INTO GEOMETRIC LATTICES
BENEDEK SKUBLICS
Abstract. A lattice is said to be finite height generated if it is complete and every element is the join of some elements of finite height. Extending former results by G. Gr¨atzer and E.W. Kiss [6] onfinitelattices, we prove that everyfinite height generated algebraiclattice that has a pseudorank function is isometrically embeddable into a geometric lattice.
1. Introduction
Given a latticeLwith a lower bound 0, the height of an elementa∈Lis defined to be the supremum of lengths of chains in [0, a]. Let a function p∶L → N∞ = {0,1, . . . ,∞}be called apseudorank function if it has the following properties:
(i) p(0) =0;
(ii) a≤bimpliesp(a) ≤p(b)for alla, b∈L;
(iii) a<bimpliesp(a) <p(b)for alla, b∈L of finite height;
(iv) p(a∧b) +p(a∨b) ≤p(a) +p(b)for alla, b∈L;
(v) p(a) < ∞ iffais of finite height.
Note that a function L → N∞ is called submodular or semimodular if it satisfies condition (iv). Also note that if Lis of finite length, our definition of pseudorank function coincides that of D.T. Finkbeiner [4] and M. Stern [9].
It is well known that the height function of a semimodular lattice of finite length is a pseudorank function. It is an easy consequence of the Jordan-H¨older Chain Condition that the elements of finite height in a semimodular lattice form a sublat- tice. Hence the height function of anarbitrarysemimodular lattice is a pseudorank function. Note that each lattice of finite length has a pseudorank function, see [4, Theorem 2.1].
Letpbe a pseudorank function on a latticeL, and letSbe a semimodular lattice.
Then an embeddingϕ∶L→S is said to beisometrical ifp=h○ϕ, wherehdenotes the height function ofS. (We compose mappings from right to left.) An embedding is said to becover-preserving, if it preserves the covering relation. E.g. ifL andS are of finite length, L is also semimodular andpdenotes the height function of L thenϕis isometrical iff it is cover-preserving.
A result of R.P. Dilworth (around 1950) states that every finite lattice can be embedded into a geometric lattice [2, Theorem 14.1]. D.T. Finkbeiner [4] showed that every lattice of finite height with a pseudorank function pcan be embedded
Date: August 30, 2012.
2000Mathematics Subject Classification. Primary: 06C10; Secondary: 06B15.
Key words and phrases. Semimodular lattice, geometric lattice, isometrical embedding, cover- preserving embedding.
This research was supported by T ´AMOP-4.2.2/B-10/1-2010-0012.
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into a semimodular lattice isometrically with respect top. Blending these results, G. Gr¨atzer and E.W. Kiss [6] managed to prove that every finite lattice with a pseudorank functionpcan be embedded into a geometric lattice isometrically with respect top.
The main goal of our paper is an extension of the result of G. Gr¨atzer and E.W.
Kiss for algebraic lattices that have sufficiently many elements of finite height. A lattice is said to befinite height generated if it is complete and every element is the join of some elements of finite height. Note that lattices of finite length are finite height generated. To show a finite height generated lattice that is not of finite length, consider e.g.N∞ with the usual ordering.
Theorem 1. Every finite height generated algebraic lattice with a pseudorank func- tion can be embedded isometrically into a geometric lattice.
Corollary 2. Every finite height generated semimodular algebraic lattice has a cover-preserving embedding into a geometric lattice.
M. Wild [10] pointed out that the proof of R.P. Dilworth [2, Theorem 14.1]
implies that every finite semimodular lattice has a cover-preserving embedding into a geometric lattice. Using the toolkit of matroid theory, he also shortened R.P.
Dilworth’s proof. Later G. Cz´edli and E.T. Schmidt [3] extended this result for lattices of finite length. Their proof is different from that of G. Gr¨atzer and E.W.
Kiss and that of R.P. Dilworth and M. Wild.
Our construction is an extension of that of R.P. Dilworth. M. Wild’s matroid theoretical approach has helped us to understand R.P. Dilworth’s proof better.
However, using the toolkit of matroid theory does not seem helpful in our case since the theory of infinite matroids is much more difficult. Even the definition of infinite matroids is not clear since there are various reasonable ways to define them, see J.G. Oxley [7, 8]. Also note that in the finite case, R.P. Dilworth’s original argument gives a completely different proof for the Gr¨atzer-Kiss Isometrical Embedding Theorem [9, Theorem 6.2.4].
Now, let us give a short outline of our paper. In Section 2, we introduce our construction. In Section 3, we prove Theorem 1 and Corollary 2. Then in Section 4, we compare the above-mentioned embedding constructions with ours.
Finally, let us overview our notation. We will use ∩ resp.∪ for set theoretical intersection resp. union and∧,∨for lattice operations. Sometimes ∧will coincide with∩. In these cases, we will usually use∩in order to emphasize this coincidence.
Let(a]resp.[a)denote the principal ideal resp. filter generated bya. For the sake of simplicity, sometimes we will writexinstead of{x}, e.g.X∪xinstead ofX∪{x}, if it is clear thatX denotes a set and xdenotes an element. X−Y will denote the set theoretical difference ofX andY.
2. Basic concepts and lemmas
Given a set S, we define a collectionL of subsets of S to be a complete lattice of subsets ofS if∅, S∈ Land L is closed under arbitrary intersection. Note that in this caseL is a complete lattice with respect to set inclusion. Also note that a collection of subsets ofSis closed under arbitrary intersection iff it is the lattice of closed sets ofS with respect to an appropriate closure operator, see e.g. S. Burris and H.P. Sankappanavar [1].
We say that a pseudorank functionr on a complete latticeL of subsets of S is arank function if
(1) r(A) −r(B) ≤ ∣A−B∣ for allA, B∈ L of finite height.
Note that ifS is finite, our rank function on a complete lattice of subsets of S are strictly increasing rank function in sense of P. Crawley and R.P. Dilworth [2].
For every finite height generated lattice with a pseudorank function, we are going to construct a complete lattice of subsets, which is isomorphic to the lattice and on which the pseudorank function becomes a rank function. Note, that our construc- tion is an extension of that of P. Crawley and R.P. Dilworth [2, Lemma 14.1.B].
For the rest of this section, let us fix a finite height generated latticeL with a pseudorank functionp. LetJ ⊆Ldenote the set of nonzero join-irreducible elements of finite height. For f ∈J, let f0 denote the unique lower cover of f. The setJ is a poset with respect to the restriction of the partial ordering of L. Since L is finite height generated, the set{(a]∩J∶a∈L}of order-ideals ofJ forms a complete lattice of subsets that is isomorphic toL. However, (1) does not necessarily hold.
To avoid this problem, we need sufficiently many elements in the ground set of the required complete lattice of subsets.
Let{Xf∶f ∈J} be a collection of pairwise disjoint sets such that∣Xf∣ =p(f) − p(f0). Set S = ⋃(Xf∶f ∈ J). For each a ∈ L, define «a= ⋃(Xf∶f ∈ J, f ≤ a). Then«0= ∅, «1=S and« ( ⋀A) = ⋂(«a∶a∈A)for all A⊆L. Consequently, the collectionL = {«a∶a∈L}forms a complete lattice of subsets ofS. SinceLis finite height generated, the map ϕ∶L→ L, a ↦ «a is an isomorphism. For each a∈ L, definer(«a) =p(a).
Lemma 3. The above definedr is a rank function onLwith p=r○ϕ.
Proof. Certainly,ris a pseudorank function. This fact will be used hereafter with- out further reference. To prove (1), observe that it suffices to show, that
(2) r(«a) −r(«b) ≤ ∣ «a− «b∣ for alla, b∈L, a≥bof finite height.
Indeed, if (2) holds then for anyc, d∈Lof finite height, we have r(«c) −r(«d) ≤ r(«c) −r(«c∩ «d) ≤ ∣ «c− («c∩ «d)∣and«c− («c∩ «d) = «c− «d.
We prove (2) by induction on h(a), where hdenotes the height function of L.
Leta, b∈L, a≥bbe arbitrary elements of finite height. The caseh(a) =0 is trivial.
Suppose thath(a) >0.
Ifa=bthen (2) holds trivially. Ifa≻bandais join-irreducible thenb=a0 and (2) holds by definition. If a≻b and a is not join-irreducible then there exists an elementf∈J such thata=b∨f andf<a. Using the induction hypothesis and the submodularity ofr, we obtainr(«a)−r(«b) ≤r(«f)−r(«b∩«f) ≤ ∣ «f−(«b∩«f)∣ ≤
∣ «a− «b∣.
If a > b and a /≻ b then there is an element c ∈ L such that b < c ≺ a. Hence the induction hypothesis and the previous paragraph yieldsr(«a)−r(«b) =r(«a)−
r(«c) +r(«c) −r(«b) ≤ ∣ «a− «c∣ + ∣ «c− «b∣ = ∣ «a− «b∣. LetF⊆Ldenote the set of elements of finite height. Since Lhas a pseudorank function,F is a sublattice. Notice that
(3) for any finiteA⊆S there isx∈F such that A⊆ «x.
For eachx∈F, we define rx to be the map
rx∶2S→N= {0,1, . . .}, A↦min{r(«y) + ∣(A∩ «x) − «y∣ ∶y∈F}.
Note that the above definition is an extension of that of P. Crawley and R.P.
Dilworth [2, Lemma 14.1.C]. Given a set A ⊆ S and an element x ∈ F, we say that y ∈ F represents rx(A) if rx(A) = r(«y) + ∣(A∩ «x) − «y∣. Some important properties ofrx can be found in the following statements.
Lemma 4.
(i) rx(A) =min{r(«y) + ∣(A∩ «x) − «y∣ ∶y∈ [0, x]} for allA⊆S.
(ii) 0≤rx(A) =rx(A∩ «x) ≤min{∣A∩ «x∣, r(«x)}for allA⊆S.
(iii) A⊆B impliesrx(A) ≤rx(B)for allA, B⊆S.
(iv) rx(A) =ry(A)for all A⊆S andx, y∈F satisfying A⊆ «x∩ «y.
(v) If x≥y then rx(«y) =r(«y)for allx, y∈F.
Proof. The first four statements follow easily from the definition. To prove (v), let x≥y be elements of F. By (iv) and (ii), we have rx(«y) =ry(«y) ≤r(«y). To prove the opposite direction, let u∈ F representry(«y). Then by (1), we obtain ry(«y) =r(«u) + ∣ «y− «u∣ ≥r(«u) +r(«y) −r(«u) =r(«y). Lemma 5. Let A⊆B ⊆ S and x∈ F. Suppose that u∈F represents rx(A)and v∈F representsrx(B). Then u∧v representsrx(A)andu∨v represents rx(B), that is
rx(A) =r(«u∩ «v) + ∣(A∩ «x) − («u∩ «v)∣ and (4)
rx(B) =r(«u∨ «v) + ∣(B∩ «x) − («u∨ «v)∣. (5)
Proof. First, we need some elementary calculations.
∣(A∩ «x) − («u∩ «v)∣ = ∣(A∩ «x) − «u∣ + ∣(A∩ «x∩ «u) − «v∣ ≤
≤ ∣(A∩ «x) − «u∣ + ∣(B∩ «x∩ «u) − «v∣ =
= ∣(A∩ «x) − «u∣ + ∣(B∩ «x) − «v∣ − ∣(B∩ «x) − («u∪ «v)∣ ≤
≤ ∣(A∩ «x) − «u∣ + ∣(B∩ «x) − «v∣ − ∣(B∩ «x) − («u∨ «v)∣.
Now, using the definition ofrx, the submodularity ofrand the above calculations, we obtain the following inequalities
rx(A) ≤r(«u∩ «v) + ∣(A∩ «x) − («u∩ «v)∣ ≤
≤r(«u) +r(«v) −r(«u∨ «v) + ∣(A∩ «x) − («u∩ «v)∣ ≤
≤r(«u) +r(«v) −r(«u∨ «v)+
+ ∣(A∩ «x) − «u∣ + ∣(B∩ «x) − «v∣ − ∣(B∩ «x) − («u∨ «v)∣ =
=rx(A) +rx(B) −r(«u∨ «v) − ∣(B∩ «x) − («u∨ «v)∣ ≤rx(A). Therefore the above inequalities are equalities. Thus the underlined part is zero,
which gives (5), while (4) is the first inequality.
Corollary 6. For any A ⊆ S and x ∈ F, there exists a smallest and a largest element in [0, x]that represents rx(A).
Lemma 7. LetA⊆Sanda∈S. Ifrx(A∪a) =rx(A)andy∈F representsrx(A∪a) then the following hold:
(i) y also representsrx(A)and (ii) a/∈ «xora∈ «y.
Proof. rx(A∪a) =r(«y) + ∣((A∪a) ∩ «x) − «y∣ ≥r(«y) + ∣(A∩ «x) − «y∣ ≥rx(A) = rx(A∪a)implies thatyalso representsrx(A)and∣((A∪a)∩«x)−«y∣ = ∣(A∩«x)−«y∣.
Hencea/∈ «xora∈ «y.
Lemma 8. Let A⊆S andB= {b∈S∶rx(A∪b) =rx(A)}. Thenrx(B) =rx(A). Proof. By the monotonicity of rx, that is Lemma 4(iii), we know that rx(A) ≤ rx(B). By Corollary 6, there exists a largest element y ∈ [0, x] that represents rx(A). Letb ∈B−A be an arbitrary element. Letz ∈ [0, x] representrx(A∪b). Then Lemma 7(i) implies that z represents rx(A), and Lemma 7(ii) implies that b /∈ «x or b ∈ «z. We also have «z ⊆ «y, because y is the largest element that represents rx(A). Hence b/∈ «xor b∈ «y. Consequently,(B−A) ∩ «x⊆ «y, which yields that
rx(A) =r(«y) + ∣(A∩ «x) − «y∣ =r(«y) + ∣(B∩ «x) − «y∣ ≥rx(B). Note that geometric lattices and closure operators are closely related. That is, a lattice is a geometric lattice iff it is the lattice of closed sets with respect to an algebraic closure operator cl that satisfies the so-called Exchange Property:
b∈cl(A∪ {a}) −cl(A)impliesa∈cl(A∪ {b}). For more details, see G. Gr¨atzer [5]
and M. Stern [9].
Usingrx, we define two kinds of closure operators onS: clx for eachx∈F and cl. Namely, for anyA⊆F,
clx(A) = {a∈S∶ry(A∪a) =ry(A)for ally∈F∩ [x)}, cl(A) = ⋃{cly(A) ∶y∈F}.
Notice that
(6) clx(A) ⊆cly(A)ifx≤y.
Lemma 9. The functions clx∶2S →2S, A↦clx(A)andcl∶2S →2S, A↦cl(A)are algebraic closure operators. Moreover, clsatisfies the Exchange Property.
Proof. The extensivity of clxis immediate from the definition. To prove the mono- tonicity, let A ⊆ B ⊆ S. By the definition of clx, it is enough to prove that ry(B∪a) =ry(B) for alla∈clx(A)and all y ∈F∩ [x). Suppose indirectly that there are elements a∈clx(A)and y∈F∩ [x)such thatry(B∪a) >ry(B). Then ry(B∪a) =ry(B) +1. By Corollary 6, there exists a smallest element u∈ [0, y] that representsry(A∪a). Letv ∈ [0, y] representry(B). Then v also represents ry(B∪a), thusa∈ «y anda/∈ «vmust hold. Using Lemma 5 forA∪a⊆B∪a, we obtain thatu∧vrepresentsry(A∪a). Thenu≤u∧vgives«u⊆ «v. By Lemma 7(ii) forry(A∪a)andu, we have thata/∈ «y ora∈ «u⊆ «v, which contradicts the fact thata∈ «y anda/∈ «v. Consequently, clxis monotone.
To prove that clx is idempotent, let A ⊆ S. For any y ∈ F ∩ [x), we have A⊆clx(A) ⊆By = {b∈S∶ry(A∪b) =ry(A)}. By Lemma 8 and the monotonicity ofry, we also havery(A) =ry(clx(A)) =ry(By). Now, for any a∈clx(clx(A))and anyy∈F∩ [x),
ry(A) ≤ry(A∪a) ≤ry(clx(A) ∪a) =ry(clx(A)) =ry(A).
Hencea∈clx(A)and clx(clx(A)) ⊆clx(A). The other direction follows immediately from the extensivity of clx. Consequently, clxis idempotent. We conclude that clx is a closure operator.
To prove that clxis algebraic, letA⊆Sanda∈clx(A). Lety= ⋀{z∈F∩[x) ∶a∈
«z}. By (3), we obtain y ∈F. First, let A0 ⊆A∩ «y be a finite subset such that ry(A0)is maximal. Thenry(A0) =ry(A). Indeed, the maximality ofry(A0)implies that ry(A0∪b) =ry(A0)for allb∈A. Using Lemma 8 forB= {b∈S∶ry(A0∪b) = ry(A0)}, we obtain ry(A0) = ry(B). Therefore ry(A0) = ry(A) = ry(B) by the monotonicity ofry. Now, we havery(A0) ≤ry(A0∪a) ≤ry(A∪a) =ry(A) =ry(A0), hencery(A0∪a) =ry(A0). Finally, letz∈F∩[x). Ifa/∈ «zthenrz(A0) =rz(A0∪a) trivially holds. If a∈ «z theny≤z by the definition ofy. Using Lemma 4(iv) for A0∪a ⊆ «y ⊆ «z, we obtain rz(A0) = ry(A0) = ry(A0∪a) = rz(A0∪a). Hence a∈clx(A0). We conclude that clx is an algebraic closure operator.
The extensivity and monotonicity of cl follow immediately from those of clx. To prove the idempotency of cl, let A ⊆ S and suppose that a ∈ cl(cl(A)). By definition, a ∈clx(cl(A)) for somex∈ F. Since clx is algebraic, there is a finite subsetA0⊆cl(A)such thata∈clx(A0). By (6) and the definition of cl,A0⊆cly(A) for some y ∈ F. By (6) and the monotonicity and idempotency of clx, we have a∈clx(cly(A)) ⊆clx∨y(clx∨y(A)) =clx∨y(A) ⊆cl(A). Hence cl(cl(A)) ⊆cl(A). The other direction follows immediately from the extensivity of cl. Consequently, cl is a closure operator. It is algebraic since clx is algebraic for allx∈F.
To prove that cl satisfies the Exchange Property, let A ⊆ S and a, b ∈ S such thata∈cl(A∪b) −cl(A). Since cl is algebraic, there is a finite subsetA0⊆Asuch that a∈ cl(A0∪b) −cl(A0). Hence a∈ clx(A0∪b) −clx(A0) for some x∈F. By (3) and (6), we can assume that A0∪ {a, b} ⊆ «x. By the definition of clx, there are u, v ∈F∩ [x)such that ru(A0∪a) =ru(A0) +1 and rv(A0∪b) =rv(A0) +1, since a, b/∈clx(A0). Using this and Lemma 4(iv) forA0∪ {a, b} ⊆ «x, we obtain that ry(A0∪a) = ry(A0∪b) = ry(A0) +1 for all y ∈ F ∩ [x). The assumption a∈clx(A0∪b)impliesry(A0∪ {a, b}) =ry(A0∪b) =ry(A0∪a)for ally∈F∩ [x). Now,b∈clx(A0∪a) ⊆cl(A∪a)follows immediately. Hence cl satisfies the Exchange
Property.
3. The main proofs
Before the proof of Theorem 1, we need a short technical lemma about finite height generatedalgebraic lattices.
Lemma 10. If L is a finite height generated algebraic lattice and the elements of finite height form a sublattice then its elements of finite height are exactly its compact elements.
Proof. Suppose thata∈Lis compact. Then a= ⋁B for some elements B⊆L of finite height, because L is finite height generated. Since a is compact, there is a finite B0⊆B with a= ⋁B0. Hence ais of finite height. Now, suppose that b∈L is of finite height. Thenb = ⋁A for some compact elements A⊆L, becauseL is algebraic. Sincebis of finite height, there is a finiteA0⊆Awithb= ⋁A0. Henceb
is compact.
Proof of Theorem 1. Given a finite height generated algebraic latticeLwith a pseu- dorank functionp, defineLandras we did in the previous section. We will also use S for the ground set ofLandF⊆Lfor the set of elements of finite height. Recall thatF is a sublattice, sinceLhas a pseudorank function. DenoteLclthe complete lattice of subsets that corresponds to the closure operator cl. By Lemma 9,Lclis a geometric lattice. It is enough to prove thatLis a sublattice ofLclsuch thatrand
the height function ofLclcoincide onL. Thenϕ∶L→ Lcl, x↦ «xis an isometrical embedding.
First, we show that L ⊆ Lcl. Letx∈ L anda∈S− «x. Suppose, for a contra- diction, that a ∈ cl(«x). Then, by definition, there is a y ∈ F with a ∈ cly(«x). By (3) and (6), we can assume that a∈ «y. Notice that a∈cly(«x)implies that ry(«x∪a) =ry(«x). Letz∈F representry(«x∪a). Then Lemma 7(ii) anda∈ «y implies thata∈ «z. Sincea∈ «z− «x, we have that «z≠ «x∩ «z= « (x∧z), hence x∧z<z. However, («x∩ «y) − « (x∧z) = («x∩ «y) − («x∩ «z) = («x∩ «y) − «z= ((«x∪a) ∩ «y) − «z. Sincer is strictly monotone for elements of finite height, we obtain thatry(«x) ≤r(« (x∧z))+∣(«x∩«y)−« (x∧z)∣ <r(«z)+∣(«x∩«y)−« (x∧z)∣ = r(«z) + ∣((«x∪a) ∩ «y) − «z∣ =ry(«x∪a), which contradictsry(«x) =ry(«x∪a). This proves that«x∈ Lcl and L ⊆ Lcl. Moreover, the meet operation both on L andLcl is the intersection, thereforeLis a meet-subsemilattice ofLcl.
To prove that L is a sublattice of Lcl, observe that the join of two elements
«x,«y∈ L in the larger latticeLclis cl(«x∪ «y). Since cl(«x∪ «y) ⊆ « (x∨y), it is enough to prove that cl(«x∪«y) ⊇ « (x∨y). Leta∈ « (x∨y). By definition, it means thata∈Xb for someb∈J∩ (x∨y]. SinceLis finite height generated,x= ⋁X and y= ⋁Y for someX, Y ⊆F. By Lemma 10,bis compact, henceb≤ ⋁X0∨ ⋁Y0 for some finiteX0⊆X, Y0⊆Y. Letx0= ⋁X0andy0= ⋁Y0. By Lemma 10,x0, y0∈F. Now,b≤x0∨y0 impliesa∈ « (x0∨y0). In order to prove thata∈cl(«x∪ «y), it is enough to show thata∈cl(«x0∪ «y0). We prove that a∈clx0∨y0(«x0∪ «y0), which yieldsa∈cl(«x0∪ «y0). As a preparation for this, we show that
(7) rz(«x0∪ «y0) =rz(« (x0∨y0))for allz∈F∩ [x0∨y0).
Since rz(«x0) = r(«x0) by Lemma 4(v), x0 represents rz(«x0). Similarly, y0
represents rz(«y0). Assume that z0 represents rz(«x0∪ «y0). Using Lemma 5 twice for «x0 ⊆ «x0∪ «y0 and «y0 ⊆ «x0∪ «y0, we obtain that x0∨y0∨z0 repre- sents rz(«x0∪ «y0). However, «x0∪ «y0 ⊆ « (x0∨y0∨z0), hencerz(«x0∪ «y0) = r(« (x0∨y0∨z0)) + ∣((«x0∪ «y0) ∩ «z) − « (x0∨y0∨z0)∣ =r(« (x0∨y0∨z0)). On the other hand, «x0∪ «y0⊆ « (x0∨y0), which impliesrz(«x0∪ «y0) ≤r(« (x0∨y0)) +
∣((«x0∪ «y0) ∩ «z) − « (x0∨y0)∣ =r(« (x0∨y0)) ≤r(« (x0∨y0∨z0)). Together with r(« (x0∨y0∨z0)) =rz(«x0∪ «y0), we obtain thatrz(«x0∪ «y0) =r(« (x0∨y0)). Lemma 4(v) yields that r(« (x0∨y0)) = rz(« (x0∨y0)), which finishes the proof of (7). Now, «x0∪ «y0 ⊆ «x0∪ «y0∪a⊆ « (x0∨y0), the monotonicity of rz and (7) implies that rz(«x0∪ «y0∪a) =rz(«x0∪ «y0)for all z∈F∩ [x0∨y0). Hence a∈clx0∨y0(«x0∪ «y0). We conclude thatL is a sublattice ofLcl.
To prove that the embedding is isometrical, we have to show thatr(«x) =h(«x) for allx∈L, wherehdenotes the height function ofLcl. By the definition of finite height generated lattices, it suffices to prove thatr(«x) =h(«x)for allx∈F. We use induction on the height ofx∈F. Ifx=0 thenr(«0) =0=h(«0).
Suppose that 0 ≤x≺ y and r(«x) = h(«x). Since r is a rank function on L, we have a set A = {a1, . . . , ak−1, ak} ⊆ «y− «x with k = r(«y) −r(«x) distinct elements. LetA0= «xandAi= «x∪a1∪ ⋅ ⋅ ⋅ ∪ai for alli∈ {1, . . . , k}. Assume that z∈F∩ [y). Clearly,rz(Ai) ≤r(«x) +i. By Lemma 4(v), rz(«x) =r(«x). Hence rz(Ai) ≤rz(«x)+i. The opposite direction is also true. Suppose, for a contradiction, that rz(Ai) <r(«x) +i for somei∈ {1, . . . , k}. Letibe minimal for this property, that isrz(Ai) <r(«x)+iandrz(Ai−1) =r(«x)+i−1. Note that suchiexists, since rz(A0) =r(«x). Then, by the monotonicity ofrz,rz(Ai) =r(«x) +i−1=rz(Ai−1). Letui representrz(Ai). Applying Lemma 7(ii) forrz(Ai)andui, we obtain that
ai∈ «ui, sinceai∈ «y⊆ «z. Notice thatxrepresentsrz(«x)by Lemma 4(v). Using Lemma 5 for «x⊆Ai, we obtain thatx∨ui represents rz(Ai). We conclude from
«x⊂Ai⊆ «y, x≺y, ai∈ «ui and the construction ofS and Lthat x∨ui≥y. Now, we have
r(«x) +i>rz(Ai) =r(« (x∨ui)) + ∣(Ai∩ «z) − « (x∨ui)∣ =
=r(« (x∨ui)) ≥r(«y) =r(«x) +k≥r(«x) +i, which is a contradiction. Therefore
(8) rz(Ai) =rz(«x) +ifor alli∈ {1, . . . , k} and allz∈F∩ [y).
Hence for everyf∈F, we have thatrz(Ai−1) ≠rz(Ai)forz=f∨y, which shows that ai/∈clf(Ai−1)for allf ∈F, that isai/∈cl(Ai−1). This gives that cl(Ai−1) ≠cl(Ai). Clearly, cl(Ai−1) ⪯cl(Ai−1∪ai) =cl(Ai). Thus
(9) «x=cl(A0) ≺cl(A1) ≺ ⋅ ⋅ ⋅ ≺cl(Ak).
We know from (8) and the definition of k that rz(Ak) = rz(«y) for all z ∈ F ∩ [y). Since rz is monotone, rz(Ak) = rz(Ak ∪b) for all b ∈ «y−Ak. Hence b ∈ cly(Ak) ⊆ cl(Ak) for all b ∈ «y−Ak, and we obtain that «y ⊆ cl(Ak). This, together with Ak ⊆ «y, yields that cl(Ak) = «y. Consequently, we conclude from (9), the semimodularity ofLcland the induction hypothesis thath(«y) =h(«x)+k=
r(«x) +k=r(«y).
Proof of Corollary 2. LetLbe a finite height generated semimodular algebraic lat- tice. Consider the height functionhL∶L→N∞. We conclude from Theorem 1 that L has an isometrical embedding ψ into a geometric latticeG with respect tohL. Assume thatx≺yinLand choose a minimal elementf of finite height in(y]−(x]. Let g be a lower cover of f. Then x= x∨g and y = x∨f. Now, ψ(f) covers ψ(g), sincehG(ψ(f))−hG(ψ(g)) =hL(f)−hL(g) =1, wherehGdenotes the height function of G. Hence the semimodularity of G implies that ψ(y) = ψ(x) ∨ψ(f) coversψ(x) =ψ(x) ∨ψ(g). Thereforeψis cover-preserving.
Remark 11. IfLis of finite length, the construction ofLclbecomes more simple:
we need onlyr1 since cl=cl1. Note that in this caser1is a rank function on Lcl. 4. Concluding remarks
As to isometrical embeddings, let L be a finite lattice. Then our construction is completely different from that of G. Gr¨atzer and E.W. Kiss [6]. First of all, we embedL≅ Ldirectly into the geometric latticeLcl, while they did their embedding in two steps: first, they embeddedLinto a semimodular lattice, and second, they embedded this semimodular lattice into a geometric lattice.
On the other hand, our L is uniquely determined, while they provided a tech- nique, where they used a so-calledsemimodular construction scheme. This scheme consists of theframeLand finitely many lattices, the so-calledblocks. These blocks can be chosen infinitely many different ways, and the geometric lattice depends on their choice.
As to cover-preserving embeddings, letLbe a semimodular lattice of finite length and let p=hbe the height function of L. Then our construction yields the same geometric lattice as the entirely different method of G. Cz´edli and E.T. Schmidt [3].
For more details about different cover-preserving embeddings of semimodular lattices, see the historical comments in G. Cz´edli and E.T. Schmidt [3].
Acknowledgment. The author is grateful to the anonymous referee, who im- proved the paper with a thorough report. Several comments and suggestions are acknowledged.
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