The lattice of monomial clones on finite fields

c 2021 The Author(s) 1420-8911/21/040001-31 published onlineAugust 31, 2021

https://doi.org/10.1007/s00012-021-00733-6 Algebra Universalis

The lattice of monomial clones on finite fields

Sebastian Kreinecker

Abstract. We investigate the lattice of clones that are generated by a set of functions that are induced on a finite field F by monomials. We study the atoms and coatoms of this lattice and investigate whether this lattice contains infinite ascending chains, or infinite descending chains, or infinite antichains.We give a connection between the lattice of these clones and semi-affine algebras. Furthermore, we show that the sublattice of idempotent clones of this lattice is finite and every idempotent monomial clone is principal.

Mathematics Subject Classification.08A40, 08A02.

Keywords. Clone, Set of functions, Monomial, Finite field, Semi-affine algebra.

1. Introduction and preliminaries

LetN:={1,2,3, . . .}, letN0:=N∪ {0}, letAbe a finite set, letF ⊆ {A^Ai | i∈N}and letn∈N. We denoteF∩A^AnbyF^[n]. LetC⊆

{A^Ai |i∈N}.Cis called aclone on Aif it is closed under composition of functions, i.e., ifn, m∈ N,f ∈C^[n],g₁, . . . , g_n∈C^[m]thenf(g₁, . . . , g_n)∈C^[m], and ifCcontains the projections, i.e., forn, j ∈N withj ≤n,πⁿ_j: Aⁿ →A, π_jⁿ(x₁, . . . , x_n) :=x_j lies inC. The characterization of clones on a two-element set by E. Post [13]

was the beginning of the study of clones. Already in the case of a three-element set, there are uncountable many clones [6], and thus a full description seems to be hard. Hence the investigation of clones led to the study of clones which contain, or which contain only, specific functions. Results for clones containing only affine mappings are given in [14], a full characterization for polynomial clones (clones containing all the constant functions) on Zp×Zp and on Zp²

for any prime pwith the operation + can be found in [4], a description for polynomial clones onZpqwith the operation + for different primesp,qis given

Presented by R. Willard.

Supported by the Austrian Science Fund (FWF): P29931.

in [2], and in [11] it is shown that the lattice of polynomial clones on a group of square-free order that contain the group operation is finite. For general notes on clone theory we refer to [12] and [15].

In this paper we study clones of the following kind: Letα(1), α(2), . . .∈ N0. We call g=

i∈Nx^α(i)_i a monomial if there is ann∈Nsuch that for all i∈Nwith i > nwe have α(i) = 0 and there is aj ∈N such thatα(j)= 0.

Then we writegalso as_n

i=1x^α(i)_i . We define mI(g) by max({i∈N|α(i)= 0}) and call mI(g) thearity ofg. Letq=p^tfor some primepand for somet∈N, and let Fq := GF(q). Let n ≥ mI(g). We call the function g^Fnq: Fⁿ_q → Fq, g^Fnq(x1, . . . , xn) :=_n

i=1x^α(i)_i then-ary induced function of g onFq. We call a function which is induced by a monomial amonomial function. We say that g^Fnq is idempotent if g^Fnq(x, . . . , x) = x for all x ∈ Fq, and call then g an idempotent monomial. This holds if and only if _mI(g)

i=1 α(i) ≡q−1 1. We call two monomialsm₁ andm₂ equivalent if the mI(m₁)-ary induced function of m1onFq is equal to the mI(m2)-ary induced function ofm2onFq. Note that m₁ =n

i=1x^α(i)_i is equivalent tom₂ =n

i=1x^β(i)_i if and only ifα(i)≡ β(i) for all 1 ≤ i ≤ n, where for a, b ∈ N0 we let a ≡ b if and only if a = b or a, b > 0 and a ≡q−1 b. Now let M be a set of monomials. We call M a monomial clone on Fq if C := {m^Fnq | m ∈ M, n ∈ N, n ≥ mI(m)} is a clone on Fq and M is closed under equivalent monomials, which means if m∈M and some monomialm is equivalent tom, thenm∈M. We say that M is idempotent if all monomials of M are idempotent. For any set M of monomials, there is a least monomial cloneMcontainingM,the monomial clone on Fq generated by M. In order to describe clones that contain only monomial functions, we will describe monomial clones. Clones that contain only monomial functions have been studied in [7], [8], [9], and [10], where special clones (e.g. generated by unary or binary monomial functions) are characterized. This was further developed in [5] where the binary part of an idempotent monomial clone on Fq which is generated by one single binary idempotent monomial is fully described.

LetMqbe the set of all monomial clones onFq. LetC, Dbe two monomial clones onFq. Then we denote by C∨D the smallest monomial clone which containsCandD, and by∩we denote the intersection of sets. Then (Mq,∨,∩) is a lattice. The smallest monomial clone Δ is the monomial clone generated by the monomial x₁ which induces the clone of projections, and the largest monomial clone∇is the monomial clone generated byx₁x₂, which induces the clone generated by the field multiplication. The set of idempotent monomial clones onFq with∨and∩forms a sublattice of the lattice of monomial clones onFq. LetCbe a monomial clone onFq such thatC= Δ andC=∇. We call Canatom if for all monomial clonesDwithD⊆Cwe haveD= Δ orD=C.

We callCa coatom if for all monomial clonesD withC⊆D we have D=∇ orD=C. LetC, Dbe two monomial clones onFq. IfC⊆D andC=D, we writeC⊂D. In this paper we investigate the general structure of the lattice of monomial clones on Fq for any prime power q: In Section 2 we start our investigation on monomials and give techniques for generating monomials. In

Section 3 we give a full description of the lattice of monomial clones on Fq

if q ∈ {2,3,4}. In Section 4 we give a connection between monomial clones and semi-affine algebras (see [15]). In Section 5 we investigate the top and the bottom of the lattice of monomial clones onFq for any prime power q. In Theorem 5.4we see that at the top of the lattice of monomial clones on Fq, there is an interval dually isomorphic to the divisor lattice ofq−1. Furthermore, we get a full description of the atoms in Corollary5.2and of the coatoms in Theorem5.10. In Section6 we prove that (Mq,∨,∩) is well-partially ordered (Theorem6.3), i.e., there are no infinite antichains and no infinite descending chains of monomial clones onFq. Furthermore, we show in Theorem6.8that infinite ascending chains of monomial clones onFq exist if and only ifq−1 is not square-free. In Section7we show that the lattice of idempotent monomial clones on Fq is finite and that every idempotent monomial clone on Fq is principal, i.e. singly generated.

1.1. Notation for monomial clones

Letqbe a prime power. LetCbe a clone onFq which contains only monomial functions, and letM be the monomial clone such thatC={m^Fnq |m∈M, n∈ N, n≥mI(m)}. Let f ∈C be induced by the monomial m=x^α(1)_i1 · · ·x^α(n)_in , where n∈N,α(1), . . . , α(n)∈Nand i₁, . . . , i_n ∈N withi₁ < i₂ <· · ·< i_n. Then m can be seen as i_n-ary function and since m has n variables with exponents unequal to 0, we say that m has width n. Since x^q = x for all x∈Fq andM is closed under equivalent monomials, we have thatM contains all these monomialsx^β(1)_i

1 · · ·x^β(n)_i

n where for allj≤i_n we haveβ(j)≡q−1α(j) andβ(j)>0. By permuting the variables we havex^α(1)₁ · · ·x^α(n)n ∈M. We see that a monomial clone is uniquely determined by those monomials_n

i=1x^α(i)_i withn∈Nsuch that for alli≤n,α(i)∈ {1, . . . , q−1}. Leti, j∈N. If we set a variablex_ito a variablex_j of a monomial, we say that weidentify the variable x_iwithx_j. SinceCis a clone andM is closed under equivalence of monomials, M is closed under substitution of monomials and thus closed under identifying variables, since a variable induces a projection. LetM be a set of monomials.

Then M is a monomial clone on Fq if and only if {xi | i ∈ N} ⊆ M, M is closed under substitution of monomials andM is closed under equivalent monomials.

Example 1.1. Letqbe a prime power. We have Δ ={x1}and∇={x1x2}.

LetC be a monomial clone onF5. Ifm(x1, x2) = x²₁x2∈C, then permuting the variables yields m1(x1, x2, x3) = m(x2, x3) = x²₂x3 ∈ C. Then we have m2(x1, x2, x3) =m(x1, m1(x2, x2, x3)) =x²₁x²₂x3∈C, and therefore we obtain m3(x1, x2, x3, x4) =m(x1, m2(x2, x3, x4)) =x²₁x²₂x²₃x4∈C. By identifying all the variables withx₁we get thatm₄(x₁) =m₃(x₁, x₁, x₁, x₁) =x⁷₁∈C. Since x⁷₁ induces the same function asx³₁, we havex³₁∈C, and alsox¹¹₁ ∈C.

Sincex^q =xfor all x∈Fq, we will describe monomial clones on Fq by investigating the arithmetic properties modulo q−1 of the exponents of the monomials.

1.2. Modulus calculation

Letqbe a prime power. Sincex^q =xfor allx∈Fq, we calculate moduloq−1 in the exponents, butx⁰ =x^q−1 for x = 0, and thus if an exponent a of a monomial has the property thata≡q−10 anda >0, we reduceatoq−1 and not to 0. Hence, we define fora∈N0, its representative with respect to≡by

a:=

q−1, ifamodq−1 = 0 anda >0, amodq−1, otherwise.

IfCis a monomial clone onFq, we denote byCthe set{n

i=1x^α(i)_i ∈C| ∀i≤ n:α(i)∈ {1, . . . , q−1}}. As we already mentioned,Cis uniquely determined byC.

2. First observations and procedures for generating monomials

Letqbe a prime power and letCbe a monomial clone onFq fixed for the rest of the section. We start with an example of a monomial clone onFq.

Lemma 2.1. Letbbe a divisor ofq−1. The setC :={_n

i=1x^α(i)_i |n∈N,∀i≤ n:α(i)∈N0,∃j≤n:α(j)= 0,_n

i=1α(i)≡b1}is a monomial clone onFq. Proof. Obviously {x₁} ⊆ C. Now we show that C is closed under substi- tution of monomials. Let n ∈ N, let m := _n

i=1x^α(i)_i ∈ C, and let m_j :=

n_j

i=1x^β_i^j(i) ∈ C for all j ≤ n. Now we get m(m1, . . . , mn) = _n

i=1

_ni

i=1

x^α(i)·β_i ⁱ⁽ⁱ⁾, and we have n i=1

n_i

i=1α(i)·β_i(i) ≡b

n

i=1α(i) ≡b 1. For all c, d ∈ N with c ≡q−1 d we have c ≡b d, since b divides q−1, and thus if n

i=1x^α(i)_i ∈C, then{n

i=1x^β(i)_i | ∀i≤n:β(i) =α(i)} ⊆C. Hence,C is a

monomial clone onFq.

Lemma 2.2. Let

i∈Ix^α(i)_i ∈C with |I| ≥2 andα(i)= 0 for all i∈ I, and letD⊂I be such that

i∈Dα(i)≡q−10. Then

i∈I\Dx^α(i)_i ∈C.

Proof. We identify allxifori∈Dwith anxjwherej∈I\D. The result holds, sinceCis closed under identifying variables andα(j) +

i∈Dα(i) =α(j).

Example 2.3. LetDbe a monomial clone onF5. Ifx³₁x²₂x²₃∈D, then 2+2≡40 and thus by Lemma2.2we havex³₁∈D.

Lemma 2.4. Let t ∈ N, let α(1), . . . , α(t)∈ N and letj ∈ N.C contains the monomial(t

i=1x^α(i)_i )(t+j

i=t+1x^q_i⁻¹)if and only if for alln∈N0we have that C contains(_t

i=1x^α(i)_i )(_t+n

i=t+1x^q_i⁻¹).

Proof. The “if”-direction obviously holds. We show the “only if”-direction. By Lemma2.2, we havet

i=1x^α(i)_i ∈C. Now we proceed by induction onn∈N.

For n = 1 we have by Lemma 2.2 that (_t

i=1x^α(i)_i )x^q_t+1⁻¹ ∈ C, since j ∈ N. Now let n ≥ 2. Let m(x1, . . . , xt+1) := (t

i=1x^α(i)_i )x^q−1_t+1. By the induction

hypothesis we have that m(x1, . . . , xt+n−1) := (_t

i=1x^α(i)_i )(_t+n−1

i=t+1 x^q−1_i ) ∈ C. Now we have

m(x1, . . . , xt+n−2, m(xt+n−1, xt+n, . . . , xt+n))

= _t

i=1

x^α(i)_i

t+n−2 i=t+1

x^q−1_i

x^α(1)_t+n−1^·(q−1)x(^(t_i=2α(i))+(q−1))·(q−1)

t+n ∈C,

since C is a monomial clone. Therefore _t

i=1x^α(i)_{i t+n}

i=t+1x^q_i⁻¹ ∈ C, since b·(q−1) =q−1 for allb∈N. This finishes the induction step.

Example 2.5. Let D be a monomial clone on F5. IfD containsx²₁x³₂x⁴₃, then we get by Lemma2.4thatx²₁x³₂x⁴₃· · ·x⁴_2+t∈D for allt∈N0.

Lemma 2.6. Let N, n ∈ N with n ≤N and let f: {1, . . . , N} →N be injec- tive. IfC contains _N

i=1x^α(i)_f(i), then there exists γ ∈N0 such thatC contains x^α(1)₁ · · ·x^α(n)n x^γ_n+1.

Proof. For alli∈ {1, . . . , n}we substitute the variablexifor the variablex_f(i), and for alli∈ {n+ 1, . . . , N}we substitute the variablexn+1 forx_f(i). Lemma 2.7. Let t∈N0 and letα(1), . . . , α(t)∈N. Ifx_1t

i=1x^α(i)_i+1 ∈C, then we have for alln∈N0 thatx1

n−1 j=0

t

i=1x^α(i)_j·t+i+1∈C.

Proof. Letm(x₁, . . . , x_t+1) := x₁t

i=1x^α(i)_i+1. We proceed by induction on n.

The statement is true for n= 0, since x₁ ∈C and for n= 1, since m ∈ C.

Let n > 1. By the induction hypothesis, we have m(x₁, . . . , x_(n−1)t+1) :=

x1

n−2 j=0

t

i=1x^α(i)_j·t+i+1∈C. Now we have

m(x1, . . . , x_n·t+1) :=m(m(x1, x(n−1)t+1+1, . . . , x_n·t+1), x2, . . . , x(n−1)t+1)

=x1

_t

i=1

x^α(i)_{(n−1)t+i+1 n−2}

j=0

t i=1

x^α(i)_j·t+i+1

=x1 n−1 j=0

t i=1

x^α(i)_j·t+i+1,

which lies in C, since C is a monomial clone. This finishes the induction

step.

We will often use Lemma2.7in the following context:

Example 2.8. Let k∈N. If x1· · ·x1+k ∈ C, then we get by Lemma 2.7that for allt∈Nwe havex₁· · ·x_1+t·k∈C.

Lemma 2.9. If C contains a monomial with two times the exponent 1, then x₁· · ·x_q ∈C.

Proof. By Lemma 2.6 we have m = x₁x₂x^α₃ ∈ C for some α ∈ N0. Now we get by Lemma 2.7 that x1(x2x^α₃)(x2+2x^α₃₊₂)· · ·(x_2+(q−2)2x^α_3+(q−2)2)∈ C.

By renaming the variables we getx₁x₂· · ·x_qx^α_q+1· · ·x^α_2(q−1)+1∈C. By iden- tifying the variables from xq+2 to x_2(q−1)+1 with the variable xq+1 we get x1x2· · ·xqx^(q_q+1^−1)·α∈C. The result follows now from Lemma2.2.

Lemma 2.10. We assume that C contains x^α(1)₁ · · ·x^α(n)n where n ∈ N with n≥2,gcd(α(1), q−1) = 1, α(1), α(2)∈N, α(i)∈N0 for i >2. Then there existsγ∈N0 such that x1x^α(2)₂ · · ·x^α(n)n x^γ_n+1∈C.

Proof. Ifα(1) = 1 the result is obvious. Now we assume that α(1) = 1. Let m(x1, . . . xn) :=x^α(1)₁ · · ·x^α(n)n ∈C. We show by induction that for allt∈N, there is aγ∈N0 such thatx^α(1)₁ ^tx^α(2)₂ · · ·x^α(n)n x^γ_n+1 ∈C. For t= 1, we have m∈C. Lett >1. By the induction hypothesis, we have m(x₁, . . . , x_n+1) = x^α(1)₁ ^(t−1)x^α(2)₂ · · ·x^α(n)n x^γ_n+1∈C for someγ∈C. Now we get

m(x1, . . . , xn+1) :=m(m(x1, xn+1. . . , xn+1), x2, . . . , xn)

=x^α(1)₁ ^tx^α(2)₂ · · ·x^α(n)_n x^γ_n+1 ∈C, whereγ=α(1)·(γ+k

i=2α(i)). This concludes the induction step. Ifq−1 = 1, lett = 1 and then α(1)^t = α(1) = 1, since α(1) >0. If q−1 >1, then let t=φ(q−1), whereφdenotes Euler’s totient function. Thenα(1)^t= 1, because gcd(α(1), q−1) = 1. Since C is closed under equivalent monomials, we get x1x^α(2)₂ · · ·x^α(n)n x^γ_n+1 ∈C.

Lemma 2.11. Let n∈N,n≥2 and letα(1), . . . , α(n)∈Nwith gcd(α(1), q− 1) = gcd(α(2), q−1) = 1. IfCcontainsx^α(1)₁ x^α(2)₂ · · ·x^α(n)n , thenx1· · ·xq ∈C.

Proof. By Lemma2.10and Lemma2.6we get x₁x^α(2)₂ x^γ₃ ∈C for some expo- nentγ ∈N0. Since C is a monomial clone, we have x^α(2)₁ x₂x^γ₃ ∈ C and thus we get by Lemma2.10and Lemma2.6thatx1x2x^δ₃∈C for someδ∈N0. The

result follows now from Lemma2.9.

Lemma 2.12. Let k∈Nand letα∈ {1, . . . , q}. We assume that x₁· · ·x_k ∈C andx^α₁ ∈C. Ifk > q−α, thenx₁· · ·x_k−(q−α)∈C.

Proof. We assume thatα < q, otherwise the claim is trivial. Lett:=q−1−α and letm(x₁, . . . , x_k) :=x₁· · ·x_k. Sincek−(q−α)>0, we have that

m(x₁, . . . , x_k−t+1) :=m(x₁, . . . , x_k−t, x_k−t+1, . . . , x_k−t+1)

=x1· · ·xk−tx^t_k−t+1

lies inC. Sincex^α₁ ∈C, we get by substitution of monomials that m(x₁, . . . , x_k−t+1) :=m(x₁, . . . , x_k−t−1, x^α_k−t, x_k−t+1)

=x1· · ·xk−t−1x^α_k−tx^t_k−t+1∈C.

Finally, we getm(x1, . . . , xk−t−1, xk−t−1, xk−t−1) =x1· · ·x^1+t+α_k−t−1 ∈C. Then x1· · ·xk−(q−α)∈C, since 1 +t+α= 1 and 1 +t=q−α, and thusk−t−1 =

k−(q−α).

As a special case we get the following corollary:

Corollary 2.13. We assume thatC contains x1· · ·xq. Let α∈ {1, . . . , q} such thatx^α₁ ∈C. Thenx1· · ·xα∈C.

Proof. By Lemma2.12we havex1· · ·x_q−(q−α)=x1· · ·xα∈C.

Lemma 2.14. {x₁· · ·x_q} is the clone of all idempotent monomials overFq. Proof. We have x1· · ·xq = x1

q−1

i=1xi+1. By Lemma 2.7 we have for each n∈Nthat

x1· · ·xqxq+1· · ·x_q+n(q−1)∈ {x1· · ·xq},

and thus we can generate all monomials with total degree 1 modulo q−1 by identifying and permuting variables. Hence,{x1· · ·xq}contains all idem- potent monomials over Fq. On the other hand, the clone of all idempotent

monomial containsx1· · ·xq.

More generally:

Lemma 2.15. Let k ∈ N\{1}. Then {x₁· · ·x_k} consists of all monomials x^α(1)₁ · · ·x^α(n)n with_n

i=1α(i)≡gcd(k−1,q−1)1.

Proof. By Lemma2.11we havex₁· · ·x_q ∈C:={x₁· · ·x_k}. By identifying the variablesx3, . . . , xkwithx2of the monomialx1· · ·xk, we get thatx1x^k−1₂ ∈ C. By Lemma 2.7 and by identifying variables, we have for all t ∈ N that x1x^t(k₂ ⁻¹⁾∈C. There aret1, t2∈Zsuch thatt1·(k−1) +t2·(q−1) = gcd(k− 1, q−1), and thus there is a t ∈Nsuch thatt·(k−1) = gcd(k−1, q−1).

This means that x₁xgcd(k−1,q−1)

2 ∈ C, and by identifying variables we get x1+gcd(k−1,q−1)

1 ∈ C. Since x₁· · ·x_q ∈ C and x1+gcd(k−1,q−1)

1 ∈C, we get by

Corollary2.13thatx₁· · ·x_{1+gcd(k−1,q−1)}∈C. By Lemma2.7we get now for all n∈Nthatx₁· · ·x_{1+n·gcd(k−1,q−1)}∈C. Hence, we can generate all monomials where the sum of the exponents is congruent to 1 modulo gcd(k−1, q−1) by identifying variables.

Finally, the set of such monomials is a monomial clone by Lemma2.1,

and it clearly containsx1· · ·xk.

3. The whole lattice of monomial clones if q ≤ 4

We start our investigation on the lattice of monomial clones for the casesF2

andF3. The following Proposition slightly generalizes the result of [9] and [7, Corollary 3.7] forF3where the lattice of monomial clones which are generated by a single monomial is given. In the case ofF2 these two lattices are equal and in the case of F3 the difference of the number of members of these two lattices is one, which means, there is exactly one monomial clone onF3 that is not singly generated.

Proposition 3.1. The lattices of monomial clones on F2 and F3 are given in Figure1.

x₁x₂

x₁

(a) Lattice of monomial clones onF2

x₁x₂

x₁x₂x₃ x²₁, x₁x²₂

x₁x²₂

x²₁x²₂

x²₁

x₁

(b)Lattice of monomial clones on F3

Figure 1. Monomial Clones on F2andF3

Proof. LetCbe a monomial clone onF2. All functions onF2can be induced by polynomials with only 0 and 1 as exponents. There is only one unary function that is induced by a monomial onF2, namely the function induced byx1. IfC contains a monomial with width t∈Nwith t >1, then C containsx1. . . xt. The result follows now from Lemma2.4.

Let us now determine the monomial clones on F3: C₁ = {x₁} is the smallest monomial clone.C₂={x²₁}contains all monomials of width 1, and thus C₁ ⊂ C₂ and there are no other monomial clones between C₁ and C₂, since a ∈ {0,1,2} for all a ∈ N0. Let C₃ := {x²₁x²₂}. By Lemma 2.2 and the fact that C2 only contains monomials of width 1 we have C2 ⊂C3. By Lemma 2.4 we have that C3 contains all monomials x²₁· · ·x²_n with n ∈ N.

By Lemma 2.4 every monomial in C3 of width greater 1 generates C3, and therefore there are no other monomial clones betweenC2 and C3. Now let C be a monomial clone which contains a monomial that does not lie in {x1} but contains 1 as exponent. ThenC containsm=x1

n

i=2x^α(i)_i forn≥2 and α(2), . . . , α(n)∈ {1,2}. First, we assume that 1 +|{j∈ {2, . . . , n} |α(j) = 1}|

is even. Then by Lemma 2.2, C contains the monomial x₁x₂ and thus all monomials. Now we assume that 1 +|{j ∈ {2, . . . , n} | α(j) = 1}| is odd and C does not contain a monomial with an even number (> 0) of 1’s as exponent. If 1 +|{j∈ {2, . . . , n} |α(j) = 1}|= 1, we havem=x_1n

i=2x²_i. By Lemma2.2 we havex₁x²₂ ∈C and by Lemma2.4we havem∈ {x₁x²₂}. We see that {x1x²₂} contains exactly all monomials with exactly one exponent with 1 and thus x²₁ ∈ {x₁x²₂}. If x²₁ ∈ C, we get x²₁x²₂ ∈ C by plugging x²₁ into x₁ of x₁x²₂. If x²₁· · ·x²_n ∈ C for somen ≥ 2 we get by Lemma 2.2 that x²₁ ∈ C and thus there are no monomial clones between {x₁x²₂} and {x²₁, x₁x²₂}, and neither between{x²₁x²₂}and{x²₁, x₁x²₂}. Now we assume 1 +|{j ∈ {2, . . . , n} | α(j) = 1}|> 1. Then, by Lemma2.2, C contains the monomialx1x2x3. We see thatx1x2x3does not generatex²₁, since{x1x2x3} contains only monomials with an odd number of variables with exponent 1.

x₁x₂

x₁

x1x³₂ x²₁ x³₁

x²₁x³₂ x²₁, x³₁

x1x2x3x4

x³₁x³₂ x1x³₂, x³₁x³₂ x²₁, x³₁x³₂ x²₁x³₂, x³₁x³₂

Figure 2. Lattice of monomial clones onF4

We also have{x1x²₂} ⊂ {x1x₂x₃}, since all monomials of{x1x²₂}contain exactly one exponent with 1. If C contains x1x2x3 and x²₁, then C contains x1x2x²₃and Lemma 2.2yieldsx1x2∈C. This finishes the proof.

The next proposition describes the whole lattice of monomial clones if q= 4.

Proposition 3.2. The lattice of monomial clones on F4 is given in Figure2.

Proof. LetCbe a monomial clone on F4. IfCcontainsx1x2, thenCcontains all monomials. First, we show that{x1x2x3x4} and {x1x2} are different and that these two monomial clones are the only monomial clones that contain a monomial which contains two exponents different from 0 modulo 3. To this end, we assume that C contains a monomial m =x^α(1)₁ x^α(2)₂ · · ·x^α(n)n where n∈N\{1} and at least 2 exponents are not equal to 0 modulo 3. We assume that α(1) ≡3 0 and α(2) ≡3 0. Then Lemma 2.11 yields x₁x₂x₃x₄ ∈ C, since gcd(α(1),3) = 1 and gcd(α(2),3) = 1. By Lemma 2.14C contains all idempotent monomials. Now we assume thatx₁· · ·x₄∈ C and C contains a monomialm which is not idempotent. By identifying all variables ofm with x₁, we get that there existsα∈ Nsuch that C contains x^α₁. Sincem is not idempotent, we have α = 1. If α = 3, we have x₁x₂x₃x₄, x³₁ ∈ C and thus x1x2x³₃x³₄∈C. Now Lemma2.2yieldsx1x2∈C. Ifα= 2, thenCcontainsx²₁. This means that x1x2x3x4, x²₁ ∈C and thus Corollary 2.13yields x1x2 ∈C.

If C contains only idempotent monomials, then we get by Lemma 2.14that C = {x1x2x3x4} and x1x2 ∈ C. By the previous analysis we have, if C contains a monomial which is not idempotent and it contains two exponents which are not 0 modulo 3, thenCcontainsx1x2.

Now we assume that all monomials ofC contain at most one exponent which is not equal to 0 modulo 3. By Lemma 2.2and Lemma2.4 we can re- strict to the following monomials as possible generators for C: x²₁, x³₁, x1x³₂,

x²₁x³₂, and x³₁x³₂. These monomials induce different functions. Now we want to find all monomial clones which we get by different combinations of these generators. We start with different combinations including x³₁x³₂. We have that x²₁ ∈ {x²₁x³₂}, x³₁ ∈ {x³₁x³₂} and x₁x³₂ ∈ {x²₁x³₂}. Hence, the mono- mial clones above {x³₁x³₂} and below {x1x2} are given by {x1x³₂, x³₁x³₂}, {x²₁, x³₁x³₂}, and{x²₁x³₂, x³₁x³₂}. We have thatx²₁∈ {x1x³₂, x³₁x³₂}, sincex²₁ does not preserve the set{1, a}whereais a generator for (F4\{0},·), but the operations onF4 given by (x1, x2)→x1x³₂ and (x1, x2)→x³₁x³₂ preserve the set{1, a}. On the other hand we have thatx1x³₂∈ {x²₁, x³₁x³₂}, since all mono- mials of{x²₁, x³₁x³₂}of width larger than 1 have the property that all expo- nents are 0 modulo 3. Therefore,{x1x³₂, x³₁x³₂} x²₁and{x²₁, x³₁x³₂} x1x³₂ are distinct and both monomial clones are strictly contained in{x²1x³₂, x³₁x³₂}.

Then the combinations of the generatorsx²₁,x³₁,x1x³₂,x²₁x³₂are left. Now we try to find different monomial clones ifx²₁x³₂is a generator. We havex²₁∈ {x²₁x³₂}, x₁x³₂ ∈ {x²₁x³₂}, and ifx³₁ andx²₁x³₂ lie inC, thenx³₁x³₂∈C. Hence, we just get the monomial clone{x²₁x³₂}as a new one. Now the combinations of the generatorsx²₁, x³₁,x₁x³₂ are left. Ifx³₁∈C andx₁x³₂∈C, thenx³₁x³₂∈C, and ifx²₁ ∈C and x₁x³₂ ∈C, then x²₁x³₂ ∈ C. Hence, we only get the monomial clones{x1x³₂}, {x²₁},{x³₁}and {x²₁, x³₁}, which we have not found yet.

The smallest monomial clone is given by{x1}. Altogether we have found 12 different monomial clones, which are ordered as given in Figure2.

4. Connection to semi-affine algebras

In this section we give a connection of monomial clones to semi-affine alge- bras. First, we recall the definition of a semi-affine algebra (cf. [15]). LetA= (A,+,−,0) be an abelian group. Letn∈N0 and letf be ann-ary operation onA. We callf affine with respect to Aiff(u+v) +f(0, . . . ,0) =f(u) +f(v) for all u, v ∈ Aⁿ, where + is the componentwise addition. We call then an algebraB= (A, F)semi-affine with respect to Aif everyf ∈F is affine with respect toA. Furthermore, we say thatf is 0-preserving iff(0, . . . ,0) = 0. Let F ⊆

{A^Ai |i∈N}. We call the algebra (A, F) 0-preserving if all term func- tions of (A, F) are 0-preserving. LetCbe a clone onA. We callC0-preserving semi-affine with respect toAif (A, C) is a 0-preserving and semi-affine algebra with respect toA. Fora∈N0, we write [a] ∈Zq−1 for the equivalence class ofamoduloq−1. Letqbe a prime power and let Gbe a set of monomials.

Then we define ϕ(G) :=

f:Zⁿq−1→Zq−1,(y1, . . . , yn)→

n

i=1

[r(i)]yi|n∈N, n i=1

x^r(i)_i ∈G

. IfGis a monomial clone onFq, then the setϕ(G) is a 0-preserving semi-affine clone with respect to (Zq−1,+,−,0). Note that for allC, D∈Mq withC⊆D we haveϕ(C)⊆ϕ(D). On the other hand we have that the monomial clones ϕ({x^q−1₁ }) and ϕ({x^q−1₁ x^q−1₂ }) are both equal to the clone on Zq−1 that is generated by{0}, but{x^q−1₁ } ⊂ {x^q−1₁ x^q−1₂ }. LetF be a set of finitary

f_{0 Z2}

f_{2 Z2} f_{1 Z2} f_{3 Z2}

Figure 3. Lattice of 0-preserving semi-affine clones with re- spect to (Z2,+,−,0)

g_{0 Z3} g_{3 Z3}

g_{1 Z3} g_{2 Z3}

g₁, g_{2 Z3} g_{4 Z3}

Figure 4. Lattice of 0-preserving semi-affine clones with re- spect to (Z3,+,−,0)

operations onZq−1. We denote the clone generated by F byFZq−1. Let us consider the difference of monomial clones and semi-affine algebras with respect to (Zq−1,+,−,0) in the exampleq = 3. Let f₀: Z2 → Z2, x → x, f₁:Z2 → Z2, x→0,f₂: Z³2→Z2,(x, y, z)→x+y+z, andf₃:Z²2→Z2,(x, y)→x+y.

Then the lattice of 0-preserving semi-affine clones with respect to (Z2,+,−,0) is given by Figure3.

In Proposition3.1we find the lattice of monomial clones on F3 and see that |M3| = 7. We will give a connection between monomial clones on Fq

and 0-preserving semi-affine algebras with respect to (Zq−1,+,−,0) via ϕin Proposition 4.3, which we use in Section 6 and Section7. If q = 4 we have given the lattice of monomial clones in Theorem 3.2. From this lattice we can easily derive the lattice of 0-preserving semi-affine clones with respect to (Z3,+,−,0) given in Figure 4. For this figure, we use g0:Z3 → Z3, x → x, g1: Z3 → Z3, x → 0,g2: Z3 → Z3, x → 2x, g3: Z⁴₃ → Z3,(x1, x2, x3, x4) → x1+x2+x3+x4, and letg4:Z²₃→Z3,(x, y)→x+y.

Now we investigate some properties of the lattice of monomial clones if q−1 is square-free.

Lemma 4.1. Let q be a prime power such that q−1 is square-free, let n ∈ N0 and let d, α(1), . . . , α(n) ∈ N . Let C be a monomial clone on Fq. If x^d₁x^d₂n

i=1x^α(i)_2+i ∈C, thenx^d₁x^d₂(n

i=1x^α(i)_2+i)x^q−1_3+n ∈C.