Monomial clones over $\mathbb {F}_q$

Abstract

The description of the poset of clones generated by a single binary idempotent monomial over $\mathbb {F}_q$ is given by purely number theoretic means.

1 Introduction

Let q be a prime power and let $\mathbb {F}_q$ denote the q element field. Every n-variable polynomial over $\mathbb {F}_q$ defines a polynomial function over $\mathbb {F}_q$, and every n-variable function is uniquely expressed as an n-variable polynomial of “low” degree. A clone is a subset of functions over $\mathbb {F}_q$ which contains all projections and closed under composition of functions. For more on clone theory, we refer the reader to [1, 2].

As substructures in general, clones over a set S can be ordered with respect to inclusion and they form a partially ordered set. In [5] all binary polynomials are given over the field $\mathbb {F}_3$ that generate a minimal clone. A polynomial will be called a minimal polynomial if it generates a minimal clone. In [5] a description of minimal linear polynomials and binary minimal monomials were given. The investigation was extended in [6] to the case of ternary majority minimal polynomials over $\mathbb {F}_3$. Recently in [3] the closed sets of binary monomials were investigated and the corresponding posets over $\mathbb {F}_2$, $\mathbb {F}_3$ and $\mathbb {F}_5$ were described. The investigation was further developed in [4], where it was shown that over the field $\mathbb {F}_q$ the poset of all closed sets of the unary and binary monomials generated by $xy^b$ is isomorphic to the lattice of divisors of $q-1$. The description of all clones generated by a single binary monomial was formulated as an open problem. In this paper we answer their question (Theorem 2.4 in Section 2).

A binary monomial over $\mathbb {F}_q$ is a polynomial of the form $x^ay^b$ for some positive integers a, b and the corresponding binary monomial function is $(s;t) \mapsto s^at^b$ for any $s,t\in \mathbb {F}_q$, as usual. In this paper we shall be interested in binary monomial functions, so for simplicity we write $x^ay^b$ for the function determined by the polynomial $x^ay^b$, as well. Note, that the function $x^ay^b$ over $\mathbb {F}_q$ is the same as $x^{a+q-1}y^b$ or $x^ay^{b+q-1}$, since $x \mapsto x^q$ is the identity function. Therefore, in the paper we mainly will be interested in the modulo $q-1$ residues of the exponents of binary monomials. The modulo $q-1$ residues will be mostly taken from the set $\left\{ \,1, \dots , q-1\,\right\}$.

A binary monomial clone $\mathcal {C}$ contains the functions x, y, and binary monomials such that $\mathcal {C}$ is closed under function composition and permutation of the variables. That is, if $x^a y^{a'}, x^by^{b'}, x^sy^{s'} \in \mathcal {C}$ for some nonnegative integers a, $a'$, b, $b'$, s, $s'$, then

$$\begin{aligned} \left( x^a y^{a'} \right) ^s \left( x^by^{b'} \right) ^{s'} = x^{as+bs'} y^{a's+b's'} \in \mathcal {C}. \end{aligned}$$

(1.1)

Furthermore, if $x^ay^{a'} \in \mathcal {C}$, then $x^{a'}y^a \in \mathcal {C}$ by permuting the variables x and y.

A binary monomial $x^ay^{a'}$ is idempotent if substituting the same variable x into every variable we obtain the identity function $x \mapsto x$, that is if $x^{a}x^{a'} \equiv x$. This happens if and only if $a+a' \equiv 1 \pmod {q-1}$. A binary idempotent monomial clone $\mathcal {C}$ is a binary monomial clone $\mathcal {C}$ containing only idempotent binary monomials. Composition of idempotent functions results an idempotent function, as if $a+a' \equiv b\,{+}\,b' \equiv s\,{+}\,s' \equiv 1 \pmod {q-1}$, then $as \,{+}\, bs' \,{+} a's \,{+}\, b's' \equiv 1 \pmod {q-1}$, as well. Hence the set of idempotent binary monomials is a clone itself.

In Section 2 we recall some preliminary results, prove some easy propositions, and state the main result (Theorem 2.4). Then in Section 3 we prove Theorem 2.4. We finish the paper by posing some open problems in Section 4.

2 Preliminaries

Let $\mathcal {C}$ be an idempotent monomial clone, that is for all $x^ay^{a'} \in \mathcal {C}$ we have $a+a' \equiv 1 \pmod {q-1}$. Let

$$\begin{aligned} H = \left\{ \,1\le a \le q-1\mid x^ay^{q-a} \in \mathcal {C}\,\right\} . \end{aligned}$$

Assume $a, b, s \in H$, that is $x^ay^{q-a}, x^by^{q-b}, x^sy^{q-s} \in \mathcal {C}$. By (1.1) we have that $x^{as+b(q-s)}y^{(q-a)s+(q-b)(q-s)} \in \mathcal {C}$. Now,

$$\begin{aligned} as+b(q-s) \equiv as + b(1-s) \pmod {q-1}, \end{aligned}$$

thus H contains the modulo $q-1$ residue class of $as + b(1-s)$. Furthermore, if $x^ay^{q-a} \in \mathcal {C}$, then by symmetry $x^{q-a}y^{a} \in \mathcal {C}$, as well. That is, if $a \in H$, then $q-a\equiv 1-a \in H$. Thus, characterizing all idempotent monomial clones translates to characterize all those subsets $H \subseteq \left\{ \,1, \dots , q-1\,\right\}$ which have the property that if $a, b, s \in H$, then

$$\begin{aligned} as+b(1-s) \pmod {q-1}&\in H, \end{aligned}$$

(2.1)

$$\begin{aligned} 1-a \pmod {q-1}&\in H. \end{aligned}$$

(2.2)

Let $S \subseteq \left\{ \,1, \dots , q-1\,\right\}$ be a subset. Then $\left\langle S \right\rangle$ denotes the smallest subset of $\left\{ \,1, \dots , q-1\,\right\}$ containing S which is closed under the operations (2.1–2.2). The problem posed in [4] was to completely characterize $\left\langle u \right\rangle$ for arbitrary $1\le u\le q-1$.

Example 2.1

Note that not every clone can be generated by one element. For example, for $q=31$ the set

$$\begin{aligned} H= \left\{ 1,6,10,15,16,21,25,30 \right\} \end{aligned}$$

is closed under the operations (2.1–2.2) modulo 30, but none of its elements generates the whole set. For every $h\in H$ we have $h^2 \equiv h \pmod {30}$, hence each element distinct from 1 and 30 generates a 4 element clone.

The smallest and largest binary monomial clones have already been determined in [4].

Proposition 2.2

[4, Proposition 5.2]. $\left\langle 2 \right\rangle = \left\{ \,1, \dots , q-1\,\right\}$.

Proposition 2.3

[4, Proposition 5.7]. For arbitrary $1\le u\le q-1$ we have $\left\{ \,1, q-1\,\right\} = \left\langle 1 \right\rangle \subseteq \left\langle u \right\rangle$.

In the following we give a complete characterization of $\left\langle u \right\rangle$ for all $1\le u\le q-1$ by pure number theoretic means. Note, that operations (2.1–2.2) make sense even if q is not a prime power. Therefore, in the following we do not assume that q is a prime power, but only that q is a positive integer and $q>1$. For convenience, from now on when we write $a \in H$ we mean that the modulo $q-1$ remainder of a from the set $\left\{ \,1, \dots , q-1\,\right\}$ is in H. For example, $q \in H$ means that 1 is in H, and $0 \in H$ means that $q-1$ is in H. Moreover, when we simply write $a \equiv b$ without specifying the module of the congruence, we mean $a \equiv b \pmod {q-1}$.

Throughout the paper we use the notation $\left( a, b\right)$ for the greatest (positive) common divisor of the integers a and b. To distinguish from the greatest common divisor, we denote the pair of a and b by putting semicolon in between a and b, i.e. $\left( a; b \right)$.

Let $q>1$ be a positive integer. For our characterization, we will need the following definition. Let $d \mid q-1$ be a divisor, and consider

$$\begin{aligned} H_d&= \left\{ \,1\le a \le q-1\mid a\equiv 0 \text { or } a \equiv 1 \pmod {d}\,\right\} . \end{aligned}$$

Then it is easy to check that $H_d$ is closed under the operations (2.1–2.2). Note, that $H_1 = H_2 = \left\{ \,1, 2, \dots , q-1\,\right\}$. Our main result is the following.

Theorem 2.4

Let $1\le u <q$, $d_1 = \left( u, q-1 \right)$, $d_2 = \left( 1-u, q-1 \right)$. Then

$$\begin{aligned} \left\langle u\right\rangle =H_{d_1}\cap H_{d_2}. \end{aligned}$$

Example 2.5

The set of all of the idempotent monomial clones over $\mathbb {F}_{13}$ is $\{\left\langle 1\right\rangle , \left\langle 2\right\rangle , \dots ,\left\langle 6\right\rangle \}$. The clones are ordered by inclusion and the structure of this lattice is presented in Figure 1.

Fig. 1

Idempotent monomial clones over $\mathbb {F}_{13}$

The following is a very useful property of sets closed under (2.1–2.2).

Proposition 2.6

Assume $s \in \left\langle u \right\rangle$ such that $1-s$ is invertible modulo ${q-1}$. Then for all $t \in \left\langle u \right\rangle$ and nonnegative integer k we have that $t + ks \in \left\langle u \right\rangle$.

Proof

Let $H = \left\langle u \right\rangle$. We prove Proposition 2.6 by induction on k. The statement holds for $k=0$. Assume that the statement holds for $(k-1)$, that is for all $t \in \left\langle u \right\rangle$ we have that $t + (k-1)s \in H$. We prove that $t+ks \in H$. Let n be the multiplicative order of $1-s \pmod {q-1}$, then $\left( 1-s\right) ^{-1} \equiv \left( 1-s \right) ^{n-1}$. Applying (2.1) with $b=q-1 \equiv 0$ we obtain that H is closed under multiplication. Hence, $\left( 1-s\right) ^{n-1} \equiv \left( 1-s \right) ^{-1} \in H$. Now, $t + (k-1)s \in H$, and therefore $\left( t + (k-1)s \right) \left( 1-s\right) ^{-1} \in H$. Applying (2.1) with $a = 1$, $b \equiv \left( t + (k-1)s \right) \left( 1-s\right) ^{-1}$ shows

$$\begin{aligned} as + b(1-s)&\equiv 1\cdot s + \left( t + (k-1)s \right) \left( 1-s\right) ^{-1} \left( 1 - s \right) \\&= s+t+(k-1)s = t+ks \in H.\\ \end{aligned}$$

$\square$

We mention the following easy consequence of Proposition 2.6, which generalizes Proposition 2.2 and is a special case of Theorem 2.4.

Corollary 2.7

Let $1\le u\le q-1$ be an integer such that both u and $1-u$ are invertible modulo $q-1$. Then $\left\langle u \right\rangle = \left\{ \,1, \dots , q-1\,\right\}$.

Proof

Let $H = \left\langle u \right\rangle$. We prove by induction that for every positive integer k we have $ku \in H$. For $k=1$ we have $u \in H$. Assume that $ku \in H$ for some positive integer k. Then applying Proposition 2.6 with $t=ku$ and $s=u$ we obtain that $H \ni t+s = ku+u = (k+1)u$.

Let x be a positive integer solution of the congruence

$$\begin{aligned} ux \equiv 2 \pmod {q-1}. \end{aligned}$$

Such x exists, because u is invertible modulo $q-1$. Then with $k=x$ we have that $ux \pmod {q-1}$ is in H, that is $2 \in H$. By Proposition 2.2 we have $H = \left\{ \,1, 2, \dots , q-1\,\right\}$. $\square$

3 Proof of Theorem 2.4

Fix $q>u\ge 1$, and let $H = \left\langle u\right\rangle$. Since $u \in H_{d_1}$ and $u \in H_{d_2}$, we have $H \subseteq H_{d_1} \cap H_{d_2}$. In the following we prove $H \supseteq H_{d_1} \cap H_{d_2}$

Note, that $\left( u, 1-u \right) = 1$, therefore

$$\begin{aligned} \left( d_1, d_2 \right) = 1. \end{aligned}$$

(3.1)

We need the following about the structure of $H_{d_1} \cap H_{d_2}$.

Lemma 3.1

Let $v \in H_{d_1} \cap H_{d_2}$ be arbitrary. Then there exists an integer m and $t \in \left\{ \,0, 1, u, 1-u\,\right\}$ such that

$$\begin{aligned} v = t + md_1d_2. \end{aligned}$$

In particular, for arbitrary integer k we have $v + kd_1d_2 \in H_{d_1} \cap H_{d_2}$.

Proof

Let $v \in H_{d_1} \cap H_{d_2}$ be arbitrary. We will apply the Chinese remainder theorem. We distinguish four cases depending on the remainder of v by $d_1$ and by $d_2$.

• $v \equiv 0 \pmod {d_1}$ and $v \equiv 0 \pmod {d_2}$. By the Chinese remainder theorem, $v \equiv 0 \pmod {d_1 d_2}$, and hence there exists an integer m such that $v = md_1d_2$.

• $v \equiv 1 \pmod {d_1}$ and $v \equiv 1 \pmod {d_2}$. By the Chinese remainder theorem, $v \equiv 1 \pmod {d_1 d_2}$, and hence there exists an integer m such that $v = 1+md_1d_2$.

• $v \equiv 0 \pmod {d_1}$ and $v \equiv 1 \pmod {d_2}$. Since $d_1 \mid u$ and $d_2 \mid 1-u$, we have $u \equiv 0 \pmod {d_1}$ and $u \equiv 1 \pmod {d_2}$. By the Chinese remainder theorem, $v \equiv u \pmod {d_1 d_2}$, and hence there exists an integer m such that $v = u+md_1d_2$.

• $v \equiv 1 \pmod {d_1}$ and $v \equiv 0 \pmod {d_2}$. Since $d_1 \mid u$ and $d_2 \mid 1-u$, we have $1-u \equiv 1 \pmod {d_1}$ and $1-u \equiv 0 \pmod {d_2}$. By the Chinese remainder theorem, $v \equiv 1-u \pmod {d_1 d_2}$, and hence there exists an integer m such that $v = 1-u+md_1d_2$.$\square$

From (3.1) we have $(d_1, d_2)=1$, hence $d_1d_2 \mid q-1$. In the following we prove $H \supseteq H_{d_1} \cap H_{d_2}$ by downward induction on $d_1d_2$. If $d_1d_2 = q-1$, then $H_{d_1} \cap H_{d_2} = \left\{ \,0,1,u,1-u\,\right\}$ by Lemma 3.1. Since $0, 1, u, 1-u \in H$, we obtain $H_{d_1} \cap H_{d_2} \subseteq H$.

Assume now, that Theorem 2.4 holds for all pairs $(q-1; v)$ for which the product $\left( v, q-1 \right) \cdot \left( 1-v, q-1 \right)$ is strictly greater than $d_1 d_2$. Applying (2.1) with $a = u$, $s=q-u \equiv 1-u$ and $b=q-1 \equiv 0$, we obtain

$$\begin{aligned} as + b(1-s) \equiv u \left( 1 - u \right) + 0 (1-s) = u-u^2 \in H. \end{aligned}$$

(3.2)

Since $(u, 1-u) = 1$, we have

$$\begin{aligned}&\displaystyle \left( u-u^2, q-1 \right) = \left( u(1-u), q-1 \right)&\nonumber \\&\displaystyle \quad = \left( u, q-1 \right) \cdot \left( 1-u, q-1 \right) = d_1d_2.&\end{aligned}$$

(3.3)

Applying (2.2) on (3.2) we obtain $1-u+u^2 \in H$. Let

$$\begin{aligned} d_3 = \left( 1-u+u^2, q-1 \right) . \end{aligned}$$

Now, $(u, 1-u+u^2)=1$, thus $(d_1, d_3)=1$. Similarly, $(1-u, 1-u+u^2)=1$, thus $(d_2, d_3)=1$. Furthermore, if $2 \not \mid q-1$, then $2 \not \mid d_3$, as well. However, if $2 \mid q-1$, then either u or $1-u$ is even, thus $2 \mid d_1 d_2$. Since $\left( d_1d_2, d_3 \right) =1$, we have $2 \not \mid d_3$. In any case, $\left( 2, d_3 \right) =1$. Thus, we have

$$\begin{aligned} \left( 2d_1d_2, d_3 \right) = 1. \end{aligned}$$

(3.4)

Lemma 3.2

If $d_3 = 1$, then $H_{d_1} \cap H_{d_2} \subseteq H$.

Proof

If $d_3 = 1$, then let m be an arbitrary nonnegative integer, and let x be a positive integer solution of the congruence

$$\begin{aligned} \left( 1-u+u^2 \right) \left( u-u^2 \right) \cdot x \equiv m d_1 d_2 \pmod {q-1}. \end{aligned}$$

Such x exists, because $\left( 1-u+u^2, q-1 \right) = 1$ and $\left( u - u^2, q-1 \right) = d_1d_2$. By Proposition 2.6 we obtain that $t + k\left( u-u^2 \right) \in H$ for any $t \in H$ and nonnegative integer k. Choosing $k = \left( 1-u+u^2 \right) x$ and $t \in \left\{ \,0, 1, u, 1-u \,\right\}$ (then $t \in H$) we obtain that $md_1d_2$, $1+md_1d_2$, $u+md_1d_2$ and $1-u+md_1d_2 \pmod {q-1}$ are all in H. Therefore, by Lemma 3.1 we have $H_{d_1} \cap H_{d_2} \subseteq H$.

$\square$

Thus, Theorem 2.4 holds if $d_3 = 1$. In the following we assume $d_3>1$. Now, applying (2.1) with $a = u$, $s=u$ and $b=q-u \equiv 1-u$ we obtain

$$\begin{aligned} as + b(1-s) \equiv u^2 + \left( 1 - u \right) ^2 = 1 -2u+2u^2 \in H. \end{aligned}$$

(3.5)

Since $(u-u^2, q-1) = d_1d_2$, we have

$$\begin{aligned} \left( 2u-2u^2, q-1 \right) \in \left\{ \, d_1d_2, 2d_1d_2\,\right\} . \end{aligned}$$

(3.6)

Applying (2.2) on (3.5) we obtain $2u-2u^2 \in H$. Let

$$\begin{aligned} d_4 = \left( 1-2u+2u^2, q-1 \right) . \end{aligned}$$

Now, $(u, 1-2u+2u^2)=1$, thus $(d_1, d_4)=1$. Furthermore, we have $(1-u, 1-2u+2u^2)=1$, thus $(d_2, d_4)=1$. Finally, from $\left( 1-u+u^2, 1-2u+2u^2 \right) = 1$ we obtain $\left( d_3, d_4 \right) = 1$. Thus, we have

$$\begin{aligned} \left( d_1d_2d_3, d_4 \right) = 1. \end{aligned}$$

(3.7)

Lemma 3.3

If $d_4 = 1$, then $H_{d_1} \cap H_{d_2} \subseteq H$.

Proof

Let $d_4 = 1$. Applying (2.1) with $a = u$, $s=u$ and $b=q-1 \equiv 0$ we obtain $as + b(1-s) \equiv u^2 + 0\cdot \left( 1 - u \right) = u^2 \in H$. Applying (2.2) we have $1-u^2 \in H$. Applying Proposition 2.6 with $s \equiv 2 \left( u - u^2 \right)$ we obtain that $t + k\left( 2u-2u^2 \right) \in H$ for any $t \in H$ and nonnegative integer k. With the choices of Table 1 we obtain that for all $t \in \left\{ \,0,1,u,1-u\,\right\}$ and for every integer l (whether l is even or odd) we have $t+l \left( u-u^2 \right) \in H$.

Table 1 $t + l(u-u^2) \in H$ for every integer l and $t \in \{ 0, 1, u, 1-u \}$

Now, let m be an arbitrary nonnegative integer, and let x be a positive integer solution of the congruence

$$\begin{aligned} \left( 1-2u+2u^2 \right) \left( u-u^2 \right) \cdot x \equiv m d_1 d_2 \pmod {q-1}. \end{aligned}$$

Such x exists, because $\left( 1-2u+2u^2, q-1 \right) = 1$ and $\left( u - u^2, q-1 \right) = d_1d_2$. Choosing $l = \left( 1-2u+2u^2 \right) x$ and $t \in \left\{ \,0, 1, u, 1-u \,\right\}$ (then $t \in H$) we obtain that $md_1d_2$, $1+md_1d_2$, $u+md_1d_2$ and $1-u+md_1d_2 \pmod {q-1}$ are all in H. Therefore, by Lemma 3.1 we have $H_{d_1} \cap H_{d_2} \subseteq H$. $\square$

Thus, Theorem 2.4 holds if $d_4 = 1$. In the following we assume $d_4>1$.

Lemma 3.4

For every $v \in H$ and for an arbitrary integer l we have $v + l \cdot 2 d_1d_2d_4 \in H$.

Proof

Let $v \in H$ be arbitrary, and let l be an arbitrary integer. By (3.6) we have that $\left( 2u-2u^2, q-1 \right)$ is either $d_1d_2$ or $2d_1d_2$. Now, if $\left( 2u-2u^2, q-1 \right) = 2d_1d_2$, then from $d_4 >1$ we obtain by induction that $H_{2d_1d_2} \cap H_{d_4} = \left\langle 2u-2u^2 \right\rangle \subseteq H$. Choosing $k=l$, Lemma 3.1 yields that $v + l\cdot 2 d_1d_2d_4 \in H$.

If $\left( 2u-2u^2, q-1 \right) = d_1d_2$, then from $d_4 >1$ we obtain by induction that $H_{d_1d_2} \cap H_{d_4} = \left\langle 2u-2u^2 \right\rangle \subseteq H$. Then choosing $k = 2l$, Lemma 3.1 yields that $v + 2l\cdot d_1d_2d_4 \in H$. $\square$

Finishing the proof of Theorem 2.4. Let $t \in \left\{ \,0, 1, u, 1-u\,\right\}$ be arbitrary, and let m be an arbitrary integer. We prove $t + md_1d_2 \in H$, which establishes $H_{d_1} \cap H_{d_2} \subseteq H$ and finishes the proof of Theorem 2.4. From (3.7) we have $\left( d_3, d_4 \right) = 1$. From (3.4) we have $\left( d_3, 2 \right) =1$. Thus $\left( d_3, 2d_4 \right) = 1$. Therefore, there exist integers x, y such that

$$\begin{aligned} x d_3 + y 2d_4 =1. \end{aligned}$$

From $d_3 > 1$ by induction we have $H_{d_1d_2} \cap H_{d_3} = \left\langle u-u^2 \right\rangle \subseteq H$. Let

$$\begin{aligned} v = t + mx\cdot d_1d_2d_3. \end{aligned}$$

By choosing $k=mx$, Lemma 3.1 yields $v \in H_{d_1d_2} \cap H_{d_3} = \left\langle u-u^2 \right\rangle \subseteq H$. By choosing $l=my$, Lemma 3.4 yields $v + my \cdot 2 d_1d_2d_4 \in H$. That is,

$$\begin{aligned} v + my\cdot 2 d_1d_2d_4= & {} t + mx\cdot d_1d_2d_3 + my \cdot 2d_1d_2d_4 \\= & {} t + \left( xd_3 + 2 y d_4 \right) \cdot md_1d_2 \\= & {} t + md_1d_2 \in H. \end{aligned}$$

Thus, for every $t \in \left\{ \,0, 1, u, 1-u\,\right\}$ and for an arbitrary integer m we have $t + md_1d_2 \in H$, establishing $H_{d_1} \cap H_{d_2} \subseteq H$. Theorem 2.4 is proved. $\square$

4 Open questions

It looks rather difficult to answer a general question on monomial clones. It does not seem feasible to continue along idempotent clones on several variables before understanding all binary monomial clones.

Problem 1

Find all binary monomial clones over $\mathbb {F}_q$.

The following conjecture could be a good start:

Conjecture 2

Every binary monomial clone can be obtained as an intersection of some $H_d$-s.

Another approach could be to omit monomiality. As every finite clone contains idempotent elements, it makes sense to look for idempotent polynomials in general.

Problem 3

Find all binary idempotent clones over $\mathbb {F}_q$.