positive primitive formulas preventing enough algebraic operations

Algebra univers. 52 (2004) 303–3120002-5240/04/030303 – 10DOI 10.1007/s00012-004-1896-7c©Birkhauser Verlag, Basel, 2004 Algebra Universalis

Positive primitive formulas preventing enough algebraicoperations

Jennifer Hyndman

In Celebration of the Sixtieth Birthday of Ralph N. McKenzie

Abstract. A finite unary algebra with a positive primitive formula defining a pp-acyclicrelation does not have enough algebraic operations and does not have a finite basis for itsquasi-equations.

1. Introduction

In [1] Bestsennyi shows that a 3-element unary algebra of finite type does not havea finite basis for its quasi-equations if and only if it has one of three bad algebras asa term reduct. In [4] Hyndman and Pitkethly provide equivalent conditions for a3-element unary algebra to have enough algebraic operations. One of the conditionswas the absence of Bestsennyi’s bad algebras as a term reduct. In this paper weextend these results by showing that a finite unary algebra with a positive primitiveformula defining a pp-acyclic relation does not have enough algebraic operationsand does not have a finite basis for its quasi-equations.

2. Not enough algebraic operations

Fix a finite algebra M. An algebraic operation of M is a homomorphismh : B → M where B ≤ Mn for some finite n. For Y ⊆ Hom(Mn,M) the ho-momorphism �Y : Mn → MY is defined for a ∈ Mn by �Y (a) = 〈y(a) | y ∈ Y 〉.Given two homomorphisms h : B → M and ψ : B → A we say that h lifts to Athrough ψ if there exists a homomorphism h∗ : A → M such that h = h∗ ◦ ψ.

In [6] Lampe, McNulty and Willard introduced the concept of enough algebraicoperations as a tool for proving a dualizable algebra is strongly dualizable. Thealgebra M has enough algebraic operations if there is a map f : ω → ω for which

Presented by R. McKenzie.Received June 17, 2003; accepted in final form August 12, 2004.2000 Mathematics Subject Classification: 08A60, 08C16.Key words and phrases: Enough algebraic operations, finite basis, quasi-equations.

303

304 J. Hyndman Algebra univers.

the following condition holds:

for all n ∈ ω \ {0}, all algebras B ≤ A ≤ Mn and all homomorphismsh : A → M, there exists a set Y ⊆ Hom(Mn,M), with |Y | ≤ f(|B|), suchthat h�B lifts to �Y (A) through �Y �B : B → �Y (A).

This definition is illustrated in Figure 1. The idea is that h�B can be lifted to�Y (A) which is a small homomorphic image of A obtained via total algebraicoperations. The map f′ : ω → ω defined by f′(n) = max{f(0), . . . , f(n)} will satisfythe conditions for enough algebraic operations when f does, as |Y | ≤ f(|B|) implies|Y | ≤ f′(|B|). Thus in the definition of enough algebraic operations we may assumef is increasing.

M

≤B

�Y �B

h�B

�Y (A)

�Y �A

A≤

Mn

Figure 1. Enough algebraic operations.

A binary relation � is acyclic if for any integer k ≥ 2 the sequence

c1 � c2 � · · · � ck � c1

implies c1 = ci for i ≤ k. A positive primitive formula is an existentially quantifiedconjunction of atomic formulas. In the case of unary algebras the atomic formulasare of the form f(x) = g(y) for variables x and y (not necessarily different) andterms (possibly the identity) f and g. A unary algebra M is pp-acyclic if thereis a positive primitive formula φ(x, y) defining an acyclic binary relation such thatthere exists 0 and 1 in M with 0 = 1, φ(0, 0), φ(0, 1) and φ(1, 1). The relationdefined by φ will be called a pp-acyclic relation.

Theorem 2.1. Suppose M is a finite unary algebra that is pp-acyclic then M doesnot have enough algebraic operations.

The remainder of this section is the proof of this theorem. For contradictionassume f : ω → ω is an increasing function that satisfies the conditions for M tohave enough algebraic operations.

Assume M = 〈M ;F〉 where F is a monoid of unary operations on M . We mayassume that the pp-acyclic relation φ is given by φ(x, y) = ∃z1∃z2 · · · ∃zk α(x, y, z)

Vol. 52, 2004 Not enough algebraic operations 305

where α(x1, x2, . . . , xk+2) is a conjunction of expressions of the form t1(u1) = t2(u2)with t1, t2 ∈ F and u1, u2 ∈ {x1, x2, . . . , xk+2}. We write a � b in B if a, b ∈ Bsatisfy φ(a, b). Choose 0, 1 ∈ M such that 0 � 0, 0 � 1 and 1 � 1 in M. Pickwitnesses c1, . . . , ck, d1, . . . , dk and e1, . . . , ek in M such that α(0, 0, c)∧α(0, 1, d)∧α(1, 1, e) holds in M.

Fix n odd greater than (|M | − 1)f(|M |3) + 5. For 1 ≤ i ≤ n− 1, define si ∈Mn

by si(u) = 0 for 1 ≤ u ≤ i and si(u) = 1 for i + 1 ≤ u ≤ n. For i ≥ j we havesi � sj in Mn. Define in Mn the elements f j

1 , . . . , fjk that witness sn−1 � sj for

2 ≤ j ≤ n−3 even and the elements gq1, . . . , g

qk that witness sq � s1 for 3 ≤ q ≤ n−2

odd as follows:

f jr (i) =

⎧⎨⎩

cr if 1 ≤ i ≤ j

dr if j + 1 ≤ i ≤ n− 1er if i = n

and

gqr(i) =

⎧⎨⎩

cr if i = 1dr if 2 ≤ i ≤ q

er if q + 1 ≤ i ≤ n.

Fix the following subalgebras of Mn:

A0 = SgMn

({sj | j even } ∪ {f jr | 2 ≤ j ≤ n− 3 even and 1 ≤ r ≤ k}),

A1 = SgMn

({sq | q odd } ∪ {gqr | 3 ≤ q ≤ n− 2 odd and 1 ≤ r ≤ k}),

A = SgMn

(A0 ∪A1), and

B = SgMn

({s1, sn−1}).

The next two lemmas describe the algebra A and the homomorphism h : A → Mthat is used to show that M does not have enough algebraic operations. For m ∈ Mthe tuple whose coordinates are all m is denoted m.

Lemma 2.2. A0 ∩ A1 ⊆ {m | m ∈ M}.

Proof. Let t1 and t2 be in F . Assume t1(f jr1

) = t2(gqr2

) with 2 ≤ j ≤ n − 3 evenand 3 ≤ q ≤ n− 2 odd. If q < j then by considering coordinates 1, 2 and j we have

t1(cr1) = t2(cr2)

t1(cr1) = t2(dr2)

t1(cr1) = t2(er2).


If q > j then consider coordinates 1, 2, j + 1 and q + 1 to obtain

t1(cr1) = t2(cr2)

t1(cr1) = t2(dr2)

t1(dr1) = t2(dr2)

t1(dr1) = t2(er2).

In either case t2(gqr2

) is constant valued. For j even and q odd, the cases t1(sj) =t2(gq

r2), t1(sj) = t2(sq), and t1(f j

r1) = t2(sq) are simpler. �

Lemma 2.3. The following are true:

(1) A = A0 ∪A1;(2) sn−1 � sj in A for j even;(3) sq � s1 in A for q odd;(4) the map h = πn�A0

∪ π1�A1is a well-defined homomorphism from A to M;

(5) h(sn−1) = 1 and h(s1) = 0.

Proof. Item (1) is true because M is a unary algebra. The witnesses f jr and gq

r

were chosen so that items (2) and (3) would hold.To show that item (4) holds it is sufficient to note that, by Lemma 2.2, A0 ∩A1

is a subset of C = {m | m ∈ M}, the constant valued elements. As πn�C = π1�C,it follows that h is well defined. �

The next lemma shows that total algebraic operations behave nicely, with respectto �, on the set {s1, s2, . . . , sn−1}.Lemma 2.4. Let r = |M | and let y : Mn→M be a homomorphism. The kernel ofy partitions {s1, s2, . . . , sn−1} into at most r intervals.

Moreover, if Y is a set of homomorphisms from Mn to M then the kernel of �Ypartitions {s1, s2, . . . , sn−1} into at most |Y |(r − 1) + 1 intervals.

Proof. By construction sn−1 � sn−2 � · · · � s2 � s1 in Mn. That is, φ(si, si−1)holds in Mn for 2 ≤ i ≤ n − 1. Assume y(si) = y(sj) with i ≥ j. Since φ is apositive primitive formula and homomorphisms from Mn to M preserve positiveprimitive formulas,

y(si) � y(si−1) � · · · � y(sj+1) � y(sj) = y(si) in M.

The relation � is acyclic on M so y(si) = y(st) for j ≤ t ≤ i. Thus each elementm of M determines at most one interval of {s1, s2, . . . , sn−1}.

Assume Y is a set of k homomorphisms. For y in Y there are at most r − 1values i ≤ n − 2 with y(si+1) = y(si). Call such an i a breakpoint of y. As�Y (si+1) = �Y (si) exactly when y(si+1) = y(si) for all y in Y , there are at most


k(r − 1) breakpoints of �Y . Thus �Y partitions {s1, s2, . . . , sn−1} into at mostk(r − 1) + 1 intervals. �

Recall B = SgMn

({s1, sn−1}) which is isomorphic to SgM3({(0, 1, 1), (0, 0, 1)}).

Thus |B| ≤ |M |3. We have h�B : B→M. Let Y ⊆ Hom(Mn,M) be such that h�B

lifts to �Y (A) through �Y �B : B → �Y (A). That is, there exists h∗ : �Y (A)→Msuch that h�B = h∗ ◦ �Y �B. By Lemma 2.3, h∗(�Y (sn−1)) = h(sn−1) = 1 andh∗(�Y (s1)) = h(s1) = 0. In order for M to have enough algebraic operations theremust be such a Y with |Y | ≤ f(|B|). If |Y | ≤ f(|B|) then |Y | ≤ f(|M |3), as f

is increasing. The integer n was chosen so that (|M | − 1)f(|M |3) + 5 < n, whichimplies (|M | − 1)|Y | + 1 < n − 4. By Lemma 2.4, the kernel of �Y partitions{s1, s2, . . . , sn−1} into at most |Y |(|M| − 1) + 1 intervals. Thus for some i with2 ≤ i ≤ n − 3, we have �Y (si) = �Y (si+1). Let {j, k} = {i, i + 1} with j

even and k odd. Thus sn−1 � sj and sk � s1 in A. That is, the witnesses toφ(sn−1, sj) and φ(sk, s1) are in A. For each y ∈ Y it follows that y(sn−1) � y(sj)and y(sk) � y(s1) in y(A). Hence �Y (sn−1) � �Y (sj) and �Y (sk) � �Y (s1) in�Y (A). Finally, applying h∗ gives

1 = h∗(�Y (sn−1)) � h∗(�Y (sj)) = h∗(�Y (sk)) � h∗(�Y (s1)) = 0 in M.

Since φ defines an acyclic relation and 0 � 1, this is a contradiction. Hence M doesnot have enough algebraic operations and the proof of Theorem 2.1 is finished.

3. Examples

V = 〈{0, 1, 2}; 010, 011〉

0 1

2

L = 〈{0, 1, 2}; 010, 110〉

0 2

1

D = 〈{0, 1, 2}; 121, 200〉

2 1

0

Figure 2

The algebras V,L,D shown in Figure 2 are pp-acyclic. The orders are definedby:

∃z x = 010(z) ∧ y = 011(z) for V, and

∃z x = 010(z) ∧ y = 110(z) for L.


For D, the formula is more complicated. Let g denote 121 and let f denote 200.Define ρ(x, y, z1, z2, z3) as

f2(x) = f(z2) ∧ g2(x) = g(z2) = g(z3)

∧ g2(y) = g(z1) ∧ f2(y) = f(z1) = f(z3).

Then η(x, y) = ∃z1∃z2∃z3 ρ(x, y, z1, z2, z3) defines the relation {(0, 0), (0, 1), (1, 1)}on D as D satisfies exactly

ρ(0, 0, 1, 1, 1),

ρ(0, 1, 0, 1, 2), and

ρ(1, 1, 0, 0, 0).

A direct application of Theorem 2.1 gives the following corollary.

Corollary 3.1. If V, L or D is a term reduct of a 3-element unary algebra M,then M does not have enough algebraic operations.

4. No finite basis

Theorem 4.1. If M is a pp-acyclic unary algebra then M does not have a finitebasis for its quasi-equations.

Proof. Fix M = 〈M ;F〉 with F a monoid. Let φ define the pp-acyclic relation anduse x � y to denote φ(x, y). Assume that

φ(x, y) = ∃z1, . . . , zk α(x, y, z1, . . . , zk)

where α is quantifier free. Let 0 and 1 satisfy 0 � 0, 0 � 1 and 1 � 1 in M. Thereare witnesses c1, c2, . . . , ck, d1, d2, . . . , dk and e1, e2, . . . , ek such that the followinghold in M

α(0, 0, c1, c2, . . . , ck)

α(0, 1, d1, d2, . . . , dk)

α(1, 1, e1, e2, . . . , ek).

Fix a positive integer λ and let n = 6λ + 11. We will define an algebra A ≤ Mn,a congruence θ of A and quasi-equations Kδ (δ ≥ 9) such that A/θ satisfies allquasi-equations in fewer than λ variables but A/θ does not satisfy Kn−2.

In Mn, for 1 ≤ i ≤ n− 1 define

si(u) ={

0 if 1 ≤ u ≤ i

1 if i+ 1 ≤ u ≤ n.


Witnesses in Mn to si+1 � si for 1 ≤ i ≤ n− 2 are

pir(u) =

⎧⎨⎩

cr if 1 ≤ u ≤ i

dr if u = i+ 1er if i+ 2 ≤ u ≤ n

where 1 ≤ r ≤ k. That is, Mn satisfies α(si+1, si, pi1, . . . , p

ik).

For 1 ≤ i ≤ n− 2 define

Ai = SgMn

({si, si+1, pi1, . . . , p

ik})

and A = SgMn

(⋃n−2

i=1 Ai) =⋃n−2

i=1 Ai. For j ≥ i+ 2 we have Ai ∩ Aj ⊆ {m | m ∈M}. To see this holds, consider t1 and t2 in F and j ≥ i+2 with t1(pi

r1) = t2(pj

r2).

Then by considering coordinates i, i+ 1 and i+ 2

t1(cr1) = t1(pir1

)(i) = t2(pjr2

)(i) = t2(cr2)

t1(dr1) = t1(pir1

)(i+ 1) = t2(pjr2

)(i+ 1) = t2(cr2)

t1(er1) = t1(pir1

)(i+ 2) = t2(pjr2

)(i+ 2) = t2(cr2)

which imply t1(pir1

) = t2(cr2). A similar argument holds for the other possibilities.Let σ1, σ2 : M2 →Mn be defined for u, v ∈M by

σ1(u, v) = 〈uvvv · · · vvv〉σ2(u, v) = 〈uuuu · · ·uuv〉.

Define θ to be the congruence on A generated by

{(σ1(u, v), σ2(u, v)) | u, v ∈ M}.It is easy to check that if w1θw2, then w1 = w2 or {w1, w2} = {σ1(u, v), σ2(u, v)}for some u, v ∈M . In fact, for a ∈ Ai with 2 ≤ i ≤ n−3 the set a/θ is the singleton{a}.

Define Kδ for δ ≥ 9 to be the quasi-equation

( δ∧i=1

α(xi, xi+1, zi1, . . . , z

ik) ∧ xδ+1 = x1

)→ (x1 = x2).

For all δ ≥ 9, if M satisfies∧δ

i=1 α(ai, ai+1, bi1, . . . , b

ik) ∧ aδ+1 = a1 then a1 � a2 �

· · · � aδ � a1 in M. Since � is acyclic we must have a1 = a2, so M satisfies Kδ.We now show that A/θ does not satisfy Kn−2. For 1 ≤ i ≤ n − 2, the algebra

A satisfiesα(si+1, si, pi

1, . . . , pik),

which impliesα(si+1/θ, si/θ, pi

1/θ, . . . , pik/θ)


holds in A/θ. Since sn−1/θ = s1/θ but s1/θ = s2/θ the algebra A/θ does notsatisfy the quasi-equation Kn−2.

As A ≤ Mn it follows that A satisfies every quasi-equation true in M. Wenow define a collection of subalgebras Bk ≤ A/θ such that there exist embeddingshk : Bk → A and if C ≤ A/θ with C generated by any set S of at most λ elements ofA/θ then C ≤ Bk for some k. This will imply that C is isomorphic to a subalgebraof a power of M and C satisfies every quasi-equation of M. Hence A/θ will satisfyevery quasi-equation of M in at most λ variables. It will follow that the λ-variablequasi-equations of M do not form a basis of the quasi-equations of M.

For 4 ≤ k ≤ n− 5 define

Bk = {a/θ ∈ A/θ | a ∈ A1 ∪ · · · ∪ Ak−3 ∪ Ak+3 ∪ · · · ∪ An−2}.The algebra Bk is generated by {a/θ | a ∈ Dr

k ∪Dlk} where

Dlk = {si, pj

r | k + 3 ≤ i ≤ n− 1, k + 3 ≤ j ≤ n− 2, 1 ≤ r ≤ k} and

Drk = {si, pj

r | 1 ≤ i ≤ k − 2, 1 ≤ j ≤ k − 3, 1 ≤ r ≤ k}.Since sk ∈ Ak−1 ∩ Ak, and sk+1 ∈ Ak ∩ Ak+1 and Ai ∩ Aj ⊆ {m | m ∈ M}

for |i− j| ≥ 2, the elements sk/θ = {sk} and sk+1/θ = {sk+1} are not in Bk. LetA∗

i = Ai \ {m | m ∈ M} for 1 ≤ i ≤ n− 2. By the comment above if A∗i ∩ A∗

j = ∅then j ∈ {i− 1, i, i+ 1}. Thus an element a of A3 ∪ · · · ∪ An−4 is in at most twoA∗

i ’s. If a is in two of these sets, they are consecutive. Given a set S of at most λelements of A3 ∪ · · · ∪An−4, the set

I = {3 ≤ i ≤ n− 4: ∃a ∈ S, a ∈ A∗i but a ∈ A∗

i−1}has size at most λ. As n− 6 = 6λ+ 5, there exists 5 ≤ j ≤ n− 4 with

S ∩ (A∗j−2 ∪A∗

j−1 ∪A∗j ∪A∗

j+1 ∪A∗j+2) = ∅.

Thus the subalgebra generated by {a/θ | a ∈ S} is a subalgebra of Bj . That is, anysubalgebra of A/θ generated by at most λ elements is a subalgebra of some Bk.

To see that Bk can be embedded in A, we must first define left and right shiftmaps lk, rk : Mn → Mn by

lk(a)(i) ={a(i+ k) if 1 ≤ i ≤ n− k

a(n) if n− k + 1 ≤ i ≤ n

and

rk(a)(i) ={a(i− k) if k + 1 ≤ i ≤ n

a(1) if 1 ≤ i ≤ k.

Notice that for a ∈ Dlk and b ∈ Dr

k the maps act like inverses in that rk+2(lk+2(a)) =a and ln−k−4(rn−k−4(b)) = b.


The first step in the construction of the embedding of Bk into A is the definitionof the map hk : {a/θ | a ∈ Dl

k ∪Drk} → A given by

hk(a/θ) ={lk+2(a) if a ∈ Dl

k

rn−k−4(a) if a ∈ Drk.

If a/θ = b/θ with a, b ∈ Dlk ∪Dr

k and a = b, then {a, b} = {σ1(u, v), σ2(u, v)} forsome u, v ∈ M where σ1(u, v) ∈ Dr

k and σ2(u, v) ∈ Dlk. Since lk+2(σ2(u, v)) =

rn−k−4(σ1(u, v)), the map hk is well defined.Extend hk to Hk : Bk → A by Hk(t(a/θ)) = t(hk(a/θ)) for a ∈ Dl

k ∪ Drk and

t ∈ F . By construction, Hk will be a homomorphism if it is well defined. Considerterms t1 and t2 in F and elements a and b in Dl

k ∪ Drk with t1(a/θ) = t2(b/θ).

The first possibility is that t1(a) = t2(b) = m for some m ∈ M. In this caseHk(t1(a/θ)) = m = Hk(t2(b/θ)) as lk+2(m) = rn−k−4(m). The second possibilityis that t1(a) = t2(b) = m for any m. Since Ai ∩ Aj ⊆ {m | m ∈ M} for |i− j| ≥ 2we may assume {t1(a), t2(b)} ⊆ Ai ∪ Ai+1 for some i. Thus either a, b ∈ Dr

k ora, b ∈ Dl

k. In the former case rn−k−4(t1(a)) = rn−k−4(t2(b)) and in the lattercase lk+2(t1(a)) = lk+2(t2(b)) so Hk(t1(a/θ)) = Hk(t2(b/θ)) as required. The lastpossibility is that t1(a) = σ1(u, v) and t2(b) = σ2(u, v) for some u, v ∈ M. Theseequations imply that a = s1 or p1

r and b = sn−1 or pn−1q for some 1 ≤ r, q ≤ k.

Thus Hk(t1(a/θ)) = t1(rn−k−4(a)) = uu · · ·uv · · · vv = t2(lk+2(b)) = Hk(t2(b/θ))where u occurs n− k − 3 times. Thus Hk is a well-defined homomorphism.

To see that Hk is an embedding assume Hk(t1(a/θ)) = Hk(t2(b/θ)) for somea, b ∈ Dl

k ∪Drk and t1, t2 ∈ F . Consider a, b ∈ Dl

k. Since lk+2(t1(a)) = Hk(t1(a/θ))= Hk(t2(b/θ)) = lk+2(t2(b)) applying rk+2 gives t1(a) = t2(b) so t1(a/θ) = t2(b/θ).A similar argument holds for a, b ∈ Dr

k. The final step is to consider a ∈ Dlk and

b ∈ Drk. Application of rk+2 and ln−k−4 to the equalities

lk+2(t1(a)) = Hk(t1(a/θ)) = Hk(t2(b/θ)) = rn−k−4(t2(b))

imply that t1(a) = rn−2(t2(b)) and ln−2(t1(a)) = t2(b). Thus there exist u and v

(possibly equal) in M with t1(a) = σ2(u, v) and t2(b) = σ1(u, v). This means thatt1(a/θ) = t2(b/θ). We have now established that Hk is an embedding.

As any subalgebra C of A/θ generated by at most λ elements is a subalgebra ofsome Bk we have C in the quasi-variety generated by M so C satisfies all n-variablequasi-equations of M. Hence the n-variable quasi-equations of M do not form abasis for the quasi-equations of M. �

References

[1] I. P. Bestsennyi, Quasiidentities of finite unary algebras, Algebra and Logic 28 (1989),327–340.

[2] D. M. Clark and B. A. Davey, Natural Dualities for the Working Algebraist, CambridgeUniversity Press, 1998.


[3] D. M. Clark, B. A. Davey and J. G. Pitkethly, The complexity of dualisability:three-element unary algebras, preprint, 2001.

[4] J. Hyndman and J. Pitkethly, How finite is a three-element unary algebra?, submitted toInternat. J. of Algebra and Comput.

[5] J. Hyndman and R. Willard, An algebra that is dualizable but not fully dualizable, J. PureAppl. Algebra 151 (2000), 31–42.

[6] W. A. Lampe, G. F. McNulty and R. Willard, Full duality among graph algebras and flatgraph algebras, Algebra Universalis 45 (2001), 311–334.

[7] J. G. Pitkethly, Strong and full dualisability: three-element unary algebras, J. Austral.Math. Soc. 73 (2002), 187–219.

Jennifer HyndmanDepartment of Mathematics and Computer Science, University of Northern BritishColumbia, Prince George, British Columbia V2N 4Z9, Canadae-mail : [email protected]

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http://www.birkhauser.ch

positive primitive formulas preventing enough algebraic operations

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