Monomial clones over \(\mathbb {F}_q\)
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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.
Keywords
Finite fields Clone Monomial clone Idempotent clone Congruences Chinese Remainder TheoremMathematics Subject Classification
08A40 11A071 Introduction
Let q be a prime power and let \(\mathbb {F}_q\) denote the q element field. Every nvariable polynomial over \(\mathbb {F}_q\) defines a polynomial function over \(\mathbb {F}_q\), and every nvariable function is uniquely expressed as an nvariable 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 \(q1\). 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 \(x^{a+q1}y^b\) or \(x^ay^{b+q1}\), since \(x \mapsto x^q\) is the identity function. Therefore, in the paper we mainly will be interested in the modulo \(q1\) residues of the exponents of binary monomials. The modulo \(q1\) residues will be mostly taken from the set \(\left\{ \,1, \dots , q1\,\right\} \).
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 \(a+a' \equiv 1 \pmod {q1}\). 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 \(a+a' \equiv b\,{+}\,b' \equiv s\,{+}\,s' \equiv 1 \pmod {q1}\), then \(as \,{+}\, bs' \,{+} a's \,{+}\, b's' \equiv 1 \pmod {q1}\), 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
Example 2.1
The smallest and largest binary monomial clones have already been determined in [4].
Proposition 2.2
[4, Proposition 5.2]. \(\left\langle 2 \right\rangle = \left\{ \,1, \dots , q1\,\right\} \).
Proposition 2.3
[4, Proposition 5.7]. For arbitrary \(1\le u\le q1\) we have \(\left\{ \,1, q1\,\right\} = \left\langle 1 \right\rangle \subseteq \left\langle u \right\rangle \).
In the following we give a complete characterization of \(\left\langle u \right\rangle \) for all \(1\le u\le q1\) 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 \(q1\) remainder of a from the set \(\left\{ \,1, \dots , q1\,\right\} \) is in H. For example, \(q \in H\) means that 1 is in H, and \(0 \in H\) means that \(q1\) is in H. Moreover, when we simply write \(a \equiv b\) without specifying the module of the congruence, we mean \(a \equiv b \pmod {q1}\).
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. \(\left( a; b \right) \).
Theorem 2.4
Example 2.5
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 \(1s\) is invertible modulo \({q1}\). 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
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 \(1\le u\le q1\) be an integer such that both u and \(1u\) are invertible modulo \(q1\). Then \(\left\langle u \right\rangle = \left\{ \,1, \dots , q1\,\right\} \).
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 \(H \ni t+s = ku+u = (k+1)u\).
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 \(H \supseteq H_{d_1} \cap H_{d_2}\)
Lemma 3.1
Proof

\(v \equiv 0 \pmod {d_1}\) 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 \(d_2 \mid 1u\), 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 \(d_2 \mid 1u\), we have \(1u \equiv 1 \pmod {d_1}\) and \(1u \equiv 0 \pmod {d_2}\). By the Chinese remainder theorem, \(v \equiv 1u \pmod {d_1 d_2}\), and hence there exists an integer m such that \(v = 1u+md_1d_2\).\(\square \)
From (3.1) we have \((d_1, d_2)=1\), hence \(d_1d_2 \mid q1\). In the following we prove \(H \supseteq H_{d_1} \cap H_{d_2}\) by downward induction on \(d_1d_2\). If \(d_1d_2 = q1\), then \(H_{d_1} \cap H_{d_2} = \left\{ \,0,1,u,1u\,\right\} \) by Lemma 3.1. Since \(0, 1, u, 1u \in H\), we obtain \(H_{d_1} \cap H_{d_2} \subseteq H\).
Lemma 3.2
If \(d_3 = 1\), then \(H_{d_1} \cap H_{d_2} \subseteq H\).
Proof
\(\square \)
Lemma 3.3
If \(d_4 = 1\), then \(H_{d_1} \cap H_{d_2} \subseteq H\).
Proof
\(t + l(uu^2) \in H\) for every integer l and \(t \in \{ 0, 1, u, 1u \}\)
t  \(\in H\) 

0  \(2k \left( u  u^2 \right) \) 
1  \(1+2k \left( u  u^2 \right) \) 
u  \(u+2k \left( u  u^2 \right) \) 
\(1u\)  \(1u+2k \left( u  u^2 \right) \) 
\(uu^2\)  \(\left( 2k+1 \right) \left( u  u^2 \right) \) 
\(1u+u^2\)  \(1+\left( 2k1 \right) \left( u  u^2 \right) \) 
\(u^2\)  \(u+\left( 2k1 \right) \left( u  u^2 \right) \) 
\(1u^2\)  \(1u+\left( 2k+1 \right) \left( u  u^2 \right) \) 
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 \(\left( 2u2u^2, q1 \right) \) is either \(d_1d_2\) or \(2d_1d_2\). Now, if \(\left( 2u2u^2, q1 \right) = 2d_1d_2\), then from \(d_4 >1\) we obtain by induction that \(H_{2d_1d_2} \cap H_{d_4} = \left\langle 2u2u^2 \right\rangle \subseteq H\). Choosing \(k=l\), Lemma 3.1 yields that \(v + l\cdot 2 d_1d_2d_4 \in H\).
If \(\left( 2u2u^2, q1 \right) = d_1d_2\), then from \(d_4 >1\) we obtain by induction that \(H_{d_1d_2} \cap H_{d_4} = \left\langle 2u2u^2 \right\rangle \subseteq H\). Then choosing \(k = 2l\), Lemma 3.1 yields that \(v + 2l\cdot d_1d_2d_4 \in H\). \(\square \)
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 \(\mathbb {F}_q\).
Notes
Acknowledgements
Open access funding provided by University of Debrecen (DE).
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