ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II

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1 PACIFIC JOURNAL OF MATHEMATICS Vol, No, ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II G GRATZER AND J PLONK A In part I of this paper a conjecture was formulated according to which, with a few obvious exceptions, the sequence (Pn(&)} of an idempotent algebra is eventually strictly increasing In this paper this conjecture is verified for idempotent algebras satisfying p QX) 0, p z QX) > 0, and p*qx) > 0 In fact, somewhat more is proved: THEOREM Let It be an idempotent algebra with no essentially binary polynomial and with essentially ternary and quaternary polynomials Then the sequence is strictly increasing, that is, for all n ^ PnQX) + Pn+l(U) The proof starts in where a lemma of K Urbanik is modified to show that the proof splits naturally into three cases and handle the first two cases In the third case is analyzed and it is proved that it splits into two further cases that are settled in and In each of these sections examples are provided that the case under consideration is not void For the undefined concepts and basic results the reader is referred to [] Examples of algebras satisfying the conditions of the Theorem abound On a two element Boolean algebra {0, } the operation (x A y) V (y A z) V (z A x) defines such an algebra * The classification* An algebra U = {A; F} is idempotent if every operation fef has type (arity) > 0, and f(a,, a) = a for all α e A All algebras considered in this paper are assumed to have more than one element An ^-ary polynomial p of II (that is, an %-ary function or A composed from functions in JP) depends on x i ( ^ i ^ n) if there exist α?, α Λ, α e A with p(a l, a {,, a n ) Φ p(a l, eel,, a n ); p is essentially n-ary, if p depends on x ly, x n For n ^, let p Λ (tt) denote the number of essentially w-ary polynomials In this paper we shall deal exclusively with idempotent algebras satisfying ftox) - 0, p s (U) Φ 0, and p (U) Φ 0 The sequence <jp w (U)> is strictly increasing because II must have

2 00 G GRATZER AND J PLONKA essentially ternary polynomials with very nice properties This will be used to classify all algebras satisfying these conditions A ternary (idempotent) polynomial p is called a minority polynomial if p is a majority polynomial, if p(x, x, y) = p(x, y, x) = p(y, x, x) = y p(x, x, y) = p(x, y, x) = p(y, x, x) = x p is a first projection polynomial, if p(x, x, y) = p(x, y, x) = p(x, y,y) = x Observe that a minority or majority ternary polynomial is essentially ternary LEMMA Let p(x, y, z) be an essentially ternary polynomial satisfying p(x, y, y) y Then one of p(z, y, x) and p(y, x, z) is an essentially ternary first projection polynomial or one of p(x, y, z) and p(p(x, y, z), y, z) is a majority polynomial This statement can be verified by easy computation, observing that p(y, x, y) = x or y, p{y, y,x) x or y, and considering the four cases separately This argument is the first half of the proof of Lemma of K Urbanik [] THEOREM Let U be an idempotent algebra satisfying p z (U) 0 and p QX) Φ 0 Then U satisfies one (or more) of the following three conditions: (a) U has a ternary majority polynomial; (b) U has an essentially ternary first projection polynomial; (c) all essentially ternary polynomials of VL are minority polynomials Proof Since p (U) Φ 0, U has an essentially ternary polynomial p Since U is idempotent and p (tt) = 0, p(x, y,y) = % or y, p(y, x,y)-x or y, and p{y, y, x) = x or y If the second alternative occurs for any essentially ternary p, say p(x, y, y) = y, then by Lemma, p(z, y, x) or p(y, χ z) is an essentially ternary first projection polynomial, or one of p(χ f y, z) and p(p(x y, z), y, z) is a majority polynomial Thus U satisfies (a) or (b) This conclusion cannot be drawn only if for any essentially ternary polynomial p we have p{x, y, y) = p(y, x, y) p(y, V, %) x, which is (c)

3 ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II 0 * Majority polynomial* Algebras satisfying condition (a) of Theorem shall be handled in this section THEOREM Let U be an ίdempotent algebra satisfying p (VL) 0 If has a ternary majority polynomial /, then for n ^ p n (U) + ^ Pn+QX) Proof For any %-ary polynomial p define an (n + l)-ary polynomial pf: pf = f(p(x l, O, p(x l, x n -u» +i), P(»i,,»»-i,»i)) Let / = / and for w ^ define recursively: Finally, we define an (n + l)-ary polynomial g: = f{fn{%u,»»), Λ(»l,,»»-l,»n+l), ^) Now we make the following claims: ( i ) For n ^, (ii) For w ^, f n (x l, a? n _ lf aji) = a? x (iii) If the polynomial p is essentially ti-ary, then pi is essentially (n + l)-ary (iv) f n is essentially w-ary (v ) pf qf implies p q (vi) g is essentially (n + l)-ary (vii) g = ^F for no polynomial p Statements (i)-(vii) easily imply the statement of Theorem Indeed, consider the set {g} U {pf\p is an essentially w-ary polynomial of U) By (iii) and (vi) all elements of this set are essentially (n + l)-ary polynomials, (vii) shows that the union is a disjoint union, and so by (v) the set has p n (U) + elements Thus, p Λ+ (U) ^ p n (U) + Proof of (i) For n / = / is a majority polynomial, hence /sfai,, #) = #i Proceeding by induction, if f n (x l, a? n «!, x,) = x lf then

4 0 G GRATZER AND J PLONKA Proo/ 0/ (ii) For w = (ii) is trivial By induction, if / J n (Xl, X, # ' ', #) ^ then = /(/»(«!,», *, ^), fn(%l, %,, O, / (»!, 0?,, X, X λ ) Proo/ 0/ (iii) Setting x n = α? %+ in pf we get p, since / is a majority polynomial Hence pf depends on x lf,x n _ and on one or both of x n and x n+ Since pf is symmetric in x n and x n+ in any two element subalgebra the first possibility cannot occur, hence pf is essentially (n + l)-ary Proof of (iv) Trivial induction using (iii) Proof of (v) pf with x n = x n+ί yields p, from which the statement follows Proof of (vi) Same as the proof of (iv) Proof of (vii) Let g = pf Setting x n x n+ we conclude that f n = p Thus g = / Λ F = / w+, in other words, f(fn(x if,» w ), / (»!,, Xn-l, Xn+l), fn(xl, '", B»-i,»i)) = f(fn(xι,,»»), Λ(»i, * * *, _l,»»+i), ^) Setting #! = a? w+ and using (i) and that / is majority we get X ι = J\Jn\Xu ' ' * y *^n)j Xly Xv Finally, setting x = x Ά = = # w and using (ii) we obtain x t = a?, a contradiction, proving (vii) An example of an algebra satisfying the conditions of Theorem was given in Further examples are easy to construct First projections polynomial* In this section the Theorem is proved, in a somewhat sharper form, for algebras having an essentially ternary first projection polynomial

5 ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II 0 THEOREM Let U be an idempotent algebra with p z (U) = 0 If U has an essentially ternary first projection polynomial f, then for n^ (n - l)p n (U) ^ p n+ί (U) REMARK Since, for n ^, (n - l)p»(tt) ^ p n (U) ^ p n (U) +, Theorem is stronger than the corresponding special case of the Theorem Proof For an w-ary polynomial p and ^ i rg n set pf { = f(p(x l, O, x i x n+ ) Then we make the following claims: (i) pfi qfi implies p = q (ii) If i Φ j, then pf t Φ qf ά (iii) pfi depends on^, x n Since substituting x = x n+ in pf { yields p, we see that if pf { is not essentially (n + l)-ary, then by (iii) pf { = p By (ii), pfi Φ pfj if i Φ j; hence for i Φ j we cannot have both pfi and pf - not essentially (n + l)-ary Thus for an essentially w-ary p {pftli^,,,w} contains at least n essentially (n + l)-ary polynomials Furthermore, by (i) and (ii) the sets {pf t \i =,,,w} and {gfji =,, -, w} are disjoint if ί) and g are distinct essentially w-ary polynomials, from which Theorem follows trivially Proof of (i) pfi with x { x n+ι yields p, hence (i) is trivial Proof of (ii) Let us assume that i Φ j and pf t = gjpy, that is, f(p(x l, a? n ), a?i, a? Λ+ ) = f(q(x l, a? n ), %,»n+i) Set α? = % for k Φ i,l -^ k -^ n, in this identity; since p (ϊl) = 0 after the substitution p =α? or a? f and g = α; or a? <# The four possibilities yield the following identities: f(x, x j x n+ι ) = f(x, x, x n+ ), /(«, x ό, x n+ ) = /(%, a?, α? n+ ), /(a?, a?y, a? Λ+ ) = f(x, x, x n+ ), J\Xf Xjj Xn+i) == J\Xji Xj Xn+l)

6 0 G GRATZER AND J PLONKA The first and third contradict that / is essentially ternary, while the second and fourth mean that / is symmetric in its first and second variable, contradicting that / is a first projection polynomial Proof of (iii) Setting x t = x n+ in pfi gives p, hence pfi depends on x l, Xi_ u x i+,,$» Assume that pf { does not depend on x { Then Substituting α; = ^ for Φ i y l^ one of f(x x i x n+ ) = a?, j ^ n and using p (U) = 0 we get i, X%y Xn+l) ^+ ' The first contradicts that / is essentially ternary, while the second is Xi x n+, a contradiction An example of an algebra satisfying the condition of Theorem can be defined on the two element set {0, } taking as operation Taking both x + (x + y)(x + z)(y + z) (u + v = (u A v') V (u' A v)) (x A y) V (y A z) V (z A x) and x + (x - y)(x + z)(y + z) as operations we get an algebra satisfying the conditions of Theorems and Note that in Theorems and p (tt) Φ 0 follows from the assumptions * The second classification* In this and the subsequent sections we consider an idempotent algebra U with p (U) = 0 in which all essentially ternary polynomials are minority polynomials LEMMA U has exactly one essentially ternary polynomial Proof Let / and g be essentially ternary polynomials, and consider the polynomial f(g(x, y, z), y, z) = h Then h(x, y, x) =f(g(x, y, x), y, x) = f(y, y, #) = # Thus h cannot be essentially ternary, because it is not

7 ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II 0 minority Due to p (VL) = 0, h(x, y z) = x, or y, or z h(x, y, x) = x eliminates h y Furthermore, h(x, x, z) = f(g(x, x, z), x, z) = f(z, x, z) = x, eliminating h z Hence, h x, that is, we proved the identity f(g(x, y,z),y,z) = x Now let α, α', b, ce A and /(α,, c) f{a\ δ, c) Then α - /(/(α, δ, c),, c) - /(/(α', δ, c), δ, c) = α', by the above identity (used with / = g) (α =)f(f(a, δ, c), δ, c) = (a =)f(g(a, δ, c), δ, c), and so by the above remark, /(α, δ, c) = ^(α, δ, c), proving that / = #, completing the proof of Lemma The only essentially ternary polynomial shall be denoted by / Keep in mind that f(f(x, y,z),y,z) = x, and that / is fully symmetric The next important step is again due to K Urbanik We call a ternary function g on A a Boolean group reduct if a Boolean group operation + can be defined on A(i e, (A; +> is an abelian group satisfying x = 0) such that g(x, y, z) = x + y + z The proof of the next lemma is identical with the proof of Lemma of K Urbanik [] LEMMA f is a Boolean group reduct if and only if f(f(x y, z), x, u) does not depend on x If this is the case + is defined by fixing an arbitrary element OeA and x + y = f(x, y, 0) Accordingly, the proof of the Theorem in the minority polynomial case splits into two completely different cases according to whether or not /(/(«, y, z), x, u) depends on x * The minority polynomial is not a Boolean group reduce THEOREM Let U be an idempotent algebra satisfying p QX) = 0 Let f be the unique essentially ternary minority polynomial of U If f(f( x, V, z )> χ y u ) depends on x, then for n ^ p n (U) + ^ p n+ (U)

8 0 G GRATZER AND J PLONKA Proof We define / = / and, inductively, for n ^ Λ+i = /(Λ(«i,, O,»i,»»+i) For an w-ary polynomial p and <Ξ ί ^ w we set pg, = f(p(x l, a; n ), x l x n+ ), pg = p(x l x,, α;^, /(»<, a?!, α? Λ+ ), & ί+,,<> Observe that pg* with ^ = a? n+ yields p We make the following claims: (i) f n is essentially n-ary, and /sθi, «, ^) = &i, Λ(a?i, a?, ^, ^ = # /»(«i, a?,», *,, α? n ) = Λ- («i,»,, «) for n- ^ (ii) If p and g are essentially w-ary polynomials and ^ ί, j ^ n, then pgi qg ό implies p q (iii) For an essentially w-ary polynomial p, at least one of pg u, pg n is essentially (w + l)-ary Using (i)-(iii) it is easy to prove Theorem Indeed, by (ii) and (in), P = {pgi\p is essentially n-arγ, i =,, n} contains at least p n {%) essentially (n + l)-ary polynomials By (i), Jn+\% *^> Mn + lj X W, *, X n ) is also essentially (n -j- l)-ary If g e P, that is, = ^Gi, for some essentially ^-ary p and < i ^ n, then the substitution»i =»»+i yields /«+i(a?, *i,»i, ^,, α») = p{x u», -,»*) By the second part of (i) the left-hand side does not depend on x x while the right-hand side does, a contradiction Thus gίp, and so P (J {g} contains at least p Λ (X) + essentially (w + l)-ary polynomials, proving Theorem Proof of (i) We start by proving the formulas in (i) Obviously, and f z {x u x, x ) = x, / (^, a?, x, x ) = /(/ fe, α?, ffj), a? lf α? ) = /fe, χ l x,) = x,

9 ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II 0 Thus, for n ;>, by induction, = f(f«-*(xi, X*,, ^~i),»i,»») (For n = interpret / w- as a^) / is essentially ternary by assumption f (χ x, x, x,) = /gfo, a?, a? ), hence / depends on x x By assumption, f depends on x l Finally, Λ(#i» χ y χ, XA) = ^> hence / depends on a? Thus / is essentially -ary Proceeding by induction for n^,f n with ^ = α? w yields f n -ι{x u,# n -i)> hence f n depends on x,, α? _ lβ Finally, / w with α; = α; gives which depends on ^ and x n, hence / % depends on x x and a? n P = Q- Proof of (ii) Obvious; by setting ^ = x n+ in pg, qg we get Proof of (iii) pgi with x = a? n+ gives pfo, -, a?»), hence ^G^ depends on x,, α? Λ Furthermore, pg ί with ^ = x = = a? n gives f(x ί x l x n+ι ) = x n+ and pg$ > ) with x ί = x n+ι gives p(x ί y, «?<_!, a? w +, a? <+,,&«), hence all pg^, ^ i ^ ^ depend on x n+ Thus if none of pg lf, pg TC is essentially (w + l)-ary then none of them depend on x λ So assume that none of pg u **,pg n depend on x x Then by substituting x x n+ in pg t we get the identity For i > we obtain Since this holds for all i >, p is symmetric (*) repeatedly we obtain Then, using the identity = f(f(p(xl, Xl, %n-, *,»l),»i,»»-l),»l, fl?») = /(' -/(P(«i,»i,, Xi), Xi, ««) ) = fn(xl X,, ff») Hence, p = f n But then (*) states that f n+ ι(x l, x n +i) does not depend on x a contradiction

10 0 G GRATZER AND J PLONKA Idempotent algebras satisfying p = 0 and having a unique ternary minority polynomial can be constructed from Steiner quadruple systems and vice versa, A Steiner quadruple system is a set A and a set S of four element subsets of A with the property that any three element subset of A belongs to one and only one member of S For such a system define an algebra (A; /) as follows: / is a minority function and for three distinct elements α, δ, c e A there is a unique member Be S with a, δ, c e B; let B = {α, δ, c, d}; set /(α, δ, c) = d Conversely, if an idempotent algebra (A; F} satisfies p = 0 and / is the unique ternary minority polynomial, then set S = {{α, δ, c,f(a, δ, c)}\a, b,cea, {α,, c} = } Then this defines a Steiner quadruple system The smallest Steiner quadruple system which is associated with an algebra satisfying the conditions of Theorem can be defined on A = {,,, 0} as follows (see []): Obviously, the associated / is not a Boolean group reduct since A = 0 is not a power of two However, an example, which is due to NS Mendelsohn, shows that even if A is a power of two, examples of algebras satisfying the conditions of Theorem can be defined on A provided that \A\ ^ Let A = {,,, } and let S be given by the following table: 0

11 ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II

12 0 G GRATZER AND J PLONKA In this example, and therefore f(j X, ίks), *i ,X>, 0 0 f(χ ιt x, x t )) defines a Boolean group operation for any fixed Qe A However, f(x, y, z) Φ x + y + z To prove this it suffices by Lemma to illustrate that f(f(x, y, z), x, u) depends on x Indeed, /(/(I,, ),, ) = and /(/(,, ),, ) - It may be of interest to note that recently C Treash [] has solved the word problem for algebras {A; /> of type <>, where / is a minority function and f(f(x, y, z), y z) = a? Boolean reducts* In this section we settle the final case of the Theorem THEOREM Let U be an idempotent algebra satisfying p (VL) = 0, p z (U) Φ 0, and p (tt) Φ 0, having a unique essentially ternary minority

13 ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA, II polynomial / If f(f(x, y, z), x, u) does not depend on x, then p n (U) + ^ p n+ (VL) for n =,, Proof By Lemma, a Boolean group operation + can be defined on A such that f(x, y, z) = x + y + z Let p be an essentially -ary polynomial of U (recall that Pi(Vί) Φ 0) It follows from Lemma of [], that there exists a ternary polynomial p 0 of <A; /> such that p{x, y, z, u) p o (x, y, z, u) whenever x, y, z> and u are not all distinct If p 0 = x, then we can conclude that p is an essentially i-ary first projection polynomial, that is, it satisfies p(x, y, z,u) = x whenever x, y, z, and u are not all distinct If p 0 = y y p 0 z, or p 0 = u, we get a first projection polynomial by permuting the variables of p If p 0 = x + y + z, then p + V + % is the first projection polynomial Observe that p + y + z is essentially -ary, since otherwise #> +?/ + z would be a polynomial of /, implying that p (p + y + z) + y + z is a polynomial of / If p 0 a? + V + u, we proceed similarly Thus there exists in an essentially -ary first projection polynomial g (This statement is a small part of Lemma in []) Now we start our constructions Let p = p(x u -,&») be an essentially w-ary polynomial, w ^ We construct an (w + l)-ary polynomial p as follows: p(x, x,, a?, /, ) is a ternary polynomial of IX, hence it is», y, z, or x + y + z; () if p(x,, x, y, z) = x, then p = g(p(x ίf -,&»), a?»_i, α», a?»+i); () if p(x, --,x,y,z) = y, then p = g(p(x l, α? n ), ^, aj w, x n+ ); () if j>(α?,, x, y, z) =, then p = g(p(x l, α? Λ ), ^, av-i,»«+i); () if p(α?,, α?, y, z) = α? + y +, then Furthermore, for n ^ we define g n by recursion: g^ g and Then we claim the following: (i) For an essentially w-ary p, the polynomial p is essentially (n + l)-ary (ii) If p and q are essentially w-ary polynomials and p Φ q then pφq (iii) For w ^, ^%(^,, x u x n u x % ) = ^fe, #, α^, -, α%) = x, (iv) flr n is essentially w-ary (v) flr» + = p for no essentially w-ary polynomial p Now Theorem is clear:

14 G GRATZER AND J PLONKA {p I p essentially w-ary} U {g n +ι} is a set of p n (U) + essentially (n + l)-ary polynomials by (i), (ii), (iv), and (v) In the subsequent proofs Case,, Case refer to the cases in the definition of p Proof of (i) Case p with x n+ = x n yields p(x lf, x n ), hence p depends on x l, x n^ The substitution x n+ = x n _ t gives that p depends on x n Setting x t = x = = x n _ γ in p (observe that p(x ίf,»!, $ _!, x n ) = a?! by assumption) yields flrfo, αv-i, a? n, x n+ ), hence p depends on x n+ Case Use the substitutions x n+ = x lt x n+ = a? Λ, and ^ = = α? n _ Case Use the substitutions x n+ = a?i, a Λ+ = &»_!, and ^ = = x n _ Case Just as in the previous cases, x n+ι = α; w and x n+ί =» _! establish that jp depends on α? x,, x n Setting x = = ^Λ_ in p we get λ = ί/ία?!, OJ!, x n, x n+ ) + ^w_! + x n Observe that h + x n ~i Λ-^n- 0(%i, %n~u χ n, x n +i) depends on x n+, therefore so does h Thus p depends on x n+ Proof of (ii) Set A λ = {n, n n + }, A = {, n, n + }, A, = {, ^, n + }, and A+= {n l n f n + } If p belongs to Case i and k, I e A if k Φ ϊ, then x k = # z substituted into p yields p Observe that I A { Π Aj I ^ for ίg i, y ^ Now if ^ = q, p belongs to Case i f q to Case j, then we can choose k, I e A { Π A j k Φ I Substituting x k = a?, into p q gives p = q Proof of (iii) For w =, ffcfo, α? lf α?, x ) = g n (x lf x, x*, a*) = a?i» since ^ = g is a first projection polynomial Assuming the identities for n, we compute: and Proof of (iv) ^ = g so the statement is true for n = Assume it for w Substituting a? w+ = α; or x n+ x % into ^r Λ+ yields g n, hence

15 ON THE NUMBER OF POLYNOMIALS OF AN IDEMPOTENT ALGEBRA II g n+ depends on x u, χ n Substituting x = x = = α? n into # %+ι gives by (iii) g(g n (x l x, a?,, ), «, «, a«+i) = 0(α?i, α?,, a? Λ+ ), hence 0 Λ+ depends on x n+ Proof of (v) Observe that g n+ι {x u &, &, a*,, aj n+ ) = a?! Hence, if g n+ = p, then pfe, a? a, α?, α?,, x n+ ) = a? x Further substituting a? w+ = x w or (Case ) x n+ = a?^^ we conclude that This is impossible if p belongs to Cases or, and it immediately yields a contradiction in Case (namely, x t = g(x a? n-, a; n, α? n+ )) and in Case REFERENCES R D Carmichael, Groups of Finite Order, Dover Publ Inc G Gratzer, Universal Algebra, The University Series in Higher Mathematics D Van Nostrand Co, Princeton, N J, # Composition of Functions, Proceedings of the Conference on Universal Algebra, October Queen's Papers in Pure and Applied Mathematics, No, Queen's University, Kingston, Ont, 0 G Gratzer and J Plonka, On the number of polynomials of an idempotent algebra I, Pacific J Math, (0), -0 C Treash, The Completion of Finite Incomplete Steiner Triple Systems with Applications to Loop Theory K Urbanik, On algebraic operations in idempotent algebras, Colloq Math, (), - Received May, The work of both authors was supported by the National Research Council of Canada THE UNIVERSITY OF MANITOBA, AND THE MATHEMATICAL RESEARCH INSTITUTE OF THE POLISH ACADEMY OF SCIENCES, WROCLAW, POLAND

REGULAR IDENTITIES IN LATTICES

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 158, Number 1, July 1971 REGULAR IDENTITIES IN LATTICES BY R. PADMANABHANC) Abstract. An algebraic system 3i = is called a quasilattice