∈L3, define u=x∨y,v =x∨z, w=y∨z. Set x1 =u∧v, y1=u∧w,z1=v∧w. Thenhx1, y1, z1iis Boolean by (ii) andhx, y, zi ≤ hx1, y1, z1i inL3. Now ifhx, y, zi ≤ hx2, y2, z2iin L3 andhx2, y2, z2iis Boolean, then
x2= (x2∨y2)∧(x2∨z2) (by (B))
≥(x∨y)∧(x∨z) (byhx, y, zi ≤ hx2, y2, z2i)
=u∧v=x1,
and similarly,y2≥y1,z2≥z1. Thushx2, y2, z2i ≥ hx1, y1, z1i, and sohx1, y1, z1iis the smallest Boolean triple containinghx, y, zi.
(iv) M3hLi 6= ∅; for instance, for all x ∈ L, the diagonal element hx, x, xi ∈ M3hLi. It is obvious from (ii) that M3hLi is meet closed. By (iii), M3hLi is a closure system inL3, from which the formulas of (iv) follow.
The proofs of (v) and (vi) are left to the reader.
4 G. GR ¨ATZER AND F. WEHRUNG
4. Proof of the theorem
Let L be a lattice with more than one element. We identify x ∈ L with the diagonal element hx, x, xi ∈ M3hLi, so we regard M3hLian extension of L. This is an embedding of L into M3hLi different from the embedding in Lemma 1.(v).
Moreover, the embedding in Lemma 1.(v) requires that L have a zero, while the embedding discussed here always works.
Note that M3hLi is a proper extension; indeed, since L has more than one element, we can choose the elements a < b in L. Then ha, a, bi ∈ M3hLi but ha, a, bi is not on the diagonal, so ha, a, bi ∈ M3hLi −L. In fact, if L = C2, the two-element chain, then this is the only type of nondiagonal element:
M3hC2i={h0,0,0i,h1,0,0i,h0,1,0i,h0,0,1i,h1,1,1i}.
For a congruence Θ ofL, let Θ3denote the congruence ofL3defined component- wise. LetM3hΘibe the restriction of Θ3to M3hLi.
Lemma 2. M3hΘi is a congruence relation ofM3hLi.
Proof. M3hΘiis obviously an equivalence relation onM3hLi. SinceM3hLiis a meet subsemilattice ofL3, it is clear thatM3hΘisatisfies the Substitution Property for meets. To verify for M3hΘi the Substitution Property for joins, let hx0, y0, z0i, hx1, y1, z1i ∈M3hLi, let
hx0, y0, z0i ≡ hx1, y1, z1i (M3hΘi), (that is,
x0≡x1 (Θ), y0≡y1 (Θ), and z0≡z1 (Θ) inL) and lethu, v, wi ∈M3hLi. Set
hx′i, yi′, zi′i=hxi, yi, zii ∨ hu, v, wi (the join formed inM3hLi), fori= 0, 1.
Then, using Lemma 1.(iii) and (iv) forx0∨u,y0∨v, andz0∨w, we obtain that x′0= (x0∨u∨y0∨v)∧(x0∨u∨z0∨w)
≡(x1∨u∨y1∨v)∧(x1∨u∨z1∨w) =x′1(M3hΘi), and similarly,y′0≡y′1 (M3hΘi),z0′ ≡z′1 (M3hΘi), hence
hx0, y0, z0i ∨ hu, v, wi ≡ hx1, y1, z1i ∨ hu, v, wi (M3hΘi).
Since L was identified with the diagonal of M3hLi, it is obvious that M3hΘi restricted toLis Θ. So to complete the proof of Theorem 2, it is sufficient to verify the following statement:
Lemma 3. Every congruence of M3hLi is of the formM3hΘi, for a suitable con- gruenceΘofL.
Proof. Let Φ be a congruence of M3hLi, and let Θ denote the congruence of L obtained by restricting Φ to the diagonal of M3hLi, that is, x ≡ y (Θ) in L iff hx, x, xi ≡ hy, y, yi (Φ) inM3hLi. We prove that Φ =M3hΘi.
To show that Φ⊆M3hΘi, let
(1) hx0, y0, z0i ≡ hx1, y1, z1i (Φ).
Define
o=x0∧x1∧y0∧y1∧z0∧z1, (2)
i=x0∨x1∨y0∨y1∨z0∨z1. (3)
Meeting the congruence (1) withhi, o, oiyields (4) hx0, o, oi ≡ hx1, o, oi (Φ).
Since
hx0, o, oi ∨ ho, o, ii=hx0, o, ii=hx0, x0, ii, joining the congruence (4) with ho, o, iiyields
(5) hx0, x0, ii ≡ hx1, x1, ii (Φ).
Similarly,
(6) hx0, i, x0i ≡ hx1, i, x1i (Φ).
Now we meet the congruences (5) and (6) to obtain (7) hx0, y0, z0i ≡ hx1, y1, z1i (Θ3) inL3, proving that Φ⊆M3hΘi.
To prove the converse,M3hΘi ⊆Φ, take
(8) hx0, y0, z0i ≡ hx1, y1, z1i (M3hΘi) inM3hLi, that is,
x0≡x1 (Θ), y0≡y1 (Θ), z0≡z1 (Θ) inL. Equivalently,
hx0, x0, x0i ≡ hx1, x1, x1i (Φ), (9)
hy0, y0, y0i ≡ hy1, y1, y1i (Φ), (10)
hz0, z0, z0i ≡ hz1, z1, z1i (Φ) (11)
inM3hLi.
Now, defineo,i as in (2) and (3). Meeting the congruence (9) withhi, o, oi, we obtain
(12) hx0, o, oi ≡ hx1, o, oi (Φ).
Similarly, from (10) and (11), we obtain the congruences ho, y0, oi ≡ ho, y1, oi (Φ), (13)
ho, o, z0i ≡ ho, o, z1i (Φ).
(14)
Finally, joining the congruences (12)–(14), we get (15) hx0, y0, z0i ≡ hx1, y1, z1i (Φ),
that is,M3hΘi ⊆Φ. This completes the proof of this lemma and of Theorem 2.
6 G. GR ¨ATZER AND F. WEHRUNG
5. Discussion
Special extensions. We can get a slightly stronger result by requiring that the extension preserve the zero and the unit, provided they exist. To state this result, we need the following concept.
An extension K of a latticeL isextensive, provided that the convex sublattice ofK generated byLisK.
Note that ifLhas a zero, 0, then an extensive extension is a{0}-extension (and similarly for the unit, 1); ifLhas a zero, 0, and unit 1, then an extensive extension is a{0,1}-extension.
Theorem 4. Every latticeLwith more than one element has a proper congruence- preserving extensive extensionK.
Proof. Indeed, every hx, y, zi ∈M3hLi is in the convex sublattice generated byL since
hx∧y∧z, x∧y∧z, x∧y∧zi ≤ hx, y, zi ≤ hx∨y∨z, x∨y∨z, x∨y∨zi.
In Theorem 3.(iii), we pointed out thatM3[D] is a congruence-preserving exten- sion of D =D, where D is an ideal of M3[D]. This raises the question whether Theorem 2 can be strengthened by requiring thatLbe an ideal inK. This is easy to do, ifLhas a zero, 0, since then we can identifyx∈Lwithhx,0,0i ∈M3hLi.
Theorem 5. Every latticeLwith more than one element has a proper congruence- preserving extension K with the property thatLis an ideal in K.
Proof. Take an element a ∈ L such that [a) (the dual ideal generated by a) has more than one element. Then by Lemma 1.(v),A=M3h[a)iis a proper congruence- preserving extension of [a) andI= [a) is an ideal inA. Now form the latticeKby gluingLwith the dual ideal [a) toAwith the idealI. It is clear thatKis a proper
congruence-preserving extension ofL.
Modularity and semimodularity. R. W. Quackenbush [8] proved that if L is a modular lattice, then M3[L] is a semimodular lattice. For our construction, the analogous result fails: M3hPiis not semimodular, whereP is a projective plane (a modular lattice). Indeed, let a,b,c be a triangle inP, with sidesl,m, n, that is, letl,m,nbe three distinct lines in the planeP, and define the points a=n∧m, b =n∧l, c =m∧l. Let pbe a point inP not on any one of these lines. Then hp,∅,∅iis an atom inM3hPi, ha, b, ci ∈M3hPibut
h{p},∅,∅i ∨ ha, b, ci=hp∨a, b, ci=hP, l, li and
ha, b, ci<hn, b, li<hP, l, li, showing thatM3hPiis not semimodular.
Now we characterize whenM3hLiis modular.
Theorem 6. Let L be a lattice with more than one element. Then M3hLi is modular iff Lis distributive.
Proof. If L is distributive, then M3hLi =M3[L], so M3hLi is modular by Theo- rem 3.
Conversely, if M3hLi is modular, then L is modular since it is a sublattice of M3hLi. Now if L is not distributive, then L contains anM3 ={o, a, b, c, i} as a sublattice. By Lemma 1.(vi), the elements
ho, o, ai,ho, c, ai,hc, c, ii,hi, i, ii,hb, o, ai belong toM3hLi. Obviously,
ho, o, ai<ho, c, ai<hc, c, ii<hi, i, ii and
ho, o, ai<hb, o, ai<hi, i, ii.
To prove that these five elements form anN5, it is enough to prove that hc, c, ii ∧ hb, o, ai=ho, o, ai
and
ho, c, ai ∨ hb, o, ai=hi, i, ii.
The meet is obvious. Now the join:
ho, c, ai ∨ hb, o, ai=hb, c, ai=hi, i, ii.
SoM3hLicontainsN5as a sublattice, contradicting the assumption thatM3hLiis
modular. Therefore,Lis distributive.
Further results. M3[L] is not a lattice for a generalL. See, however, G. Gr¨atzer and F. Wehrung [6], where a new concept of n-modularity is introduced, for any natural numbern. Modularity is the same as 1-modularity.
By definition, n-modularity is an identity; for larger n, a weaker identity. For ann-modular latticeL,M3[L] is a lattice, a congruence-preserving extension ofL.
For distributive lattices (in fact, forn-modular lattices), the constructionM3[L]
is a special case of the tensor product construction of two semilattices with zero, see, for instance, G. Gr¨atzer, H. Lakser, and R. W. Quackenbush [2] and R. W.
Quackenbush [8]. The M3hLi construction is generalized in G. Gr¨atzer and F.
Wehrung [7] to two bounded lattices; the new construction is called box product.
Some of the arguments of this paper carry over to box products.
Problems
Lattices. As usual, let us denote byT, D,M, andL the variety of one-element, distributive, modular, and all lattices, respectively. A variety V is nontrivial if V6=T.
Let us say that a varietyVof lattices has theCongruence Preserving Extension Property (CPEP, for short), if every lattice inV with more than one element has a proper congruence-preserving extension in V. It is easy to see that no finitely generated lattice variety has CPEP. (Indeed, by J´onsson’s lemma, a nontrivial finitely generated lattice variety V has a finite maximal subdirectly irreducible member L; if K is a proper congruence-preserving extension ofL, thenK is also subdirectly irreducible and |L|>|K|, a contradiction.) In particular, Ddoes not have CPEP.
Theorem 2 can be restated as follows: Lhas CPEP.
Problem 1. Find all lattice varietiesV with CPEP. In particular, doesM have CPEP?
8 G. GR ¨ATZER AND F. WEHRUNG
Groups. Let us say that a varietyVof groups has theNormal Subgroup Preserving Extension Property (NSPEP, for short), if every groupGinVwith more than one element has a proper supergroupGinVwith the following property: every normal subgroup N in G can be uniquely represented in the form N ∩G, where N is a normal subgroup ofG.
Not every group variety V has NSPEP, for instance, the variety A of Abelian groups does not have NSPEP.
Problem 2. Does the varietyGof all groups have NSPEP? Find all group varieties having NSPEP?
Rings. For ring varieties, we can similarly introduce theIdeal Preserving Extension Property(IPEP, for short). The varietyRof all (not necessarily commutative) rings has IPEP. Indeed, if R is a ring with more than one element, then embedR into M2(R) (the ring of 2×2 matrices overR) with the diagonal map. The two-sided ideals of M2(R) are of the form M2(I), where I is a two-sided ideal of R, and I=M2(I)∩R.
Problem 3. Find all ring varieties having IPEP? In particular, does the variety of all commutative rings have IPEP?
The second author found a positive answer for Dedekind domains: every Dedekind domain with more than one element has a proper ideal-preserving extension that is, in addition, a principal ideal domain.
Acknowledgment
This work was partially completed while the second author was visiting the University of Manitoba. The excellent conditions provided by the Mathematics Department, and, in particular, a quite lively seminar, were greatly appreciated.
References
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[7] ,A new lattice construction: the box product, manuscript.
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Department of Mathematics, University of Manitoba, Winnipeg MN, R3T 2N2, Canada E-mail address: gratzer@cc.umanitoba.ca
URL:http://server.maths.umanitoba.ca/homepages/gratzer.html/
C.N.R.S., E.S.A. 6081, D´epartment de Math´ematiques, Universit´e de Caen, 14032 Caen Cedex, France
E-mail address: gremlin@math.unicaen.fr URL:http://www.math.unicaen.fr/~wehrung
