Algebra universalis

, 80:23

Monomial clones over $$\mathbb {F}_q$$

Open Access
Article

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

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  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  a description of minimal linear polynomials and binary minimal monomials were given. The investigation was extended in  to the case of ternary majority minimal polynomials over $$\mathbb {F}_3$$. Recently in  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 , 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  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 .

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 $$\mathbb {F}_q$$.

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Machida, H., Pantović, J.: Three classes of closed sets of monomials. In: 2017 IEEE 47th International Symposium on Multiple-Valued Logic, pp. 100–105. IEEE Computer Society, Los Alamitos (2017)Google Scholar
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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