# On the number of universal algebraic geometries

Erhard Aichinger  Institut für Algebra, Johannes Kepler Universität Linz,
Altenberger Straße 69, 4040 Linz, Austria<EMAIL_ADDRESS>and
Bernardo Rossi  Institut für Algebra, Johannes Kepler Universität Linz,
Altenberger Straße 69, 4040 Linz, Austria<EMAIL_ADDRESS>
###### Abstract.

The _algebraic geometry_ of a universal algebra $\mathbf{A}$ is defined as the
collection of solution sets of systems of term equations. Two algebras
$\mathbf{A}_{1}$ and $\mathbf{A}_{2}$ are called _algebraically equivalent_ if
they have the same algebraic geometry. We prove that on a finite set $A$ with
$\lvert A\rvert>3$ there are countably many algebraically inequivalent Mal’cev
algebras and that on a finite set $A$ with $\lvert A\rvert>2$ there are
continuously many algebraically inequivalent algebras.

###### Key words and phrases:

Universal algebraic geometry, Clones, Algebraic sets

###### 2010 Mathematics Subject Classification:

08A40, (08A62, 08B05, 03C05)

Supported by the Austrian Science Fund (FWF): P33878.

##  1\. Introduction

Universal algebraic geometry, introduced in [18, 10], is based on the notion
of _algebraic sets_. Given a universal algebra ${\mathbf{A}}=(A;(f_{i})_{i\in
I})$, a subset $B$ of $A^{n}$ is _algebraic_ if it is the solution set of a
system of (possibly infinitely many) term equations in the language of
${\mathbf{A}}$. Following [17], we denote the collection of the algebraic
subsets of $A^{n}$ by $\operatorname{Alg}_{n}{\mathbf{A}}$ and we define the
_universal algebraic geometry_ of ${\mathbf{A}}$ by
$\operatorname{Alg}{\mathbf{A}}:=\bigcup_{n\in\mathbb{N}}\operatorname{Alg}_{n}{\mathbf{A}}$.
Clearly, for a universal algebra ${\mathbf{A}}$,
$\operatorname{Alg}{\mathbf{A}}$ is completely determined by its clone of term
functions (cf. [8, 19]). We define the universal algebraic geometry of a clone
$\mathcal{C}$ on a set $A$ by
$\operatorname{Alg}\mathcal{C}:=\operatorname{Alg}(A;\mathcal{C})$. Following
[17], we say that two clones $\mathcal{C}$ and $\mathcal{D}$ on the same set
$A$ are algebraically equivalent,
$\mathcal{C}\operatorname{\sim_{\text{alg}}}\mathcal{D}$, if
$\operatorname{Alg}\mathcal{C}=\operatorname{Alg}\mathcal{D}$. In [17, 21] one
finds examples of different clones that are algebraically equivalent; in fact
Tóth and Waldhauser [21] proved that on the two element set there are only
finitely many algebraically inequivalent clones. Pinus showed that on each
finite set there are only finitely many universal algebraic geometries closed
under taking unions [17, 6]. The aim of this note is to provide infinite
families of algebraically inequivalent clones on a finite set.

Analysing the construction of $2^{\aleph_{0}}$ clones on $\\{0,1,2\\}$ in
[13], we obtain:

######  Theorem 1.1.

Let $A$ be a finite set with $\lvert A\rvert\geq 3$. Then there are
$2^{\aleph_{0}}$ algebraically inequivalent clones on $A$.

The proof will be given in Section 5.

A clone $\mathcal{C}$ is called a _Mal’cev clone_ if it contains a ternary
function such that for all $a,b\in A$,

$d(a,b,b)=d(b,b,a)=a,$

and it is called _constantive_ if every unary constant function on $A$ is in
$\mathcal{C}$. Using Idziak’s construction from [12], we prove:

######  Theorem 1.2.

Let $A$ be a finite set with $\lvert A\rvert\geq 4$. Then there are exactly
$\aleph_{0}$ algebraically inequivalent constantive Mal’cev clones on $A$.

The proof will be given in Section 4. It relies on the following fact:

######  Theorem 1.3.

Let $p$ be a prime, and let $n\in\mathbb{N}$. Then there are exactly
$\aleph_{0}$ algebraically inequivalent clones that contain
$\operatorname{Pol}(\mathbb{Z}_{np^{2}},+)$.

The proof will be given in Section 3.

In Table 1 we summarize our current knowledge on the number of algebraically
inequivalent clones on a finite set.

Property of the clone on an $n$-element set |  Number of clones |  Number of clones modulo $\operatorname{\sim_{\text{alg}}}$
---|---|---
$n=2$ |  $\aleph_{0}$ [20] |  finite [21]
$n\geq 3$ |  $2^{\aleph_{0}}$ [13] |  $2^{\aleph_{0}}$ (Theorem 1.1)
constantive, $n\geq 3$ |  $2^{\aleph_{0}}$ [1] |  ?
equationally additive |  ? |  finite [17, 6]
Mal’cev, $n\leq 3$ |  finite [7] |  finite
constantive, Mal’cev, $n\geq 4$ |  $\aleph_{0}$ [2, 12] |  $\aleph_{0}$ (Theorem 1.2)
contains $\operatorname{Clo}(\mathbb{Z}_{n},+)$, $n$ squarefree |  finite [11, 4, 15] |  finite
contains $\operatorname{Pol}(\mathbb{Z}_{mp^{2}},+)$, $m\geq 1$ |  $\aleph_{0}$ [2, 12] |  $\aleph_{0}$ (Theorem 1.3)
Table 1. The first column lists the properties of the clones we consider, the
second column reports the number of such clones and the third column lists the
number of such clones modulo $\operatorname{\sim_{\text{alg}}}$. The number
$p$ is assumed to be prime. Question marks represent open problems.

##  2\. Notation

We write $\mathbb{N}$ for the set of positive integers and for
$n\in\mathbb{N}$, $[{n}]:=\\{1,\dots,n\\}$. For a set $A$ the $i$-th component
of $\mathbfsl{a}\in A^{n}$ is denoted by $a_{i}$ and $\mathbfsl{a}(i)$. We
will use the notions _clone_ , $\operatorname{Pol}_{n}({\mathbf{A}})$ and
$\operatorname{Clo}_{n}({\mathbf{A}})$ as they are commonly used in universal
algebra [8, 16]. For $k\in\mathbb{N}$ with $k\leq n$, we define the $k$-th
$n$-ary projection $\pi_{k}^{(n)}:A^{n}\to A$ by
$\pi_{k}^{(n)}(a_{1},\dots,a_{n})=a_{k}$ for all $\mathbfsl{a}\in A^{n}$. We
set $\mathcal{J}_{A}:=\\{\pi_{k}^{(n)}\mid n\in\mathbb{N}\text{ and }k\leq
n\\}$. For a clone $\mathcal{C}$ on $A$, $\mathcal{C}^{[n]}$ is the set of
$n$-ary functions in $\mathcal{C}$. For $k\in\mathbb{N}$, for
$\sigma\colon[{k}]\to[{n}]$ and for $f\in\mathcal{C}^{[k]}$, we define
$f_{\sigma}\in\mathcal{C}^{[n]}$ by
$f_{\sigma}(x_{1},\dots,x_{n}):=f(x_{\sigma(1)},\dots,x_{\sigma(k)})$ for all
$x_{1},\dots,x_{n}\in A$, and we call $f_{\sigma}$ a _minor_ of $f$. If $t$ is
a $k$-ary term in the language of a universal algebra ${\mathbf{A}}$, we write
$t_{\sigma}$ for the term $t(x_{\sigma(1)},\dots,x_{\sigma(k)})$. Observe that
in this case $(t_{\sigma})^{\mathbf{A}}=(t^{\mathbf{A}})_{\sigma}$.

Let $A$ be a set and let $\mathcal{C}$ be a clone on $A$. For $n\in\mathbb{N}$
and for $X\subseteq A^{n}$, we define $\text{V}_{\mathcal{C}}(X)$ to be the
intersection of all the elements of $\operatorname{Alg}_{n}\mathcal{C}$ that
contain $X$ as a subset. Since the intersection of a collection of elements of
$\operatorname{Alg}_{n}\mathcal{C}$ is an element of
$\operatorname{Alg}_{n}\mathcal{C}$, we infer that
$\text{V}_{\mathcal{C}}(X)\in\operatorname{Alg}_{n}\mathcal{C}$. The following
lemma will be used to assess whether a set $X$ is algebraic with respect to a
clone $\mathcal{C}$, which is equivalent to $X=\text{V}_{\mathcal{C}}(X)$.

######  Lemma 2.1.

Let $A$ be a set, let $\mathcal{C}$ be a clone on $A$, let $n\in\mathbb{N}$,
let $X\subseteq A^{n}$, and let $\mathbfsl{a}\in A^{n}$. Then we have

$\mathbfsl{a}\in\text{V}_{\mathcal{C}}(X)\Longleftrightarrow\Bigl{(}\forall
f,g\in\mathcal{C}^{[n]}:f|_{X}=g|_{X}\Rightarrow
f(\mathbfsl{a})=g(\mathbfsl{a})\Bigr{)}.$

###### Proof.

“$\Leftarrow$”: We assume that for all $f,g\in\mathcal{C}^{[n]}$ with
$f|_{X}=g|_{X}$, we have $f(\mathbfsl{a})=g(\mathbfsl{a})$. We show that
$\mathbfsl{a}\in\text{V}_{\mathcal{C}}(X)$. To this end, we show that for each
$B\in\operatorname{Alg}_{n}\mathcal{C}$ with $X\subseteq B$, we have
$\mathbfsl{a}\in B$. Since $B$ is algebraic with respect to $\mathcal{C}$,
there exists a system of $\mathcal{C}$-equations $\\{p_{i}\approx q_{i}\mid
i\in I\\}$ such that $B=\\{\mathbfsl{x}\in A^{n}\mid\forall i\in I\colon
p_{i}(\mathbfsl{x})=q_{i}(\mathbfsl{x})\\}$. Since $X\subseteq B$, we have
$p_{i}|_{X}=q_{i}|_{X}$ for all $i\in I$. Therefore, the assumption yields
$p_{i}(\mathbfsl{a})=q_{i}(\mathbfsl{a})$ for all $i\in I$, and thus
$\mathbfsl{a}\in B$.

“$\Rightarrow$”: We assume that there exist $f,g\in\mathcal{C}^{[n]}$ such
that $f|_{X}=g|_{X}$ and $f(\mathbfsl{a})\neq g(\mathbfsl{a})$. Then
$B:=\\{\mathbfsl{x}\in A^{n}\mid f(\mathbfsl{x})=g(\mathbfsl{x})\\}$ satisfies
$B\in\operatorname{Alg}_{n}\mathcal{C}$, $X\subseteq B$, and
$\mathbfsl{a}\notin B$. Thus, $\mathbfsl{a}\notin\text{V}_{\mathcal{C}}(X)$. ∎

##  3\. Countably many algebraically inequivalent constantive expansions of
$\mathbb{Z}_{np^{2}}$

Let ${\mathbf{A}}=(A;+,-,0,(f_{i})_{i\in I})$ be an expanded group. Following
[3], a function $f\colon A^{n}\to A$ is _absorbing_ if
$f(a_{1},\dots,a_{n})=0$ for all $a_{1},\dots,a_{n}\in A$ with
$0\in\\{a_{1},\dots,a_{n}\\}$.

######  Lemma 3.1.

Let $p$ be a prime, let $n,d,k\in\mathbb{N}$ with $d\geq 2$ and $k\geq d+1$,
let ${\mathbf{A}}_{d}$ be the algebra $(\mathbb{Z}_{np^{2}};+,-,0,f_{d})$,
where $f_{d}\colon\mathbb{Z}_{np^{2}}^{d}\to\mathbb{Z}_{np^{2}}$ is defined by

$f_{d}(x_{1},\dots,x_{d})=np\prod_{j=1}^{d}x_{j}\text{ for
}x_{1},\dots,x_{d}\in\mathbb{Z}_{np^{2}},$

and let $g\in\operatorname{Pol}_{k}({\mathbf{A}}_{d})$. If $g$ is absorbing,
then $g$ is the constant zero function.

Lemma 3.1 states that the algebra ${\mathbf{A}}_{d}$ is $d$-supernilpotent
(cf. [5]). We provide a proof that makes no use of the notion of
supernilpotency.

###### Proof.

Let $\mathcal{C}_{d}$ be the set of all operations on $\mathbb{Z}_{np^{2}}$
that are induced by a sum of monomials of the form $ax_{i_{1}}^{n_{1}}\dots
x_{i_{r}}^{n_{r}}$, where $n_{1},\dots,n_{r}\in\mathbb{N}$,
$x_{i_{1}},\dots,x_{i_{r}}$ are pairwise distinct variables,
$a\in\mathbb{Z}_{np^{2}}$, and the following property is satisfied:

(3.1) $(r=0)\text{ or }(r=n_{1}=1)\text{ or }(2\leq n_{1}+\dots+n_{r}\leq
d\text{ and }np\text{ divides }a).$

It is clear that $\\{+,-,0,f_{d}\\}\subseteq\mathcal{C}_{d}$, that the
projection clone satisfies
$\mathcal{J}_{\mathbb{Z}_{np^{2}}}\subseteq\mathcal{C}_{d}$, and that
$\mathcal{C}_{d}$ contains the constant functions. Furthermore, it is
straightforward to verify that for $t_{1},t_{2},\dots,t_{d}\in\mathcal{C}_{d}$
we have that $t_{1}+t_{2}\in\mathcal{C}_{d}$, $t_{1}-t_{2}\in\mathcal{C}_{d}$
and $f_{d}(t_{1},\dots,t_{d})\in\mathcal{C}_{d}$. Hence
$\operatorname{Pol}({\mathbf{A}}_{d})\subseteq\mathcal{C}_{d}$. We assume that
$g\in\operatorname{Pol}_{k}({\mathbf{A}}_{d})$ is absorbing, and consider the
representation of $g$ as a sum of monomials, each one satisfying (3.1). Since
$g$ is absorbing, $g(0,x_{2},\dots,x_{k})$ is the constant $0$-function. Thus,
the sum of those monomials of $g$ that do not involve $x_{1}$ induces the
constant zero function. Therefore, there exists a representation of $g$ as a
sum of monomials of the above form, each of them involving the variable
$x_{1}$. A similar argument applies to all the other variables
$x_{2},\dots,x_{k}$. Therefore, there exists a (possibly empty) sum of
monomials that induces the same polynomial function as $g$ such that each
monomial involves all the variables $x_{1},\dots,x_{k}$ and satisfies (3.1).
If $x_{i_{1}}^{n_{1}}\dots x_{i_{r}}^{n_{r}}$ is such a monomial, then $r\geq
k$, and therefore $n_{1}+\cdots+n_{r}\geq k$. Since $k\geq 3$, (3.1) implies
$d\geq n_{1}+\cdots+n_{r}\geq k$, contradicting the assumption $k\geq d+1$.
Hence $g$ is induced by the empty sum of monomials, and therefore it is the
constant $0$-function. ∎

###### Proof of Theorem 1.3.

For each $d\in\mathbb{N}\setminus\\{1\\}$, let ${\mathbf{A}}_{d}$ and $f_{d}$
be as in the statement of Lemma 3.1. For $d\in\mathbb{N}\setminus\\{1\\}$, for
$m\in\mathbb{N}$ and for $X\subseteq\mathbb{Z}_{np^{2}}^{m}$, we use
$\text{V}_{d}(X)$ as a shorthand for
$\text{V}_{\operatorname{Pol}({\mathbf{A}}_{d})}(X)$. Let $l,i\in\mathbb{N}$
with $l>i\geq 2$. We claim that
$\operatorname{Pol}({\mathbf{A}}_{i})\not\sim_{\text{alg}}\operatorname{Pol}({\mathbf{A}}_{l})$.
For proving this, we let

$Q:=\\{\mathbfsl{a}\in\mathbb{Z}_{np^{2}}^{l}\mid\exists
s\in[{l}]:\mathbfsl{a}(s)=0\\}$

and show that $\text{V}_{i}(Q)\neq\text{V}_{l}(Q)$; as a consequence
$\operatorname{Pol}({\mathbf{A}}_{i})\not\sim_{\text{alg}}\operatorname{Pol}({\mathbf{A}}_{l})$.

First we prove that $(1,\dots,1)\notin\text{V}_{l}(Q)$. To this end, we
consider the term equation $f_{l}(x_{1},\dots,x_{l})\approx 0$. We have that
$f_{l}(1,\dots,1)=np$ and $f_{l}|_{Q}=0$. Hence Lemma 2.1 yields
$(1,\dots,1)\not\in\text{V}_{l}(Q)$.

We now prove that $\text{V}_{i}(Q)=\mathbb{Z}_{np^{2}}^{l}$. To this end, let
us consider an equation of the form $t_{1}(x_{1},\dots,x_{l})\approx
t_{2}(x_{1},\dots,x_{l})$ with
$t_{1},t_{2}\in\operatorname{Pol}_{l}({\mathbf{A}}_{i})$. This equation is
equivalent to $(t_{1}-t_{2})\,(x_{1},\dots,x_{l})\approx 0$. Hence it suffices
to consider equations of the form $t(x_{1},\dots,x_{l})\approx 0$. If
$t(x_{1},\dots,x_{l})\approx 0$ is satisfied by all elements of $Q$, then the
function $t$ is absorbing and has arity greater than $i$. Lemma 3.1 implies
that $t$ is the constant zero function. Therefore, the equation
$t(x_{1},\dots,x_{l})\approx 0$ is satisfied by all elements in
$\mathbb{Z}_{np^{2}}^{l}$. Hence Lemma 2.1 yields $(1,\dots
1)\in\text{V}_{i}(Q)$ and therefore $\text{V}_{i}(Q)\neq\text{V}_{l}(Q)$.

We conclude that
$(\operatorname{Pol}({\mathbf{A}}_{i}))_{i\in\mathbb{N}\setminus\\{1\\}}$ is
an infinite family of algebraically inequivalent clones. ∎

##  4\. Countably many algebraically inequivalent constantive Mal’cev clones
on finite sets with at least 4 elements

We report a construction from [12, Section 3]: Let $A$ be a finite set such
that $\lvert A\rvert\geq 4$ and let $\mathcal{C}$ be a constantive clone on
$A$ with a Mal’cev term $p$. The construction picks an element $1\in A$ and an
element $0\notin A$ and constructs a new clone $\mathcal{C}^{\oplus}$ on the
set $A^{\oplus}:=A\cup\\{0\\}$.

For all $k\in\mathbb{N}$ and for all $f\in\mathcal{C}$, the operation
$f^{\oplus}\colon(A^{\oplus})^{k}\to A^{\oplus}$ is defined by

$f^{\oplus}(x_{1},\dots,x_{k}):=\begin{cases}f(x_{1},\dots,x_{k})&\text{ if
}(x_{1},\dots,x_{k})\in A^{k},\\\ 0&\text{ otherwise.}\end{cases}$

A binary operation $\cdot$ on $A^{\oplus}$ is defined by

$x\cdot y:=\begin{cases}0&\text{ if }(x,y)\in A^{2},\\\ 1&\text{ if }x=y=0,\\\
x&\text{ if }x\in A\text{ and }y=0,\\\ y&\text{ if }x=0\text{ and }y\in
A.\end{cases}$

Finally, $\mathcal{C}^{\oplus}$ is defined as the clone on $A^{\oplus}$
generated by $\\{f^{\oplus}\mid f\in\mathcal{C}\\}\cup\\{\cdot\\}$ and all
unary constant operations.

######  Lemma 4.1.

Let $A$ be a finite set with $\lvert A\rvert\geq 4$, let $\mathcal{C}$ be a
constantive clone on $A$ and let $f\in\mathcal{C}^{\oplus}$. Then

(4.1) $f|_{A^{k}}=0\text{ or there exists }\hat{f}\in\mathcal{C}\text{ such
that }f|_{A^{k}}=\hat{f}.$

###### Proof.

Let $K$ be the set of all unary constant functions on $A^{\oplus}$. We will
show that all projections on $A^{\oplus}$ satisfy (4.1), and that for all
$k,n\in\mathbb{N}$, for all $n$-ary

$s\in\\{f^{\oplus}\mid f\in\mathcal{C}\\}\cup\\{\cdot\\}\cup K$

and for all $k$-ary $t_{1},\ldots,t_{n}$ that satisfy (4.1), also
$s(t_{1},\ldots,t_{n})$ satisfies (4.1). From this, we conclude that every
function in $\mathcal{C}^{\oplus}$ satisfies (4.1).

Clearly, all the projections satisfy (4.1).

Let $k,n\in\mathbb{N}$, let $t_{1},\ldots,t_{n}$ be $k$-ary functions on
$A^{\oplus}$ that satisfy (4.1), and let $s$ be an $n$-ary function on
$A^{\oplus}$ from $\\{f^{\oplus}\mid f\in\mathcal{C}\\}\cup\\{\cdot\\}\cup K$.
We will show that $s(t_{1},\ldots,t_{n})$ satisfies (4.1).

We first consider the case that $s=f^{\oplus}$ with $f\in\mathcal{C}^{[n]}$.
If there exists $i\in[{n}]$ such that $t_{i}|_{A^{k}}=0$, then by the
definition of $f^{\oplus}$, $f^{\oplus}(t_{1},\dots,t_{n})|_{A^{k}}=0$, and
hence $s(t_{1},\ldots,t_{n})$ satisfies (4.1). If for all $i\in[{n}]$, there
exists $\hat{t_{i}}\in\mathcal{C}^{[k]}$ such that
$t_{i}|_{A^{k}}=\hat{t_{i}}$, then by the definition of $f^{\oplus}$ we have

$s(t_{1},\ldots,t_{n})|_{A^{k}}=f^{\oplus}(t_{1},\dots,t_{n})|_{A^{k}}=f(\hat{g_{1}},\dots,\hat{g_{n}})\in\mathcal{C}.$

Therefore $s(t_{1},\ldots,t_{n})$ satisfies (4.1).

Now we consider the case that $s$ is the function $\cdot$ (and $n=2$). Let
$t_{1},t_{2}$ be $k$-ary functions on $A^{\oplus}$ satisfying (4.1). We show
that $t_{1}\cdot t_{2}$ satisfies (4.1). Since $t_{1},t_{2}$ satisfy (4.1), we
have to consider the following cases.

Case 1: $t_{1}|_{A^{k}}=t_{2}|_{A^{k}}=0$: In this case, the definition of
$\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=1$. Since $\mathcal{C}$ is
constantive, $t_{1}\cdot t_{2}|_{A^{k}}\in\mathcal{C}$, and thus $t_{1}\cdot
t_{2}$ satisfies (4.1).

Case 2: _$t_{1}|_{A^{k}}=0$ and there exists $\hat{t_{2}}\in\mathcal{C}^{[k]}$
such that $t_{2}|_{A^{k}}=\hat{t_{2}}$_: In this case, the definition of
$\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=\hat{t_{2}}$. This implies that
$t_{1}\cdot t_{2}|_{A^{k}}\in\mathcal{C}$, and thus $t_{1}\cdot t_{2}$
satisfies (4.1).

Case 3: _$t_{2}|_{A^{k}}=0$ and there exists $\hat{t_{1}}\in\mathcal{C}^{[k]}$
such that $t_{1}|_{A^{k}}=\hat{t_{1}}$_: In this case, the definition of
$\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=\hat{t_{1}}$. This implies that
$t_{1}\cdot t_{2}|_{A^{k}}\in\mathcal{C}$, and thus $t_{1}\cdot t_{2}$
satisfies (4.1).

Case 4: _There exist $\hat{g_{1}},\hat{g_{2}}\in\mathcal{C}^{[k]}$ such that
$g_{1}|_{A^{k}}=\hat{g_{1}}$ and $g_{2}|_{A^{k}}=\hat{g_{2}}$_: In this case,
the definition of $\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=0$, thus
$t_{1}\cdot t_{2}$ satisfies (4.1).

Finally, let us consider the case $s\in K$ (and $n=1$). Let $a\in A^{\oplus}$
be the unique function value of $s$, and let $t$ be a $k$-ary function on
$A^{\oplus}$ that satisfies (4.1). Clearly, $s(t)(x_{1},\dots,x_{k})=a$ for
all $x_{1},\dots,x_{k}\in A^{\oplus}$. If $a\in A$, since $\mathcal{C}$ is
constantive, we have that $s(t)$ satisfies (4.1). If $a=0$, then
$s(t)(x_{1},\dots,x_{k})=0$ for all $(x_{1},\dots,x_{k})\in A^{k}$. Thus,
$s(t)$ satisfies (4.1). ∎

######  Lemma 4.2.

Let $A$ be a finite set with $\lvert A\rvert\geq 4$, let $\mathcal{C}$ be a
constantive clone on $A$, let $f\in(\mathcal{C}^{\oplus})^{[k]}$, and let
$\mathbfsl{a},\mathbfsl{b}\in A^{k}$. Then $f(\mathbfsl{a})=0$ if and only if
$f(\mathbfsl{b})=0$.

###### Proof.

If $f(\mathbfsl{a})\neq 0$, then $f|_{A^{k}}\neq 0$. Thus by Lemma 4.1, there
exists $\hat{f}\in\mathcal{C}$ such that $f|_{A^{k}}=\hat{f}$, and therefore
$f(\mathbfsl{b})=\hat{f}(\mathbfsl{b})\in A$, which implies
$f(\mathbfsl{b})\neq 0$. ∎

######  Lemma 4.3.

[12, Section 3] Let $A$ be a finite set with $\lvert A\rvert\geq 4$ and let
$\mathcal{C}$ be a constantive Mal’cev clone on $A$. Then
$\mathcal{C}^{\oplus}$ is a constantive Mal’cev clone on a set of cardinality
$\lvert A\rvert+1$.

######  Proposition 4.4.

Let $A$ be a finite set with $\lvert A\rvert\geq 4$, and let
$\mathcal{C}_{1},\mathcal{C}_{2}$ be two constantive Mal’cev clones on $A$. If
$\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are not algebraically equivalent, then
$\mathcal{C}_{1}^{\oplus}$ and $\mathcal{C}_{2}^{\oplus}$ are not
algebraically equivalent.

###### Proof.

Without loss of generality, we assume that there exists $B\subseteq A^{k}$
that is algebraic with respect to $\mathcal{C}_{1}$ but not with respect to
$\mathcal{C}_{2}$. Note that then $B\neq A^{k}$. Since $\mathcal{C}_{2}$ is
constantive, $B$ is not empty. Then there exists $\mathbfsl{a}\in
A^{k}\setminus B$ such that $\mathbfsl{a}\in\text{V}_{\mathcal{C}_{2}}(B)$.

Let $B_{0}:=B\cup((A^{\oplus})^{k}\setminus A^{k})$. We claim that

(4.2) $B_{0}\in\operatorname{Alg}\mathcal{C}_{1}^{\oplus}.$

To prove (4.2), let $\mathbfsl{c}\notin B_{0}$. By Lemma 2.1 it suffices to
show that there are two functions $p_{1},p_{2}\in\mathcal{C}_{1}^{\oplus}$
such that $p_{1}|_{B_{0}}=p_{2}|_{B_{0}}$ and $p_{1}(\mathbfsl{c})\neq
p_{2}(\mathbfsl{c})$. Since $\mathbfsl{c}\in A^{k}\setminus B$ and $B$ is
algebraic with respect to $\mathcal{C}_{1}$, Lemma 2.1 implies that there are
$f_{1},f_{2}\in\mathcal{C}_{1}^{[k]}$ such that $f_{1}|_{B}=f_{2}|_{B}$ and
$f_{1}(\mathbfsl{c})\neq f_{2}(\mathbfsl{c})$. We set $p_{1}=f_{1}^{\oplus}$
and $p_{2}=f_{2}^{\oplus}$. Then $p_{1}|_{B_{0}}=p_{2}|_{B_{0}}$ and
$p_{1}(\mathbfsl{c})\neq p_{2}(\mathbfsl{c})$, which concludes the proof of
(4.2).

We now claim that

(4.3) $\mathbfsl{a}\in\text{V}_{\mathcal{C}_{2}^{\oplus}}(B_{0}).$

Suppose that $\mathbfsl{a}\notin\text{V}_{\mathcal{C}_{2}^{\oplus}}(B_{0})$.
Then Lemma 2.1 implies that there are
$f_{1},f_{2}\in(\mathcal{C}_{2}^{\oplus})^{[k]}$ such that
$f_{1}|_{B_{0}}=f_{2}|_{B_{0}}$ and $f_{1}(\mathbfsl{a})\neq
f_{2}(\mathbfsl{a})$. Since $B\subseteq B_{0}$ we infer that
$f_{1}|_{B}=f_{2}|_{B}$. From Lemma 4.2 and $B\neq\emptyset$, we deduce that
$f_{1}|_{B}=f_{2}|_{B}\neq 0$. Moreover, by Lemma 4.1, there are
$\hat{f_{1}},\hat{f_{2}}\in\mathcal{C}_{2}^{[k]}$ such that
$f_{1}|_{A^{k}}=\hat{f_{1}}$ and $f_{2}|_{A^{k}}=\hat{f_{2}}$. Therefore,
$\hat{f_{1}}|_{B}=\hat{f_{2}}|_{B}$ and
$\hat{f_{1}}(\mathbfsl{a})\neq\hat{f_{2}}(\mathbfsl{a})$. Hence by Lemma 2.1,
$\mathbfsl{a}\notin\text{V}_{\mathcal{C}_{2}}(B)$, contradicting the choice of
$\mathbfsl{a}$.

From (4.2) and (4.3) we obtain
$\operatorname{Alg}\mathcal{C}_{1}^{\oplus}\neq\operatorname{Alg}\mathcal{C}_{2}^{\oplus}$.
∎

###### Proof of Theorem 1.2.

We proceed by induction on $n=\lvert A\rvert$. For the base case $n=4$, we use
Theorem 1.3 with $n=1$ and $p=2$. For the induction step we let $n\geq 4$ and
$\\{\mathcal{C}_{i}\mid i\in\mathbb{N}\\}$ be a collection of pairwise
algebraically inequivalent constantive Mal’cev clones on a set of cardinality
$n$. Then Proposition 4.4 and Lemma 4.3 imply that
$\\{\mathcal{C}_{i}^{\oplus}\mid i\in\mathbb{N}\\}$ is a collection of
pairwise algebraically inequivalent constantive Mal’cev clones on a set of
cardinality $n+1$. ∎

##  5\. Continuously many algebraically inequivalent clones on sets with at
least 3 elements

Let $Z=\\{0,1,2\\}$. For $n\in\mathbb{N}$ and for $i\leq n$, a function
$f:Z^{n}\to Z$ _depends on its $i$-th argument_ if there exist
$\mathbfsl{a}\in Z^{n}$ and $b\in Z$ such that $f(\mathbfsl{a})\neq
f(a_{1},\dots,a_{i-1},b,a_{i+1},\dots,a_{n})$. If $f$ depends on exactly one
of its arguments, then $f$ is called _essentially unary_. The essential arity
of $f$ is defined by $\text{essArity}(f):=\lvert\\{i\in[{n}]:f\text{ depends
on its }i\text{-th argument}\\}\rvert$. Let $\mathcal{F}$ be a set of finitary
functions on $Z$. We say that _$f$ belongs to $\mathcal{F}$ up to inessential
arguments_ if there exist $l\in\mathbb{N}$, $1\leq i_{1}<\dots<i_{l}\leq n$
and a function $g\in\mathcal{F}^{[l]}$ such that
$f(x_{1},\dots,x_{n})=g(x_{i_{1}},\dots,x_{i_{l}})$ for all $\mathbfsl{x}\in
Z^{n}$.

Let $D_{1}=\emptyset$ and for each $m\in\mathbb{N}\setminus\\{1\\}$, let
$D_{m}:=\\{(1,2,\dots,2),\dots,(2,\dots,2,1)\\}$. For each $n\in\mathbb{N}$,
let $F_{n}$ be the set consisting of all the functions $f\colon Z^{n}\to Z$
such that

(5.1) $\displaystyle f(Z^{n})\subseteq\\{0,1\\},\text{ and}$ (5.2)
$\displaystyle f^{-1}(\\{1\\})\subseteq D_{n}.$

We set $F:=\bigcup_{n\in\mathbb{N}}F_{n}$ and

$F^{\prime}:=\bigcup_{n\in\mathbb{N}}\\{f\colon Z^{n}\to Z\mid f\text{ belongs
to }F\text{ up to inessential arguments}\\}\cup\mathcal{J}_{Z}.$

######  Lemma 5.1.

The sets $F$ and $F^{\prime}$ satisfy the following properties:

  1. (1)

For all $f\in F$ with $\text{essArity}(f)\leq 1$, we have $f=0$.

  2. (2)

For all $n\in\mathbb{N}$ and for all $f\in F_{n}$ with $\text{essArity}(f)>0$,
we have that $\text{essArity}(f)=n$ and $n\geq 2$.

  3. (3)

For all $f\in F^{\prime}$ with $\text{essArity}(f)=1$, we have
$f\in\mathcal{J}_{Z}$.

  4. (4)

For all $l,n\in\mathbb{N}$, for all $g\in F_{l}$ and for all
$\rho\colon[{l}]\to[{n}]$, we have $g_{\rho}\in F^{\prime}_{n}$.

  5. (5)

For all $k,n\in\mathbb{N}$, for all $f\in F^{\prime}_{k}$ and for all
$\sigma\colon[{k}]\to[{n}]$, we have $f_{\sigma}\in F^{\prime}_{n}$.

###### Proof.

(1) Let $n\in\mathbb{N}$, let $f\in F_{n}$ with $\text{essArity}(f)\leq 1$ and
let $\mathbfsl{x}\in Z^{n}$. We prove that $f(\mathbfsl{x})=0$. If
$\text{essArity}(f)\leq 1$, then there exists $i\in[{n}]$ such that for all
$j\in[{n}]\setminus\\{i\\}$ the function $f$ does not depend on its $j$-th
argument. Let us define $f^{\prime}\in Z^{Z}$ by $f^{\prime}(y)=f(y,\dots,y)$
for all $y\in Z$. We prove that $f^{\prime}=0$. To this end, let $a\in Z$.
Since $(a,\dots,a)\notin D_{n}$, we have $f^{\prime}(a)=f(a,\dots,a)=0$.
Moreover, since for all $j\neq i$, $f$ does not depend on its $j$-th argument,
$f(\mathbfsl{x})=f(x_{i},\dots,x_{i})=f^{\prime}(x_{i})=0$. This concludes the
proof of (1).

(2) Let $n\in\mathbb{N}$ and let $f\in F_{n}$. If $f$ is not constantly zero,
then there exists $\mathbfsl{a}\in D_{n}$ such that $f(\mathbfsl{a})=1$.
Changing any one of the coordinates of $\mathbfsl{a}$ to $0$, the value of $f$
changes from $1$ to $0$. Hence $f$ depends on all of its arguments and (1)
implies that it cannot be essentially unary. This concludes the proof of (2).

(3) Let $f\in F^{\prime}_{n}$ with $\text{essArity}(f)=1$. Seeking a
contradiction, we suppose that $f$ is not a projection. Then $f$ belongs to
$F$ up to inessential arguments. Thus, there exist $1\leq
j_{1}<\dots<j_{k}\leq n$ and $\hat{f}\in F$ such that
$f(x_{1},\dots,x_{n})=\hat{f}(x_{j_{1}},\dots,x_{j_{k}})$ for all
$\mathbfsl{x}\in Z^{n}$. If $\text{essArity}(\hat{f})\leq 1$, then (1) yields
$\hat{f}=0$, and so $f$ is the constant $0$-function, contradicting
$\text{essArity}(f)=1$. If $\text{essArity}(\hat{f})>1$, then
$\text{essArity}(f)>1$, contradicting $\text{essArity}(f)=1$. This concludes
the proof of (3).

(4) Let us assume that the image of $\rho$ is
$\\{j_{1},\dots,j_{k}\\}\subseteq[{n}]$, with $j_{1}<\dots<j_{k}$, and let us
define

$h:=\Set{\bigl{(}(x_{j_{1}},\dots,x_{j_{k}}),g(x_{\rho(1)},\dots,x_{\rho(l)})\bigr{)}\mid(x_{1},\dots,x_{n})\in
Z^{n}}.$

We first prove that $h$ is a functional relation. To this end, let
$\mathbfsl{x},\mathbfsl{y}\in Z^{n}$ be such that
$(x_{j_{1}},\dots,x_{j_{k}})=(y_{j_{1}},\dots,y_{j_{k}})$. We show that
$g(x_{\rho(1)},\dots,x_{\rho(l)})=g(y_{\rho(1)},\dots,y_{\rho(l)})$. Clearly,
if $(x_{j_{1}},\dots,x_{j_{k}})=(y_{j_{1}},\dots,y_{j_{k}})$, then
$(x_{\rho(1)},\ldots,x_{\rho(l)})=(y_{\rho(1)},\ldots,y_{\rho(l)})$, and
therefore $g(x_{\rho(1)},\dots,x_{\rho(l)})=g(y_{\rho(1)},\dots,y_{\rho(l)})$.
This concludes the proof that $h$ is a functional relation. Now we prove that
$h\in F_{k}$. Condition (5.1) is clearly satisfied since $g\in F_{l}$. We
prove that $h$ satisfies (5.2). To this end, let $\mathbfsl{x}\in A^{n}$ with
$g(x_{\rho(1)},\dots,x_{\rho(l)})=1$. We show that
$(x_{j_{1}},\dots,x_{j_{k}})\in D_{k}$. Since $g\in F_{l}$ and
$g(x_{\rho(1)},\dots,x_{\rho(l)})=1$, there exists $a\in[{l}]$ such that
$x_{\rho(a)}=1$ and $x_{\rho(b)}=2$ for all $b\in[{l}]\setminus\\{a\\}$. Since
$j_{1}<\dots<j_{k}$ we have $\lvert\\{r\in[{k}]\colon
j_{r}=\rho(a)\\}\rvert\leq 1$. Moreover, since $\\{j_{1},\dots,j_{k}\\}$ is
the image of $\rho$, we have $\lvert\\{r\in[{k}]\colon
j_{r}=\rho(a)\\}\rvert\geq 1$. Thus, $\lvert\\{r\in[{k}]\colon
j_{r}=\rho(a)\\}\rvert=1$. Hence there exists $i\in[{k}]$ such that
$x_{j_{i}}=x_{\rho(a)}=1$. Then for each $r\in[{k}]\setminus\\{i\\}$ there
exists $b\in[{l}]\setminus\\{a\\}$ such that $j_{r}=\rho(b)$, and then
$x_{j_{r}}=x_{\rho(b)}=2$. Thus, $(x_{j_{1}},\dots,x_{j_{k}})\in D_{k}$, and
so $h$ satisfies (5.2). This proves that $h\in F_{k}$. Since
$j_{1}<\dots<j_{k}$ and $g_{\rho}(\mathbfsl{a})=h(a_{j_{1}},\dots,a_{j_{k}})$
for all $\mathbfsl{a}\in A^{n}$, $g_{\rho}$ belongs to $F$ up to inessential
arguments. Hence $g_{\rho}\in F_{n}^{\prime}$. This concludes the proof of
(4).

(5) Let $k,n\in\mathbb{N}$, let $f\in F^{\prime}_{k}$ and let
$\sigma\colon[{k}]\to[{n}]$. We prove that $f_{\sigma}\in F^{\prime}_{n}$. By
the definition of $F^{\prime}$, there exist $l\in\mathbb{N}$, $g\in F_{l}$ and
$\tau\colon[{l}]\to[{k}]$ injective and increasing such that
$f(y_{1},\dots,y_{k})=g(y_{\tau(1)},\dots,y_{\tau(l)})$ for all
$\mathbfsl{y}\in Z^{k}$. Therefore, for all $\mathbfsl{x}\in Z^{n}$ we have

$f_{\sigma}(x_{1},\dots,x_{n})=f(x_{\sigma(1)},\dots,x_{\sigma(k)})=g(x_{\sigma(\tau(1))},\dots,x_{\sigma(\tau(l))}).$

Since $g\in F_{l}$, (4) implies that $g_{\sigma\circ\tau}\in F^{\prime}_{n}$,
and thus $f_{\sigma}\in F^{\prime}_{n}$. This concludes the proof of (5). ∎

######  Lemma 5.2.

The set $F^{\prime}$ is a clone.

###### Proof.

We prove that $F^{\prime}$ is closed under $\zeta$, $\tau$, $\Delta$, $\nabla$
and $*$ as they are defined in [14] (cf. [9, 19]). To this end, let $f\in
F^{\prime}_{n}$. Lemma 5.1(5) yields that $\mathop{\zeta}f\in F^{\prime}_{n}$,
$\mathop{\tau}f\in F^{\prime}_{n}$, $\mathop{\Delta}f\in F^{\prime}_{n-1}$,
and $\mathop{\nabla}f\in F^{\prime}_{n+1}$. Let $g\in F^{\prime}_{m}$ and let
$h$ be the function defined by
$h(\mathbfsl{x})=(f*g)(\mathbfsl{x})=f(g(x_{1},\dots,x_{m}),x_{m+1},\dots,x_{m+n-1})$
for all $\mathbfsl{x}\in Z^{m+n-1}$. We prove that $h$ belongs to
$F^{\prime}$. If $f$ is constant, then $h$ is a constant mapping with the same
function value as $f$. Hence $h$ is a minor of $f$, and thus by Lemma 5.1(5),
$h\in F^{\prime}$. If $f$ is a projection, then either $h$ is a projection or
$h=g$. In both cases $h$ belongs to $F^{\prime}$. If $g$ is a projection, then
$h$ is a minor of $f$. Hence Lemma 5.1(5) yields $h\in F^{\prime}$. Finally,
let us assume that both $f$ and $g$ belong to $F$ up to inessential arguments
and $f$ is not constant. Then there exist $1\leq
j_{1}<\dots<j_{n^{\prime}}\leq n$, $1\leq i_{1}<\dots<i_{m^{\prime}}\leq m$,
$\hat{f}\in F_{n^{\prime}}$ and $\hat{g}\in F_{m^{\prime}}$ such that for all
$\mathbfsl{x}\in Z^{n}$ and for all $\mathbfsl{y}\in Z^{m}$ we have
$f(\mathbfsl{x})=\hat{f}(x_{j_{1}},\dots,x_{j_{n^{\prime}}})$ and
$g(\mathbfsl{y})=\hat{g}(y_{i_{1}},\dots,y_{i_{m^{\prime}}})$. If $j_{1}\geq
2$, then for all $\mathbfsl{x}\in Z^{m+n-1}$ we have
$h(\mathbfsl{x})=\hat{f}(x_{m+j_{1}-1},\dots,x_{m+j_{n^{\prime}}-1})$. Thus,
$h$ is a minor of $\hat{f}$ and Lemma 5.1(4) yields $h\in F^{\prime}$. Let us
now assume that $j_{1}=1$ and let $\hat{h}$ be defined by
$\hat{h}(x_{1},\dots,x_{m^{\prime}+n^{\prime}-1})=\hat{f}(\hat{g}(x_{1},\dots,x_{m^{\prime}}),x_{m^{\prime}+1},\dots,x_{m^{\prime}+n^{\prime}-1})$
for all $\mathbfsl{x}\in Z^{m^{\prime}+n^{\prime}-1}$. We now prove that
$\hat{h}\in F$. Since $\hat{f}$ belongs to $F$, $\hat{h}$ satisfies (5.1). We
now prove that $\hat{h}$ satisfies (5.2). To this end, let $\mathbfsl{a}\in
Z^{m^{\prime}+n^{\prime}-1}$ be such that $\hat{h}(\mathbfsl{a})=1$. Then,
since $\hat{f}$ satisfies (5.2), we infer that
$(g(a_{1},\dots,a_{m^{\prime}}),a_{m^{\prime}+1},\dots,a_{m^{\prime}+n^{\prime}-1})\in
D_{n^{\prime}}$. Moreover, since $g$ satisfies (5.1) and (5.2), we have that
$(a_{1},\dots,a_{m^{\prime}})\in D_{m^{\prime}}$ and
$(a_{m^{\prime}+1},\dots,a_{m^{\prime}+n^{\prime}-1})=(2,\dots,2)$. Thus,
there exists $r\in[{m^{\prime}}]$ such that $a_{r}=1$ and for all
$k\in[{m^{\prime}+n^{\prime}-1}]\setminus\\{r\\}$ we have $a_{k}=2$. This
proves that $\mathbfsl{a}\in D_{m^{\prime}+n^{\prime}-1}$. Thus, $\hat{h}$
satisfies (5.2), and so $\hat{h}\in F$. Moreover, we have that for all
$\mathbfsl{x}\in Z^{m+n-1}$

$h(x_{1},\dots,x_{m+n-1})=\hat{f}(\hat{g}(x_{i_{1}},\dots,x_{i_{m^{\prime}}}),x_{m+j_{2}-1},\dots,x_{m+j_{n^{\prime}}-1}).$

Therefore $h$ belongs to $F$ up to inessential arguments, and so $h\in
F^{\prime}$. Therefore, $F^{\prime}$ is closed under $*$. ∎

We make use of the following construction from [13] (cf. [19, Chapter 3]). For
each $i\in\mathbb{N}\setminus\\{1\\}$, the function $g_{i}\colon Z^{i}\to Z$
is defined by

$g_{i}(x_{1},\dots,x_{i})=\begin{cases}1&\text{ if }(x_{1},\dots,x_{i})\in
D_{i},\\\ 0&\text{ otherwise}.\end{cases}$

For each $I\subseteq\mathbb{N}\setminus\\{1\\}$, ${\mathbf{Z}}_{I}$ is defined
as $(Z;(g_{i})_{i\in I})$.

######  Lemma 5.3.

Let $I\subseteq\mathbb{N}\setminus\\{1\\}$, let $k\in\mathbb{N}$ and let
$g\in\operatorname{Clo}_{k}({\mathbf{Z}}_{I})$. Then we have:

  1. (1)

If $g$ is constant, then $g=0$.

  2. (2)

If $g$ is essentially unary, then $g$ is a projection.

  3. (3)

If $g$ depends exactly on the arguments $i_{1}<\dots<i_{l}$ with $l\geq 2$,
then the following two conditions are satisfied:

    1. (a)

For all $\mathbfsl{b}\in Z^{k}$ we have $g(\mathbfsl{b})\neq 2$.

    2. (b)

For all $\mathbfsl{a}\in Z^{k}$ with $(a_{i_{1}},\dots,a_{i_{l}})\notin
D_{l}$, we have $g(\mathbfsl{a})=0$.

###### Proof.

Equations (5.1) and (5.2) imply that for all $i\in\mathbb{N}\setminus\\{1\\}$,
$g_{i}\in F$. Thus, Lemma 5.2 yields that for all
$I\subseteq\mathbb{N}\setminus\\{1\\}$,
$\operatorname{Clo}({\mathbf{Z}}_{I})\subseteq F^{\prime}$. Since
$g\in\operatorname{Clo}_{k}({\mathbf{Z}}_{I})\subseteq F^{\prime}$, then
either $g\in\mathcal{J}_{Z}$ or $g$ belongs to $F$ up to inessential
arguments. If $g$ is a constant, then it cannot be a projection. Hence it
belongs to $F$ up to inessential arguments. Thus, by Lemma 5.1(1), is the
constant zero. This proves (1). We now prove (2). Let us assume that $g$ is
essentially unary. Then by Lemma 5.1(3), $g$ is a projection. This proves (2).
We now prove (3). Let us assume that $g$ depends on its arguments
$i_{1}<\dots<i_{l}$, let $\mathbfsl{b}\in Z^{k}$ and let $\mathbfsl{a}\in
Z^{k}$ be such that $(a_{i_{1}},\dots,a_{i_{l}})\notin D_{l}$. Since $l\geq
2$, $g$ is not a projection. Since $g\in F^{\prime}$, $g$ belongs to $F$ up to
inessential arguments. This, together with Lemma 5.1(2), implies that there
exists $\hat{g}\in F_{l}$ such that
$g(x_{1},\dots,x_{k})=\hat{g}(x_{i_{1}},\dots,x_{i_{l}})$ for all
$\mathbfsl{x}\in Z^{k}$, and $\hat{g}$ depends on all of its arguments. Since
$\hat{g}\in F$, (5.1) implies that
$g(\mathbfsl{b})=\hat{g}(b_{i_{1}},\dots,b_{i_{l}})\in\\{0,1\\}$. Moreover,
since $(a_{i_{1}},\dots,a_{i_{l}})\notin D_{l}$, (5.2) and (5.1) yield
$g(\mathbfsl{a})=\hat{g}(a_{i_{1}},\dots,a_{i_{l}})=0$. This proves (3). ∎

######  Proposition 5.4.

Let $I\subseteq\mathbb{N}\setminus\\{1\\}$, let
$i\in\mathbb{N}\setminus\\{1\\}$ and let
$g_{i}^{\bullet}=\\{(x_{1},\dots,x_{i},x_{i+1})\in Z^{i+1}\mid
x_{i+1}=g(x_{1},\dots,x_{i})\\}$. Then
$g_{i}^{\bullet}\in\operatorname{Alg}{\mathbf{Z}}_{I}$ if and only if $i\in
I$.

###### Proof.

If $i\in I$, then $g_{i}^{\bullet}\in\operatorname{Alg}{\mathbf{Z}}_{I}$ by
definition. We now assume that
$g_{i}^{\bullet}\in\operatorname{Alg}{\mathbf{Z}}_{I}$ and prove that $i\in
I$.

If $g_{i}^{\bullet}\in\operatorname{Alg}{\mathbf{Z}}_{I}$, since
$(1,\dots,1)\notin g_{i}^{\bullet}$, Lemma 2.1 implies that there exist
$f_{1},f_{2}\in\operatorname{Clo}_{i+1}({\mathbf{Z}}_{I})$ such that
$f_{1}|_{g_{i}^{\bullet}}=f_{2}|_{g_{i}^{\bullet}}$ and $f_{1}(1,\dots,1)\neq
f_{2}(1,\dots,1)$. We now prove that one of the two is a projection and the
other depends on at least two of its arguments. Seeking contradictions, let us
suppose that this is not the case. We distinguish cases accordingly to the
essential arity of $f_{1}$ and $f_{2}$.

Case 1: _$f_{1}$ and $f_{2}$ are constant_: In this case
$f_{1}(1,\dots,1)=f_{2}(1,\dots,1)$.

Case 2: _$f_{1}$ is constant and $f_{2}$ is essentially unary_: In this case,
Lemma 5.3 yields that $f_{1}$ is the constant $0$ function and $f_{2}$ is a
projection. Let $\mathbfsl{a}=(1,2,\dots,2,1)$. Since
$g(a_{1},\dots,a_{i})=1$, $\mathbfsl{a}\in g_{i}^{\bullet}$. Lemma 5.3 yields
$f_{1}(\mathbfsl{a})=0$ and $f_{2}(\mathbfsl{a})\in\\{1,2\\}$. Hence
$f_{1}(\mathbfsl{a})\neq f_{2}(\mathbfsl{a})$. Thus,
$f_{1}|_{g_{i}^{\bullet}}\neq f_{2}|_{g_{i}^{\bullet}}$.

Case 3: _$f_{1}$ is essentially unary and $f_{2}$ is constant_: This case is
symmetric to Case 2.

Case 4: _$f_{1}$ is constant and $f_{2}$ depends on at least two of its
arguments_: In this case Lemma 5.3 yields
$f_{1}(1,\dots,1)=0=f_{2}(1,\dots,1)$.

Case 5: _$f_{1}$ is depends on at least two of its arguments and $f_{2}$ is
constant_: This case is symmetric to Case 4.

Case 6: _$f_{1}$ and $f_{2}$ are both essentially unary_: In this case Lemma
5.3(2) implies that both $f_{1}$ and $f_{2}$ are projections. Thus
$f_{1}(1,\dots,1)=1=f_{2}(1,\dots,1)$.

Case 7: _$f_{1}$ and $f_{2}$ both depend on at least two of their arguments_:
In this case Lemma 5.3(3) yields $f_{1}(1,\dots,1)=0=f_{2}(1,\dots,1)$.

Thus, we have proved that one among $f_{1}$ and $f_{2}$ is a projection, while
the other depends on at least two of its arguments. Without loss of
generality, let us assume that $f_{2}$ is a projection and that $f_{1}$
depends on at least two of its arguments. Let

$T:=\\{\mathbfsl{a}\in Z^{i+1}\mid a_{i+1}=1\text{ and }(a_{1},\dots,a_{i})\in
D_{i}\\}.$

Note that $T\subseteq g_{i}^{\bullet}$. Let $j\in\\{1,\dots,i,i+1\\}$ be such
that $f_{2}(x_{1},\dots,x_{i+1})=x_{j}$ for all $\mathbfsl{x}\in Z^{i+1}$. We
claim $j=i+1$. Seeking a contradiction, let us suppose that $j\leq i$. Let
$\mathbfsl{x}\in T$ be such that $\mathbfsl{x}(j)=2$. Then, since
$f_{1}|_{T}=f_{2}|_{T}$, we have $f_{1}(\mathbfsl{x})=2$. On the other hand,
Lemma 5.3(3) implies that $f_{1}(\mathbfsl{x})\neq 2$. Thus we get the desired
contradiction and deduce that $f_{2}$ is the $(i+1)$-th projection. Next, we
prove that $f_{1}$ does not depend on its $(i+1)$-th argument. Seeking a
contradiction, let us suppose that $f_{1}$ depends on its $(i+1)$-th argument.
Then, since $f_{1}$ depends on at least two of its arguments, there exists
$j\in[{i}]$ such that $f_{1}$ depends also on its $j$-th argument. Let
$\mathbfsl{a}\in T$ be such that $\mathbfsl{a}(j)=1$. Since
$f_{1}|_{T}=f_{2}|_{T}$, we have that
$f_{1}(\mathbfsl{a})=f_{2}(\mathbfsl{a})=a_{i+1}=1$. On the other hand, Lemma
5.3 implies that $f_{1}(\mathbfsl{a})=0$ because $a_{j}=a_{i+1}=1$. From this
contradiction, we conclude that $f_{1}$ does not depend on its $(i+1)$-th
argument.

Let $f\colon Z^{i}\to Z$ be defined by $f(x_{1},\dots
x_{i}):=f_{1}(x_{1},\dots,x_{i},x_{1})$. Since $f_{1}$ does not depend on its
$(i+1)$-th argument, for each $\mathbfsl{a}\in g_{i}^{\bullet}$, we have

(5.3)
$f(a_{1},\dots,a_{i})=f_{1}(a_{1},\dots,a_{i},a_{1})=f_{1}(a_{1},\dots,a_{i+1})=f_{2}(a_{1},\dots,a_{i+1})=a_{i+1}.$

We show that $g_{i}=f$. To this end, let $\mathbfsl{a}\in Z^{i}$. Then
$(a_{1},\dots,a_{i},g_{i}(\mathbfsl{a}))\in g_{i}^{\bullet}$, and thus (5.3)
yields $f(\mathbfsl{a})=g_{i}(\mathbfsl{a})$. Since
$f\in\operatorname{Clo}_{i}({\mathbf{Z}}_{I})$, we deduce that
$g_{i}\in\operatorname{Clo}_{i}({\mathbf{Z}}_{I})$. We now apply [19, Theorem
3.1.4] and deduce that $i\in I$. ∎

######  Corollary 5.5.

On the set $Z=\\{0,1,2\\}$ there are exactly $2^{\aleph_{0}}$ algebraically
inequivalent clones.

###### Proof.

Let $\mathcal{L}_{Z}$ be the set of all clones on $Z$ and let
$\psi\colon\mathcal{P}(\mathbb{N}\setminus\\{1\\})\to\mathcal{L}_{Z}/{\operatorname{\sim_{\text{alg}}}}$
be defined by
$I\mapsto\operatorname{Clo}({\mathbf{Z}}_{I})/{\operatorname{\sim_{\text{alg}}}}$.
Proposition 5.4 yields that for all $I\neq
J\in\mathcal{P}(\mathbb{N}\setminus\\{1\\})$ we have
$\operatorname{Clo}({\mathbf{Z}}_{I})\operatorname{\sim_{\text{alg}}}\operatorname{Clo}({\mathbf{Z}}_{J})$
if and only if $I=J$. Therefore, $\psi$ is injective. Hence there are at least
$2^{\aleph_{0}}$ algebraically inequivalent clones on $Z$. Since by [19,
Theorem 3.1.4] there are at most $2^{\aleph_{0}}$ clones on $Z$, we can
conclude that there are exactly $2^{\aleph_{0}}$ algebraically inequivalent
clones on $Z$. ∎

###### Proof of Theorem 1.1.

We first show that for a set $A$ and an element $u\notin A$, $A\cup\\{u\\}$
has at least as many algebraically inequivalent clones as $A$. To this end,
let $A$ be a finite set with $\lvert A\rvert\geq 3$, let $\mathcal{L}_{A}$ be
the set of all clones on $A$, let $u\notin A$, let $A^{\oplus}=A\cup\\{u\\}$,
let $\mathcal{L}_{A^{\oplus}}$ be the set of all clones on $A^{\oplus}$, and
let $\Phi\colon\mathcal{L}_{A}\to\mathcal{L}_{A^{\oplus}}$ be defined by

$\Phi(\mathcal{C})=\underset{n\in\mathbb{N}}{\bigcup}\\{f:(A^{\oplus})^{n}\to
A^{\oplus}\,\mid\,f|_{A^{n}}\in\mathcal{C}\\}.$

We first prove that for all $\mathcal{C}\in\mathcal{L}_{A}$,
$\Phi(\mathcal{C})$ is a clone on $A^{\oplus}$. Clearly, all the projections
belong to $\Phi(\mathcal{C})$. Let $f\in\Phi(\mathcal{C})^{[n]}$, let
$t_{1},\dots,t_{n}\in\Phi(\mathcal{C})^{[k]}$. Then

$f(t_{1},\dots,t_{n})|_{A^{k}}=f(t_{1}|_{A^{k}},\dots,t_{n}|_{A^{k}})=f|_{A^{n}}(t_{1}|_{A^{k}},\dots,t_{n}|_{A^{k}}).$

The definition of $\Phi$ yields
$f|_{A^{n}},t_{1}|_{A^{k}},\dots,t_{n}|_{A^{k}}\in\mathcal{C}$. Thus,
$f|_{A^{n}}(t_{1}|_{A^{k}},\dots,t_{n}|_{A^{k}})$ belongs to $\mathcal{C}$ and
$f(t_{1},\dots,t_{n})\in\Phi(\mathcal{C})$. Therefore, $\Phi(\mathcal{C})$ is
a clone on $A^{\oplus}$.

Next, we prove that $\Phi$ satisfies

(5.4) $\forall\mathcal{C}\in\mathcal{L}_{A},\forall n\in\mathbb{N},\forall
B\subseteq(A^{\oplus})^{n}\colon
B\in\operatorname{Alg}\Phi(\mathcal{C})\Leftrightarrow B\cap
A^{n}\in\operatorname{Alg}\mathcal{C}.$

For proving (5.4), let $\mathcal{C}\in\mathcal{L}_{A}$, let $n\in\mathbb{N}$,
let $B\subseteq(A^{\oplus})^{n}$ and let us assume that
$B\in\operatorname{Alg}_{n}\Phi(\mathcal{C})$. We prove that $B\cap
A^{n}\in\operatorname{Alg}_{n}\mathcal{C}$. To this end, let $\mathbfsl{b}\in
A^{n}\setminus B$. By Lemma 2.1 it suffices to prove that there are
$\hat{f_{1}},\hat{f_{2}}\in\mathcal{C}$ such that $\hat{f_{1}}|_{A^{n}\cap
B}=\hat{f_{2}}|_{A^{n}\cap B}$ and
$\hat{f_{1}}(\mathbfsl{b})\neq\hat{f_{2}}(\mathbfsl{b})$. Since
$B\in\operatorname{Alg}_{n}\Phi(\mathcal{C})$ and $\mathbfsl{b}\notin B$,
Lemma 2.1 yields that there exist $f_{1},f_{2}\in\Phi(\mathcal{C})$ such that
$f_{1}|_{B}=f_{2}|_{B}$ and $f_{1}(\mathbfsl{b})\neq f_{2}(\mathbfsl{b})$. Let
$\hat{f_{1}}=f_{1}|_{A^{n}}$ and let $\hat{f_{2}}=f_{2}|_{A^{n}}$. Then
clearly $\hat{f_{1}},\hat{f_{2}}\in\mathcal{C}$, $\hat{f_{1}}|_{A^{n}\cap
B}=\hat{f_{2}}|_{A^{n}\cap B}$ and
$\hat{f_{1}}(\mathbfsl{b})\neq\hat{f_{2}}(\mathbfsl{b})$. Thus, $B\cap
A^{n}\in\operatorname{Alg}\mathcal{C}$.

Let us now assume that $B\cap A^{n}\in\operatorname{Alg}\mathcal{C}$. We prove
that $B\in\operatorname{Alg}_{n}\Phi(\mathcal{C})$. Let
$\mathbfsl{a}\in(A^{\oplus})^{n}\setminus B$. By Lemma 2.1 it suffices to show
that there are $f_{1},f_{2}\in\Phi(\mathcal{C})$ such that
$f_{1}|_{B}=f_{2}|_{B}$ and $f_{1}(\mathbfsl{a})\neq f_{2}(\mathbfsl{a})$. We
split the proof into two cases.

Case 1: $\mathbfsl{a}\in A^{n}$: In this case Lemma 2.1 implies that there are
$g_{1},g_{2}\in\mathcal{C}$ such that $g_{1}|_{B\cap A^{n}}=g_{2}|_{B\cap
A^{n}}$ and $g_{1}(\mathbfsl{a})\neq g_{2}(\mathbfsl{a})$. Then for
$i\in\\{1,2\\}$, we set

$f_{i}(\mathbfsl{x})=\begin{cases}g_{i}(\mathbfsl{x})&\text{ if
}\mathbfsl{x}\in A^{n},\\\ u&\text{ otherwise}.\end{cases}$

By the definition of $\Phi(\mathcal{C})$, both $f_{1}$ and $f_{2}$ belong to
$\Phi(\mathcal{C})$, $f_{1}|_{B}=f_{2}|_{B}$ and, since $\mathbfsl{a}\in
A^{n}$, $f_{1}(\mathbfsl{a})\neq f_{2}(\mathbfsl{a})$.

Case 2: $\mathbfsl{a}\notin A^{n}$: Let $v_{1},v_{2}\in A$ with $v_{1}\neq
v_{2}$. For $i\in\\{1,2\\}$, we define

$f_{i}(\mathbfsl{x})=\begin{cases}u&\text{ if
}\mathbfsl{x}\in(A^{\oplus})^{n}\setminus(A^{n}\cup\\{\mathbfsl{a}\\}),\\\
x_{1}&\text{ if }\mathbfsl{x}\in A^{n},\\\ v_{i}&\text{ if
}\mathbfsl{x}\in\\{\mathbfsl{a}\\}.\end{cases}$

By construction, we have that $f_{1}|_{A^{n}},f_{2}|_{A^{n}}\in\mathcal{C}$.
Hence $f_{1},f_{2}\in\Phi(\mathcal{C})$. Moreover, $f_{1}|_{B}=f_{2}|_{B}$ and
$f_{1}(\mathbfsl{a})\neq f_{2}(\mathbfsl{a})$. Thus, we can conclude that
$B\in\operatorname{Alg}\Phi(\mathcal{C})$. Equation (5.4) implies that for all
$\mathcal{C},\mathcal{D}\in\mathcal{L}_{A}$, if
$\mathcal{D}\not\sim_{\text{alg}}\mathcal{C}$ then
$\Phi(\mathcal{C})\not\sim_{\text{alg}}\Phi(\mathcal{D})$.

Since by Corollary 5.5 there are $2^{\aleph_{0}}$ algebraically inequivalent
clones on the set $\\{0,1,2\\}$ and since there are at most $2^{\aleph_{0}}$
clones on a finite set, we deduce that on each finite set $A$ with $\lvert
A\rvert\geq 3$ there are exactly $2^{\aleph_{0}}$ algebraically inequivalent
clones. ∎

## Acknowledgements

The authors thank Tamás Waldhauser for discussions on the questions addressed
in the present note and the referee, who has suggested the content of Lemmas
3.1 and 5.3 to replace our originally more involved argument for proving
Theorems 1.3 and 1.1.

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