# Clonoids between modules

Peter Mayr  Department of Mathematics
University of Colorado
Boulder CO 80309
USA<EMAIL_ADDRESS>
<EMAIL_ADDRESS>and  Patrick Wynne

###### Abstract.

Clonoids are sets of finitary functions from an algebra $\mathbf{A}$ to an
algebra $\mathbf{B}$ that are closed under composition with term functions of
$\mathbf{A}$ on the domain side and with term functions of $\mathbf{B}$ on the
codomain side. For $\mathbf{A},\mathbf{B}$ (polynomially equivalent to) finite
modules we show: If $\mathbf{A},\mathbf{B}$ have coprime order and the
congruence lattice of $\mathbf{A}$ is distributive, then there are only
finitely many clonoids from $\mathbf{A}$ to $\mathbf{B}$. This is proved by
establishing for every natural number $k$ a particular linear equation that
all $k$-ary functions from $\mathbf{A}$ to $\mathbf{B}$ satisfy. Else if
$\mathbf{A},\mathbf{B}$ do not have coprime order, then there are infinitely
many clonoids from $\mathbf{A}$ to $\mathbf{B}$ and not all of them are
finitely generated. Consequently any extension of $\mathbf{A}$ by $\mathbf{B}$
has countably infinitely many $2$-nilpotent expansions up to term equivalence.

###### Key words and phrases:

closed function classes, clones, abelian Mal’cev algebras, linear functional
equations

###### 1991 Mathematics Subject Classification:

08A40, 08A02

The research of the first author was partially supported by the NSF under
grant no. DMS-1500254 and by the Austrian Science Fund (FWF): P33878.

##  1\. Introduction

Clones are sets of finitary operations on a set that are closed under
composition and contain all projection maps. They have been used for studying
completeness of Boolean operations (Post [15]), classifying algebraic
structures (see the books [9, 12, 16]), and capturing the computational
complexity of Constraint Satisfaction Problems (see the survey [2]) for many
decades.

More recently and more generally, sets of finitary functions from one set to
another that satisfy certain closure properties have found applications, e.g.,
for studying equational theories of finite algebras [1, 13], classifying
expansions of algebras [7, 11], and investigating the computational complexity
of Promise Constraint Satisfaction Problems (see the survey [3]). These
function classes fit into the general framework of _clonoids_ : sets of
finitary functions from an algebra $\mathbf{A}$ to an algebra $\mathbf{B}$
that are closed under composition with term functions of $\mathbf{A}$ on one
side and composition with term functions of $\mathbf{B}$ on the other.

######  Definition 1.1.

[1, cf. Definition 4.1] Let $\mathbf{A}$ and $\mathbf{B}$ be algebras and let
$C\subseteq\bigcup_{n\in\mathbb{N}}B^{A^{n}}$ be a set of finitary functions
from $A$ to $B$. We say that $C$ is a _clonoid from $\mathbf{A}$ to
$\mathbf{B}$_ if

$C\mathrm{Clo}(\mathbf{A})\subseteq C\,\text{ and
}\,\mathrm{Clo}(\mathbf{B})C\subseteq C.$

Here $\mathrm{Clo}(\mathbf{A})$ and $\mathrm{Clo}(\mathbf{B})$ are the clones
of term functions of $\mathbf{A}$ and $\mathbf{B}$, respectively, and
juxtaposition represents function class composition, i.e.

$\displaystyle C\mathrm{Clo}(\mathbf{A})=\\{f(s_{1},\dots,s_{k})\ :\
k,m\in\mathbb{N},f\,k\text{-ary in }C,s_{1},\dots,s_{k}\ $ $\displaystyle
m\text{-ary term functions on }\mathbf{A}\\},$
$\displaystyle\mathrm{Clo}(\mathbf{B})C=\\{t(f_{1},\dots,f_{n})\ :\
n,k\in\mathbb{N},t\,n\text{-ary term function on }\mathbf{B}$ $\displaystyle
f_{1},\dots,f_{n}\,k\text{-ary in }C\\}.$

For example, for abelian groups $\mathbf{A}$ and $\mathbf{B}$, the set
$\bigcup_{k\in\mathbb{N}}\mathrm{Hom}(\mathbf{A}^{k},\mathbf{B})$ of
homomorphisms from finite powers of $\mathbf{A}$ to $\mathbf{B}$ forms a
clonoid from $\mathbf{A}$ to $\mathbf{B}$.

The name ‘clonoid’ was originally introduced by Aichinger and the first author
of this paper in [1] for the case that $\mathbf{A}$ is a set with no
operations. There clonoids were used to represented equational theories and to
show that every subvariety of a finitely generated variety with cube term is
finitely generated. For subsequent applications the slight generalization that
allows for $\mathbf{A}$ an algebra as in Definition 1.1 now appears to be more
desirable. We mention some other contexts in which clonoids occurred.

Already in 2002, Pippenger developed a Galois correspondence for clonoids
between sets, which he called _minor-closed classes_ , and pairs of relational
structures in [14]. The theory was extended by Couceiro and Foldes in [5] to
clonoids from an algebra $\mathbf{A}$ to an algebra $\mathbf{B}$, which they
called classes of functions that are _stable_ under right composition with
$\mathrm{Clo}(\mathbf{A})$ and left composition with
$\mathrm{Clo}(\mathbf{B})$ (see Section 2.3 below).

In 2018, Brakensiek and Guruswami observed that Pippenger’s Galois
correspondence can be used to analyze Promise Constraint Satisfaction Problems
via their finitary symmetries, i.e., sets of polymorphisms between relational
structures. In this context, clonoids from sets $A$ to $B$ have been called
_minions_ (see the survey by Barto, Bulín, Krokhin, and Opršal [3]).

Kreinecker [11] and Fioravanti [6, 7] studied clonoids between modules
$\mathbf{A}$ and $\mathbf{B}$, which they called _linearly-closed clonoids_ ,
and used them to classify expansions of $\mathbf{A}\times\mathbf{B}$ up to
term equivalence. For algebras $\mathbf{A}_{1},\mathbf{A}_{2}$ on the same
universe, $\mathbf{A}_{1}$ is an _expansion_ of $\mathbf{A}_{2}$ if
$\mathrm{Clo}(\mathbf{A}_{1})\supseteq\mathrm{Clo}(\mathbf{A}_{2})$; we say
$\mathbf{A}_{1}$ and $\mathbf{A}_{2}$ are _term equivalent_ if
$\mathrm{Clo}(\mathbf{A}_{1})=\mathrm{Clo}(\mathbf{A}_{2})$.

All of the function classes mentioned above are instances of clonoids from
$\mathbf{A}$ to $\mathbf{B}$ in the sense of Definition 1.1 for the
appropriate choice of algebras.

In this paper we collect basic properties and background on clonoids in
Section 2. Then we mainly investigate clonoids between algebras that are
polynomially equivalent to modules in the vein of [6, 11]. Recall that an
algebra $\mathbf{A}$ is polynomially equivalent to a module if all basic
operations of $\mathbf{A}$ are affine functions on that module (see Section
2.4). In particular we consider the following problem:

######  Question 1.2.

Given finite (algebras that are polynomially equivalent to) modules
$\mathbf{A}$ and $\mathbf{B}$, how many clonoids are there from $\mathbf{A}$
to $\mathbf{B}$? Are they all finitely generated?

From work of Aichinger and the first author of this paper in [1] it follows
that for $\mathbf{B}$ a finite algebra with few subpowers (in particular, for
$\mathbf{B}$ polynomially equivalent to a module), all clonoids from a finite
algebra $\mathbf{A}$ to $\mathbf{B}$ are finitely related and hence there are
at most countably infinitely many (see Theorem 2.9).

Kreinecker [11] showed that there are countably infinitely many clonoids from
$(\mathbb{Z}_{p},+)$ to $(\mathbb{Z}_{p},+)$ for any prime $p$, which he then
used to construct infinitely many non-finitely generated clones on
$\mathbb{Z}_{p}^{2}$ that contain $+$. We generalize his result to obtain the
following in Section 5.

######  Theorem 1.3.

Let $\mathbf{A},\mathbf{B}$ be finite modules whose orders are not coprime.
Then there are countably infinitely many clonoids from $\mathbf{A}$ to
$\mathbf{B}$ and not all of them are finitely generated.

Together with Lemma 2.6 below and the subsequent remark, this yields that any
finite module $\mathbf{A}$ with a submodule $B$ and $\gcd(|A/B|,|B|)>1$ has
countably infinitely many $2$-nilpotent expansions up to term equivalence.

Fioravanti proved for $\mathbf{A}$ and $\mathbf{B}$ finite coprime regular
modules over products of finite fields that every clonoid from $\mathbf{A}$ to
$\mathbf{B}$ is generated by its unary functions (and so there are finitely
many of them) [6, Theorem 1.2]. Note that under his assumptions the lattices
of submodules of $\mathbf{A}$ and $\mathbf{B}$ are distributive (in fact,
Boolean). As a consequence, he then showed that a finite abelian group has
finitely many expansions up to term equivalence if and only if the order of
that group is squarefree in [7].

Recall that a module is _distributive_ if its lattice of submodules is
distributive. Our main result generalizes Fioravanti’s result and settles
Question 1.2 for $\mathbf{A}$ (polynomially equivalent to) a distributive
module.

######  Theorem 1.4.

Let $\mathbf{A}$ be polynomially equivalent to a finite distributive
$\mathbf{R}$-module, let $n$ be the nilpotence degree of the Jacobson radical
of $\mathbf{R}$, and let $\mathbf{B}$ be polynomially equivalent to an
$\mathbf{S}$-module such that $|A|$ is invertible in $\mathbf{S}$.

  1. (1)

Then every clonoid from $\mathbf{A}$ to $\mathbf{B}$ is generated by its
$n+1$-ary functions (by its $n$-ary functions if $\mathbf{A}$ is an
$\mathbf{R}$-module).

  2. (2)

If $\mathbf{B}$ is finite, then there are only finitely many clonoids from
$\mathbf{A}$ to $\mathbf{B}$.

For example, let $m\in\mathbb{N}$ and let $n$ be the exponent of the largest
power of a prime $p$ such that $p^{n}$ divides $m$. Then
$\mathbf{A}:=(\mathbb{Z}_{m},x-y+z)$ is polynomially equivalent to
$\mathbf{A}_{0}:=(\mathbb{Z}_{m},+)$, the regular module of the ring
$\mathbf{R}:=(\mathbb{Z}_{m},+,\cdot,1)$. Clearly $\mathbf{A}_{0}$ is
distributive and the Jacobson radical $J(\mathbf{R})$ satisfies
$J(\mathbf{R})^{n}=0$. Let $\mathbf{B}$ be a finite module of order coprime to
$m$. By Theorem 1.4 every clonoid from $\mathbf{A}_{0}$ into $\mathbf{B}$ is
generated by $n$-ary functions and every clonoid from $\mathbf{A}$ to
$\mathbf{B}$ is generated by $n+1$-ary functions.

We first prove Theorem 1.4 for modules $\mathbf{A},\mathbf{B}$ in Section 3.
Our main tool for this is that under the given assumptions, for every
$k\in\mathbb{N}$ we find a specific linear equation that is satisfied by every
$k$-ary function from $\mathbf{A}$ to $\mathbf{B}$ (Theorem 3.1). As it turns
out, every finitary function from $\mathbf{A}$ to $\mathbf{B}$ can be
uniformly generated by its $n$-ary $\mathbf{A},\mathbf{B}$-minors (see
Sections 2.1, 2.7). In Section 4 we then extend these results from modules to
algebras that are polynomially equivalent to modules. The main difference
there is that if $\mathbf{A}$ does not have a constant term function $0$, we
have to replace that by a projection, i.e., an additional variable $z$, in our
interpolation arguments. This additional variable yields that the clonoids in
the general case are generated by $n+1$-ary functions instead of $n$-ary as
for modules.

Finally we investigate necessary conditions such that every clonoid from
algebras $\mathbf{A}$ to $\mathbf{B}$ is generated by $n$-ary functions in
Section 5. We show that every subalgebra of $\mathbf{A}$ has to be generated
by $n$ elements in Lemma 5.1. Further, if every clonoid from a regular
$\mathbf{R}$-module $\mathbf{A}$ to $\mathbf{B}$ is generated by unary
functions, then $\mathbf{R}$ is semisimple (Lemma 5.2). Hence for a ring
$\mathbf{R}$ with $J(\mathbf{R})\neq 0$ and $J(\mathbf{R})^{2}=0$, the
assertion of Theorem 1.4 that all clonoids from distributive
$\mathbf{R}$-modules $\mathbf{A}$ to coprime modules $\mathbf{B}$ are
generated by binary functions cannot be strengthened to unary functions.

As consequence and partial converse to Theorem 1.4 we can characterize the
pairs of finite modules $\mathbf{A}$ over commutative rings and modules
$\mathbf{B}$ such that every clonoid from $\mathbf{A}$ to $\mathbf{B}$ is
generated by unary functions.

######  Theorem 1.5.

For a finite faithful $\mathbf{R}$-module $\mathbf{A}$ over a commutative ring
$\mathbf{R}$ and a finite $\mathbf{S}$-module $\mathbf{B}$ the following are
equivalent:

  1. (1)

Every clonoid from $\mathbf{A}$ to $\mathbf{B}$ is generated by unary
functions.

  2. (2)

The orders of $\mathbf{A}$ and $\mathbf{B}$ are coprime, $\mathbf{R}$ is a
direct product of finite fields and $\mathbf{A}$ is isomorphic to the regular
$\mathbf{R}$-module.

The proof is given in Section 5. Note that item (2) in Theorem 1.5 implies
that $\mathbf{A}$ is a distributive module.

In general it remains open whether the condition that $\mathbf{A}$ is
distributive is necessary so that the number of clonoids from $\mathbf{A}$
into any coprime module is finite. We do not even know the answer to the
following.

######  Question 1.6.

For a prime $p$, how many clonoids are there from the group
$(\mathbb{Z}_{p}^{2},+)$ into a finite module of order coprime to $p$?

##  2\. Preliminaries and Notation

###  2.1. General

We write $\mathbb{N}:=\\{1,2,\dots\\}$ for the set of positive integers and
$[n]:=\\{1,\dots,n\\}$ for $n\in\mathbb{N}$.

Algebras are pairs $\mathbf{A}:=(A,F)$ with _universe_ (domain) $A$ and a set
$F$ of finitary operations on $A$ (the _basic operations_ of $\mathbf{A}$).

######  Definition 2.1.

Let $\mathbf{A},\mathbf{B}$ be algebras with universes $A,B$, respectively.

  1. (1)

Let $F(A,B):=\bigcup_{k\in\mathbb{N}}B^{A^{k}}$ denote the clonoid of finitary
functions from $\mathbf{A}$ to $\mathbf{B}$.

  2. (2)

For $F\subseteq F(A,B)$, we denote the smallest clonoid from $\mathbf{A}$ to
$\mathbf{B}$ that contains $F$ by $\langle F\rangle_{\mathbf{A},\mathbf{B}}$
and call it the clonoid _generated_ by $F$. For $f\in F(A,B)$, we also write
$\langle f\rangle_{\mathbf{A},\mathbf{B}}$ instead of
$\langle\\{f\\}\rangle_{\mathbf{A},\mathbf{B}}$.

  3. (3)

A clonoid $C$ from $\mathbf{A}$ to $\mathbf{B}$ is _finitely generated_ if it
is generated by some finite set $F\subseteq F(A,B)$

  4. (4)

If $g\in\langle f\rangle_{\mathbf{A},\mathbf{B}}$ for $f\in F(A,B)$, we call
$g$ an $\mathbf{A},\mathbf{B}$-_minor_ of $f$.

If $\mathbf{A}$ and $\mathbf{B}$ are clear from context, we also write
$\langle F\rangle$ and $\langle f\rangle$ instead of $\langle
F\rangle_{\mathbf{A},\mathbf{B}}$ and $\langle
f\rangle_{\mathbf{A},\mathbf{B}}$.

######  Definition 2.2.

For $C\subseteq F(A,B)$ and $k\in\mathbb{N}$, let $C^{(k)}:=C\cap B^{A^{k}}$
denote the set of $k$-ary functions in $C$.

###  2.2. The lattice of clonoids

For algebras $\mathbf{A}$ and $\mathbf{B}$, we denote the set of all clonoids
from $\mathbf{A}$ to $\mathbf{B}$ by $\mathcal{C}_{\mathbf{A},\mathbf{B}}$.
Then $\mathcal{C}_{\mathbf{A},\mathbf{B}}$ forms a lattice with meet given by
intersection and the join of clonoids $C,D$ given by the clonoid generated by
the union $\langle C\cup D\rangle_{\mathbf{A},\mathbf{B}}$. This is a bounded
lattice with bottom element $\emptyset$ and top element $F(A,B)$.

We note some straightforward relations for clonoids between algebras, their
expansions, quotients and subalgebras.

######  Lemma 2.3.

Let $\mathbf{A},\mathbf{B}$ be a algebras with expansions
$\mathbf{A}^{+},\mathbf{B}^{+}$, respectively. Then
$\mathcal{C}_{\mathbf{A}^{+},\mathbf{B}^{+}}$ is a meet-subsemilattice of
$\mathcal{C}_{\mathbf{A},\mathbf{B}}$.

###### Proof.

Every clonoid from $\mathbf{A}^{+}$ to $\mathbf{B}^{+}$ is also a clonoid from
$\mathbf{A}$ to $\mathbf{B}$. Clearly the intersection of clonoids in
$\mathcal{C}_{\mathbf{A}^{+},\mathbf{B}^{+}}$ is the same as in
$\mathcal{C}_{\mathbf{A},\mathbf{B}}$. Since joins in
$\mathcal{C}_{\mathbf{A}^{+},\mathbf{B}^{+}}$ require closure under
$\mathrm{Clo}(\mathbf{A}^{+})$ and $\mathrm{Clo}(\mathbf{B}^{+})$, they may
properly contain the joins in $\mathcal{C}_{\mathbf{A},\mathbf{B}}$. ∎

For any equivalence relation $\alpha$ on $A$, any function
$f\colon(A/\alpha)^{k}\to B$ can be lifted to a function $f^{\prime}\colon
A^{k}\to B$ that is constant on $\alpha$-blocks via

$f^{\prime}(x_{1},\ldots,x_{k}):=f(x_{1}/\alpha,\ldots,x_{k}/\alpha).$

######  Lemma 2.4.

Let $\mathbf{A}$ and $\mathbf{B}$ be algebras, let $\alpha$ be a congruence on
$\mathbf{A}$ and let $\mathbf{B}^{\prime}$ be a subalgebra of $\mathbf{B}$.
Then

$\varphi\colon\mathcal{C}_{\mathbf{A}/\alpha,\mathbf{B}^{\prime}}\to\mathcal{C}_{\mathbf{A},\mathbf{B}},\
C\mapsto\\{f^{\prime}\ :\ f\in C\\},$

is a lattice embedding.

###### Proof.

Let $C\in\mathcal{C}_{\mathbf{A}/\alpha,\mathbf{B}^{\prime}}$. To show
$\varphi(C)\in\mathcal{C}_{\mathbf{A},\mathbf{B}}$, let $f\in C^{(k)}$,
$s_{1},\ldots,s_{k}\in\text{Clo}(\mathbf{A})^{(m)}$, and
$(x_{1},\dots,x_{m})\in A^{m}$. Then

$\displaystyle[f^{\prime}(s_{1},\ldots,s_{k})](x_{1},\ldots,x_{m})$
$\displaystyle=f^{\prime}(s_{1}(x_{1},\ldots,x_{m}),\ldots,s_{k}(x_{1},\ldots,x_{m}))$
$\displaystyle=f(s_{1}(x_{1},\ldots,x_{m})/\alpha,\ldots,s_{k}(x_{1},\ldots,x_{m})/\alpha)$
$\displaystyle=f(s_{1}(x_{1}/\alpha,\ldots,x_{m}/\alpha),\ldots,s_{k}(x_{1}/\alpha,\ldots,x_{m}/\alpha))$
$\displaystyle=[f(s_{1},\ldots,s_{k})]^{\prime}(x_{1},\ldots,x_{m})\in\varphi(C).$

Also for $f_{1},\ldots,f_{k}\in C^{(m)}$, $t\in\mathrm{Clo}(\mathbf{B})^{(k)}$
and $(x_{1},\dots,x_{m})\in A^{m}$,

$\displaystyle[t(f_{1}^{\prime},\ldots,f_{k}^{\prime})](x_{1},\ldots,x_{m})$
$\displaystyle=t(f_{1}^{\prime}(x_{1},\ldots,x_{m}),\ldots,f_{k}^{\prime}(x_{1},\ldots,x_{m}))$
$\displaystyle=t(f_{1}(x_{1}/\alpha,\ldots,x_{m}/\alpha),\ldots,f_{k}(x_{1}/\alpha,\ldots,x_{m}/\alpha))$
$\displaystyle=[t(f_{1},\ldots,f_{k})]^{\prime}(x_{1},\ldots,x_{m})\in\varphi(C).$

Hence $\varphi(C)$ is a clonoid from $\mathbf{A}$ to $\mathbf{B}$.

That $\varphi$ is injective and preserves meets and joins is immediate. ∎

######  Lemma 2.5.

Let $\mathbf{A}$ and $\mathbf{B}$ be algebras and let $\mathbf{A}^{\prime}$ be
a subalgebra of $\mathbf{A}$. Then restriction of functions from $A$ to
$A^{\prime}$ yields a lattice homomorphism from
$\mathcal{C}_{\mathbf{A},\mathbf{B}}$ to
$\mathcal{C}_{\mathbf{A}^{\prime},\mathbf{B}}$.

###### Proof.

Straightforward since term functions of $\mathbf{A}^{\prime}$ are restrictions
of term functions of $\mathbf{A}$. ∎

We give one easy example of how clonoids arise in the description of
expansions of algebras. Let $\mathbf{A}$ be a module with a submodule $B$. For
$f\colon A^{k}\to A$ and $g\colon(A/B)^{k}\to B$ write

$f+g^{\prime}\colon A^{k}\to A,\ x\mapsto f(x)+g(x+B^{k}).$

For a clonoid $C$ from $\mathbf{A}/B$ to $\mathbf{B}$ let

$\mathrm{Clo}(\mathbf{A})+C:=\\{f+g^{\prime}\ :\
k\in\mathbb{N},f\in\mathrm{Clo}(\mathbf{A})^{(k)},g\in C^{(k)}\\}.$

######  Lemma 2.6.

Let $\mathbf{A}$ be a module with a submodule $\mathbf{B}$ and let
$\mathcal{D}$ be the lattice of clones on $A$ that contain
$\mathrm{Clo}(\mathbf{A})$. Then

$\varphi\colon\mathcal{C}_{\mathbf{A}/B,\mathbf{B}}\to\mathcal{D},\
C\mapsto\mathrm{Clo}(\mathbf{A})+C,$

is a lattice embedding.

###### Proof.

Let $C$ be a clonoid from $\mathbf{A}/B$ to $\mathbf{B}$. Since $C$ contains
the constant $0$ function,
$\mathrm{Clo}(\mathbf{A})\subseteq\mathrm{Clo}(\mathbf{A})+C$ and the latter
contains all projections on $A$. To see that $\mathrm{Clo}(\mathbf{A})+C$ is
closed under composition, let $u\in\mathrm{Clo}(\mathbf{A}),v\in C$ be $n$-ary
and let $f_{1},\dots,f_{n}\in\mathrm{Clo}(\mathbf{A}),g_{1},\dots,g_{n}\in C$
be $k$-ary. Then

$\displaystyle\left[u+v^{\prime}\right](f_{1}+g^{\prime}_{1},\dots,f_{n}+g^{\prime}_{n})$
$\displaystyle=$ $\displaystyle
u(f_{1}+g^{\prime}_{1},\dots,f_{n}+g^{\prime}_{n})+v^{\prime}(f_{1}+g^{\prime}_{1},\dots,f_{n}+g^{\prime}_{n})$
$\displaystyle=$ $\displaystyle
u(f_{1},\dots,f_{n})+[u(g_{1},\dots,g_{n})]^{\prime}+[v(f_{1},\dots,f_{n})]^{\prime}$

since $u$ is linear on $\mathbf{A}$ and $v^{\prime}$ is constant on
$B$-cosets. Note that $u(f_{1},\dots,f_{n})\in\mathrm{Clo}(\mathbf{A})$ and
$u(g_{1},\dots,g_{n})+v(f_{1},\dots,f_{n})\in C$ by the respective closure
properties. Hence
$\left[u+v^{\prime}\right](f_{1}+g^{\prime}_{1},\dots,f_{n}+g^{\prime}_{n})$
is in $\mathrm{Clo}(\mathbf{A})+C$ and the latter is a clone.

That $\varphi$ is a lattice embedding is straightforward. ∎

We note that all the expansions
$\mathbf{A}_{C}:=(A,\mathrm{Clo}(\mathbf{A})+C)$ in Lemma 2.6 are
$2$-nilpotent with respect to the term condition commutator [8, Chapter 7]. In
particular the congruence modulo $B$ is central in $\mathbf{A}_{C}$.

###  2.3. Relational description of clonoids

Pippenger [14] extended the Galois theory between clones and relations to the
setting of clonoids between sets and pairs of relations. Couceiro and Foldes
[5] further expanded this to our setting of clonoids between algebras.

######  Definition 2.7.

Let $A$ and $B$ be nonempty sets and let $n\in\mathbb{N}$. For $R\subseteq
A^{n},S\subseteq B^{n}$, let

$\mathrm{Pol}(R,S):=\bigcup_{k\in\mathbb{N}}\\{f\colon A^{k}\to B\mid
f(R,\dots,R)\subseteq S\\}$

denote the set of _polymorphisms_ of the relational pair $(R,S)$.

######  Theorem 2.8.

[5, cf. Theorem 2] Let $\mathbf{A}$ and $\mathbf{B}$ be algebras with $|A|$
finite. For $C\subseteq\bigcup_{n\in\mathbb{N}}B^{A^{n}}$ the following are
equivalent:

  1. (1)

$C$ is a clonoid from $\mathbf{A}$ to $\mathbf{B}$.

  2. (2)

There exist subalgebras $R_{i},S_{i}$ of
$\mathbf{A}^{m_{i}},\mathbf{B}^{m_{i}}$, respectively, for
$m_{i}\in\mathbb{N}$ and $i$ in a set $I$ such that $C=\bigcap_{i\in
I}\mathrm{Pol}(R_{i},S_{i})$.

In [1] Aichinger and the first author of this paper showed that in certain
settings the number of relational pairs needed to determine a clonoid is
finite.

A finite algebra $\mathbf{B}$ has _few subpowers_ if there exists a polynomial
$p$ such that for every $n\in\mathbb{N}$ the number of subalgebras of
$\mathbf{B}^{n}$ is at most $2^{p(n)}$. All finite algebras with (quasi)group
or lattice operations have few subpowers. Hence the following holds in
particular for $\mathbf{B}$ a group or module.

######  Theorem 2.9.

[1, Theorem 5.3] Let $\mathbf{A},\mathbf{B}$ be finite algebras such that
$\mathbf{B}$ has few subpowers.

  1. (1)

Then the lattice of clonoids from $\mathbf{A}$ to $\mathbf{B}$ satisfies the
descending chain condition.

  2. (2)

Every clonoid from $\mathbf{A}$ to $\mathbf{B}$ is finitely related (i.e. the
polymorphism clonoid of a single relational pair).

  3. (3)

The number of clonoids from $\mathbf{A}$ to $\mathbf{B}$ is finite or
countably infinite.

We note that between finite algebras there are only finitely many clonoids if
and only if they are all generated by functions of some bounded arity.

######  Lemma 2.10.

For finite algebras $\mathbf{A},\mathbf{B}$ the following are equivalent:

  1. (1)

The number of clonoids from $\mathbf{A}$ to $\mathbf{B}$ is finite.

  2. (2)

There exists some $n\in\mathbb{N}$ such that every clonoid from $\mathbf{A}$
to $\mathbf{B}$ is generated by $n$-ary functions.

  3. (3)

There exists some $n\in\mathbb{N}$ such that for all $k\in\mathbb{N}$ every
function $f\colon A^{k}\to B$ is generated by its $n$-ary
$\mathbf{A},\mathbf{B}$-minors.

###### Proof.

Straighforward. ∎

###  2.4. Abelian Mal’cev algebras

Algebras $(A,F_{1})$ and $(A,F_{2})$ on the same universe are _polynomially
equivalent_ if they have the same clone of polynomial functions.

A ternary operation $m$ on a set $A$ is _Mal’cev_ if

$m(x,y,y)=x=m(y,y,x)\text{ for all }x,y\in A.$

An algebra $\mathbf{A}$ is _Mal’cev_ if it has a term operation which is
Mal’cev.

Every expansion of a group $(A,+,-,0)$ has a Mal’cev term operation
$m(x,y,z)=x-y+z$. A Mal’cev algebra is abelian (with respect to the term
condition commutator) if and only if it is polynomially equivalent to a module
[8]. We will need the following explicit characterization of abelian Mal’cev
algebras in Section 4.

######  Lemma 2.11.

[8, Chapters 5, 9] Let $\mathbf{A}$ be an abelian algebra with Mal’cev term
operation $m$. Then

  1. (1)

the set of rank-$1$-commutator terms
$R_{\mathbf{A}}:=\\{r\in\mathrm{Clo}(\mathbf{A})^{(2)}\ :\ r(z,z)=z\ \forall
z\in A\\}$ of $\mathbf{A}$ forms a ring $\mathbf{R}_{\mathbf{A}}$ under

$\displaystyle r(x,z)+s(x,z):=m(r(x,z),z,s(x,z)),$
$\displaystyle-r(x,z):=m(z,r(x,z),z)$ $\displaystyle r(x,z)\cdot
s(x,z):=r(s(x,z),z)$

for $r,s\in R$ and with neutral elements $z,x$ for $+,\cdot$, respectively.

  2. (2)

Fixing any $0\in A$ as neutral element, $A$ forms an
$\mathbf{R}_{\mathbf{A}}$-module $\mathbf{A}_{0}$ under

$\displaystyle a+b:=m(a,0,b),\quad-a:=m(0,a,0),$ $\displaystyle ra:=r(a,0)$

for $a,b\in A,r\in R_{\mathbf{A}}$.

The expansion of $\mathbf{A}$ with the constant operation $0$ is term
equivalent to the expansion of $\mathbf{A}_{0}$ with
$\mathrm{Clo}(\mathbf{A})^{(1)}$. In particular, $\mathbf{A}$ and
$\mathbf{A}_{0}$ are polynomially equivalent.

  3. (3)

For $0,z\in A$,

$h\colon A\to A,\ x\mapsto m(x,0,z),$

is an $\mathbf{R}_{\mathbf{A}}$-module isomorphism from $\mathbf{A}_{0}$ to
$\mathbf{A}_{z}$.

Let $\mathbf{A}$ be an abelian Mal’cev algebra. For $r=(r_{ij})_{i,j\in[k]}$ a
$k\times k$-matrix over $R_{\mathbf{A}}$, $x=(x_{1},\dots,x_{k})$ a $k$-tuple
of variables and $z$ another variable, define

$r*_{z}x:=\left(\sum_{j=1}^{k}r_{ij}(x_{j},z)\right)_{i\in[k]}$

with term functions added by $s(x,z)+t(x,z):=m(s(x,z),z,t(x,z))$. Then
$r*_{z}x$ is a $k$-tuple of $k+1$-ary term functions over $\mathbf{A}$.

Since the ring $\mathbf{R}_{\mathbf{A}}$ acts faithfully on any abelian
Mal’cev algebra $\mathbf{A}$, the nilpotence degree of the Jacobson radical of
$\mathbf{R}_{\mathbf{A}}$ is at most the height of the congruence lattice of
$\mathbf{A}$.

###  2.5. Distributive modules

A ring $\mathbf{R}$ is _semiperfect_ if it has a set $e_{1},...,e_{\ell}$ of
orthogonal idempotents with $e_{1}+\dots+e_{\ell}=1$ such that $e_{i}Re_{i}$
is a local ring for all $i\in[\ell]$. Then the idempotents
$e_{1},\dots,e_{\ell}$ are called local. All left Artinian (in particular all
finite) rings are semiperfect.

A module is _uniserial_ if its submodules are linearly ordered by inclusion.
In particular, simple modules and $\mathbb{Z}_{p^{n}}$ for any prime $p$ and
$n\in\mathbb{N}$ are uniserial. For an example over a noncommutative ring, let
$F$ be a field and

$R:=\left\\{\begin{pmatrix}a&b\\\ 0&a\end{pmatrix}\ :\ a,b\in F\right\\}.$

Then $A:=F^{2}$ with the usual action of matrices on column vectors forms a
uniserial $\mathbf{R}$-module with submodules $0\leq F\times 0\leq F^{2}$.

Distributive modules over semiperfect rings can be decomposed as sums of
uniserial modules over local subrings as follows.

######  Lemma 2.12.

Let $\mathbf{R}$ be a semiperfect ring with $1=\sum_{i=1}^{\ell}e_{i}$ for
local orthogonal idempotents $e_{1},\dots,e_{\ell}$ and let $\mathbf{A}$ be a
distributive $\mathbf{R}$-module. Then

  1. (1)

[10, Lemma 4] [17, 1.28] $e_{i}A$ is a uniserial $e_{i}Re_{i}$-module for all
$i\in[\ell]$.

  2. (2)

$A=\sum_{i=1}^{\ell}e_{i}A$ is a distributive module over the subring
$R^{\prime}:=\sum_{i=1}^{\ell}e_{i}Re_{i}$ of $\mathbf{R}$ with the induced
action from $\mathbf{R}$.

  3. (3)

$J(\mathbf{R}^{\prime})\subseteq J(\mathbf{R})$.

###### Proof.

(1) We include the short proof from [10, Lemma 4] for the convenience of the
reader. For $i\in[\ell]$, write $e:=e_{i}$ and $J:=J(\mathbf{R})$. We first
show that the set of $\mathbf{R}$-modules $\\{Rex\ :\ x\in A\\}$ is linearly
ordered under inclusion. Seeking a contradiction, suppose that there are
$x,y\in A$ such that $Rex$ and $Rey$ are not comparable. Since $Je$ is the
unique maximal $\mathbf{R}$-submodule of $Re$ by the assumptions, $Rex$ and
$Rey$ have unique maximal submodules $Jex$ and $Jey$, respectively. Hence
$Rex\cap Rey=Jex\cap Jey$. It follows that

$(Rex+Rey)/(Jex+Jey)\cong Re/Je\times Re/Je,$

which contradicts the distributivity of $\mathbf{A}$. Thus $\\{Rex\ :\ x\in
A\\}$ is linearly ordered.

So for each $x,y\in A$ either $ex\in eRey$ or $ey\in eRex$. Therefore $eA$ is
a uniserial $eRe$-module.

(2) Using the orthogonality of the idempotents, we see that every
$\mathbf{R}^{\prime}$-submodule of $\mathbf{A}$ is a direct sum of
$\mathbf{R}^{\prime}$-submodules of $e_{i}A$ for $i\in[\ell]$. Further the
$\mathbf{R}^{\prime}$-submodules of $e_{i}A$ coincide with the
$e_{i}Re_{i}$-submodules. Since the latter are linearly ordered by (1), it
follows that $\sum_{i=1}^{\ell}e_{i}A$ is a distributive
$\mathbf{R}^{\prime}$-module.

For (3), let $s\in J(e_{1}Re_{1})$ and $r\in R$. Then $s=e_{1}se_{1}$ and so

$1-rs=1-\sum_{i=1}^{\ell}e_{i}re_{1}se_{1}=1-e_{1}re_{1}se_{1}-\sum_{i>1}e_{i}re_{1}se_{1}.$

Since $e_{1}re_{1}se_{1}$ is in the Jacobson radical of the local ring
$e_{1}Re_{1}$ with identity $e_{1}$, we see that $e_{1}-e_{1}re_{1}se_{1}$ is
a unit in $e_{i}Re_{i}$. Since the quotient of $e_{1}Re_{1}$ by
$J(e_{1}Re_{1})$ is a division ring, the inverse of $e_{1}-e_{1}re_{1}se_{1}$
has the form $e_{1}-u$ for some $u\in J(e_{1}Re_{1})$. By the orthogonality of
idempotents again,

$\displaystyle(1-u)(1-rs)$
$\displaystyle=\underbrace{(1-u)(1-e_{1}re_{1}se_{1})}_{=1}-(1-\underbrace{u}_{=ue_{1}})(\sum_{i>1}e_{i}re_{1}se_{1})$
$\displaystyle=1-\sum_{i>1}e_{i}re_{1}se_{1}.$

Since $(\sum_{i>1}e_{i}re_{1}se_{1})^{2}=0$, we see that
$1-\sum_{i>1}e_{i}re_{1}se_{1}$ has inverse $1+\sum_{i>1}e_{i}re_{1}se_{1}.$
Hence $1-rs$ is a unit for every $r\in R$, so $s\in J(\mathbf{R})$.

A similar argument shows that $J(e_{i}Re_{i})\subseteq J(\mathbf{R})$ for each
$i\in[\ell]$. Hence
$J(\mathbf{R}^{\prime})=\sum_{i=1}^{\ell}J(e_{i}Re_{i})\subseteq
J(\mathbf{R})$. ∎

Let $\mathbf{A}$ be a module over a ring $\mathbf{R}$, let $k\in\mathbb{N}$
and $T=(t_{ij})_{1\leq i,j\leq k}$ a $k\times k$-matrix over $\mathbf{R}$.
Then $T$ acts on $x:=(x_{1},\dots,x_{k})$ in $\mathbf{A}^{k}$ by

$Tx:=(\sum_{j=1}^{k}t_{1j}x_{j},\dots,\sum_{j=1}^{k}t_{kj}x_{j}).$

Here $A^{k}\to A^{k},\ x\mapsto Tx,$ can be considered as a $k$-tuple of
$k$-ary term functions of $\mathbf{A}$ but is not necessarily an
$\mathbf{R}$-homomorphism if $\mathbf{R}$ is noncommutative.

For interpolating functions with domain $\mathbf{A}^{k}$ in Section 3, it will
be useful to cover $\mathbf{A}^{k}$ by its $\mathbf{R}$-submodules that are
isomorphic to $\mathbf{A}\times(J(\mathbf{R})A)^{k-1}$. We collect some
information about these submodules in the following.

######  Lemma 2.13.

Let $\mathbf{R}$ be a ring with Jacobson radical $J:=J(\mathbf{R})$, let
$\mathbf{A}$ be a finite distributive $\mathbf{R}$-module, let
$k\in\mathbb{N}$ and let

$V:=\\{M\leq\mathbf{A}^{k}\ :\ (JA)^{k}\leq
M,M/(JA)^{k}\cong\mathbf{A}/JA\\}.$

Then

  1. (1)

for every $M\leq\mathbf{A}^{k}$ such that $(JA)^{k}\leq M$ and $M/(JA)^{k}$
embeds into $\mathbf{A}/JA$ there exists an invertible $k\times k$-matrix $T$
over $\mathbf{R}$ such that $TM\leq A\times(JA)^{k-1}$; in particular,
$\mathrm{GL}_{k}(\mathbf{R})$ acts transitively on $V$.

  2. (2)

For all distinct $M,N\in V$ there exists $L<\mathbf{A}$ such that $M\cap N\leq
L^{k}$.

  3. (3)

$A^{k}=\bigcup V\cup\bigcup\\{L^{k}\ :\ L<\mathbf{A}\\}$.

###### Proof.

Factoring $\mathbf{R}$ by the annihilator of $\mathbf{A}$ if necessary, we may
assume that $\mathbf{R}$ is faithful on $\mathbf{A}$, in particular that
$\mathbf{R}$ is finite. We first prove (1) for the case that $\mathbf{R}$ is
commutative and semisimple. Then $J=0$ and $\mathbf{R}$ is the direct sum of
finite fields $\mathbf{K}_{1},\dots,\mathbf{K}_{\ell}$ for some
$\ell\in\mathbb{N}$ by the Wedderburn-Artin Theorem. Since $\mathbf{A}$ is
distributive and semisimple, it is a direct sum of pairwise non-isomorphic
simple $\mathbf{R}$-modules. It follows that $\mathbf{A}$ is isomorphic to the
regular $\mathbf{R}$-module $\sum_{i=1}^{\ell}K_{i}$.

Let $M\leq\mathbf{A}^{k}$ such that $M$ embeds into $\mathbf{A}$. Then we have
$S\subseteq[\ell]$ such that $M\cong\sum_{i\in S}K_{i}$. More precisely, we
also have $j_{i}\in[k]$ for $i\in S$ such that

$M=\sum_{i\in S}0\times\dots\times 0\times\underset{j_{i}}{K_{i}}\times
0\times\dots\times 0\leq\mathbf{A}^{k}.$

Clearly there exists an invertible $k\times k$ matrix $T$ over $\mathbf{R}$
that moves $K_{i}$ from the $j_{i}$-th component of $A^{k}$ to the first
component for all $i\in S$. Hence $TM=\sum_{i\in S}K_{i}\times
0^{k-1}\leq\mathbf{A}\times 0^{k-1}$. This proves (1) for $\mathbf{R}$
commutative and semisimple.

Next assume that $\mathbf{R}$ is semisimple. By the Wedderburn-Artin Theorem
$\mathbf{R}\cong\sum_{i=1}^{\ell}\mathbf{F}_{i}^{n_{i}\times n_{i}}$ for some
$\ell,n_{1},\dots,n_{\ell}\in\mathbb{N}$ and finite fields
$\mathbf{F}_{1},\dots,\mathbf{F}_{\ell}$. Since $\mathbf{A}$ is distributive
and semisimple, it is a direct sum of pairwise non-isomorphic simple
$\mathbf{R}$-modules, hence $\mathbf{A}\cong\sum_{i=1}^{\ell}F_{i}^{n_{i}}$.
Note that $\mathbf{F}_{i}^{n_{i}\times n_{i}}$ (and hence $\mathbf{R}$) has a
subring $\mathbf{K}_{i}$, which is a field of order $|F_{i}|^{n_{i}}$ for
every $i\in[\ell]$. Let $\hat{\mathbf{R}}:=\sum_{i=1}^{\ell}\mathbf{K}_{i}$
and let $\hat{\mathbf{A}}$ denote the $\hat{\mathbf{R}}$-module reduct of
$\mathbf{A}$. Since $|A|=|\hat{R}|$, we see that $\hat{\mathbf{A}}$ is
isomorphic to the regular $\hat{\mathbf{R}}$-module. Further
$\mathbf{R}$-submodules of $\mathbf{A}$ and $\hat{\mathbf{R}}$-submodules of
$\hat{\mathbf{A}}$ coincide and $\hat{\mathbf{A}}$ is distributive as well.
Applying (1) for $\hat{A}$ over the commutative ring $\hat{\mathbf{R}}$, we
obtain that for every $M\leq\mathbf{A}^{k}$ that embeds into $\mathbf{A}$,
there exists some
$T\in\mathrm{GL}_{k}(\hat{\mathbf{R}})\leq\mathrm{GL}_{k}(\mathbf{R})$ such
that $TM\leq\mathbf{A}\times 0^{k-1}$. This proves (1) for semisimple
$\mathbf{R}$.

Finally, for the general case let $\mathbf{R}$ be arbitrary, let
$M\leq\mathbf{A}^{k}$ such that $(JA)^{k}\leq M$ and
$\overline{M}:=M/(JA)^{k}$ embeds into $\overline{\mathbf{A}}:=\mathbf{A}/JA$.
Applying (1) for the distributive module $\overline{\mathbf{A}}$ over the
semisimple ring $\overline{\mathbf{R}}:=\mathbf{R}/J$, we obtain $T\in
R^{k\times k}$ whose projection $\overline{T}$ in $\overline{R}^{k\times k}$
is invertible and satisfies
$\overline{T}\overline{M}\leq\overline{\mathbf{A}}\times 0^{k-1}$. Since
$\overline{T}$ is invertible over $\overline{\mathbf{R}}$, we have $S\in
R^{k\times k}$ such that $ST=I_{k}-U$ for $I_{k}$ the $k\times k$ identity
matrix over $\mathbf{R}$ and some $U\in J^{k\times k}$. Since $J$ is
nilpotent, it follows that $ST\in\mathrm{GL}_{k}(\mathbf{R})$ and further
$T\in\mathrm{GL}_{k}(\mathbf{R})$. Since $\overline{T}\overline{M}=N/J\times
0^{k-1}$ for some $JA\leq N\leq\mathbf{A}$, we obtain $TM\subseteq
N\times(JA)^{k-1}$. Equality follows since $|M|=|N\times(JA)^{k-1}|$ and $T$
is invertible. Thus (1) is finally proved in full generality.

For (2) let $M,N\in V$ be distinct. By (1) we have
$T\in\mathrm{GL}_{k}(\mathbf{R})$ such that $T(M\cap N)=L\times(JA)^{k-1}$ for
some $JA\leq L<\mathbf{A}$. Hence $M\cap N\leq T^{-1}L^{k}\leq L^{k}$.

For (3) let $a\in A^{k}$ and $M:=Ra+(JA)^{k}$. Again we use the notation
$\overline{\mathbf{R}}:=\mathbf{R}/J$, $\overline{\mathbf{A}}:=\mathbf{A}/JA$
and $\overline{M}:=M/(JA)^{k}$. Then $\overline{M}$ is a cyclic
$\overline{\mathbf{R}}$-submodule of $\overline{\mathbf{A}}^{k}$. Hence
$\overline{M}$ is a sum of pairwise non-isomorphic simple
$\overline{\mathbf{R}}$-modules and embeds into $\overline{\mathbf{A}}$. By
(1) we have $T\in\mathrm{GL}_{k}(\mathbf{R})$ such that $TM=L\times(JA)^{k-1}$
for some $JA\leq L\leq\mathbf{A}$. If $L=A$, then $M\in V$ and $a\in\bigcup
V$. Else if $L<\mathbf{A}$, then $M\leq T^{-1}L^{k}\leq L^{k}$ and
$a\in\bigcup\\{L^{k}\ :\ L<\mathbf{A}\\}$. Thus $A^{k}\subseteq\bigcup
V\cup\bigcup\\{L^{k}\ :\ L<\mathbf{A}\\}$. The converse inclusion is trivial.
∎

We note that Lemma 2.13 does not generalize to arbitrary cyclic
$\mathbf{R}$-modules. For example, let $\mathbf{R}:=\mathbf{F}^{2\times 2}$ be
the $2\times 2$-matrix ring over a field $\mathbf{F}$ and let $\mathbf{A}$ be
the regular $\mathbf{R}$-module. Then

$M:=\left\\{\begin{bmatrix}a&0\\\ b&0\end{bmatrix}\ :\ a,b\in
F\right\\}^{2}\leq\mathbf{A}^{2}$

is isomorphic to $\mathbf{A}$ and fixed by $\mathbf{R}^{2\times 2}$. Hence
there exists no $T\in\mathbf{R}^{2\times 2}$ such that $TM=A\times 0$.

###  2.6. Inner rank of matrices

The rank of matrices over fields can be generalized to arbitrary rings as
follows (see [4, Proposition 5.4.3]). Let $\mathbf{R}$ be a ring,
$m,n\in\mathbb{N}$ and $A\in R^{m\times n}$. The _inner rank_ of $A\neq 0$ is
defined as

$\mathrm{rk}(A):=\min\\{r\in\mathbb{N}\ :\ A=BC\text{ for some }B\in
R^{m\times r},C\in R^{r\times n}\\}$

and $\mathrm{rk}(0):=0$.

Then $\mathrm{rk}(A)$ is the least $r$ such that the right
$\mathbf{R}$-submodule of $R^{m}$ that is generated by the columns of $A$ is
contained in an $r$-generated module (equivalently, the least $r$ such that
the left $\mathbf{R}$-submodule of $R^{n}$ that is generated by the rows of
$A$ is contained in an $r$-generated module). Note that in particular
$\mathrm{rk}(A)\leq\min(m,n)$.

If $\mathbf{R}$ is a field, the inner rank is just the usual row or column
rank.

###  2.7. Uniformly generated functions

Let $\mathbf{A},\mathbf{B}$ be algebras. For $k,\ell,n\in\mathbb{N}$, let

$R^{\ell,k}_{n}:=\\{r\in(\mathrm{Clo}(\mathbf{A})^{(\ell)})^{k}\ :\ \\\
r(x)=(v_{1}(w_{1}(x),\dots,w_{n}(x)),\dots,v_{k}(w_{1}(x),\dots,w_{n}(x)))\\\
\text{ for
}v_{1},\dots,v_{k}\in\mathrm{Clo}(\mathbf{A})^{(n)},w_{1},\dots,w_{n}\in\mathrm{Clo}(\mathbf{A})^{(\ell)}\\}$

denote the set of $k$-tuples of $\ell$-ary term functions on $\mathbf{A}$ that
factor through $n$-ary term functions.

For an $\mathbf{R}$-module $\mathbf{A}$ we simply have

$R^{\ell,k}_{n}=\\{A^{\ell}\to A^{k},x\mapsto ax\ :\ a\in
R^{k\times\ell},\mathrm{rk}(a)\leq n\\}.$

For $U\subseteq F(A,B)^{(k)}$, $o\in\mathbb{N},r_{1},\dots,r_{o}\in
R^{\ell,k}_{n}$ and $s\in\mathrm{Clo}(\mathbf{B})^{(o)}$, the operator

$d\colon U\to F(A,B)^{(\ell)},\ f(x)\mapsto s(f(r_{1}(x)),\dots,f(r_{o}(x))),$

is defined by composing $n$-ary $\mathbf{A},\mathbf{B}$-minors with term
functions of $\mathbf{A}$ and $\mathbf{B}$ on the respective sides. We will
express this more shortly by saying that the $\ell$-ary function
$f^{\prime}(x):=s(f(r_{1}(x)),\dots,f(r_{o}(x)))$ is _uniformly generated_ by
$n$-ary $\mathbf{A},\mathbf{B}$-minors for the $k$-ary functions $f\in U$.

For example, for every $f\in F(A,B)^{(2)}$ the minor
$f^{\prime}(x_{1}):=f(x_{1},x_{1})$ is uniformly generated by the unary
$\mathbf{A},\mathbf{B}$-minors of $f$ (in fact by $f(x_{1},x_{1})$ itself).

Less trivially, for groups $\mathbf{A}=\mathbf{B}=(\mathbb{Z},+,-,0)$ and
$k=\ell=2,n=1$ the binary function

$f^{\prime}(x_{1},x_{2}):=f(0,0)-f(x_{1},-x_{1})+2f(x_{1}+x_{2},2(x_{1}+x_{2}))$

is uniformly generated by the unary $\mathbf{A},\mathbf{B}$-minors
$f(0x,0x),f(x,-x),f(x,2x)$ of $f\in F(A,B)^{(2)}$.

We will use the following properties of uniformly generated functions between
modules in Section 3.

######  Lemma 2.14.

Let $\mathbf{A}$ be an $\mathbf{R}$-module, $\mathbf{B}$ an
$\mathbf{S}$-module, $k,\ell,n\in\mathbb{N}$, $U\subseteq F(A,B)^{(k)}$ and
$U\to F(A,B)^{(\ell)},\ f\mapsto f^{\prime}$.

  1. (1)

Then $f^{\prime}$ is uniformly generated by $n$-ary
$\mathbf{A},\mathbf{B}$-minors of $f\in U$ if and only if there exists
$s\colon\\{r\in R^{k\times\ell}\ :\ \mathrm{rk}(r)\leq n\\}\to S$ with finite
support such that for all $f\in U$ and all $x\in A^{\ell}$

$f^{\prime}(x)=\sum_{r\in R^{k\times\ell},\mathrm{rk}(r)\leq n}s(r)f(rx).$

  2. (2)

Assume $k=\ell$, that $f^{\prime}$ is uniformly generated by $n$-ary
$\mathbf{A},\mathbf{B}$-minors of $f\in U$ and that $f-f^{\prime}$ is
uniformly generated by its $n$-ary $\mathbf{A},\mathbf{B}$-minors for every
$f\in U$. Then every $f\in U$ is uniformly generated by its $n$-ary
$\mathbf{A},\mathbf{B}$-minors.

###### Proof.

(1) is straightforward from the definition. For (2) let $s,t\colon\\{r\in
R^{k\times k}\ :\ \mathrm{rk}(r)\leq n\\}\to S$ such that for all $f\in U$ and
all $x\in A^{k}$

$\displaystyle f^{\prime}(x)$ $\displaystyle=\sum_{r\in R^{k\times
k},\mathrm{rk}(r)\leq n}s(r)f(rx),$ $\displaystyle(f-f^{\prime})(x)$
$\displaystyle=\sum_{u\in R^{k\times k},\mathrm{rk}(u)\leq
n}t(u)(f-f^{\prime})(ux).$

Since $\mathrm{rk}(ru)\leq\min(\mathrm{rk}(r),\mathrm{rk}(u))$, it follows
that

$\displaystyle(f-f^{\prime})(x)=$ $\displaystyle\sum_{u\in R^{k\times
k},\mathrm{rk}(u)\leq n}t(u)\left(f(ux)-\sum_{r\in R^{k\times
k},\mathrm{rk}(r)\leq n}s(r)f(rux)\right)$

is uniformly generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$.
Together with the assumption on $f^{\prime}$ we see that
$f=(f-f^{\prime})+f^{\prime}$ is also uniformly generated by $n$-ary
$\mathbf{A},\mathbf{B}$-minors of $f$. ∎

###  2.8. Sums

Let $S$ be a subset of an abelian group $(A,+)$. To simplify notation we will
also write $\sum S$ for $\sum_{s\in S}s$.

##  3\. Clonoids from distributive modules

This section consists of the proof that every function from a finite
distributive $\mathbf{R}$-module $\mathbf{A}$ into a coprime module
$\mathbf{B}$ is uniformly generated by its $n$-ary
$\mathbf{A},\mathbf{B}$-minors, where $n$ is the nilpotence degree of the
Jacobson radical of $\mathbf{R}$. This is the basis of all our finite
generation results for clonoids between coprime abelian Mal’cev algebras. More
precisely we show the following.

######  Theorem 3.1.

Let $\mathbf{A}$ be a finite distributive $\mathbf{R}$-module, let
$n\in\mathbb{N}$ such that $J(\mathbf{R})^{n}=0$, and let $\mathbf{B}$ be an
$\mathbf{S}$-module such that $|A|$ is invertible in $\mathbf{S}$.

For all $k\in\mathbb{N}$ there exists $s\colon\\{r\in R^{k\times k}\ :\
\mathrm{rk}(r)\leq n\\}\to S$ such that for all $f\colon A^{k}\to B$ and all
$x\in A^{k}$

$f(x)=\sum_{r\in R^{k\times k},\mathrm{rk}(r)\leq n}s(r)f(rx).$

######  Example 3.2.

We illustrate Theorem 3.1 for binary functions from
$\mathbf{A}=(\mathbb{Z}_{2},+)$ to an $\mathbf{S}$-module $\mathbf{B}$ such
that $2$ a unit in $\mathbf{S}$. Then every $f\colon A^{2}\to B$ is uniformly
generated by its unary $\mathbf{A},\mathbf{B}$-minors $f(0,0)$, $f(x,0)$,
$f(0,x),f(x,x)$ via

$\displaystyle f(x_{1},x_{2})=$ $\displaystyle f(0,0)$
$\displaystyle+2^{-1}[f(x_{1},0)+f(x_{1}+x_{2},0)-f(0,0)-f(x_{2},0)$
$\displaystyle+f(0,x_{2})+f(0,x_{1}+x_{2})-f(0,0)-f(0,x_{1})$
$\displaystyle+f(x_{1},x_{1})+f(x_{2},x_{2})-f(0,0)-f(x_{1}+x_{2},x_{1}+x_{2})].$

This identity can be derived from the interpolation arguments for Theorem 3.1
below but is more elementary verified by the following case analysis. If
$x_{2}=0$, then lines 1 and 2 on the right hand side add up to $f(x_{1},0)$
while lines 3 and 4 each cancel. If $x_{1}=0$, then similarly lines 1 and 3
add up to $f(0,x_{2})$ while lines 1 and 4 each cancel. Finally if
$x_{1}=x_{2}$, then lines 1 and 4 add up to $f(x_{1},x_{2})$ while lines 2 and
3 each cancel. Hence in each case the right hand side of the formula yields
$f(x_{1},x_{2})$ and the claim is proved.

By Lemma 2.12 it suffices to prove Theorem 3.1 for
$\mathbf{R}=\sum_{i=1}^{\ell}R_{i}$ a direct product of finite local rings
$R_{i}$ with greatest ideal $J_{i}$ and uniserial module $A_{i}$ for
$i\in[\ell]$. Further $\mathbf{A}=\sum_{i=1}^{\ell}A_{i}$ is a faithful
distributive $\mathbf{R}$-module and
$J(\mathbf{R})=\sum_{i=1}^{\ell}J_{i}=:J$. Because of its faithfulness,
$\mathbf{R}$ is finite and $|R|$ has the same prime divisors as $|A|$. Since
$|A|$ is invertible in $\mathbf{S}$, so is $|R|$. We fix these assumptions and
notation throughout this section for the proof of Theorem 3.1.

We will show that $f$ is generated in a uniform way by its $n$-ary
$\mathbf{A},\mathbf{B}$-minors via induction on $\mathbf{A}$ and a series of
reductions and interpolations.

The base case of Theorem 3.1 for trivial $\mathbf{A}$ is clear. Next the
induction assumption yields the following.

######  Lemma 3.3.

Let $M$ be a proper submodule of $\mathbf{A}$. There exists $f^{\prime}\colon
A^{k}\to B$ such that $f^{\prime}|_{M^{k}}=f|_{M^{k}}$ and $f^{\prime}$ is
uniformly generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$.

Note that in Lemma 3.3 the difference of $f$ and $f^{\prime}$ is $0$ on
$M^{k}$. So by Lemma 2.14(2) it suffices to prove Theorem 3.1 for functions
satisfying $f(M^{k})=0$. If $f(N^{k})=0$ for some submodule $N$ of
$\mathbf{A}$, then also $f^{\prime}(N)=0$. Hence by repeating this reduction
for all maximal submodules $M$ of $\mathbf{A}$, we may assume that $f$
vanishes on all proper submodules of $\mathbf{A}$.

First we show that for any maximal $M$ every such $f$ uniformly generates a
function that is equal to $f$ on $A\times M^{k-1}$ and $0$ else.

######  Lemma 3.4.

Let $M$ be a maximal submodule of $\mathbf{A}$ and assume $f(M^{k})=0$. Then

$g_{M}\colon A^{k}\to B,\ x\mapsto\begin{cases}f(x)&\text{ if }x\in A\times
M^{k-1},\\\ 0&\text{ else},\end{cases}$

is uniformly generated by $\mathbf{A},\mathbf{B}$-minors of $f$.

###### Proof.

From the decomposition of $\mathbf{A}$ we may assume that
$M=J_{1}A_{1}+\sum_{i=2}^{\ell}A_{i}$. We use induction on the number of
submodules of $\mathbf{A}_{1}$.

For the base case, $\mathbf{A}_{1}$ is simple, $J_{1}=0$ and $\mathbf{R}_{1}$
is a simple ring. Let $e$ be the multiplicative identity of
$\sum_{i=2}^{\ell}R_{i}$. We claim that for all $x_{1},\dots,x_{k}\in A$

(3.1)
$\begin{array}[]{ll}g_{M}(x_{1},\dots,x_{k})=&|R_{1}|^{1-k}\left(\sum_{a_{2},\dots,a_{k}\in
R_{1}}f(x_{1}+\sum_{i=2}^{k}a_{i}x_{i},x_{2},\dots,x_{k})\right.\\\
&\left.-\sum_{a_{2},\dots,a_{k}\in
R_{1}}f(ex_{1}+\sum_{i=2}^{k}a_{i}x_{i},x_{2},\dots,x_{k})\right).\end{array}$

If $x_{2},\dots,x_{k}\in M$, this follows since $\sum_{i=2}^{k}a_{i}x_{i}=0$
and the right hand side of (3.1) simplifies to

$f(x_{1},x_{2},\dots,x_{k})-\underbrace{f(ex_{1},x_{2},\dots,x_{k})}_{=0\text{
since }f(M^{k})=0}=f(x_{1},x_{2},\dots,x_{k}).$

Else if $(x_{2},\dots,x_{k})\not\in M^{k-1}$, then the maximality of $M$ in
$\mathbf{A}$ implies

$\left\\{\sum_{i=2}^{k}a_{i}x_{i}\ :\ a_{2},\dots,a_{k}\in
R_{1}\right\\}=\sum_{i=2}^{k}R_{1}x_{i}=A_{1}$

where each element in $A_{1}$ is attained with the same multiplicity
$\frac{|R_{1}|^{k-1}}{|A_{1}|}$. So the right hand side of (3.1) yields

$|R_{1}|^{-1}\left(\sum f(x_{1}+A_{1},x_{2},\dots,x_{k})-\sum
f(\underbrace{ex_{1}+A_{1}}_{=x_{1}+A_{1}\text{ since }A=A_{1}\times
M},x_{2},\dots,x_{k})\right)=0.$

This shows (3.1) and the base case of the induction.

For the induction step let $M=J_{1}A_{1}+\sum_{i=2}^{\ell}A_{i}$ and
$J_{1}A_{1}\neq 0$ in the following. Let $N$ be the smallest
$\mathbf{R}$-submodule of $\mathbf{A}_{1}$. Then $N\leq M$ and $N=IA_{1}$ for
$I$ a power of $J_{1}$. In order to apply the induction assumption, we will
take averages of functions over arguments $\sum Ix$. For this we note that by
the minimality of $N$,

$Ix=\begin{cases}0&\text{for all }x\in M,\\\ N&\text{for all }x\in A\setminus
M.\end{cases}$

First we claim that

$\bar{f}(x_{1},\dots,x_{k}):=|I|^{-k^{2}}\sum_{a_{1},\dots,a_{k}\in
I^{k}}f(x_{1}+\sum_{j=1}^{k}a_{1j}x_{j},x_{2}+\sum_{j=1}^{k}a_{2j}x_{j},\dots,x_{k}+\sum_{j=1}^{k}a_{kj}x_{j})$

satisfies

$\bar{f}(x)=|N|^{-k}\sum f(x+N^{k})\text{ for all }x\in A^{k}.$

For $x\in M^{k}$, this follows from the assumption that $f(M^{k})=0$ and
$N\subseteq M$. Else if $x\not\in M^{k}$, it follows from

$\left\\{\sum_{j=1}^{k}a_{ij}x_{j}\ :\ a_{i}\in
I^{k}\right\\}=\sum_{j=1}^{k}Ix_{j}=N,$

where each element in $N$ is attained with the same multiplicity
$\frac{|I|^{k}}{|N|}$. Since $\bar{f}$ is constant on blocks modulo $N$, we
may view it as a function from $\mathbf{A}/N$ to $\mathbf{B}$. By the
induction assumption on $\mathbf{A}/N$ with maximal submodule $M/N$,

$\bar{g}_{M}(x):=\begin{cases}\bar{f}(x)&\text{ if }x\in A\times M^{k-1},\\\
0&\text{ else},\end{cases}$

is uniformly generated by $\mathbf{A}/N,\mathbf{B}$-minors of $\bar{f}$,
equivalently uniformly generated by $\mathbf{A},\mathbf{B}$-minors of $f$.

Similarly as for $\bar{f}$ we see that

$\displaystyle\hat{f}(x_{1},\dots,x_{k}):=|I|^{-k(k-1)}\sum_{a_{1},\dots,a_{k}\in
I^{k-1}}f(x_{1}+\sum_{j=2}^{k}a_{1,j-1}x_{j},x_{2}+\sum_{j=2}^{k}a_{2,j-1}x_{j},$
$\displaystyle\dots,x_{k}+\sum_{j=2}^{k}a_{k,j-1}x_{j})$

satisfies

$\hat{f}(x)=\begin{cases}f(x)&\text{if }x\in A\times M^{k-1},\\\ |N|^{-k}\sum
f(x+N^{k})=\bar{f}(x)&\text{else}.\end{cases}$

Now $g_{M}=\hat{f}-(\bar{f}-\bar{g}_{M})$ is an $\mathbf{A},\mathbf{B}$-minor
of $f$. We note that the formulas for $\hat{f},\bar{f},\bar{g}_{M}$ and
$g_{M}$ are independent of $f$. Hence the induction step and the lemma is
proved. ∎

Next we obtain a function that coincides with $f$ on $A\times(JA)^{k-1}$ and
is uniformly generated by the $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$.

######  Lemma 3.5.

Assume $f(M^{k})=0$ for all proper submodules $M$ of $\mathbf{A}$. Then

$g\colon A^{k}\to B,\ x\mapsto\begin{cases}f(x)&\text{ if }x\in
A\times(JA)^{k-1},\\\ 0&\text{ else},\end{cases}$

is uniformly generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$.

###### Proof.

That $g$ can be uniformly generated from $\mathbf{A},\mathbf{B}$-minors of $f$
follows by repeated application of Lemma 3.4 for all maximal submodules $M$ of
$\mathbf{A}$ since $JA$ is the intersection of these submodules. It remains to
show that $n$-ary $\mathbf{A},\mathbf{B}$-minors suffice. For this we induct
on the nilpotence degree $n$ of $J$.

The base case $n=1$ and $J=0$ holds since $g$ can clearly be uniformly
generated by $f(x_{1},0,\dots,0)$. Assume $J>0$ and let $I$ be the annihilator
of $J$ in $\mathbf{R}$ in the following. Then the Jacobson radical of
$\mathbf{R}/I$ has nilpotence degree $n-1$.

For $x_{1}\in A$, define

$f_{x_{1}}\colon A^{k-1}\to B,\ (x_{2},\dots,x_{k})\mapsto
f(x_{1},x_{2},\dots,x_{k}).$

Since $JA$ is a distributive $\mathbf{R}/I$-module, the induction assumption
of Theorem 3.1 yields that the restriction of $f_{x_{1}}$ to $(JA)^{k-1}$ can
be interpolated by its $n-1$-ary $JA,\mathbf{B}$-minors in a uniform way
(independently of $x_{1}$). More precisely we have $s\colon\\{r\in
R^{k-1\times k-1}\ :\ \mathrm{rk}(r)\leq n-1\\}\to S$ such that

$f(x_{1},x_{2},\dots,x_{k})=\sum_{r\in R^{(k-1)\times(k-1)},\mathrm{rk}(r)\leq
n-1}s(r)f(x_{1},r(x_{2},\dots x_{k}))$

for all $x_{1}\in A,x_{2},\dots,x_{k}\in JA.$ Here
$(x_{1},r(x_{2},\dots,x_{k}))$ denotes the concatenation of the $1$-tuple
$x_{1}$ and the $k-1$-tuple $r(x_{2},\dots,x_{k})$. For $r\in R^{k-1\times
k-1}$ the $k\times k$ block diagonal matrix $r^{\prime}:=\begin{pmatrix}1&0\\\
0&r\end{pmatrix}$ has inner rank at most $\mathrm{rk}(r)+1$ and satisfies
$r^{\prime}(x_{1},\dots,x_{k})=(x_{1},r(x_{2},\dots,x_{k}))$. Then

$f^{\prime}(x):=\sum_{r\in R^{(k-1)\times(k-1)},\mathrm{rk}(r)\leq
n-1}s(r)f(r^{\prime}x)$

is equal to $f$ on $A\times(JA)^{k}$ and uniformly generated by $n$-ary
$\mathbf{A},\mathbf{B}$-minors of $f$.

By the argument in the first paragraph of this proof, $g$ is uniformly
generated by $\mathbf{A},\mathbf{B}$-minor of $f^{\prime}$. Hence $g$ is
uniformly generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$. ∎

Extending Lemma 3.5 we show that $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$
generate functions that are equal to $f$ on any submodule $N$ isomorphic to
$A\times(JA)^{k-1}$ and $0$ else.

######  Lemma 3.6.

Assume $f(M^{k})=0$ for all proper submodules $M$ of $\mathbf{A}$. Let
$(JA)^{k}\leq N\leq\mathbf{A}^{k}$ such that $N/(JA)^{k}\cong\mathbf{A}/JA$.
Then

$f_{N}\colon A^{k}\to B,\ x\mapsto\begin{cases}f(x)&\text{if }x\in N,\\\
0&\text{else},\end{cases}$

is uniformly generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$.

###### Proof.

By Lemma 2.13 we have some invertible $k\times k$-matrix $T$ over $\mathbf{R}$
such that $TN=A\times(JA)^{k-1}$. Let $g$ as in Lemma 3.5 be uniformly
generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f(T^{-1}x)$, that is,

$g\colon A^{k}\to B,\ x\mapsto\begin{cases}f(T^{-1}x)&\text{ if }x\in A\times
J^{k-1},\\\ 0&\text{ else.}\end{cases}$

We claim that

(3.2) $f_{N}(x)=g(Tx)\text{ for all }x\in A^{k}.$

If $x\in N$, then $Tx\in A\times(JA)^{k-1}$ and

$g(Tx)=f(T^{-1}Tx)=f(x).$

Else if $x\not\in N$, then $Tx\not\in A\times(JA)^{k-1}$ and so $g(Tx)=0$.
This proves (3.2) and in particular that $f_{N}$ is uniformly generated by
$n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$. ∎

Summing up the restrictions $f_{N}$ of $f$ from the previous lemma we now
obtain the main result of this section, which we repeat in its short form.

######  Theorem 3.1.

$f$ is uniformly generated by its $n$-ary $\mathbf{A},\mathbf{B}$-minors.

###### Proof.

Recall that by repeated application of Lemma 3.3 we have $f^{\prime}$
generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$ such that
$f-f^{\prime}$ is $0$ on $M^{k}$ for all proper submodules $M$ of
$\mathbf{A}$. By Lemma 2.14(2) we may assume $f(M^{k})=0$ for all
$M<\mathbf{A}$ in the following.

Let

$V:=\\{N\leq\mathbf{A}^{k}\ :\ (JA)^{k}\leq N,N/(JA)^{k}\cong\mathbf{A}/JA\\}$

and let $N\in V$. By Lemma 3.6 $n$-ary $\mathbf{A},\mathbf{B}$-minors of $f$
uniformly generate $f_{N}$, which is $f$ on $N$ and $0$ else. Lemma 2.13 and
$f(M^{k})=0$ for $M<\mathbf{A}$ imply that the support of the functions
$f_{N}$ partitions the support of $f$. Hence

$f=\sum_{N\in V}f_{N}.$

Thus $f$ is uniformly generated by $n$-ary $\mathbf{A},\mathbf{B}$-minors of
$f$. ∎

Theorem 1.4 for modules is an easy consequence of Theorem 3.1.

###### Proof of Theorem 1.4 for modules $\mathbf{A},\mathbf{B}$.

Let $C$ be a clonoid from $\mathbf{A}$ to $\mathbf{B}$. Then $C=\langle
C^{(n)}\rangle$ by Theorem 3.1.

If $\mathbf{B}$ is finite, then there are finitely many $n$-ary functions from
$A$ to $B$ and so there are finitely many clonoids from $\mathbf{A}$ to
$\mathbf{B}$. ∎

##  4\. Clonoids from distributive abelian Mal’cev algebras

Extending Theorem 3.1 we show that every function from a finite abelian
Mal’cev algebra $\mathbf{A}$ with distributive congruence lattice into some
coprime abelian Mal’cev algebra $\mathbf{B}$ is generated by its $n+1$-ary
$\mathbf{A},\mathbf{B}$-minors where $n$ is the nilpotence degree of the
Jacobson radical of the ring $\mathbf{R}_{\mathbf{A}}$ (see the definition in
Lemma 2.11). The increase in the arity compared to the result for modules is
due to the fact that we have to compensate for the missing constant term
function $0$ with a projection, i.e., an additional variable $z$. More
precisely we have the following.

######  Theorem 4.1.

Let $\mathbf{A}$ be a finite abelian Mal’cev algebra with distributive
congruence lattice, let $n\in\mathbb{N}$ be the nilpotence degree of the
Jacobson radical of $\mathbf{R}_{\mathbf{A}}$, and let $\mathbf{B}$ be an
abelian Mal’cev algebra such that $|A|$ is invertible in
$\mathbf{R}_{\mathbf{B}}$.

For all $k\in\mathbb{N}$ there exists $s\colon\\{r\in R_{\mathbf{A}}^{k\times
k}\ :\ \mathrm{rk}(r)\leq n\\}\to R_{\mathbf{B}}$ such that for all $f,b\colon
A^{k+1}\to B$ and all $x\in A^{k},z\in A$

(4.1) $f(x,z)=\sum_{r\in R_{\mathbf{A}}^{k\times k},\mathrm{rk}(r)\leq
n}s(r)*_{b(x,z)}f(r*_{z}x,z)$

where the sum is taken pointwise with respect to $+_{b(x,z)}$ in $\mathbf{B}$.

###### Proof.

Let $k\in\mathbb{N},a\in A,b\in B$. Since $\mathbf{A}$ and the
$\mathbf{R}_{\mathbf{A}}$-module $\mathbf{A}_{a}$ are polynomially equivalent
by Lemma 2.11(2), they have the same congruences. Since the congruence lattice
of $\mathbf{A}$ is distributive by assumption, so is the congruence lattice of
$\mathbf{A}_{a}$. Now the distributive $\mathbf{R}_{\mathbf{A}}$-module
$\mathbf{A}_{a}$ and the $\mathbf{R}_{\mathbf{B}}$-module $\mathbf{B}_{b}$
satisfy the assumptions of Theorem 3.1. Thus we have some $s\colon\\{r\in
R_{\mathbf{A}}^{k\times k}\ :\ \mathrm{rk}(r)\leq n\\}\to R_{\mathbf{B}}$ such
that for all $g\colon A^{k}\to B$ and all $x\in A$

(4.2) $g(x)=\sum_{r\in R_{\mathbf{A}}^{k\times k},\mathrm{rk}(r)\leq
n}s(r)*_{b}g(r*_{a}x).$

Note that this formula depends on the constants $a,b$ via the action $*_{a}$
of $R_{\mathbf{A}}^{k\times k}$ on $A^{k}$ and the underlying addition
$u+_{a}v=m(u,a,v)$ of $\mathbf{A}_{a}$ as well as the operations $*_{b},+_{b}$
of $\mathbf{B}_{b}$ for the outer sum.

We claim that the function $s$ is actually independent of the choices of $a,b$
since the modules $\mathbf{A}_{a},\mathbf{B}_{b}$ are isomorphic for all $a\in
A,b\in B$, respectively. To see that let first $g\colon A^{k}\to B$, $z\in A$
and $p\colon\mathbf{A}_{a}^{k}\to\mathbf{A}_{z}^{k}$ be the
$\mathbf{R}_{\mathbf{A}}$-module isomorphism which exists by Lemma 2.11(3).
Applying (4.2) to $gp$ and using that $p$ is a homomorphism, we obtain for all
$x\in A^{k}$

$g(p(x))=\sum_{r\in R_{\mathbf{A}}^{k\times k},\mathrm{rk}(r)\leq
n}s(r)*_{b}g(p(r*_{a}x))=\sum_{r\in R_{\mathbf{A}}^{k\times
k},\mathrm{rk}(r)\leq n}s(r)*_{b}g(r*_{z}p(x))).$

Since $p$ is a bijection on $A^{k}$, this yields

(4.3) $g(x)=\sum_{r\in R_{\mathbf{A}}^{k\times k},\mathrm{rk}(r)\leq
n}s(r)*_{b}g(r*_{z}x)$

for all $x\in A^{k}$. Hence $s$ is independent of the choice of the neutral
element in the module $\mathbf{A}_{z}$.

For $f\colon A^{k+1}\to B$ and $z\in A$, we set
$g(x_{1},\dots,x_{k}):=f(x_{1},\dots,x_{k},z)$ in (4.3) to obtain

(4.4) $f(x,z)=\sum_{r\in R_{\mathbf{A}}^{k\times k},\mathrm{rk}(r)\leq
n}s(r)*_{b}f(r*_{z}x,z)$

for all $x\in A^{k},z\in A$.

Next let $c\in B$ and let $h\colon\mathbf{B}_{c}\to\mathbf{B}_{b}$ be the
$\mathbf{R}_{\mathbf{B}}$-module isomorphism given in Lemma 2.11(3). Applying
(4.4) to $hf$ and using that $h$ is a homomorphism, we obtain for all $x\in
A^{k},z\in A$ that

$\displaystyle hf(x,z)$ $\displaystyle=\sum_{r\in R_{\mathbf{A}}^{k\times
k},\mathrm{rk}(r)\leq n}s(r)*_{b}hf(r*_{z}x,z)$
$\displaystyle=h\left(\sum_{r\in R_{\mathbf{A}}^{k\times k},\mathrm{rk}(r)\leq
n}s(r)*_{c}f(r*_{z}x,z)\right).$

Note that here the sum outside of $h$ is computed in $\mathbf{B}_{b}$ and
inside of $h$ in $\mathbf{B}_{c}$. Since $h$ is a bijection on $B$, it follows
that

(4.5) $f(x,z)=\sum_{r\in R_{\mathbf{A}}^{k\times k},\mathrm{rk}(r)\leq
n}s(r)*_{c}f(r*_{z}x,z)$

for all $x\in A^{k},z\in A,c\in B$. Hence $s$ is also independent of the
choice of the neutral element in the module $\mathbf{B}_{c}$. In particular by
choosing $c=b(x,z)$ independently for any point $x\in A^{k},z\in A$ in
equation (4.5), we obtain (4.1). ∎

Theorem 1.4 is an easy consequence of Theorem 4.1.

###### Proof of Theorem 1.4.

Let $\mathbf{A}$ be polynomially equivalent to a distributive
$\mathbf{R}_{\mathbf{A}}$-module with $n$ the nilpotence degree of
$J(\mathbf{R}_{\mathbf{A}})$, let $\mathbf{B}$ be polynomially equivalent to
an $\mathbf{R}_{\mathbf{B}}$-module such that $|A|$ is invertible in
$\mathbf{R}_{\mathbf{B}}$. Then every clonoid from $\mathbf{A}$ to
$\mathbf{B}$ is generated by its $n+1$-ary functions by Theorem 4.1.

Hence, if $\mathbf{B}$ is finite, then there are only finitely many clonoids
from $\mathbf{A}$ to $\mathbf{B}$. ∎

We point out one important special case of Theorem 1.4.

######  Corollary 4.2.

Let $\mathbf{A}=\mathbf{A}_{1}\times\dots\times\mathbf{A}_{\ell}$ for pairwise
non-isomorphic finite abelian simple Mal’cev algebras
$\mathbf{A}_{1},\dots,\mathbf{A}_{\ell}$ and let $\mathbf{B}$ be a finite
abelian Mal’cev algebra such that $|A|$ and $|B|$ are coprime.

Then every clonoid from $\mathbf{A}$ to $\mathbf{B}$ is generated by binary
functions.

###### Proof.

Since the factors $\mathbf{A}_{i}$ are pairwise non-isomorphic and simple,
every congruence of $\mathbf{A}$ is a product congruence. That is, the
congruence lattice of $\mathbf{A}$ is Boolean, in particular, distributive.
Since $\mathbf{A}$ is polynomially equivalent to a direct product of simple
$\mathbf{R}_{\mathbf{A}}$-modules, $\mathbf{R}_{\mathbf{A}}$ is semisimple.
Thus $\mathbf{A}$ satisfies the assumptions of Theorem 1.4 with $n=1$ and the
result follows. ∎

##  5\. Lower bounds on the number of clonoids from $\mathbf{A}$ to
$\mathbf{B}$

In this section we give necessary conditions on the algebras $\mathbf{A}$ and
$\mathbf{B}$ such that there are only finitely many clonoids between them. We
start by showing that if there exists $n\in\mathbb{N}$ such that every clonoid
from $\mathbf{A}$ to a nontrivial $\mathbf{B}$ is generated by $n$-ary
functions, then all subalgebras of $\mathbf{A}$ are generated by $n$ elements.

######  Lemma 5.1.

Let $n\in\mathbb{N}$, let $\mathbf{A}$ be an algebra not all of whose
subalgebras are generated by $n$ elements, and let $\mathbf{B}$ be a
nontrivial algebra. Then not every clonoid from $\mathbf{A}$ to $\mathbf{B}$
is generated by $n$-ary functions.

###### Proof.

By assumption $\mathbf{A}$ has some subalgebra $U$ that is generated by $n+1$
elements but not by $n$.

We construct clonoids $C$ and $D$ from $\mathbf{A}$ to $\mathbf{B}$ with
$C^{(k)}=D^{(k)}$ for all $k\leq n$ but $C^{(n+1)}\neq D^{(n+1)}$. For any
subalgebra $M$ of $\mathbf{A}$, let $\rho_{M}:=M^{2}$. Then

$C:=\mathrm{Pol}(\rho_{U},=_{B})$

is the clonoid of all functions from $A$ to $B$ that are constant on $U$. For
$a,b\in A^{n}$, let $\sigma_{a,b}$ be the subalgebra of $\mathbf{A}^{2}$ that
is generated by $(a_{1},b_{1}),\dots,(a_{n},b_{n})$. Let

$D:=\bigcap_{a,b\in U^{n}}\mathrm{Pol}(\sigma_{a,b},=_{B}).$

Then $f\colon A^{k}\to B$ is in $D$ iff
$f(g_{1}(a),\dots,g_{k}(a))=f(g_{1}(b),\dots,g_{k}(b))$ for all $a,b\in U^{n}$
and all $n$-ary term operations $g_{1},\dots,g_{k}$ of $\mathbf{A}$. Clearly
$C\subseteq D$.

To show $D^{(k)}\subseteq C^{(k)}$ for $k\leq n$, let $f\in D^{(k)}$ and
$a,b\in U^{k}$. Repeating the last entry of $a,b$, respectively, we obtain
$n$-tuples $a^{\prime}:=(a_{1},\dots,a_{k},\dots
a_{k}),b^{\prime}:=(b_{1},\dots,b_{k},\dots b_{k})$ over $U$. Since
$(a_{1},b_{1}),\dots,(a_{k},b_{k})$ are in $\sigma_{a^{\prime},b^{\prime}}$
and $f\in\mathrm{Pol}(\sigma_{a^{\prime},b^{\prime}},=_{B})$, we have
$f(a)=f(b)$. Thus $f\in C$ and $C^{(k)}=D^{(k)}$ for all $k\leq n$.

To show that $C^{(n+1)}\neq D^{(n+1)}$ let $0,1$ be distinct elements in $B$
and define

$g\colon A^{n+1}\to B,\ (x_{1},\dots,x_{n+1})\mapsto\begin{cases}1&\text{ if
}x_{1},\ldots,x_{n+1}\text{ generate }U,\\\ 0&\text{ else.}\end{cases}$

Since there exist $n+1$ elements that generate $U$ by assumption, $g$ is not
constant on $U$, hence not in $C$. However $g$ is constant $0$ on all proper
subalgebras of $U$ and in particular on all $n$-generated subalgebras. Thus
$g\in D^{(n+1)}\setminus C^{(n+1)}$. ∎

Even if $\mathbf{A}$ is a cyclic module, not all clonoids from $\mathbf{A}$ to
$\mathbf{B}$ may be generated by unary functions by the following.

######  Lemma 5.2.

Let $\mathbf{R}$ be a finite ring with nonzero Jacobson radical, let
$\mathbf{A}$ be the left regular module over $\mathbf{R}$, and let
$\mathbf{B}$ be a nontrivial algebra. Then not every clonoid from $\mathbf{A}$
to $\mathbf{B}$ is generated by unary functions.

###### Proof.

Since $\mathbf{R}$ is a finite ring, its Jacobson radical $J:=J(\mathbf{R})$
is nilpotent. It follows that $\mathbf{R}$ has a nonzero ideal $I\leq J$ such
that $I^{2}=0$. Note that $I$ is contained in every maximal left ideal of
$\mathbf{R}$ since $J$ is the intersection of all maximal left ideals of
$\mathbf{R}$.

We construct clonoids $C,D$ from $\mathbf{A}$ to $\mathbf{B}$ such that
$C^{(1)}=D^{(1)}$ but $C^{(2)}\neq D^{(2)}$. For any submodule $M$ of
$\mathbf{A}$, let $\rho_{M}:=M^{2}$. Then

$C:=\mathrm{Pol}(\equiv_{I},=)\cap\bigcap_{M<\mathbf{A}}\mathrm{Pol}(\rho_{M},=)$

is the clonoid of functions from $\mathbf{A}$ to $\mathbf{B}$ that are
constant modulo $I$ and constant on every proper submodule of $\mathbf{A}$.

For $a\in I$, let $\sigma_{a}:=R(1,1+a)$ be a submodule of $\mathbf{A}^{2}$.
Note that $\sigma_{a}$ is contained in $\equiv_{I}$. Let

$D:=\bigcap_{a\in
I}\mathrm{Pol}(\sigma_{a},=)\cap\bigcap_{M<\mathbf{A}}\mathrm{Pol}(\rho_{M},=).$

Then $f\in D^{(k)}$ iff
$f(x_{1},\dots,x_{k})=f(x_{1}+x_{1}a,\dots,x_{k}+x_{k}a)$ for all
$x_{1},\dots,x_{k}\in R$ and all $a\in I$ and $f$ is constant on every proper
submodule of $\mathbf{A}$. Since all elements $x_{i}a$ in the previous
identity are in $I$, it follows that

(5.1) $C\subseteq D.$

We claim that

(5.2) $C^{(1)}=D^{(1)}.$

For the proof let $f\in D^{(1)}$ and $x\in A$. First consider the case that
$Rx\neq R$. Then $x$ is contained in some maximal submodule $M$ of
$\mathbf{A}$ and $I\leq M$ since $I$ is contained in every maximal submodule.
Hence $x+I\subseteq M$ implies $f(x+I)=f(x)$.

Next suppose that $Rx=R$. Then $x$ has a multiplicative left inverse in $R$.
Since $I$ is finite, $xI=I$. Since $f(x)=f(x+xI)$ by assumption, we see that
$f(x)=f(x+I)$.

Hence $f$ is constant on $I$-cosets and $f\in C$. Together with (5.1) this
implies (5.2).

Next we claim that

(5.3) $C^{(2)}\neq D^{(2)}.$

To see this let $0,1$ be distinct elements in $B$ and define

$g\colon A^{2}\rightarrow B,\,(x,y)\mapsto\begin{cases}1&\text{ if
}x=y\equiv_{I}1,\\\ 0&\text{ else.}\end{cases}$

Clearly $g\notin C$ as for nonzero $a\in I$ we have $(1,1)\equiv_{I}(1+a,1)$
but $g(1,1)=1\neq 0=g(1+a,1)$.

For $g\in D$ we first show that $g\in\mathrm{Pol}(\sigma_{a},=)$ for all $a\in
I$. Let $(x,y)\in A^{2}$ and $a\in I$.

If $x\not\equiv_{I}1$, then also $x(1+a)=x+xa\not\equiv_{I}1$ and
$g(x,y)=0=g(x(1+a),y(1+a))$. The symmetric argument works if $y\neq_{I}1$.
Hence we assume $x,y\in 1+I$ in the following, say $x=1+u$ and $y=1+v$ for
some $u,v\in I$. If $u\neq v$, then $g(x,y)=0$ and also

$g(x(1+a),y(1+a))=g(1+u+a+\underbrace{ua}_{=0},1+v+a+\underbrace{va}_{=0})=0.$

Here we used the assumption that $I^{2}=0$. If $u=v$, then similarly
$g(x,y)=1$ and

$g(x(1+a),y(1+a))=g(1+u+a,1+u+a)=1.$

Hence $g\in\mathrm{Pol}(\sigma_{a},=)$.

Now let $M$ be a proper submodule of $\mathbf{A}$ and show
$g\in\mathrm{Pol}(\rho_{M},=)$. Note that by the nilpotence of $J$, every
element in $1+J$ is a unit in $\mathbf{R}$. Hence in particular $1+I$ and $M$
are disjoint. Thus $g(M,M)=0$.

Therefore $g\in D^{(2)}\setminus C^{(2)}$ proving (5.3) and that $D$ is not
generated by its unary functions. ∎

Next we prove Theorem 1.5 which characterizes the finite modules over
commutative rings for which all clonoids are generated by their unary
functions.

###### Proof of Theorem 1.5..

The implication $(2)\Rightarrow(1)$ follows from Theorem 1.4.

For $(1)\Rightarrow(2)$, assume that every clonoid from the
$\mathbf{R}$-module $\mathbf{A}$ to $\mathbf{B}$ is generated by unary
functions. By Lemma 5.1 $\mathbf{A}$ is a cyclic module. So $\mathbf{A}$ is
isomorphic to $R/L$ for some left ideal $L$ of $\mathbf{R}$. Since
$\mathbf{R}$ is commutative, $L$ is a (two-sided) ideal of $\mathbf{R}$. Now
$L(\mathbf{R}/L)=0$ yields $LA=0$ and $L=0$ since $\mathbf{A}$ is a faithful
$\mathbf{R}$-module. Hence $\mathbf{A}$ is isomorphic to the regular
$\mathbf{R}$-module. Since the Jacobson radical $J(\mathbf{R})=0$ by Lemma 5.2
and $\mathbf{R}$ is finite, the Wedderburn-Artin Theorem yields that
$\mathbf{R}$ is isomorphic to a direct product of matrix rings over finite
fields. Since $\mathbf{R}$ is commutative, each matrix ring has dimension
$1\times 1$ and $\mathbf{R}$ is a direct product of finite fields. ∎

Finally we construct an infinite ascending chain of clonoids between any two
finite modules whose orders have a nontrivial common divisor to prove Theorem
1.3. So in this case we attain the upper bound guaranteed by Theorem 2.9.

###### Proof of Theorem 1.3..

Let $p$ be a common prime divisor of $|A|$ and $|B|$. Then $\mathbf{A}$ has a
simple quotient $\mathbf{A}^{\prime}$ of $p$-power order and $\mathbf{B}$ has
a simple subalgebra $\mathbf{B}^{\prime}$ of $p$-power order. By the Jacobson
Density Theorem $\mathbf{A}^{\prime}$ and $\mathbf{B}^{\prime}$ are modules
over full matrix rings over (not necessarily the same) finite fields of
characteristic $p$. In particular we can expand $\mathbf{A}^{\prime}$ to a
module $\mathbf{A}^{+}=\mathbb{Z}_{p}^{m}$ over the matrix ring
$\mathbf{R}:=\mathbb{Z}_{p}^{m\times m}$ and $\mathbf{B}^{\prime}$ to a module
$\mathbf{B}^{+}=\mathbb{Z}_{p}^{n}$ over $\mathbf{S}:=\mathbb{Z}_{p}^{n\times
n}$ for some natural numbers $m$ and $n$.

By Lemmas 2.3 and 2.4 it suffices to construct infinitely many clonoids from
$\mathbf{A}^{+}$ to $\mathbf{B}^{+}$. For ease of notation assume
$\mathbf{A}=\mathbf{A}^{+}$ and $\mathbf{B}=\mathbf{B}^{+}$ in the following.
For $x=(x_{1},\dots,x_{m})$ and $y=(y_{1},\dots,y_{m})$ in $A$ define

$xy:=(x_{1}y_{1},\underbrace{0,\dots,0}_{m-1})\in A\quad\text{ and }\quad
x^{\prime}:=(x_{1},\underbrace{0,\dots,0}_{n-1})\in B.$

Here $x_{1}y_{1}$ denotes the usual product in $\mathbb{Z}_{p}$. For
$k\in\mathbb{N}$ let

$f_{k}\colon A^{k}\to B,\ (x_{1},\dots,x_{k})\mapsto(x_{1}\cdots
x_{k})^{\prime}.$

We claim that

(5.4) $\langle f_{1}\rangle\subsetneq\langle
f_{1},f_{2}\rangle\subsetneq\langle f_{1},f_{2},f_{3}\rangle\subsetneq\cdots$

is an infinite strictly ascending sequence of clonoids from $\mathbf{A}$ to
$\mathbf{B}$.

For the proof first note that $f_{k}$ is linear in each argument and in
particular zero-absorbing for all $k\in\mathbb{N}$. We show that $\langle
f_{1},\ldots,f_{k}\rangle$ does not contain any zero-absorbing functions of
arity $k+1$ except for the constant zero function. To this end let
$g\in\langle f_{1},\ldots,f_{k}\rangle^{(k+1)}$. By the multilinearity of the
$f_{i}$ we can expand $g$ into the form

$g(x_{1},\dots,x_{k+1})=\sum_{\emptyset\neq
I\subsetneq[k+1]}\left(\sum_{j=1}^{\ell_{I}}s_{I,j}\left(\prod_{i\in
I}x_{i}^{e_{I,j,i}}\right)^{\prime}\right)$

for some $\ell_{I}\in\mathbb{N},s_{I,j}\in\mathbb{Z}_{p}^{n\times n}$ and
$e_{I,j,i}\in[p]$. Note that no product in $A$ has more than $k$ factors. We
rewrite the summands of the outer sum of $g$ as

(5.5) $g(x_{1},\dots,x_{k+1})=\sum_{\emptyset\neq
I\subsetneq[k+1]}a_{I}(x_{1},\dots,x_{k+1})\left(\prod_{i\in
I}x_{i}\right)^{\prime}$

for some $a_{I}\colon A^{k+1}\to\mathbb{Z}_{p}^{n\times n}$ that do not depend
on $x_{j}$ for any $j\in[k+1]\setminus I$.

Assume that $g$ is zero-absorbing. We show that then every summand
$a_{I}(x_{1},\dots,x_{k+1})\left(\prod_{i\in I}x_{i}\right)^{\prime}$ in (5.5)
is $0$ by induction on $I\subsetneq[k+1]$. The base case $a_{\emptyset}(x)=0$
holds by the very form of $g$. For $I\subsetneq[k+1]$ and for $x\in A^{k+1}$,
define $x^{I}=(x^{I}_{1},\ldots,x^{I}_{k+1})$ in $A^{k+1}$ by

$x_{i}^{I}:=\begin{cases}x_{i}&\text{if }i\in I,\\\ 0&\text{else.}\end{cases}$

Then

$\displaystyle 0$ $\displaystyle=g(x^{I})\quad\text{since }g\text{ is zero-
absorbing and }|I|\leq k,$ $\displaystyle=\sum_{\emptyset\neq J\subseteq
I}a_{J}(x^{I})\left(\prod_{i\in J}x_{i}\right)^{\prime}\quad\text{since
}\prod_{i\in J}x^{I}_{i}=0\text{ unless }J\subseteq I,$
$\displaystyle=\sum_{\emptyset\neq J\subseteq I}a_{J}(x)\left(\prod_{i\in
J}x_{i}\right)^{\prime}\quad\text{since }a_{J}\text{ does not depend on
}x_{i}\text{ for }i\not\in I,$ $\displaystyle=a_{I}(x)\left(\prod_{i\in
I}x_{i}\right)^{\prime}\quad\text{by induction assumption.}$

Thus the induction step is proved and $g=0$ is the only $k+1$-ary zero-
absorbing function in $\langle f_{1},\ldots,f_{k}\rangle$. In particular
$f_{k+1}$ is not in $\langle f_{1},\ldots,f_{k}\rangle$. This proves (5.4).

Clearly the union of clonoids in (5.4) is not finitely generated. ∎

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## Appendices

###  A Algebra for Sensitivity Analysis

Assuming a positive $\beta_{1}$, we can solve for the point at which the
treatment effect becomes non-significant:

$\displaystyle\widehat{\beta_{1}}-1.96\sqrt{\widehat{V}(\beta_{1})_{\text{CTE}}+\frac{\sigma^{2}_{\zeta}}{I}}$
$\displaystyle=0$ (31)
$\displaystyle-1.96\sqrt{\widehat{V}(\beta_{1})_{\text{CTE}}+\frac{\sigma^{2}_{\zeta}}{I}}$
$\displaystyle=-\widehat{\beta}_{1}$ (32)
$\displaystyle\sqrt{\widehat{V}(\beta_{1})_{\text{CTE}}+\frac{\sigma^{2}_{\zeta}}{I}}$
$\displaystyle=\frac{\widehat{\beta}_{1}}{1.96}$ (33)
$\displaystyle\widehat{V}(\beta_{1})_{\text{CTE}}+\frac{\sigma^{2}_{\zeta}}{I}$
$\displaystyle=(\frac{\widehat{\beta}_{1}}{1.96})^{2}$ (34)
$\displaystyle\frac{\sigma^{2}_{\zeta}}{I}$
$\displaystyle=(\frac{\widehat{\beta}_{1}}{1.96})^{2}-\widehat{V}(\beta_{1})_{\text{CTE}}$
(35) $\displaystyle{\sigma^{2}_{\zeta}}$
$\displaystyle=I((\frac{\widehat{\beta}_{1}}{1.96})^{2}-\widehat{V}(\beta_{1})_{\text{CTE}})$
(36) $\displaystyle\sigma_{\zeta}$
$\displaystyle=\sqrt{I((\frac{\widehat{\beta}_{1}}{1.96})^{2}-\widehat{V}(\beta_{1})_{\text{CTE}})}$
(37)

###  B Additional Mathematical Detail for the Identification Problem in
Section 3.4

We can formalize the intuition provided by Figures 4 and 5 by noting that the
flattening of the sum score curves (i.e., the test response functions, or
TRFs) with respect to $X_{j}$ can result from two distinct processes. First,
through a direct flattening of the slopes of the individual item curves, as in
the interaction case, and second, through an increase in the variance of the
item locations in the treatment group that arises from the combination of
$\sigma_{\zeta}$ and $\rho$. That is, $\sigma_{b}$ provides the SD of item
locations in the control group, and the presence of IL-HTE changes the SD of
item locations in the treatment group according to the following formula [279,
pp. 29-32], where we define $\sigma_{b}^{*}$ as the SD of item locations in
the treatment group:

$\displaystyle\sigma_{b}^{*}$
$\displaystyle=\sqrt{\sigma^{2}_{b}+\sigma^{2}_{\zeta}+2\rho\sigma_{b}\sigma_{\zeta}}.$
(38)

When $\sigma_{\zeta}=0$, $\sigma_{b}=\sigma_{b}^{*}$ and the identification
problem is resolved. Furthermore, because $\sigma_{b}$ is typically larger
than $\sigma_{\zeta}$, and $\sigma_{\zeta}$ is typically less than 1 (relative
to $\sigma_{\theta}$, as is the case in our data, as shown in Figure 2), when
$\rho=0$, the final term drops out and the resulting inflation of
$\sigma^{*}_{b}$ is much less severe. For example, consider the simple case
where the SD of the item locations in the control group ($\sigma_{b}$), the SD
of item-specific treatment effects ($\sigma_{\zeta}$), and the location-
treatment effect correlation ($\rho$) are all 1, so that $\sigma_{b}^{*}$ will
be twice as large as $\sigma_{b}$ (i.e., the treatment doubles the SD of the
item locations):

$\displaystyle\sigma_{b}^{*}$
$\displaystyle=\sqrt{\sigma^{2}_{b}+\sigma^{2}_{\zeta}+2\rho\sigma_{b}\sigma_{\zeta}}$
(39) $\displaystyle=\sqrt{1+1+2}$ (40) $\displaystyle=\sqrt{4}$ (41)
$\displaystyle=2.$ (42)

This increase in $\sigma_{b}^{*}$ relative to $\sigma_{b}$ will attenuate the
slope of the sum score curve. We can demonstrate this with the following
formula that allows us to convert the slopes of the item curves to the slope
of the sum score curve [218, p. 116]:

$\displaystyle\beta_{S}=\frac{\beta_{I}}{\sqrt{.346\sigma^{2}_{b}+1}},$ (43)

where $\beta_{S}$ is the slope of the sum score curve (on the logit scale) and
$\beta_{I}$ is the slope of the item curve.

To illustrate, we apply Equation 43 to both the treatment and control groups,
continuing with the simple case where $\sigma_{b}=1$ and $\sigma_{b}^{*}=2$ as
described above. In the control group, if we assume $\beta_{I}=1$, then
$\beta_{S}$ is

$\displaystyle\beta_{S}\text{ if }(T_{j}=0)$
$\displaystyle=\frac{1}{\sqrt{.346\times 1+1}}$ (44) $\displaystyle=.86,$ (45)

and in the treatment group,

$\displaystyle\beta_{S}\text{ if }(T_{j}=1)$
$\displaystyle=\frac{1}{\sqrt{.346\sigma^{2*}_{b}+1}}$ (46)
$\displaystyle=\frac{1}{\sqrt{.346\times 4+1}}$ (47) $\displaystyle=.65.$ (48)

Thus, even though the slopes of the item curves in the numerator are
identical, the difference in the slopes of the sum scores is $.65-.86=-.21$,
which could easily be misinterpreted as an interaction effect between
treatment and the baseline covariate in an analysis of the sum scores.

Similarly, $\rho<0$ will produce a spurious interaction in the other direction
by “shrinking” the scale inward. Assuming $\rho=-1$,

$\displaystyle\beta_{S}\text{ if }(T_{j}=1)$
$\displaystyle=\frac{1}{\sqrt{.346\sigma^{2*}_{b}+1}}$ (49)
$\displaystyle=\frac{1}{\sqrt{.346\times(1+1-2)+1}}$ (50) $\displaystyle=1.$
(51)

Here, the difference in slopes of the sum scores is $1-.86=+.14$. Thus, for
the person-dependent and item-dependent processes to produce the same pattern
of sum scores, the following equality must hold:

$\displaystyle\frac{\beta_{2}+\beta_{3}}{\sqrt{.346\sigma_{b}^{2}+1}}=\frac{\beta_{2}}{\sqrt{.346\times(\sigma_{b}^{2}+\sigma_{\zeta}^{2}+\rho\sigma_{b}\sigma_{\zeta})+1}}.$
(52)

In the control group, these quantities will clearly be the same as both
$\beta_{3}$ and $\sigma_{\zeta}$ are 0. For a given set of parameters
$\beta_{2},\beta_{3},\sigma_{b}$ we can choose infinitely many values of
$\sigma_{\zeta}$ and $\rho$ that produce the equivalent pattern in the total
score (to a point; $\rho=-1$ can only compress the scale such that the
individual item slope is not attenuated, when $\sigma^{2*}_{b}=0$). Clearly,
therefore, an observed treatment by covariate interaction could result from
either a true interaction, or a correlation of item-specific treatment effects
and item location.111111Simulations in [208] show that this phenomenon is not
an artifact of sum scores. The results apply equally to IRT-based scores and
EIRMs that simultaneously estimate the measurement and structural models. We
emphasize $\rho$ as the primary cause of the identification problem because,
while theoretically a large enough value of $\sigma^{2}_{\zeta}$ could
generate the same problem when $\rho=0$, $\sigma^{2}_{\zeta}$ is typically
less than $\sigma^{2}_{b}$ and less than 1, so that
$\sigma_{\zeta}>\sigma^{2}_{\zeta}$. Furthermore, $\sigma_{\zeta}$ must be
positive, whereas a negative value of $\rho$ can stretch the scale in either
direction. As a final point, we note that any bias in interaction terms will
necessarily bias main effects in models that omit the interaction, because the
main effects are weighted averages of the effects in each subgroup.

###  C Sample R Code to Fit Various IL-HTE Models

The R code below illustrates how to fit various EIRMs to a data set dat with
0/1 outcome score, 0/1 treatment indicator treat, baseline covariate
std_baseline, person identifier s_id, and item identifier item using the glmer
function (generalized linear mixed effects models in R). For clarity, we omit
data = dataset, family = binomial from each function call. For further
resources, see the replication materials in our supplement or the various EIRM
R tutorials listed in the references [203, 186].

    # R Code
    # load library
    library(lme4)

    # treatment indicator only
    glmer(score ~ treat + (1|s_id) + (1|item))

    # constant effects model with pretest
    glmer(score ~ treat + std_baseline + (1|s_id) + (1|item))

    # IL-HTE model
    glmer(score ~ treat + std_baseline + (1|s_id) + (treat|item))

    # person interaction model
    glmer(score ~ treat*std_baseline + (1|s_id) + (1|item))

    # flexible model
    glmer(score ~ treat*std_baseline + (1|s_id) + (treat|item))

###  D Directed Acyclic Graph of the IL-HTE Model

Figure 8: Directed Acyclic Graph for Equation 22

Notes: Squares indicate observed variables, hollow circles indicate latent
variables, and solid circles represent cross product interaction terms.
$I_{n}$ are item responses, $T_{j}$ is the treatment indicator, and $X_{j}$ is
the covariate. $\beta_{1}$ represents the average treatment effect. $\rho$
represents the correlation between item location and item-specific treatment
effect size. Path coefficients are fixed at 1 unless otherwise indicated. The
$\sigma$ terms represent residual standard deviations.