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Algebra universalis

, 80:23 | Cite as

Monomial clones over \(\mathbb {F}_q\)

  • Gábor HorváthEmail author
  • Kamilla Kátai-Urbán
  • Csaba Szabó
Open Access
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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 Theorem 

Mathematics Subject Classification

08A40 11A07 

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 ab 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 (2.12.2). 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 (2.12.2) 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 (2.12.2) 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 (2.12.2). 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 (2.12.2).

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 \(t \in \{ 0, 1, u, 1-u \}\)

t

\(\in H\)

0

\(2k \left( u - u^2 \right) \)

1

\(1+2k \left( u - u^2 \right) \)

u

\(u+2k \left( u - u^2 \right) \)

\(1-u\)

\(1-u+2k \left( u - u^2 \right) \)

\(u-u^2\)

\(\left( 2k+1 \right) \left( u - u^2 \right) \)

\(1-u+u^2\)

\(1+\left( 2k-1 \right) \left( u - u^2 \right) \)

\(u^2\)

\(u+\left( 2k-1 \right) \left( u - u^2 \right) \)

\(1-u^2\)

\(1-u+\left( 2k+1 \right) \left( u - u^2 \right) \)

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 xy 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 \(\mathbb {F}_q\).

Notes

Acknowledgements

Open access funding provided by University of Debrecen (DE).

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Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  • Gábor Horváth
    • 1
    Email author
  • Kamilla Kátai-Urbán
    • 2
  • Csaba Szabó
    • 3
  1. 1.Institute of MathematicsUniversity of DebrecenDebrecenHungary
  2. 2.Bolyai InstituteUniversity of SzegedSzegedHungary
  3. 3.Department of Algebra and Number TheoryEötvös Loránd UniversityBudapestHungary

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