Algebra universalis

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

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


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.


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
$$\begin{aligned} \left( x^a y^{a'} \right) ^s \left( x^by^{b'} \right) ^{s'} = x^{as+bs'} y^{a's+b's'} \in \mathcal {C}. \end{aligned}$$
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
$$\begin{aligned} as+b(1-s) \pmod {q-1}&\in H, \end{aligned}$$
$$\begin{aligned} 1-a \pmod {q-1}&\in H. \end{aligned}$$
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 \(t + ks \in \left\langle u \right\rangle \).


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 \(\left\langle u \right\rangle = \left\{ \,1, \dots , q-1\,\right\} \).


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
$$\begin{aligned} \left( d_1, d_2 \right) = 1. \end{aligned}$$
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 \(v + kd_1d_2 \in H_{d_1} \cap H_{d_2}\).


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
$$\begin{aligned} as + b(1-s) \equiv u \left( 1 - u \right) + 0 (1-s) = u-u^2 \in H. \end{aligned}$$
Since \((u, 1-u) = 1\), we have
$$\begin{aligned}&\displaystyle \left( u-u^2, q-1 \right) = \left( u(1-u), q-1 \right)&\nonumber \\&\displaystyle \quad = \left( u, q-1 \right) \cdot \left( 1-u, q-1 \right) = d_1d_2.&\end{aligned}$$
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
$$\begin{aligned} \left( 2d_1d_2, d_3 \right) = 1. \end{aligned}$$

Lemma 3.2

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


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
$$\begin{aligned} as + b(1-s) \equiv u^2 + \left( 1 - u \right) ^2 = 1 -2u+2u^2 \in H. \end{aligned}$$
Since \((u-u^2, q-1) = d_1d_2\), we have
$$\begin{aligned} \left( 2u-2u^2, q-1 \right) \in \left\{ \, d_1d_2, 2d_1d_2\,\right\} . \end{aligned}$$
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
$$\begin{aligned} \left( d_1d_2d_3, d_4 \right) = 1. \end{aligned}$$

Lemma 3.3

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


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


\(\in H\)


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


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


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


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


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


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


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


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

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 \(v + l \cdot 2 d_1d_2d_4 \in H\).


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 \(\mathbb {F}_q\).



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 (, 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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