On the variety generated by planar modular lattices

On the variety generated by planar modular lattices G. Grätzer and R. W. Quackenbush Abstract. We investigate the variety generated by the class of planar modular lattices. The main result is a structure theorem describing the subdirectly irreducible members of this variety. 1. Introduction 1.1. Planar modular lattices. In this paper, we consider the variety Var(PM) generated by the class PM of planar modular lattices. We give a structure theorem, Theorem 5.3, describing the subdirectly irreducible members of this variety. 1.2. Outline. In Section 2, we state and prove Baker s Dimension Theorem, on which the whole paper is based. We tackle planar subdirectly irreducible modular lattices in Section 3. We describe them utilizing the frame of a planar subdirectly irreducible modular lattice, the distributive lattice on which it is built. We characterize the finite distributive lattices that can occur as frames. The infinite case starts in Section 4, where we prove, in Theorem 4.9, that the distributive lattices that occur as frames of subdirectly irreducible modular lattices of dimension 2 are subdirect products of two weakly atomic chains. In Section 5, we state the Structure Theorem of Subdirectly Irreducible Modular Lattices of Order- Dimension 2. The variety Var(PM) is locally finite see Lemma 6.1 so every lattice in it is a direct limit of its finite sublattices. In Section 6, we provide an example of a subdirectly irreducible lattice in Var(PM) that is not the direct limit of its finite subdirectly irreducible sublattices. 1.3. Background. It follows from Jónsson s Lemma (see B. Jónsson [12] and [8, p. 328]) that for a finite lattice L, we find the subdirectly irreducible lattices in Var(L), the variety generated by L, as homomorphic images of sublattices of L; therefore up to isomorphism they are finite in number, of size L, and we can regard them as known. Date: June 9, 2008; revised November 24, 2008. 2000 Mathematics Subject Classification: Primary: 06C05; Secondary: 06B20. Key words and phrases: Modular lattice, planar, order-dimension, subdirectly irreducible, variety, weakly atomic. Research of the first author is supported by the NSERC of Canada. 1

2 G. GRÄTZER AND R. W. QUACKENBUSH There are a number of other similar examples. For instance, take the class K of finite lattices of length two. Then Si Var(K), the subdirectly irreducible members of the variety generated by K, is the class of all lattices of length two (with the obvious adjustments: including C 2 and excluding C 3 and C2). 2 There are many deeper results of this type. See the papers K. A. Baker [2] and [3], R. Freese [5] and [6], W. Poguntke and B. Sands [14], and the book P. Jipsen and H. Rose [11]. The deepest known example of this type is in R. Freese [6]. Freese takes the class K of modular lattices of width four and he describes the class Si Var(K), the class of subdirectly irreducible modular lattices of width at most four. They are all either finite or countably infinite, 2 ℵ0 in number, and half of them are of order-dimension 2 (the finite ones, planar). So Freese provides 2 ℵ0 subvarieties of Var(PM). For a new proof of this classical result, see G. Grätzer and H. Lakser [10]. Notation and terminology. For basic concepts and terminology, see [9]. The Glossary of Notation of [9] is available as a pdf file at http://mirror.ctan.org/info/examples/math_into_latex-4/notation.pdf In addition, M 3 + is the modular lattice generated by adding an element into a side of an M 3, see Figure 1. (This lattice probably first appeared in B. Jónsson [12]; it was generalized from adding one element to adding any chain in E. T. Schmidt [15].) The ordered set of prime ideals of L is P(L). Let C n = {0, 1,..., n 1} be the n-element chain, C ω = {0, 1,... } the chain isomorphic to ω, and C = {..., 1, 0, 1,... } the chain isomorphic to the integers. Figure 1. The lattice M + 3 By definition, a planar lattice is finite. The corresponding concept for infinite lattices is order-dimension at most 2. So a finite lattice of order-dimension at most 2 is planar. Acknowledgement. We would like to thank D. Kelly for his early input. H. Lakser read the manuscript and suggested improvements. K. Baker advised us while doing this research and offered advice on revising the paper. We also received useful

ON THE VARIETY GENERATED BY PLANAR MODULAR LATTICES 3 observations from R. Freese and J. B. Nation. Finally, special thanks to the two referees for unusually insightful reports and constructive criticisms. 2. Baker s Dimension Theorem The following result appeared in K. A. Baker [1] for finite lattices. Theorem 2.1 (Baker s Dimension Theorem). Let L be a distributive lattice. Then L is of order-dimension n iff n is the least integer such that L can be embedded as a sublattice into the n-th power of a chain. I. Rival rephrased this result: If a distributive lattice L has an order embedding into the n-th power of a chain, then it has a lattice embedding into the n-th power of some chain. Baker s proof for the finite case is unpublished. As K. Baker pointed out, there is an easy way of passing from the finite to the infinite case. An infinite structure is embeddable in the ultraproduct of its finitely generated substructures, which are finite for a distributive lattice L. Thus if each finite sublattice of L is embeddable in a product of n chains, L itself is lattice-embeddable in the ultraproduct of these products of chains, which is isomorphic to the product of n ultraproducts of chains but ultraproducts of chains are chains. In this section, we are offering a proof of Baker s result not utilizing the unpublished finite case. We first need a result of R. P. Dilworth [4]: Lemma 2.2. Let L be a distributive lattice. Then L can be embedded as a sublattice into the n-th power of a chain iff P(L) is of width n. Proof of Theorem 2.1. The other direction being obvious, let L be of order-dimension n and let us assume that P(L) is of width n + 1, that is, there are prime ideals P 1,..., P n+1 of L that are pairwise incomparable. Then we can pick an element a i,j in P i but not in P j, for j i. Set a i = ( a i,j j i ). Clearly, a i is in P i but not in any of the other prime ideals. Define b i = ( a j j i ); b i is not in P i but is in P j, for j i. The elements b 1,..., b n+1 are distinct and the pairwise meet of any two distinct b i -s is the same: ( aj 1 j n + 1 ). Since L is distributive, b 1,..., b n+1 generate B n+1 as a sublattice of L, contradicting that L is of order-dimension n. 3. Planar subdirectly irreducible modular lattices Let L be a finite modular lattice. It is well-known that L is subdirectly irreducible iff it is simple. Now let L be planar (we fix the planar diagram P of L); then every sublattice isomorphic to M 3 is a cover-preserving sublattice; indeed, otherwise L would contain M + 3 as a sublattice and M + 3 is not planar. So if we remove the interior elements of the covering M n -s, we get a distributive lattice, Frame P L. More formally:

4 G. GRÄTZER AND R. W. QUACKENBUSH Definition 3.1. Let L be a planar modular lattice with a planar diagram P, and let M = [b, t] be an interval of L isomorphic to M n (we call these a convex M n ), with exterior atoms {a 1, a n } and interior atoms {a 2,..., a n 1 } (that is, for 2 i n 1, a i is between a 1 and a n in the planar diagram P ). Notice that the interior atoms must all be doubly irreducible and that a doubly irreducible element of L appears in at most one convex M n of L. We say a L is an internal element (with respect to P ) if a is an internal atom of some convex M n of L; otherwise, we say that a is an external element. Let Frame P L be the sublattice of L consisting of all external elements of L; thus Frame P L contains all reducible elements of L. Since Frame P L is obtained from by omitting a set of doubly irreducible elements (the interior elements), it is indeed a sublattice; since it contains no M 3, it is distributive. The distributive lattice Frame P L as a sublattice of L depends on the planar diagram P we fixed. If we utilize a different planar diagram Q, the resulting frame Frame Q L may be a different sublattice of L. (For instance, if we start with L = M 3 = {0, a, b, c, 1}, then depending on which planar representation we chose, we get {0, a, b, 1}, {0, a, c, 1}, {0, b, c, 1} as frames.) However, we shall prove in Theorem 3.4 that Frame P L and Frame Q L are isomorphic lattices. Definition 3.2. A narrows of L is an element a L {0, 1} (that is, a is not the zero or unit of L) such that L = a a (that is, for every b L, b a or b a). Thus, a L {0, 1} is a narrows iff x/con(a, 0) = {x}, for every x L a (since we do not insist that L have a zero, con(a, 0) is the congruence generated by collapsing all of a to a). Lemma 3.3. Let L be a planar modular lattice and let M = [b, t] be a (convex) M n of L such that each atom of M is doubly irreducible in L. Then both b and t are a narrows of L. Proof. By duality, it is enough to prove that t is a narrows. Let a be an atom of M. If t were not a narrows, then there would be u L with u incomparable to t. We cannot have u a < t; if u > a, then a would be meet-reducible. Hence, u is incomparable to a. But then u a > a so that a = (u a) t, again making a meet-reducible. This proves the lemma. Theorem 3.4. Let L and K be isomorphic planar modular lattices, with planar diagrams P and Q, respectively. Then Frame P L and Frame Q K are also isomorphic. Proof. Let ϕ: L K be an isomorphism; then ϕ induces a planar diagram Q on L from Q on K such that Frame Q L is isomorphic to Frame Q K. Thus, we may assume that L = K. We define ψ : Frame P L Frame Q L. If a Frame P L Frame Q L, then define ψ(a) = a. Now let a Frame P L Frame Q L. Thus, a is external with respect to P and internal with respect to Q and so is an atom in a unique convex M n ; denote it M. Let the other P -external atom of M be b. If b is also Q-external, then we have already defined ψ(b) = b; in this case, we must define ψ(a) to be the other Q-external atom of M. Now let b be Q-internal. Then there

ON THE VARIETY GENERATED BY PLANAR MODULAR LATTICES 5 are two ways of mapping the P -external atoms of M onto the Q-external atoms of M; choose one. Thus, ψ is a bijection, and using Lemma 3.3, it is readily seen that ψ is a lattice isomorphism. So with a planar modular lattice L, we associate a planar distributive lattice Frame L, which is unique up to isomorphism. With the converse process, starting with a finite planar distributive lattice D and adding interior elements to covering squares, forming M n -s, we construct all planar modular lattices L. How many covering M n -s, n 3, do we have to form to make the lattice simple? Let L be a planar modular lattice. Then L is simple iff any two prime intervals are projective (modularity implies that one way perspectivity works both ways). We can visualize this using the following terminology. Definition 3.5. Let p and q be prime intervals in a modular lattice L. A modular zig-zag connecting p and q is a sequence S 1,..., S m, m 1, of covering M 3 -s such that p is perspective to a prime interval in S 1, q is perspective to a prime interval in S m, and any two adjacent members of this series of covering M 3 -s are attached, that is, they form a sublattice isomorphic to one of the two lattices in Figure 2. If p q, then we regard the empty sequence as connecting p and q. By G. Grätzer [7], we have the following result. Lemma 3.6. Let p and q be prime intervals in a modular lattice L. Then p and q are projective iff they are connected by a modular zig-zag. An example of a simple planar modular lattice is shown in Figure 3. In this lattice, the prime intervals p and q are projective. The modular zig-zag connecting p and q is of length 3. The elements of the three covering M 3 -s are black-filled. Figure 2. Attached covering M 3 -s In actual practice, it may be enough to check modular zig-zag connectivity only for some special pairs of prime intervals in L. In the planar case, every prime interval is projective to a prime interval on the left boundary; it follows that it is enough to check this condition for p and q on the left boundary. Moreover, if p and q are connected and q and r are connected, then obviously p and r are connected, further reducing the computations.

6 G. GRÄTZER AND R. W. QUACKENBUSH p q Figure 3. A subdirectly irreducible planar modular lattice L Which finite distributive lattices D occur as Frame L for a finite subdirectly irreducible planar modular lattice L? Definition 3.7. The lattice L is narrows free, if it has no narrows. Equivalently, L is narrows free iff every element of L {0, 1} is in a two-element antichain. Now we characterize the frames of subdirectly irreducible planar modular lattices. Theorem 3.8. Let D be a planar distributive lattice with more than two elements. Then D is isomorphic to Frame L, for a subdirectly irreducible planar modular lattice L, iff D is narrows free. Proof. If the element a is a narrows of D, then by Definition 3.2 and its dual, con(0, a) con(a, 1) = 0, in any planar modular lattice L with Frame L = D, contradicting that L is subdirectly irreducible. Conversely, if D is narrows free, then fill each covering square to form an M 3 to construct a modular lattice L with Frame L = D. The lattice D cannot have a unique atom a, because then a would be a narrows. So D has more than one atom. Since D cannot have three atoms, because they would generate C2, 3 which is not planar, we conclude that D has exactly two atoms. Therefore, L has a unique sublattice M 3 containing 0. Let α be the congruence of L collapsing the unique M 3 containing 0. Let u be the largest element in L with u 0 (α). If u = 1, then α = 1 in L and L is simple. Let us assume that u < 1. By assumption, u is not a narrows. By Definition 3.7, D has an element v with u v. It follows easily that there is a covering square in D with u as an atom. Let w be the unit element of this covering square. Then w 0 (α), contradicting the maximality of u.

ON THE VARIETY GENERATED BY PLANAR MODULAR LATTICES 7 4. Weakly atomic distributive lattices of order-dimension 2 By Baker s Dimension Theorem, every subdirectly irreducible modular lattice of order-dimension 2 is a subdirect product of two chains. In this section, we prove that the two chains themselves are also weakly atomic. We start with two easy lemmas. Lemma 4.1. Let A and B be lattices and let L be a sublattice of A B. Then L is a subdirect product of A and B iff for every a 1 a 2 in A, there is b 1 b 2 in B such that (a 1, b 1 ), (a 2, b 2 ) L and symmetrically. Proof. Of course, if L has the property stated in the lemma, then L is a subdirect product of A and B. Conversely, assume that L is a subdirect product of A and B and let a 1 a 2 in A. Since L is a subdirect product of A and B, there are elements x 1, x 2 B such that (a 1, x 1 ), (a 2, x 2 ) L. Then (a 2, x 1 x 2 ) = (a 1, x 1 ) (a 2, x 2 ) L, so we may take b 1 = x 1 and b 2 = x 1 x 2. Definition 4.2. We shall use the following terminology. Let the lattice L be a subdirect product of two chains, E 1 and E 2. Let a 1, a 2 E 1 and b 1, b 2 E 2, with a 1 a 2, b 1 b 2. We call the interval I = [(a 1, b 1 ), (a 2, b 2 )] of L parallel with E 1, if b 1 = b 2 and call the interval [a 1, a 2 ] = I E1 the E 1 -projection of I. We define, similarly, parallel with E 2 and the E 2 -projection. Lemma 4.3. Let D be a subdirect product of two chains, E 1 and E 2. Let p be a prime interval in D. Then p E1 is either a singleton or a prime interval and the same for E 2. Proof. Obviously, because the projections of D to E 1 and E 2 are onto homomorphisms. In the finite case, we used the condition that D is narrows free. In the infinite case, we use the following related condition. Definition 4.4. Let L be a lattice and let p = [a, b] be a prime interval of L. We call p a prime-narrows of L, if L = a b, that is, both a and b are narrows. The lattice L is prime-narrows free, if it has no prime-narrows. The following equivalent formulation of this concept will be useful ( denotes perspectivity). Lemma 4.5. Let D be a distributive lattice. Then a prime interval p is a primenarrows in D iff there does not exist a prime interval q in D such that p q and p q. Proof. If the prime interval p = [a, b] of D is not a prime-narrows, then D a b. Let x D ( a b). Since D is distributive, either x a x b or x a x b. By duality, we can assume that x a x b. Define q = [x a, x b]. Then q is prime, the relation p q holds (in fact, p q), and p q because x / a b. Conversely, if the prime interval p = [a, b] of D is a prime-narrows,

8 G. GRÄTZER AND R. W. QUACKENBUSH then D = a b, so a prime interval q in D such that p q must lie in a or in b; therefore, p q must fail. Lemma 4.6. Let D be a distributive lattice with more than 2 elements. Let us further assume that D satisfies the following conditions: (i) D is a subdirect product of two chains, E 1 and E 2 ; (ii) D is prime-narrows free. Let p be a prime interval in D. Then either p is parallel with E 1 and p E1 is a prime interval in E 1 or p is parallel with E 2 and p E2 is a prime interval in E 2, but not both. Proof. By Condition (i), the lattice D is a subdirect product of two chains, E 1 and E 2. Let p = [(a 1, a 2 ), (b 1, b 2 )] be a prime interval in D. It follows immediately from Lemma 4.3 that either a 1 = b 1 or [a 1, b 1 ] is a prime interval in E 1 and either a 2 = b 2 or [a 2, b 2 ] is a prime interval in E 2. Now assume that a 1 < b 1 ; it is enough to show that a 2 = b 2. By Condition (ii), the lattice D is prime-narrows free. Applying Lemma 4.5, we conclude that p is perspective to some other prime interval q = [(c 1, c 2 ), (d 1, d 2 )] in D. By duality, we can assume that the perspectivity is up, so (c 1, c 2 ) > (a 1, a 2 ), (a 1, a 2 ) = (b 1, b 2 ) (c 1, c 2 ) = (b 1 c 1, b 2 c 2 ). Since b 1 c 1 = a 1 < b 1 in the chain E 1, it follows that a 1 = c 1. Hence, c 2 > a 2, so a 2 = b 2 c 2 forces that a 2 = b 2, proving the lemma. We now define a condition that is stronger than prime-narrows free. But first, we rephrase the definition of a modular zig-zag for distributive lattices, Definition 4.7. Let p and q be prime intervals in a planar distributive lattice L. A distributive zig-zag connecting p and q is a sequence T 1,..., T m, m 1, of covering squares such that p is perspective to a prime interval in T 1, q is perspective to a prime interval in T m, and any two adjacent members of this series of covering squares are attached, that is, they form a sublattice isomorphic to C 2 C 3 or C 2 C 4. If p q, then we regard the empty sequence as connecting p and q. Lemma 4.8. Let D be a weakly atomic distributive lattice with more than 2 elements, and let D be a subdirect product of two chains, E 1 and E 2. Let us assume that D satisfies the following condition: (ii ) any two prime intervals of D are distributive zig-zag connected. Then D satisfies Condition (ii) of Lemma 4.6, that is, D is prime-narrows free. Proof. We verify that D is prime-narrows free. Let p = [a, b] be a prime interval of D. By assumption, D has more than 2 elements, so either a is not the zero or b is not the unit of D. Let us assume that a 0. (If a = 0 and b 1, then we proceed similarly.) Since D is weakly atomic, there is a prime interval q in [0, a]. By Condition (ii ), the p and q are distributive zig-zag connected, and therefore, p is perspective to another prime interval, and so it is not a prime-narrows by Lemma 4.5.

ON THE VARIETY GENERATED BY PLANAR MODULAR LATTICES 9 Now we can state and prove the main result of this section: Theorem 4.9. Let D be a weakly atomic distributive lattice with more than 2 elements. Let us further assume that D is a subdirect product of two chains, E 1 and E 2 and that any two prime intervals of D are distributive zig-zag connected. Then the chains E 1 and E 2 are weakly atomic. Proof. We show that E 1 is weakly atomic. (The proof for E 2 is similar.) Choose a 1 < b 1 in E 1. In the interval [a 1, b 1 ], we find a prime interval. By Lemma 4.1, there are a 2, b 2 in E 2 with a 2 b 2 and (a 1, a 2 ), (b 1, b 2 ) D. We consider the interval I = [(a 1, a 2 ), (b 1, b 2 )] of D. Case 1: I is not a chain. Then D has two incomparable elements (c 1, c 2 ) and (d 1, d 2 ) I. We can assume that d 1 < c 1 and c 2 < d 2, by symmetry. So (c 1, c 2 ) (d 1, d 2 ) = (d 1, c 2 ) I and the interval [(d 1, c 2 ), (c 1, c 2 )] of D is parallel with E 1. Since D is weakly atomic, we can pick a prime interval [(d 1, c 2 ), (c 1, c 2 )] in D. By Lemma 4.8, [d 1, c 1] is a prime interval in E 1, the one we require for [a 1, b 1 ]. Case 2: I is a chain. We can make the following assumptions: a 2 < b 2 (otherwise, I is parallel with E 1, and we can argue, as in Case 1, using Lemma 4.3). All prime intervals in I are parallel with E 2 (otherwise, Lemma 4.3 finds us a prime interval in [a 1, b 1 ]). I is infinite (otherwise, by the previous item, I would be parallel with E 2, contradicting a 1 < b 1 ). Using the above assumptions and using twice the weak atomicity of D, we find prime intervals p = [(c 1, c 2 ), (c 1, c 2)] and q = [(d 1, d 2 ), (d 1, d 2)] of I satisfying a 1 < c 1 < d 1 < b 1, a 2 c 2 c 2 d 2 d 2 b 2. By assumption, p and q are distributive zig-zag connected. Since p and q are in a chain, I, they are not perspective. So p is perspective to a covering square S = {e 1, e 1} {c 2, c 2}, where e 1 e 1 in E 1. If S I, then [e 1, e 1] is a prime interval in [a 1, b 1 ], and we are done. If S I, then e 1 a 1 or e 1 b 1. Let e 1 a 1. Then (a 1, a 2 ) (e 1, c 2) = (a 1, c 2) D. Since (a 1, a 2 ) (a 1, c 2) < (c 1, c 2) (a 2, b 2 ) in I, we find a proper interval J = [(a 1, c 2), (c 1, c 2)] I parallel with E 1, yielding by Lemma 4.3 a prime interval in [a 1, b 1 ]. A similar computation for e 1 b 1 also yields a prime interval in [a 1, b 1 ]. This completes the proof of the theorem. 5. Subdirectly irreducible modular lattices of order-dimension 2 Let L be a subdirectly irreducible modular lattice of order-dimension 2. Then L is not distributive, so it contains a sublattice M = {o, a, b, c, i} isomorphic to M 3. As in the finite case, the edges of M are prime intervals. It follows that

10 G. GRÄTZER AND R. W. QUACKENBUSH Lemma 5.1. Let L be a subdirectly irreducible modular lattice of order-dimension 2. Then L is weakly atomic. Proof. Let α be the base congruence of L. Since L is not distributive, it contains a sublattice M = {o, a, b, c, i} = M 3. As discussed above, p = [o, a] is prime, so L contains a prime interval. By the modularity of L, the congruence con(p) is an atom of Con L, so α = con(p). Now let x < y L. Since L is subdirectly irreducible, we get the inequality α con(x, y). By modularity, there is a prime interval q in [x, y] with p projective to q. So L is weakly atomic. We define D = Frame L similar to the definition in the finite case: the convex hull of an M 3 is an M α, in which at most two atoms are not doubly irreducible. Thus we may remove enough doubly irreducible elements to get a covering square. D is what is left after all the removals are simultaneously carried out. This lattice D is unique only up to isomorphism. Obviously, D is a distributive lattice of orderdimension 2 and it is also weakly atomic. So by Baker s Dimension Theorem, D is a subdirect product of two chains, E 1 and E 2. Now we can characterize the distributive frame of a subdirectly irreducible modular lattice of order-dimension 2. Theorem 5.2. A distributive lattice D of order-dimension 2 is the frame, Frame L, of a subdirectly irreducible modular lattice L of order-dimension 2 iff it is weakly atomic and any two prime intervals of D are distributive zig-zag connected. Proof. Let us assume that D = Frame L, where L is a subdirectly irreducible modular lattice of order-dimension 2. By Lemma 5.1, L is weakly atomic hence D is weakly atomic and con(p) is the base congruence, for any prime interval p in L. We can clearly assume that p is a prime interval in D. Now Proposition 2 in G. Grätzer [7] (see, especially, Figure 5) provides that any two prime intervals of D are distributive zig-zag connected. Conversely, let D be a weakly atomic distributive lattice of order-dimension 2 with the property that any two prime intervals of D are distributive zig-zag connected. Form the modular lattice L by adding an interior element to all covering squares in D. Clearly, L is a modular lattice of order-dimension 2. Given two prime intervals, p and q of D, it follows that con(p) = con(q) in L since they are distributive zig-zag connected. By the construction of L utilizing that D is weakly atomic the same holds for any two prime intervals of L, hence, L is subdirectly irreducible. Now we combine Baker s Dimension Theorem, Theorem 4.9, and Theorem 5.2, to obtain the Structure Theorem of Subdirectly Irreducible Modular Lattices of Order- Dimension 2. Theorem 5.3. We can construct all subdirectly irreducible modular lattices L of order-dimension 2 as follows: (i) Take two weakly atomic chains E 1 and E 2. (ii) Let D be a weakly atomic subdirect product of E 1 and E 2 satisfying the condition that any two prime intervals of D are distributive zig-zag connected.

ON THE VARIETY GENERATED BY PLANAR MODULAR LATTICES 11 (iii) Form the lattice L by adding elements to D to turn enough covering squares into covering M α -s (α is a cardinal, α > 2, and α varies with the covering square), so that in L any two prime intervals of D that are distributive zig-zag connected become projective. Of course, the lattices we construct are arbitrarily large. Using the terminology of W. Taylor [12], Var(PM) is residually large. Indeed, already M α Var(PM) is arbitrarily large, if the cardinal α is large. But more is true. Theorem 5.4. There are arbitrarily large subdirectly irreducible modular lattices of order-dimension 2 not containing M 4 as a sublattice. Proof. Indeed, let E 1 and E 2 be well-ordered chains. Form D = E 1 E 2 and add an interior element to all covering squares in D, to form the modular lattice L. Then L is a subdirectly irreducible modular lattice of order-dimension 2. Note that this lattice L is simple iff both chains are isomorphic to subintervals of C. Corollary. There are countably infinitely many frames of simple modular lattices of order-dimension 2. 6. Direct limits of finite subdirectly irreducible lattices We start with this observation from the folklore: Lemma 6.1. The variety Var(PM) is locally finite. Proof. Let L be a planar modular lattice L with frame F. For x F, define x = x = x, and let x = x, x = x, for x L F. Let X L with X = n and define X = { x x X }, X = { x x X }. The sublattice D generated in the distributive lattice F by X X has at most F D (2n) elements, where n = X. The sublattice in L generated by X is contained in D X, so it has at most F D (2n) +n elements. This is a first order statement, so by Jónsson s Lemma, it holds in Var(PM). It follows from Lemma 6.1 that every lattice in Var(PM) is a direct limit of its finite sublattices. Is a subdirectly irreducible lattice in Var(PM) always a direct limit of its finite subdirectly irreducible sublattices? The next result shows that this is not the case. Theorem 6.2. In the variety, Var(PM), generated by planar modular lattices, there exists a subdirectly irreducible lattice S that is not a direct limit of its finite subdirectly irreducible sublattices. Proof. To define the lattice S, we start with its frame D. We define D as the sublattice { (i, j) i j + 1 } of Cω. 2 We add an interior element to the covering square S(i, j) with top element (i, j), for (i, j) ( {(n, n), (1, 2n + 2), (n + 1, 2n + 2)} n 1 ).

12 G. GRÄTZER AND R. W. QUACKENBUSH Let S denote the resulting lattice, illustrated in Figure 4. Clearly, S Var(PM). We define the sublattice M 3 (i, j) as the the sublattice S(i, j) with the additional element inside, for the (i, j) in the displayed formula above. The spine of S (black-filled in Figure 4) is the sublattice of S consisting of the union of these M 3 -s: ( M3 (i, i) i 1 ). We divide S into files and ranks. For n > 0, For n > 1, File(n) = { s S (n 1, n 1) s (n, m), for some m n }. Rank(n) = { s S (n, 0) s (n, n) }. Both File(n) and Rank(n) are convex sublattices of S. We marked File(1) and Rank(4) in Figure 4. Every file is infinite and every rank is finite; each contains a unique M 3 from the spine. File(1) contains infinitely many copies of M 3 : M 3 (1, 1), M 3 (1, 4),..., M 3 (1, 2k),..., for k 2. Rank(4) File(1) Figure 4. The lattices S for Theorem 6.2

ON THE VARIETY GENERATED BY PLANAR MODULAR LATTICES 13 For n > 1, File(n) contains just two copies of M 3 : M 3 (n, n) and M 3 (n, 2n). For n = 2 or n odd, Rank(n) contains exactly one copy of M 3 : M 3 (n, n). For n 2, Rank(2n) contains just three copies of M 3 : M 3 (2n, 2n), M 3 (n, 2n), M 3 (1, 2n). Claim 1. S is subdirectly irreducible, in fact, simple. Let p be a prime interval of S. We prove that con(p) collapses all of S. Every prime interval of S is perspective to some prime interval in some M 3 (n, n) in the spine in the same rank or file, causing it to collapse. In turn, this causes the collapse of M 3 (n, 2n), then of M 3 (1, 2), and finally of M 3 (1, 1). Applying this to any two prime intervals, we conclude using the modularity of S that they are projective. Since any interval in S is finite, the last statement shows that S is simple, proving the claim. Claim 2. S is not the direct limit of its finite subdirectly irreducible that is, simple sublattices. Assume, to the contrary, that S is the direct limit of its simple sublattices. The interval A = [(0, 0), (2, 2)] is not simple, so there must be a finite simple sublattice B of S containing A. We verify that B is a cover-preserving sublattice of S. Let p be a prime interval in B. Since B is simple, con(p) collapses all of B, so p is perspective to an edge of an M 3 in B. But an M 3 in B is also an M 3 in S, so M 3 is a cover-preserving sublattice of S, and therefore, of B. Since p is perspective to an edge of this M 3, it follows that p is prime in S, proving B is a cover-preserving sublattice of S. Let (m, n) be the top element of B. Since B is simple, it must contain the square S(m, n). If n 2, then B cannot be simple, so n 3. We distinguish two cases. Case 1: n 3 is odd. In the interval [(0, 0), (n, n)] of S, collapsing M 3 (n, n) and each interval [(i 1, n), (i, n)] is a congruence relation, α, since Rank(n) contains only one M 3, which is already collapsed by α and since File(n) [(0, 0), (n, n)] = M 3 (n, n). Since S(m, n) B, this congruence α restricted to B is nontrivial, hence B cannot be simple. This contradiction shows that this case cannot occur. Case 2: n 3 is even. Clearly, n 1 3. We verified that B is a covering sublattice of S, so it must contain a prime interval of the form [(m, n 2), (m, n 1)], for some m m. We repeat the argument of Case 1 with the interval [(0, 0), (n, n)] of S, collapsing M 3 (n 1, n 1), the interval [(n 2, n), (n 1, n)], and each interval [(i, n 2), (i, n 1)], to obtain a congruence relation, β, of [(0, 0), (n, n)]. This congruence β restricted to B is nontrivial because it collapses [(m, n 2), (m, n 1)], so again B is not simple. This verifies Claim 2 and the theorem. References [1] K. A. Baker, Honors Thesis, Harvard College, 1961, unpublished. [2], Primitive satisfaction and equational problems for lattices and other algebras, Trans. Amer. Math. Soc. 190 (1974), 125 150. [3], Some non-finitely-based varieties of lattices, Universal algebra (Esztergom, 1977), pp. 53 59, Colloq. Math. Soc. Jnos Bolyai, 29, North-Holland, Amsterdam-New York, 1982. [4] R. P. Dilworth, A decomposition theorem for partially ordered sets, Ann. of Math. (2) 51 (1950), 161 166. [5] R. Freese, Breadth two modular lattices, Proceedings of the University of Houston Lattice Theory Conference (Houston, Tex., 1973), pp. 409 451.

14 G. GRÄTZER AND R. W. QUACKENBUSH [6], The structure of modular lattices of width four with applications to varieties of lattices, Memoirs of the AMS 181, 1977. vii+91 pp. ISBN: 0-8218-2181-4. [7] G. Grätzer, Equational classes of lattices, Duke Math. J. 33 (1966), 613 622. [8], General Lattice Theory, second edition. New appendices by the author with B. A. Davey, R. Freese, B. Ganter, M. Greferath, P. Jipsen, H. A. Priestley, H. Rose, E. T. Schmidt, S. E. Schmidt, F. Wehrung, and R. Wille. Birkhäuser Verlag, Basel, 1998. xx+663 pp. ISBN: 0-12-295750-4; 3-7643-5239-6. Softcover edition, Birkhäuser Verlag, Basel Boston Berlin, 2003. ISBN: 3-7643-6996-5. Reprinted March, 2007. [9], The Congruences of a Finite Lattice, A Proof-by-Picture Approach, Birkhäuser Boston, 2005. xxiii+281 pp. ISBN: 0-8176-3224-7. [10] G. Grätzer and H. Lakser, Subdirectly irreducible modular lattices of width at most 4, Acta Sci. Math. (Szeged) 73 (2007), 3 30. [11] P. Jipsen and H. Rose, Varieties of lattices. Lecture Notes in Mathematics, 1533. Springer-Verlag, Berlin, 1992. x+162 pp. ISBN: 3-540-56314-8. [12] B. Jónsson, Algebras whose congruence lattices are distributive, Math. Scand. 21 (1967), 110 121. [13], Equational classes of lattices, Math. Scand. 22 (1968), 187 196. [14] W. Poguntke and B. Sands, On finitely generated lattices of finite width, Canad. J. Math. 33 (1981), 28 48. [15] E. T. Schmidt, Every finite distributive lattice is the congruence lattice of some modular lattice, Algebra Universalis 4 (1974), 49 57. [16] W. Taylor, Residually small varieties, Algebra Universalis 2 (1972), 33 53. Department of Mathematics, University of Manitoba, Winnipeg, MB R3T 2N2, Canada E-mail address, G. Grätzer: gratzer@ms.umanitoba.ca URL, G. Grätzer: http://server.math.umanitoba.ca/homepages/gratzer/ E-mail address, R. W. Quackenbush: qbush@cc.umanitoba.ca