# Polynomial identities satisfied by generalized polynomials

Eszter Gselmann

###### Abstract

The main purpose of this paper is solve polynomial equations that are
satisfied by (generalized) polynomials. More exactly, we deal with the
following problem: let $\mathbb{F}$ be a field with
$\mathrm{char}(\mathbb{F})=0$ and $P\in\mathbb{F}[x]$ and $Q\in\mathbb{C}[x]$
be polynomials. Our aim is to prove characterization theorems for generalized
polynomials $f\colon\mathbb{F}\to\mathbb{C}$ of degree two that also fulfill
equation

$f(P(x))=Q(f(x))$

for each $x\in\mathbb{F}$. As it turns out, the difficulty of such problems
heavily depends on that we consider the above equation for generalized
polynomials or for (normal) polynomials. Therefore, firstly we study the
connection between these two notions.

> This paper is dedicated to Professors _György Gát_ and _Zsolt Páles_ on the
> occasion of their $60$th and $65$th birthday, respectively.

##  1 Introduction and preliminaries

###  1.1 Homomorphisms and derivations

The study of additive mappings from a ring into another ring which preserve
squares was initiated by G. Ancochea in [3] in connection with problems
arising in projective geometry. Later, these results were strengthened by
(among others) Kaplansky [11] and Jacobson–Rickart [10].

Let $R,R^{\prime}$ be rings, the mapping $\varphi:R\rightarrow R^{\prime}$ is
called a _homomorphism_ if

$\varphi(a+b)=\varphi(a)+\varphi(b)\qquad\left(a,b\in R\right)$

and

$\varphi(ab)=\varphi(a)\varphi(b)\qquad\left(a,b\in R\right).$

Furthermore, the function $\varphi:R\to R^{\prime}$ is an _anti-homomorphism_
if

$\varphi(a+b)=\varphi(a)+\varphi(b)\qquad\left(a,b\in R\right)$

and

$\varphi(ab)=\varphi(b)\varphi(a)\qquad\left(a,b\in R\right).$

Henceforth, $\mathbb{N}$ will denote the set of the positive integers. Let
$n\in\mathbb{N},n\geq 2$ be fixed. The function $\varphi:R\rightarrow
R^{\prime}$ is called an $n$-Jordan homomorphism if

$\varphi(a+b)=\varphi(a)+\varphi(b)\qquad\left(a,b\in R\right)$

and

$\varphi(a^{n})=\varphi(a)^{n}\qquad\left(a\in R\right).$

Finally, we remark that in case $n=2$ we speak about homomorphisms and Jordan
homomorphisms, respectively. It was G. Ancochea who firstly dealt with the
connection of Jordan homomorphisms and homomorphisms, see [3]. These results
were generalized and extended in several ways, see for instance [10], [11],
[17].

Let $n\in\mathbb{N}$, we say that a ring $R$ is _of characteristic larger than
$n$_ if $n!x=0$ implies that $x=0$. The ring $R$ is termed to be a _prime
ring_ if $a,b\in R$ and $aRb=\left\\{0\right\\}$ imply that either $a=0$ or
$b=0$. In [9] I.N. Herstein proved that if $\varphi$ is a Jordan homomorphism
of a ring $R$ _onto_ a prime ring $R^{\prime}$ of characteristic different
from $2$ and $3$, then either $\varphi$ is a homomorphism or an anti-
homomorphism.

Furthermore, in the above-mentioned paper [9], not only Jordan homomorphisms
but also $n$-Jordan mappings were considered and he proved the following
statement.

######  Theorem 1.

Let $\varphi$ be an $n$-Jordan homomorphism from a ring $R$ onto a prime ring
$R^{\prime}$ and assume that $R^{\prime}$ has characteristic larger than $n$,
suppose further that $R$ has a unit element. Then $\varphi=\varepsilon\tau$
where $\tau$ is either a homomorphism or an anti-homomorphism and
$\varepsilon$ is an $(n-1)$st root of unity lying in the center of
$R^{\prime}$.

Besides homomorphisms, derivations also play a key role in the theory of rings
and fields. Concerning this notion, we will follow [13, Chapter 14].

Let $Q$ be a ring and let $P$ be a subring of $Q$. A function $d\colon
P\rightarrow Q$ is called a _derivation_ if it is additive, i.e.

$d(x+y)=d(x)+d(y)\quad\left(x,y\in P\right)$

and also satisfies the so-called _Leibniz rule_ , i.e. equation

$d(xy)=d(x)y+xd(y)\quad\left(x,y\in P\right).$

Fundamental examples for derivations are the following ones.

Let $\mathbb{F}$ be a field, and let in the above definition
$P=Q=\mathbb{F}[x]$ be the ring of polynomials with coefficients from
$\mathbb{F}$. For a polynomial $p\in\mathbb{F}[x]$,
$p(x)=\sum_{k=0}^{n}a_{k}x^{k}$, define the function
$f\colon\mathbb{F}[x]\rightarrow\mathbb{F}[x]$ as

$f(p)=p^{\prime},$

where $p^{\prime}(x)=\sum_{k=1}^{n}ka_{k}x^{k-1}$ is the derivative of the
polynomial $p$. Then the function $f$ clearly fulfills

$f(p+q)=f(p)+f(q)$

and

$f(pq)=pf(q)+qf(p)$

for all $p,q\in\mathbb{F}[x]$. Hence $f$ is a derivation.

Let $(\mathbb{F},+,\cdot)$ be a field, and suppose that we are given a
derivation $f\colon\mathbb{F}\to\mathbb{F}$. We define the mapping
$f_{0}\colon\mathbb{F}[x]\to\mathbb{F}[x]$ in the following way. If
$p\in\mathbb{F}[x]$ has the form

$p(x)=\sum_{k=0}^{n}a_{k}x^{k},$

then let

$f_{0}(p)=p^{f}(x)=\sum_{k=0}^{n}f(a_{k})x^{k}.$

Then $f_{0}\colon\mathbb{F}[x]\to\mathbb{F}[x]$ is a derivation.

It is well-known that in case of additive functions, Hamel bases play an
important role. As [13, Theorem 14.2.1] shows in case of derivations,
algebraic bases are fundamental.

######  Theorem 2.

Let $(\mathbb{K},+,\cdot)$ be a field of characteristic zero, let
$(\mathbb{F},+,\cdot)$ be a subfield of $(\mathbb{K},+,\cdot)$, let $S$ be an
algebraic base of $\mathbb{K}$ over $\mathbb{F}$, if it exists, and let
$S=\emptyset$ otherwise. Let $f\colon\mathbb{F}\to\mathbb{K}$ be a derivation.
Then, for every function $u\colon S\to\mathbb{K}$, there exists a unique
derivation $g\colon\mathbb{K}\to\mathbb{K}$ such that $g|_{\mathbb{F}}=f$ and
$g|_{S}=u$.

###  1.2 The symmetrization method

While proving our results, the so-called Polarization formula for multi-
additive functions and the symmetrization method will play a key role. In this
subsection the most important notations and statements are summarized. Here we
follow the monograph [15].

######  Definition 1.

Let $G,S$ be commutative semigroups, $n\in\mathbb{N}$ and let $A\colon
G^{n}\to S$ be a function. We say that $A$ is _$n$ -additive_ if it is a
homomorphism of $G$ into $S$ in each variable. If $n=1$ or $n=2$ then the
function $A$ is simply termed to be _additive_ or _bi-additive_ ,
respectively.

The _diagonalization_ or _trace_ of an $n$-additive function $A\colon G^{n}\to
S$ is defined as

$A^{\ast}(x)=A\left(x,\ldots,x\right)\qquad\left(x\in G\right).$

As a direct consequence of the definition each $n$-additive function $A\colon
G^{n}\to S$ satisfies

$A(x_{1},\ldots,x_{i-1},kx_{i},x_{i+1},\ldots,x_{n})=kA(x_{1},\ldots,x_{i-1},x_{i},x_{i+1},\ldots,x_{n})\qquad\left(x_{1},\ldots,x_{n}\in
G\right)$

for all $i=1,\ldots,n$, where $k\in\mathbb{N}$ is arbitrary. The same identity
holds for any $k\in\mathbb{Z}$ provided that $G$ and $S$ are groups, and for
$k\in\mathbb{Q}$, provided that $G$ and $S$ are linear spaces over the
rationals. For the diagonalization of $A$ we have

$A^{\ast}(kx)=k^{n}A^{\ast}(x)\qquad\left(x\in G\right).$

The above notion can also be extended for the case $n=0$ by letting $G^{0}=G$
and by calling $0$-additive any constant function from $G$ to $S$.

One of the most important theoretical results concerning multiadditive
functions is the so-called _Polarization formula_ , that briefly expresses
that every $n$-additive symmetric function is _uniquely_ determined by its
diagonalization under some conditions on the domain as well as on the range.
Suppose that $G$ is a commutative semigroup and $S$ is a commutative group.
The action of the _difference operator_ $\Delta$ on a function $f\colon G\to
S$ is defined by the formula

$\Delta_{y}f(x)=f(x+y)-f(x)\qquad\left(x,y\in G\right).$

Note that the addition in the argument of the function is the operation of the
semigroup $G$ and the subtraction means the inverse of the operation of the
group $S$.

######  Theorem 3 (Polarization formula).

Suppose that $G$ is a commutative semigroup, $S$ is a commutative group,
$n\in\mathbb{N}$. If $A\colon G^{n}\to S$ is a symmetric, $n$-additive
function, then for all $x,y_{1},\ldots,y_{m}\in G$ we have

$\Delta_{y_{1},\ldots,y_{m}}A^{\ast}(x)=\left\\{\begin{array}[]{rcl}0&\text{
if}&m>n\\\ n!A(y_{1},\ldots,y_{m})&\text{ if}&m=n.\end{array}\right.$

######  Corollary 1.

Suppose that $G$ is a commutative semigroup, $S$ is a commutative group,
$n\in\mathbb{N}$. If $A\colon G^{n}\to S$ is a symmetric, $n$-additive
function, then for all $x,y\in G$

$\Delta^{n}_{y}A^{\ast}(x)=n!A^{\ast}(y).$

######  Lemma 1.

Let $n\in\mathbb{N}$ and suppose that the multiplication by $n!$ is surjective
in the commutative semigroup $G$ or injective in the commutative group $S$.
Then for any symmetric, $n$-additive function $A\colon G^{n}\to S$,
$A^{\ast}\equiv 0$ implies that $A$ is identically zero, as well.

######  Definition 2.

Let $G$ and $S$ be commutative semigroups, a function $p\colon G\to S$ is
called a _generalized polynomial_ from $G$ to $S$ if it has a representation
as the sum of diagonalizations of symmetric multi-additive functions from $G$
to $S$. In other words, a function $p\colon G\to S$ is a generalized
polynomial if and only if, it has a representation

$p=\sum_{k=0}^{n}A^{\ast}_{k},$

where $n$ is a nonnegative integer and $A_{k}\colon G^{k}\to S$ is a
symmetric, $k$-additive function for each $k=0,1,\ldots,n$. In this case we
also say that $p$ is a generalized polynomial _of degree at most $n$_.

Let $n$ be a nonnegative integer, functions $p_{n}\colon G\to S$ of the form

$p_{n}=A_{n}^{\ast},$

where $A_{n}\colon G^{n}\to S$ are the so-called _generalized monomials of
degree $n$_.

######  Remark 1.

Obviously, generalized monomials of degree $0$ are constant functions and
generalized monomials of degree $1$ are additive functions.

Furthermore, generalized monomials of degree $2$ will be termed _quadratic
functions_.

###  1.3 Polynomial functions

As Laczkovich [14] enlightens, on groups there are several polynomial notions.
One of them is that we introduced in subsection 1.2, that is the notion of
generalized polynomials. As we will see in the forthcoming sections, not only
this notion, but also that of _(normal) polynomials_ will be important. The
definitions and results recalled here can be found in [15].

Throughout this subsection $G$ is assumed to be a commutative group (written
additively).

######  Definition 3.

Polynomials are elements of the algebra generated by additive functions over
$G$. Namely, if $n$ is a positive integer,
$P\colon\mathbb{C}^{n}\to\mathbb{C}$ is a (classical) complex polynomial in
$n$ variables and $a_{k}\colon G\to\mathbb{C}\;(k=1,\ldots,n)$ are additive
functions, then the function

$x\longmapsto P(a_{1}(x),\ldots,a_{n}(x))$

is a polynomial and, also conversely, every polynomial can be represented in
such a form.

######  Remark 2.

For the sake of easier distinction, at some places polynomials will be called
normal polynomials.

######  Remark 3.

We recall that the elements of $\mathbb{N}^{n}$ for any positive integer $n$
are called ($n$-dimensional) _multi-indices_. Addition, multiplication and
inequalities between multi-indices of the same dimension are defined
component-wise. Further, we define $x^{\alpha}$ for any $n$-dimensional multi-
index $\alpha$ and for any $x=(x_{1},\ldots,x_{n})$ in $\mathbb{C}^{n}$ by

$x^{\alpha}=\prod_{i=1}^{n}x_{i}^{\alpha_{i}}$

where we always adopt the convention $0^{0}=1$. We also use the notation
$\left|\alpha\right|=\alpha_{1}+\cdots+\alpha_{n}$. With these notations any
polynomial of degree at most $N$ on the commutative semigroup $G$ has the form

$p(x)=\sum_{\left|\alpha\right|\leq N}c_{\alpha}a(x)^{\alpha}\qquad\left(x\in
G\right),$

where $c_{\alpha}\in\mathbb{C}$ and $a=(a_{1},\dots,a_{n})\colon
G\to\mathbb{C}^{n}$ is an additive function. Furthermore, the _homogeneous
term of degree $k$_ of $p$ is

$\sum_{\left|\alpha\right|=k}c_{\alpha}a(x)^{\alpha}.$

######  Lemma 2 (Lemma 2.7 of [15]).

Let $G$ be a commutative group, $n$ be a positive integer and let

$a=\left(a_{1},\ldots,a_{n}\right),$

where $a_{1},\ldots,a_{n}$ are linearly independent complex valued additive
functions defined on $G$. Then the monomials $\left\\{a^{\alpha}\right\\}$ for
different multi-indices are linearly independent.

######  Definition 4.

A function $m\colon G\to\mathbb{C}$ is called an _exponential_ function if it
satisfies

$m(x+y)=m(x)m(y)\qquad\left(x,y\in G\right).$

Furthermore, on an _exponential polynomial_ we mean a linear combination of
functions of the form $p\cdot m$, where $p$ is a polynomial and $m$ is an
exponential function.

######  Definition 5.

Let $G$ be an Abelian group and $V\subseteq\mathbb{C}^{G}$ a set of functions.
We say that $V$ is translation invariant if for every $f\in V$ the function
$\tau_{g}f\in V$ also holds for all $g\in G$, where

$\tau_{g}f(h)=f(h+g)\qquad\left(h\in G\right).$

In view of Theorem 10.1 of Székelyhidi [15], any finite dimensional
translation invariant linear space of complex valued functions on a
commutative group consists of exponential polynomials. This implies that if
$G$ is a commutative group, then any function $f\colon G\to\mathbb{C}$,
satisfying the functional equation

$f(x+y)=\sum_{i=1}^{n}g_{i}(x)h_{i}(y)\qquad\left(x,y\in G\right)$

for some positive integer $n$ and functions $g_{i},h_{i}\colon G\to\mathbb{C}$
($i=1,\ldots,n$), is an exponential polynomial of degree at most $n$.

This enlightens the connection between generalized polynomials and
polynomials. It is easy to see that each polynomial, that is, any function of
the form

$x\longmapsto P(a_{1}(x),\ldots,a_{n}(x)),$

where $n$ is a positive integer, $P\colon\mathbb{C}^{n}\to\mathbb{C}$ is a
(classical) complex polynomial in $n$ variables and $a_{k}\colon
G\to\mathbb{C}\;(k=1,\ldots,n)$ are additive functions, is a generalized
polynomial. The converse however is in general not true. A complex-valued
generalized polynomial $p$ defined on a commutative group $G$ is a polynomial
_if and only if_ its variety (the linear space spanned by its translates) is
of _finite_ dimension. To make the situation more clear, here we also recall
Theorem 13.4 from Székelyhidi [16].

######  Theorem 4.

The torsion free rank of a commutative group is finite _if and only if_ every
generalized polynomial on the group is a polynomial.

##  2 Results

In this section $\mathbb{F}$ is assumed to be a field with
$\mathrm{char}(\mathbb{F})=0$. Let further $n$ be a positive integer and
$P\in\mathbb{F}[x]$ and $Q\in\mathbb{C}[x]$ be polynomials. Our aim is to
prove characterization theorems for generalized polynomials
$f\colon\mathbb{F}\to\mathbb{C}$ of degree at most $n$ that also fulfill
equation

$f(P(x))=Q(f(x))$

for each $x\in\mathbb{F}$.

Before presenting the results of this paper, we note that related problems
have already been considered by Z. Boros and E. Garda–Mátyás in [4, 5, 6] and
also by M. Amou in [2]. In these papers the authors consider monomial
functions $f,g\colon\mathbb{R}\to\mathbb{R}$ of degree $n$, where
$n\in\mathbb{N}$, $n\geq 2$, which satisfy the conditional equation

$x^{n}f(y)=y^{n}g(x)$

for all points $(x,y)$ on a specified planar curve.

Roughly speaking, the following lemmata tell us that the problem investigated
in this paper is meaningful in the sense that for any positive integer $n$,
there _do exist_ generalized polynomials of degree at most $n$ that satisfy
the above identity.

######  Lemma 3.

Let $\mathbb{F}$ be a field with $\mathrm{char}(\mathbb{F})=0$, $n$ be a
positive integer and
$\varphi_{1},\ldots,\varphi_{n}\colon\mathbb{F}\to\mathbb{C}$ be
homomorphisms. Define the function $f$ on $\mathbb{F}$ by

$f(x)=\varphi_{1}(x)\cdots\varphi_{n}(x)\qquad\left(x\in\mathbb{F}\right).$

Then the following statements hold true.

  1. (i)

The function $f\colon\mathbb{F}\to\mathbb{C}$ is a generalized polynomial of
degree $n$.

  2. (ii)

The function $f\colon\mathbb{F}\to\mathbb{C}$ is a polynomial of degree $n$.

  3. (iii)

For any positive integer $k$, we have

$f(x^{k})=f(x)^{k}\qquad\left(x\in\mathbb{F}\right).$

###### Proof.

Firstly, observe that the function $f$ defined on $\mathbb{F}$ by

$f(x)=\varphi_{1}(x)\cdots\varphi_{n}(x)\qquad\left(x\in\mathbb{F}\right)$

is the trace of the symmetric $n$-additive mapping $F$ defined by

$F(x_{1},\ldots,x_{n})=\frac{1}{n!}\sum_{\sigma\in\mathscr{S}_{n}}\varphi_{1}(x_{\sigma(1)})\cdots\varphi_{n}(x_{\sigma(n)})\qquad\left(x\in\mathbb{F}\right).$

Thus $f$ is a generalized monomial of degree $n$. Secondly, in view of the
definition of the function $f$ we immediately get that it is also a monomial
of degree $n$. Indeed, let

$P(x_{1},\ldots,x_{n})=x_{1}\cdots x_{n}\qquad\left(x\in\mathbb{C}\right)$

and then $f$ can be written as $f=P\circ\varphi$, where
$\varphi\colon\mathbb{F}\to\mathbb{C}^{n}$ is

$\varphi(x)=\left(\varphi_{1}(x),\ldots,\varphi_{n}(x)\right)\qquad\left(x\in\mathbb{F}\right).$

Thirdly, recall that in case $\varphi$ is a homomorphism between $\mathbb{F}$
and $\mathbb{C}$, then we also have

$\varphi(x^{k})=\varphi(x)^{k}$

for each $x\in\mathbb{K}$. Therefore,

$f(x^{k})=\varphi_{1}(x^{k})\cdots\varphi_{n}(x^{k})=\varphi_{1}(x)^{k}\cdots\varphi_{n}(x)^{k}=\left(\varphi_{1}(x)\cdots\varphi_{n}(x)\right)^{k}=f(x)^{k}\qquad\left(x\in\mathbb{F}\right).$

∎

######  Lemma 4.

Let $\mathbb{F}\subset\mathbb{C}$ be a field and
$d\colon\mathbb{F}\to\mathbb{C}$ be a derivation and consider the function
$f\colon\mathbb{F}\to\mathbb{C}$ defined by

$f(x)=d(x^{n})\qquad\left(x\in\mathbb{F}\right).$

Then the following statements are satisfied.

  1. (i)

The function $f\colon\mathbb{F}\to\mathbb{C}$ is a generalized polynomial of
degree $n$.

  2. (ii)

The function $f\colon\mathbb{F}\to\mathbb{C}$ is a polynomial of degree $n$.

  3. (iii)

For any positive integer $k\geq 2$, we have

$f(x^{k})=kx^{(k-1)n}f(x)\qquad\left(x\in\mathbb{F}\right).$

###### Proof.

Let $\mathbb{F}\subset\mathbb{C}$ be a field and
$d\colon\mathbb{F}\to\mathbb{C}$ be a derivation and let us consider the
function $f\colon\mathbb{F}\to\mathbb{C}$ defined through

$f(x)=d(x^{n})\qquad\left(x\in\mathbb{F}\right).$

Observe that in this case the function
$F_{n}\colon\mathbb{F}^{n}\to\mathbb{C}$ defined by

$F_{n}(x_{1},\ldots,x_{n})=d\left(x_{1}\cdots
x_{n}\right)\qquad\left(x_{1},\ldots,x_{n}\in\mathbb{F}\right)$

is a symmetric and $n$-additive function. Furthermore, its trace is $f$,
showing that $f$ is a generalized monomial of degree $n$.

At the same time, since $d$ is a derivation, we also have that

$d(x^{n})=nx^{n-1}d(x)\qquad\left(x\in\mathbb{F}\right),$

yielding that

$f(x)=nx^{n-1}d(x)=P(x,d(x))\qquad\left(x\in\mathbb{F}\right),$

with the two-variable complex polynomial

$P(x,y)=nx^{n-1}y\qquad\left(x,y\in\mathbb{C}\right).$

Therefore, $f$ is a monomial of degree $n$.

Finally, let $k\geq 2$ be a positive integer. Then

$f(x^{k})=d\left((x^{k})^{n}\right)=d(x^{kn})=knx^{kn-1}d(x)=kx^{(k-1)n}\cdot\left(nx^{n-1}d(x)\right)=kx^{(k-1)n}f(x)\qquad\left(x\in\mathbb{F}\right).$

∎

######  Lemma 5.

Let $n$ and $k$ be positive integers, $\alpha\in\mathbb{N}^{n}$ be an
n-dimensional multi-index, $\mathbb{F}$ be a field with
$\mathrm{char}(\mathbb{F})=0$ and
$a_{1},\ldots,a_{n}\colon\mathbb{F}\allowbreak\to\mathbb{C}$ be additive
functions and $a=(a_{1},\ldots,a_{n})$. Then the mapping

$\mathbb{F}\ni x\longmapsto a^{\alpha}(x^{k})$

is a generalized monomial of degree $|\alpha|\cdot k$.

###### Proof.

Let $n$ and $k$ be positive integers, $\alpha\in\mathbb{N}^{n}$ be an
$n$-dimensional multi-index and
$a_{1},\ldots,a_{n}\colon\mathbb{F}\allowbreak\to\mathbb{C}$ be additive
functions. If $\alpha=(\alpha_{1},\ldots,\alpha_{n})$, then the function

$\mathbb{F}\ni x\longmapsto a_{1}^{\alpha_{1}}(x^{k})\cdots
a_{n}^{\alpha_{n}}(x^{k})$

is the trace of the $|\alpha|\cdot k$-additive function $F$ defined by

$F(x_{1,1},\ldots,x_{\alpha_{n},k})=\prod_{i=1}^{n}\prod_{j=1}^{\alpha_{i}}a_{i}(x_{j,1}\cdots
x_{j,k}).$

∎

Let $\mathbb{F}$ be a field and denote $\mathbb{F}^{\times}$ the
multiplicative group of the nonzero elements of $\mathbb{F}$. Obviously, every
(normal) polynomial $p\colon\mathbb{F}^{\times}\to\mathbb{C}$ is a generalized
polynomial, too. Furthermore, in view of Theorem 4 we know that these are the
only generalized polynomials if the torsion free rank of $\mathbb{F}^{\times}$
is finite. For example if we take $\mathbb{F}=\mathbb{Q}$ (the field of the
rationals), then this is the case. While for larger fields, the same does not
hold in general. Despite Theorem 4 gives an elegant criterion for the above
problem, with the aid of this result it is not too easy to imagine how
generalized polynomials look like. From one hand, this is the purpose of the
remark below.

######  Remark 4.

Notice that the above lemma cannot be in general strengthened. To see this let
$a\colon\mathbb{F}\to\mathbb{C}$ be a non-identically zero additive function.
As the following proposition shows, the mapping

$\mathbb{F}\ni x\longmapsto a(x^{2})$

is a generalized monomial of degree two that is not necessarily a (normal)
monomial.

######  Proposition 1.

Let $\mathbb{F}\subset\mathbb{C}$ be a field and
$a\colon\mathbb{F}\to\mathbb{C}$ be a non-identically zero additive function.
The mapping

$\mathbb{F}\ni x\longmapsto a(x^{2})$

is a monomial of degree two if and only if

$a(x)=\varphi(d(x))+a(1)\cdot\varphi(x)\qquad\left(x\in\mathbb{F}\right)$

or

$a(x)=\alpha\varphi_{1}(x)+\beta\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right),$

where $\alpha,\beta$ are complex constants, $d\colon\mathbb{F}\to\mathbb{C}$
is a non-identically zero derivation and
$\varphi,\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ are homorphisms
such that $\varphi_{1}$ and $\varphi_{2}$ are linearly independent.

###### Proof.

Let $\mathbb{F}\subset\mathbb{C}$ be a field and
$a\colon\mathbb{F}\to\mathbb{C}$ be a non-identically zero additive function.
Assume that the mapping appearing in the proposition is a monomial of degree
two. Then there exist linearly independent additive functions
$a_{1},a_{2}\colon\mathbb{F}\to\mathbb{C}$ and complex constants
$\alpha_{i,j}$, $i,j=1,2$ such that

$a(x^{2})=\alpha_{1,1}a_{1}(x)^{2}+(\alpha_{1,2}+\alpha_{2,1})a_{1}(x)a_{2}(x)+\alpha_{2,2}a_{2}(x)^{2}$

for each $x\in\mathbb{F}$. Since both sides of the above identity are traces
of symmetric bi-additive functions, we can use the Polarization Formula to get
that

$a(xy)=\alpha_{1,1}a_{1}(x)a_{1}(y)+\frac{\alpha_{1,2}+\alpha_{2,1}}{2}\left(a_{1}(x)a_{2}(y)+a_{1}(y)a_{2}(x)\right)+\alpha_{2,2}a_{2}(x)a_{2}(y)$

is fulfilled by any $x,y\in\mathbb{F}$. After some rearrangement, we have

$a(xy)=a_{1}(x)\cdot\left(\alpha_{1,1}a_{1}(y)+\frac{\alpha_{1,2}+\alpha_{2,1}}{2}a_{2}(y)+\right)+a_{2}(x)\cdot\left(\frac{\alpha_{1,2}+\alpha_{2,1}}{2}a_{1}(y)+\alpha_{2,2}a_{2}(y)\right)$

for all $x,y\in\mathbb{F}$, which is a Levi-Civita equation on
$\mathbb{F}^{\times}$ (on the multiplicative group of the non-zero elements of
$\mathbb{F}$). Applying the result of [15, pages 93–94], we deduce that there
exist complex constants $\alpha,\beta$, an additive function
$l\colon\mathbb{F}^{\times}\to\mathbb{C}$ and exponentials
$m,m_{1},m_{2}\colon\mathbb{F}^{\times}\to\mathbb{C}$, $m_{1}$ and $m_{2}$ are
linearly independent such that

$a(x)=(\alpha l(x)+\beta)m(x)\qquad\left(x\in\mathbb{F}^{\times}\right)$

or

$a(x)=m_{1}(x)+m_{2}(x)\qquad\left(x\in\mathbb{F}^{\times}\right).$

Recall that on the group $\mathbb{F}^{\times}$ multiplication as group
operation is considered. Therefore these functions fulfill the following
identities:

$l(xy)=l(x)+l(y)\qquad\left(x,y\in\mathbb{F}^{\times}\right)$

and

$m(xy)=m(x)m(y)\qquad\left(x,y\in\mathbb{F}^{\times}\right)$

as well as

$m_{i}(xy)=m_{i}(x)m_{i}(y)\qquad\left(x,y\in\mathbb{F}^{\times},i=1,2\right).$

At the same time, the mapping $a$ was assumed to be additive, thus in the
above representation we have

$a(x)=\varphi(d(x))+a(1)\cdot\varphi(x)\qquad\left(x\in\mathbb{F}\right)$

or

$a(x)=\alpha\varphi_{1}(x)+\beta\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right),$

where $\alpha,\beta$ are complex constants, $d\colon\mathbb{F}\to\mathbb{C}$
is a non-identically zero derivation and
$\varphi,\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ are
homomorphisms such that $\varphi_{1}$ and $\varphi_{2}$ are linearly
independent.

The converse is obvious. ∎

######  Remark 5.

The proof of the previous proposition can be a starting point of further
investigations. More precisely, if $n$ is a positive integer,
$a\colon\mathbb{F}\to\mathbb{C}$ is a non-identically zero additive function
and the mapping

$\mathbb{F}\ni x\longmapsto a(x^{2})$

is a monomial of degree $n$, then there exist linearly independent additive
functions $a_{1},\ldots,a_{n}\colon\mathbb{F}\to\mathbb{C}$ and complex
constants $\alpha_{i,j}$, $i,j=1,\ldots,n$ such that

$a(x^{2})=\sum_{i,j=1}^{n}\alpha_{i,j}a_{i}(x)a_{j}(x)\qquad\left(x\in\mathbb{F}\right).$

Since both sides of the above identity are traces of symmetric bi-additive
functions, we can use the Polarization Formula to get that

$a(xy)=\sum_{i,j=1}^{n}\frac{\alpha_{i,j}+\alpha_{j,i}}{2}\left(a_{i}(x)a_{j}(y)+a_{j}(x)a_{i}(y)\right)=\sum_{i=1}^{n}\widetilde{a_{i}}(x)\widetilde{a_{j}}(y)$

for all $x,y\in\mathbb{F}^{\times}$ which is a Levi-Civita equation on the
group $\mathbb{F}^{\times}$. In other words, the mapping $a$ as a function
restricted to the multiplicative group $\mathbb{F}^{\times}$, is a normal
exponential polynomial of degree $n$. In view of the results of [15, page 43
and page 79],

$a(x)=\sum_{j=1}^{k}P_{j}\left(l_{j,1}(x),l_{j,2}(x),\ldots,l_{j,n_{j}-1}(x)\right)m_{j}(x)\qquad\left(x\in\mathbb{F}^{\times}\right),$

where $k,n_{1},\ldots,n_{k}$ are positive integers, $m_{1},\ldots,m_{k}$ are
different, non-zero complex-valued exponentials on the group
$\mathbb{F}^{\times}$, further $\left\\{l_{j,1},\ldots,l_{j,n_{j}-1}\right\\}$
are sets of linearly independent, complex-valued additive functions defined on
$\mathbb{F}^{\times}$ for $j=1,\ldots,k$ and
$P_{j}\colon\mathbb{C}^{n_{j}-1}\to\mathbb{C}$ are complex polynomials of
degree at most $n_{j}-1$ and in $n_{j}-1$ variables for each $j=1,\ldots,k$.

We conjecture that as a continuation, a result of Kiss–Laczkovich [12] might
be useful. According to Theorem 1.1 of this paper,
$a\colon\mathbb{F}\to\mathbb{C}$ is an additive function with $a(1)=0$ and
$D/j$, as a map from the group $\mathbb{F}^{\times}$ to $\mathbb{C}$, is a
generalized polynomial of degree at most $n$ if and only if $a$ is a
derivation of order at most $n$. Here $j$ denotes the identity map from
$\mathbb{F}$ to $\mathbb{C}$.

######  Example 1.

Let $d\colon\mathbb{C}\to\mathbb{C}$ be a non-identically zero derivation. Due
to Theorem 2 such a mapping does exist. Furthermore, due to Theorem 14.5.1 of
[13], there exist non-trivial endomorphisms of $\mathbb{C}$ (trivial
endomorphisms are meant the identically zero and the identity mapping, resp.).
Let $\varphi\colon\mathbb{C}\to\mathbb{C}$ be a non-trivial endomorphism. Then
the mapping $a$ defined on $\mathbb{C}$ by

$a(x)=d(x)+\varphi(x)\qquad\left(C\right)$

is clearly additive and we have

$a(x^{2})=d(x^{2})+\varphi(x^{2})=2xd(x)+\varphi(x)^{2}\qquad\left(x\in\mathbb{F}\right),$

showing that the mapping

$\mathbb{C}\ni x\longmapsto a(x^{2})$

is a generalized polynomial of degree exactly two and it is a (normal)
polynomial of degree exactly three.

Note that the situation changes if the domain is the real field. Indeed, all
homomorphisms $\varphi\colon\mathbb{R}\to\mathbb{C}$ are trivial, i.e. they
are either identically zero or they coincide with the identity map. Therefore
if $d\colon\mathbb{R}\to\mathbb{C}$ is a non-identically zero derivation and
we define

$a(x)=d(x)+x\qquad\left(x\in\mathbb{R}\right),$

then the mapping

$\mathbb{C}\ni x\longmapsto a(x^{2})$

is a polynomial of degree exactly two.

######  Remark 6.

From the above thoughts we obtain that if $a\colon\mathbb{F}\to\mathbb{C}$ is
an additive function such that the mapping

$\mathbb{F}\ni x\longmapsto a(x^{2})$

is a generalized monomial which is not a monomial, then for all positive
integer $k\geq 2$, the mapping

$\mathbb{F}\ni x\longmapsto a(x^{k})$

is also a generalized monomial which is not a monomial.

Indeed, observe that the above mapping is the trace of the symmetric and
$k$-additive function $A_{k}$ defined on $\mathbb{F}^{k}$ by

$A_{k}(x_{1},\ldots,x_{k})=a(x_{1}\cdots
x_{k})\qquad\left(x_{1},\ldots,x_{k}\in\mathbb{F}\right).$

If there would exist a positive integer $k>2$ such that the mapping

$\mathbb{F}\ni x\longmapsto a(x^{k})$

would be a monomial, then all the translates of this mapping would also be
generalized monomials, as well. By the Polarization formula, we have

$\Delta_{y}^{2}\Delta^{k-2}_{1}a(x^{k})=k!A(y,y,1,\ldots,1)=k!a(y^{2})\qquad\left(x,y\in\mathbb{F}\right).$

From this we would deduce that the mapping

$\mathbb{F}\ni y\longmapsto a(y^{2})$

is a monomial of degree two, which is a contradiction.

######  Lemma 6.

Let $k$ and $n$ be positive integers and $f\colon\mathbb{F}\to\mathbb{C}$ be a
generalized monomial of degree n, where $\mathbb{F}$ is assumed to be a field
with $\mathrm{char}(\mathbb{F})=0$. Then the mapping

$\mathbb{F}\ni x\longmapsto f(x^{k})$

is a generalized monomial of degree $n\cdot k$.

###### Proof.

Since $f\colon\mathbb{F}\to\mathbb{C}$ is a generalized monomial of degree
$n$, there exists an $n$-additive function
$F_{n}\colon\mathbb{F}^{n}\to\mathbb{C}$ such that its trace is the function
$f$. In this case, the mapping $F_{nk}\colon\mathbb{F}^{nk}\to\mathbb{C}$
defined by

$F_{nk}(x_{1,1},\ldots,x_{n,k})=F_{n}(x_{1,1}\cdots x_{1,k},x_{2,1}\cdots
x_{2,k},\ldots,x_{n,1}\cdots x_{n,k})\\\
\left(x_{i,j}\in\mathbb{F},i=1,\ldots,n,j=1,\ldots,k\right)$

is an $(n\cdot k)$-additive function whose trace is

$F_{nk}(x,\ldots,x)=F_{n}(x^{k},\ldots,x^{k})=f(x^{k})\qquad\left(x\in\mathbb{F}\right).$

∎

######  Remark 7.

The above lemma with a different proof can be found among others in [1].

Now we turn to deal with the main problem of this paper, which is the
following. Assume $\mathbb{F}$ to be a field. Let further $P\in\mathbb{F}[x]$
and $Q\in\mathbb{C}[x]$ be polynomials. Our aim is to prove characterization
theorems for quadratic functions $f\colon\mathbb{F}\to\mathbb{C}$ that also
fulfill equation

$f(P(x))=Q(f(x))$

for each $x\in\mathbb{F}$.

As for the difficulty of the problem, it is an important condition that the
function $f$ is supposed to be quadratic (that is, it is a _generalized_
monomial of degree two). As the statement below shows, although the result is
the same, the proof is much easier for normal monomials of degree two.

######  Proposition 2.

Let $\mathbb{F}$ be a field with $\mathrm{char}(\mathbb{F})=0$ and
$f\colon\mathbb{F}\to\mathbb{C}$ be a quadratic function that can be
represented as

$f(x)=a_{1}(x)a_{2}(x)\qquad\left(x\in\mathbb{F}\right)$

with the aid of the additive functions
$a_{1},a_{2}\colon\mathbb{F}\to\mathbb{C}$. Then equation

$f(x^{2})=f(x)^{2}$ (1)

holds for each $x\in\mathbb{F}$ _if and only if_ there exist complex-valued
non-trivial homomorphisms $\varphi_{1},\varphi_{2}$ defined on $\mathbb{F}$
such that

$f(x)=f(1)\cdot\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right).$

Furthermore,

  1. (A)

either $f(1)=0$ and then $f$ is identically zero;

  2. (B)

or $f(1)=1$.

###### Proof.

Assume $\mathbb{F}$ be a field and $f\colon\mathbb{F}\to\mathbb{C}$ be a
quadratic function that can be represented as

$f(x)=a_{1}(x)a_{2}(x)\qquad\left(x\in\mathbb{F}\right)$

with the aid of the additive functions
$a_{1},a_{2}\colon\mathbb{F}\to\mathbb{C}$. Then equation (1) in terms of the
additive functions $a_{1}$ and $a_{2}$ is

$a_{1}(x^{2})a_{2}(x^{2})=a_{1}(x)^{2}a_{2}(x)^{2}\qquad\left(x\in\mathbb{F}\right).$

Since both sides of this identity are traces of symmetric, $4$-additive
functions, a simple application of the Polarization Formula leads to

$a_{1}(xy)a_{2}(uv)+a_{1}(xu)a_{2}(yv)+a_{1}(xv)a_{2}(yu)\\\
+a_{1}(yu)a_{2}(xv)+a_{1}(yv)a_{2}(xu)+a_{1}(ux)a_{2}(xy)\\\
=a_{1}(x)a_{1}(y)a_{2}(u)a_{2}(v)+a_{1}(x)a_{1}(u)a_{2}(y)a_{2}(v)+a_{1}(x)a_{1}(v)a_{2}(y)a_{2}(u)\\\
+a_{1}(y)a_{1}(u)a_{2}(x)a_{2}(v)+a_{1}(y)a_{1}(v)a_{2}(x)a_{2}(u)+a_{1}(u)a_{1}(x)a_{2}(x)a_{2}(y)\\\
(x,y,u,v\in\mathbb{F}).$ (2)

From this, with the substitution $x=y=u=v=1$, we immediately get that

$a_{1}(1)\,a_{2}(1)\,\left(a_{1}(1)\,a_{2}(1)-1\right)=0$

and with the substitution $y=u=v=1$,

$\left(a_{1}(1)\,a_{2}(1)-1\right)\,\left(a_{1}(1)\,a_{2}(x)+a_{2}(1)\,a_{1}(x)\right)=0$

can be deduced for all $x\in\mathbb{F}$. Assume first that
$a_{1}(1)\,a_{2}(1)-1\neq 0$, then $a_{1}(1)=0$ or $a_{2}(1)=0$. If
$a_{1}(1)=0$, then

  1. (A)

either $a_{2}(1)\neq 0$ and the above identity reduces to

$a_{2}(1)a_{1}(x)=0\qquad\left(x\in\mathbb{F}\right).$

In other words $a_{1}$ is the identically zero function. In this case $f$ is
identically zero, too.

  2. (B)

or $a_{2}(1)=0$ and from (2) we get that

$a_{1}(1)\,a_{2}(x\,y)+a_{2}(1)\,a_{1}(x\,y)+\left(\left(2-2\,a_{1}(1)\,a_{2}(1)\right)\,a_{1}(x)-a_{1}(1)^{2}\,a_{2}(x)\right)\,a_{2}(y)\\\
+\left(\left(2-2\,a_{1}(1)\,a_{2}(1)\right)\,a_{2}(x)-a_{2}(1)^{2}\,a_{1}(x)\right)\,a_{1}(y)=0$

for all $x,y\in\mathbb{F}$, that is,

$a_{1}(x)\,a_{2}(y)+a_{2}(x)\,a_{1}(y)=0\qquad\left(x,y\in\mathbb{F}\right).$

From this,

$a_{1}(x)a_{2}(x)=0$

follows for all $x\in\mathbb{F}$, that is, $f$ is the identically zero
function.

Therefore, from now on $a_{1}(1)\,a_{2}(1)-1=0$ can be supposed. Note that
without the loss of generality we can (and we do) assume that
$a_{1}(1)=a_{2}(1)=1$. In this case equation (2) implies that

$a_{2}(x\,y)+a_{1}(x\,y)=a_{2}(x)\,a_{2}(y)+a_{1}(x)\,a_{1}(y)$

for all $x,y\in\mathbb{F}$. From this we deduce that either

  1. (A)

$\left\\{a_{1},a_{2}\right\\}$ is linearly dependent, that is

$a_{2}=ca_{1}(x)\qquad\left(x\in\mathbb{F}\right),$

from which we get that $a_{1}\equiv a_{2}$, since $a_{1}(1)=a_{2}(1)$. If so,
then equation (1) reduces to

$a(x^{2})^{2}=\kappa a(x)^{4}\qquad\left(x\in\mathbb{F}\right),$

that is $a$ is a homomorphism and

$f(x)=f(1)a(x)^{2}\qquad\left(x\in\mathbb{F}\right).$

  2. (B)

or $\left\\{a_{1},a_{2}\right\\}$ is linearly independent. Then due to [15,
pages 93-94]

$a_{i}(x)=(\alpha_{i}l(x)+\beta_{i})m(x)\qquad\left(x\in\mathbb{F}^{\times},i=1,2\right)$

or

$a_{i}(x)=\alpha_{i}m_{1}(x)+\beta_{i}m_{2}(x)\qquad\left(x\in\mathbb{F}^{\times},i=1,2\right),$

where $l\colon\mathbb{F}^{\times}\to\mathbb{C}$ is a logarithmic function,
$m,m_{1},m_{2}\colon\mathbb{F}^{\times}\to\mathbb{C}$ are multiplicative
functions and $\alpha_{1},\alpha_{2}$ and $\beta_{1},\beta_{2}$ are complex
constants.

Substituting the first form into equation (1), $\alpha_{1}=0$ and
$\alpha_{2}=0$. That is, $a_{1}$ and $a_{2}$ are constant multiples of a
(nonzero) homomorphism. This means that there exists linearly independent
homomorphisms $\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=f(1)\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right).$

Finally, if we substitute the second form into (1) then we get (among others)
that

$\begin{array}[]{rcl}\alpha_{1}\alpha_{2}(1-\alpha_{1}\alpha_{2})&=&0\\\
\beta_{1}\beta_{2}(1-\beta_{1}\beta_{2})&=&0\end{array}$

If we would have $\beta_{1}\beta_{2}=1$, then the remaining equations, that
is,

$\begin{array}[]{rcl}\alpha_{1}\beta_{1}\beta_{2}^{2}+\alpha_{2}\beta_{1}^{2}\beta_{2}&=&0\\\
\alpha_{1}^{2}\alpha_{2}\beta_{2}+\alpha_{1}\beta_{2}^{2}\beta_{1}&=&0\\\
\alpha_{1}\beta_{2}+\beta_{1}\alpha_{2}-\alpha_{1}^{2}\beta_{2}^{2}-\beta_{1}^{2}\alpha_{2}-4\alpha_{1}\alpha_{2}\beta_{1}\beta_{2}&=&0\end{array}$

would lead to a contradiction. The same concerns the case
$\alpha_{1}\alpha_{2}=1$.

This means however that $\alpha_{1}\alpha_{2}=0$ and $\beta_{1}\beta_{2}=0$,
that is, either $f$ is the identically zero function, or there exists
homomorphisms $\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=f(1)\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right).$

∎

######  Remark 8.

Roughly speaking the above statement says that among polynomials whose variety
is at most two-dimensional, the solutions $f\colon\mathbb{F}\to\mathbb{C}$ of
equation (1) are of the form

$f(x)=f(1)\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right),$

with appropriate homomorphisms
$\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$.

Let now $n$ be a fixed positive integer. In the general case, that is, if
$f\colon\mathbb{F}\to\mathbb{C}$ is a quadratic function whose variety is at
most $n$-dimensional, then we have

$f(x)=\sum_{p,q=1}^{n}a_{p}(x)a_{q}(x)\qquad\left(x\in\mathbb{F}\right),$

with certain additive functions
$a_{1},\ldots,a_{n}\colon\mathbb{F}\to\mathbb{C}$. In this situation equation
(1) can be investigated with an analogous argument as in the proof of
Proposition 2.

The starting point of study of equation (3) among quadratic functions is the
theorem below that can be found in [7]. Here we also recall its proof to
enlighten the way to the general case.

######  Theorem 5.

Let $\mathbb{F}$ be a field with $\mathrm{char}(\mathbb{F})=0$ and
$f\colon\mathbb{F}\to\mathbb{C}$ be a quadratic function. Then equation

$f(x^{2})=f(x)^{2}$ (3)

holds for each $x\in\mathbb{F}$ _if and only if_ there exist non-trivial
homomorphisms
$\varphi_{1},\varphi_{2}\colon\allowbreak\mathbb{F}\to\mathbb{C}$ such that

$f(x)=f(1)\cdot\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right).$

Furthermore,

  1. (A)

either $f(1)=0$ and then $f$ is identically zero;

  2. (B)

or $f(1)=1$.

###### Proof.

Since $f$ is a generalized monomial of degree $2$, there exists a symmetric
bi-additive function $F_{2}\colon\mathbb{F}^{2}\allowbreak\to\mathbb{C}$ so
that

$F_{2}(x,x)=f(x)\qquad\left(x\in\mathbb{F}\right).$

Define the symmetric $4$-additive mapping
$F_{4}\colon\mathbb{F}^{4}\to\mathbb{C}$ through

$F_{4}(x_{1},x_{2},x_{3},x_{4})=F_{2}(x_{1}x_{2},x_{3}x_{4})+F_{2}(x_{1}x_{3},x_{2}x_{4})+F_{2}(x_{1}x_{4},x_{2}x_{3})\\\
-F_{2}(x_{1},x_{2})F(x_{3},x_{4})-F_{2}(x_{1},x_{3})F(x_{2},x_{4})-F_{2}(x_{1},x_{4})F(x_{2},x_{3})\\\
\left(x_{1},x_{2},x_{3},x_{4}\in\mathbb{F}\right).$

Since

$F_{4}(x,x,x,x)=3\left(F_{2}(x^{2},x^{2})-F_{2}(x,x)^{2}\right)=3\left(f(x^{2})-f(x)^{2}\right)=0\qquad\left(x\in\mathbb{F}\right),$

the mapping $F_{4}$ has to be identically zero on $\mathbb{F}^{4}$. Therefore,
especially

$0=F_{4}(1,1,1,1)=3F_{2}(1,1)-3F_{2}(1,1)^{2},$

yielding that either $F_{2}(1,1)=0$ or $F_{2}(1,1)=1$. Moreover,

$0=F_{4}(x,1,1,1)=3F_{2}\left(x,1\right)-3F_{2}\left(1,1\right)F_{2}\left(x,1\right)\qquad\left(x\in\mathbb{F}\right),$

from which either $F_{2}(1,1)=1$ or $F_{2}(x,1)=0$ follows for any
$x\in\mathbb{F}$.

Using that

$0=F_{4}(x,x,1,1)=F_{2}(x^{2},1)-F_{2}\left(1,1\right)\,F_{2}\left(x,x\right)+2F_{2}\left(x,x\right)-2F^{2}_{2}\left(x,1\right)\qquad\left(x\in\mathbb{F}\right),$

we obtain that

$\left(F_{2}(1,1)-2\right)F_{2}\left(x,x\right)=F_{2}(x^{2},1)-2F^{2}_{2}\left(x,1\right)\qquad\left(x\in\mathbb{F}\right).$

Now, if $F_{2}(1,1)=0$, then according to the above identities $F_{2}(x,1)=0$
would follow for all $x\in\mathbb{F}$. Since $F_{4}(x,x,1,1)=0$ is also
fulfilled by any $x\in\mathbb{F}$, this immediately implies that

$-2f(x)=-2F_{2}(x,x)=F_{2}(x^{2},1)-F_{2}(x,1)^{2}=0\qquad\left(x\in\mathbb{F}\right),$

i.e., $f$ is identically zero.

In case $F_{2}(1,1)\neq 0$, then necessarily $F_{2}(1,1)=1$ from which

$-F_{2}(x,x)=F_{2}(x^{2},1)-2F_{2}(x,1)^{2}\qquad\left(x\in\mathbb{F}\right).$

Define the non-identically zero additive function
$a\colon\mathbb{F}\to\mathbb{C}$ by

$a(x)=F_{2}(x,1)\qquad\left(x\in\mathbb{F}\right)$

to get that

$f(x)=F_{2}(x,x)=-F_{2}(x^{2},1)+2F_{2}(x,1)^{2}=2a(x)^{2}-a(x^{2})\qquad\left(x\in\mathbb{F}\right).$

Since $F_{4}(x,x,x,x)=0$ has to hold, the additive function
$a\colon\mathbb{F}\to\mathbb{C}$ has to fulfill identity

$-a(x^{4})+a^{2}(x^{2})+4a^{2}(x)a(x^{2})-4a^{4}(x)=0\qquad\left(x\in\mathbb{F}\right)$
(4)

too.

In what follows, we will show that the additive function $a$ is of a rather
special form.

Indeed,

$0=F_{4}(x,y,z,1)\qquad\left(x,y,z\in\mathbb{F}\right)$

means that $a$ has to fulfill equation

$a(x)a(yz)+a(y)a(xz)+a(z)a(xy)=2a(x)a(y)a(z)+a(xyz)\qquad\left(x,y,z\in\mathbb{F}\right)$

Let now $z^{\ast}\in\mathbb{F}$ be arbitrarily fixed to have

$a(x)a(yz^{\ast})+a(y)a(xz^{\ast})+a(z^{\ast})a(xy)=2a(x)a(y)a(z^{\ast})+a(xyz^{\ast})\qquad\left(x,y,z\in\mathbb{F}\right).$

Define the additive function $A\colon\mathbb{F}\to\mathbb{C}$ by

$A(x)=a(xz^{\ast})-a(z^{\ast})a(x)\qquad\left(x\in\mathbb{F}\right)$

to receive that

$A(xy)=a(x)A(y)+a(y)A(x)\qquad\left(x,y\in\mathbb{F}\right),$

which is a special convolution type functional equation. Due to Theorem 12.2
of [15], we get that

  1. (a)

the function $A$ is identically zero under any choice of $z^{\ast}$, implying
that $a$ has to be multiplicative. Note that $a$ is additive, too. Thus, for
the quadratic mapping $f\colon\mathbb{F}\to\mathbb{C}$ there exists a
homomorphism $\varphi\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=\varphi(x)^{2}\qquad\left(x\in\mathbb{F}\right).$

  2. (b)

or there exists multiplicative functions
$m_{1},m_{2}\colon\mathbb{F}\to\mathbb{C}$ and a complex constant $\alpha$
such that

$a(x)=\frac{m_{1}(x)+m_{2}(x)}{2}\qquad\left(x\in\mathbb{F}\right)$

and

$A(x)=\alpha\left(m_{1}(x)-m_{2}(x)\right)\qquad\left(x\in\mathbb{F}\right).$

Due to the additivity of $a$, in view of the definition of the mapping $A$, we
get that $A$ is additive, too.

This however means that both the maps $m_{1}+m_{2}$ and $m_{1}-m_{2}$ are
additive, from which the additivity of $m_{1}$ and $m_{2}$ follows, yielding
that they are in fact homomorphisms.

Since

$F_{2}(x,x)=f(x)=2a(x)^{2}-a(x^{2})\qquad\left(x\in\mathbb{F}\right),$

we obtain for the quadratic function $f\colon\mathbb{F}\to\mathbb{C}$ that
there exist homomorphisms
$\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right).$

Summing up, we received the following: identity

$f(x^{2})=f(x)^{2}\qquad\left(x\in\mathbb{F}\right)$

holds for the quadratic function $f\colon\mathbb{F}\to\mathbb{C}$ if and only
if there exist homomorphisms
$\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=f(1)\cdot\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right).$

∎

######  Corollary 2.

Let $n\geq 2$ be a positive integer, $\mathbb{F}$ be a field with
$\mathrm{char}(\mathbb{F})=0$ and $f\colon\mathbb{F}\to\mathbb{C}$ be a
quadratic function. Then equation

$f(x^{n})=f(x)^{n}$ (5)

holds for each $x\in\mathbb{F}$ _if and only if_ there exist homomorphisms
$\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=f(1)\cdot\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right),$

furthermore either $f(1)=0$ and then $f$ is identically zero, or $f(1)$ is an
$(n-1)$st root of unity.

###### Proof.

In case $n=2$, then in view of the previous theorem, there is nothing to
prove. Thus $n>2$ can be supposed subsequently.

Let $\mathbb{F}$ be a field with $\mathrm{char}(\mathbb{F})=0$,
$\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ be homomorphisms and
$\lambda\in\mathbb{C}$ be such that either $\lambda$ is zero, or it is an
$(n-1)$st root of unity. Define the function $f\colon\mathbb{F}\to\mathbb{C}$
by

$f(x)=\lambda\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right).$

Since every homomorphism is additive, Lemma 3 immediately yields that $f$ is a
quadratic function and we also have

$f(x^{n})=\lambda\varphi_{1}(x^{n})\varphi_{2}(x^{n})=\lambda\varphi_{1}(x)^{n}\varphi_{2}(x)^{n}=\lambda^{n}\varphi_{1}(x)^{n}\varphi_{2}(x)^{n}=\left(\lambda\varphi_{1}(x)\varphi_{2}(x)\right)^{n}=f(x)^{n}$

for each $x\in\mathbb{F}$, since $\lambda=\lambda^{n}$.

Conversely, let $f\colon\mathbb{F}\to\mathbb{C}$ be a quadratic function such
that we additionally have that

$f(x^{n})=f(x)^{n}$

for all $x\in\mathbb{F}$. Since $f$ is a quadratic function, there exists a
symmetric, bi-additive function $F_{2}\colon\mathbb{F}^{2}\to\mathbb{C}$ such
that

$f(x)=F_{2}(x,x)\qquad\left(x\in\mathbb{F}\right).$

Equation (5) with $x=1$ immediately yields that

$f(1)=f(1)^{n},$

that is, $F(1,1)=f(1)$ is either zero, or it is an $(n-1)$st root of unity.

Furthermore, equation (5) in terms of the function $F_{2}$ is

$F_{2}(x^{n},x^{n})=F_{2}(x,x)^{n}\qquad\left(x\in\mathbb{F}\right).$

Observe that both the sides of the above identity are traces of symmetric and
$2n$-additive functions, namely we have

$\frac{1}{(2n)!}\sum_{\sigma\in\mathscr{S}_{2n}}F_{2}(x_{\sigma(1)}\cdots
x_{\sigma(n)},x_{\sigma(n+1)}\cdots
x_{\sigma(2n)})=\frac{1}{(2n!)}\sum_{\sigma\in\mathscr{S}_{2n}}F_{2}(x_{\sigma(1)},x_{\sigma(2)})\cdots
F_{2}(x_{\sigma(2n-1)},x_{\sigma(2n)})$

for all $x_{1},\ldots,x_{2n}\in\mathbb{F}$. This identity with the
substitution

$x_{1}=x,\;x_{2}=x\quad x_{i}=1\quad\text{for }i=3,\ldots,2n$

yields that there are complex constants $\alpha$ and $\beta$ depending only on
$F_{2}(1,1)$ such that

$f(x)=F_{2}(x,x)=\alpha a(x^{2})+\beta
a(x)^{2}\qquad\left(x\in\mathbb{F}\right),$

where the additive function $a\colon\mathbb{F}\to\mathbb{C}$ is defined by

$a(x)=F_{2}(x,1)\qquad\left(x\in\mathbb{F}\right).$

Writing this form back into equation (5), we deduce

$\alpha a(x^{2n})+\beta a(x^{n})^{2}=\left(\alpha a(x^{2})+\beta
a(x)^{2}\right)^{n}\qquad\left(x\in\mathbb{F}\right).$

Again, both the sides of this identity are traces of symmetric and
$2n$-additive functions, therefore

$\frac{1}{(2n)!}\sum_{\sigma\in\mathscr{S}_{2n}}\left[\alpha
a(x_{\sigma(1)}\cdots x_{\sigma_{(2n)}})+\beta a(x_{\sigma_{(1)}}\cdots
x_{\sigma_{(n)}})a(x_{\sigma(n+1)}\cdots x_{\sigma(n)})\right]\\\
=\frac{1}{(2n)!}\sum_{\sigma\in\mathscr{S}_{2n}}\prod_{k=0}^{n-1}(\alpha
a(x_{\sigma(2k+1)}x_{\sigma(2k+2)})+\beta
a(x_{\sigma(2k+1)})a(x_{\sigma(2k+2)}))$

for each $x_{1},\ldots,x_{2n}\in\mathbb{F}$. Let now $x,y,z\in\mathbb{F}$ be
arbitrary, then this identity with the substitutions

$x_{1}=x,\,x_{2}=y,\,x_{3}=z\quad\text{and}\quad x_{i}=0\quad\text{for}\quad
4\leq i\leq 2n$

leads to

$Aa(xyz)+Ba(xy)a(z)+Ca(xz)a(y)+Da(yz)a(x)+Ea(x)a(y)a(z)=0\qquad\left(x,y,z\in\mathbb{F}\right),$

that is a similar equation that appear in the proof of Theorem 5. With an
analogous thread we get that $a$ can be written as a sum of two homomorphisms
that finally implies for the function $f$ that there are homomorphisms
$\varphi_{1},\varphi_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=f(1)\varphi_{1}(x)\varphi_{2}(x)\qquad\left(x\in\mathbb{F}\right),$

where $f(1)$ is either zero, or it is an $(n-1)$st root of unity. ∎

##  3 Open problems and further perspectives

As it is written at the beginning of the second section, the main aim of this
paper was to prove characterization theorems for generalized polynomials
$f\colon\mathbb{F}\to\mathbb{C}$ of degree at most $n$ that also fulfill
equation

$f(P(x))=Q(f(x))$

for each $x\in\mathbb{F}$, where $n$ is a positive integer and
$P\in\mathbb{F}[x]$ and $Q\in\mathbb{C}[x]$ are polynomials. The results
presented in connection with this problem can be considered as initial steps.
Thus we close this paper with several open questions and we would like to give
some perspectives, too.

######  Remark 9.

Clearly, it is enough to consider the case $\deg(P)=\deg(Q)$. Indeed, if
$f\colon\mathbb{F}\to\mathbb{C}$ is a generalized monomial of degree $n$, the
due to Lemma 6, the mappings

$\mathbb{F}\ni x\longmapsto f(P(x))\quad\text{and}\quad\mathbb{F}\ni
x\longmapsto Q(f(x))$

are generalized polynomials of degree $n\cdot\deg(P)$ and $n\cdot\deg(Q)$,
respectively. Furthermore, they coincide at each point $x\in\mathbb{F}$.
However, this is only possible if $\deg(P)=\deg(Q)$.

In the second section, we studied only quadratic functions, i.e., generalized
monomials of degree two. Therefore, we formulate the following.

######  Open Problem 1.

Let $n\in\mathbb{N}$, $n\geq 2$ and $P\in\mathbb{F}[x]$ and
$Q\in\mathbb{C}[x]$ be polynomials of degree at least two and
$f\colon\mathbb{F}\to\mathbb{C}$ be a generalized monomial of degree $n$.
Prove or disprove that if

$f(P(x))=Q(f(x))\qquad\left(x\in\mathbb{F}\right),$

then and only then there exist homomorphisms
$\varphi_{1},\ldots,\varphi_{n}\colon\mathbb{F}\to\mathbb{C}$ such that

$f(x)=f(1)\cdot\varphi_{1}(x)\cdots\varphi_{n}(x)\qquad\left(x\in\mathbb{F}\right).$

######  Open Problem 2.

It might be promising to consider firstly the case

$P(x)=Q(x)=x^{2}\qquad\left(x\in\mathbb{F}\right),$

because as the results of the previous section show, hopefully, the case

$P(x)=Q(x)=x^{n}\qquad\left(x\in\mathbb{F}\right)$

where $n>2$, leads back to the case $n=2$.

######  Open Problem 3.

In the special case we considered the above general problem, it turned our
that the solutions of the functional equations are always (normal) monomials.
Additionally, we also showed that the proof is much easier if we know this,
see Proposition 2. We conjecture that this is true in general, too. Thus,
prove or disprove that if the generalized monomial
$f\colon\mathbb{F}\to\mathbb{C}$ solves equation

$f(P(x))=Q(f(x))\qquad\left(x\in\mathbb{F}\right),$

then $f$ is a monomial.

######  Remark 10.

The above problem is clearly meaningful for polynomials $P$ and $Q$ with
degree one. At the same time, in this case we cannot expect representations
similar to that ones that appeared in the statements proved in the second
section. Here we consider only the case quadratic functions. Accordingly,
assume that for the quadratic function $f\colon\mathbb{F}\to\mathbb{C}$
equation

$f(ax+b)=Af(x)+B\qquad\left(x\in\mathbb{F}\right),$

with some $a,b\in\mathbb{F}$ and $A,B\in\mathbb{C}$. Let us denote
$F\colon\mathbb{F}^{2}\to\mathbb{C}$ the uniquely determined symmetric, bi-
additive function whose trace is $f$. With $x=0$ we immediately get that
$f(b)=F(b,b)=B$. Furthermore,

$F(ax+b,ax+b)=AF(x,x)+F(b,b)\qquad\left(x,\in\mathbb{F}\right),$

that is,

$F(ax,ax)+2F(ax,b)+F(b,b)=AF(x,x)+F(b,b)\qquad\left(x,\in\mathbb{F}\right),$

or after some simplification

$\left(F(ax,ax)-AF(x,x)\right)+2F(ax,b)=0$

for all $x\in\mathbb{F}$. Since the left hand side of this equation is a
generalized polynomial of degree two which has to be identically zero, all of
its monomial terms should vanish. This means from one hand that

$F(ax,b)=0\qquad\left(x\in\mathbb{F}\right),$

especially, $f(b,b)=B=0$. On the other hand, we also have

$F(ax,ax)=AF(x,x)\qquad\left(x\in\mathbb{F}\right),$

from this however

$F(ax,ay)=AF(x,y)\qquad\left(x,y\in\mathbb{F}\right),$

follows, that is, the symmetric, bi-additive function is _semi-homogeneous_.
From [8, Theorem 3] it is known that a non-identically zero, symmetric and bi-
additive function $F$ fulfilling the above semi-homogeneity exists if and only
if there are injective homomorphism
$\delta_{1},\delta_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$\delta_{1}(a)\delta_{2}(a)=A.$

Summing up, if $f\colon\mathbb{F}\to\mathbb{C}$ is a non-identically zero
quadratic function such that

$f(ax+b)=Af(x)+B\qquad\left(x\in\mathbb{F}\right),$

with some $a,b\in\mathbb{F}$ and $A,B\in\mathbb{C}$, then

  1. (i)

$B=0$

  2. (ii)

for the uniquely determined symmetric and bi-additive function
$F\colon\mathbb{F}^{2}\to\mathbb{C}$, identity

$F(ax,b)=0$

is satisfied for all $x\in\mathbb{F}$.

  3. (iii)

there are injective homomorphism
$\delta_{1},\delta_{2}\colon\mathbb{F}\to\mathbb{C}$ such that

$\delta_{1}(a)\delta_{2}(a)=A.$

Observe that for instance with $b=0$, with arbitrary fixed $a\in\mathbb{Q}$
and with $A=a^{2}$ the above identity is fulfilled by _any_ quadratic function
$f\colon\mathbb{F}\to\mathbb{C}$ (this is obviously consistent with the fact
that quadratic functions are $\mathbb{Q}$-homogeneous of degree two). This
shows that in this case we do not get in general any information for the form
of the involved quadratic function $f$. With an analogous method, we obtain
the same for higher order generalized monomials.

######  Open Problem 4.

In this paper only equations with one unknown function were considered. At the
same time, the investigated problem can clearly be extended: let
$P\in\mathbb{F}[x]$ and $Q\in\mathbb{C}[x]$ be polynomials,
$f,g\colon\mathbb{F}\to\mathbb{C}$ be generalized polynomials (of possibly
_different order_) such that $\deg(f)\deg(P)=\deg(g)\deg(Q)$. Prove or
disprove that if

$f(P(x))=Q(g(x))$

holds for all $x\in\mathbb{F}$, then $f$ and $g$ can be represented as
products of homomorphisms.

######  Remark 11.

A particularly interesting and presumably the simplest case of the following
problem is when $\deg(g)=1$, i.e., when $g\colon\mathbb{F}\to\mathbb{C}$ is an
additive function.

In connection to this problem we prove the following special case, which is
expected to be successfully applied in the general case as well.

######  Proposition 3.

Let $\mathbb{F}$ be a field with $\mathrm{char}(\mathbb{F})=0$,
$f\colon\mathbb{F}\to\mathbb{C}$ be a quadratic function and
$a\colon\mathbb{F}\to\mathbb{C}$ be an additive function. Then equation

$f(x^{2})=a(x)^{4}\qquad\left(x\in\mathbb{F}\right)$

holds if and only if here exists a homomorphism
$\varphi\colon\mathbb{F}\to\mathbb{C}$ such that

$a(x)=a(1)\varphi(x)\qquad\left(x\in\mathbb{F}\right)$

and

$f(x)=a(1)^{4}\varphi(x)^{2}\qquad\left(x\in\mathbb{F}\right).$

###### Proof.

Assume that $f\colon\mathbb{F}\to\mathbb{C}$ is a quadratic, while
$a\colon\mathbb{F}\to\mathbb{C}$ is an additive function such that

$f(x^{2})=a(x)^{4}\qquad\left(x\in\mathbb{F}\right).$

Since both the sides of this equation are traces of symmetric, $4$-additive
functions, we obtain that

$\frac{1}{3}\left[F(x_{1}x_{2},x_{3}x_{4})+F(x_{1}x_{3},x_{2}x_{4})+F(x_{1}x_{4},x_{2}x_{3})\right]=a(x_{1})a(x_{2})a(x_{3})a(x_{4})\qquad\left(x_{1},x_{2},x_{3},x_{4}\in\mathbb{F}\right).$

Here $F\colon\mathbb{F}\times\mathbb{F}\to\mathbb{C}$ is the uniquely
determined symmetric, bi-additive function for which $F(x,x)=f(x)$ is
satisfied for all $x\in\mathbb{F}$.

This identity implies especially that

$f(1)=F(1,1)=a(1)^{4}\qquad\text{and}\qquad
F(x,1)=a(1)^{3}a(x)\qquad\left(x\in\mathbb{F}\right).$

Furthermore, we also have

$2F(x,y)=3a(1)^{2}a(x)a(y)-F(xy,1)\qquad\left(x,y\in\mathbb{F}\right).$

Thus,

$2F(x,y)=3a(1)^{2}a(x)a(y)-a(1)^{3}a(xy)\qquad\left(x\in\mathbb{F}\right).$

In other words,

$f(x)=\frac{3}{2}a(1)^{2}a(x)^{2}-\frac{1}{2}a(1)^{3}a(x^{2})\qquad\left(x\in\mathbb{F}\right).$

Substituting this back into the original equation we get that the additive
function $a\colon\mathbb{F}\to\mathbb{C}$ has to fulfill

$a(x)^{4}=\frac{3}{2}a(1)^{2}a(x^{2})^{2}-\frac{1}{2}a(1)^{3}a(x^{4})\qquad\left(x\in\mathbb{F}\right).$

Again, after symmetrization or after applying [7, Theorem 15], we derive that
there exists a homomorphism $\varphi\colon\mathbb{F}\to\mathbb{C}$ such that

$a(x)=a(1)\varphi(x)\qquad\left(x\in\mathbb{F}\right)$

and

$f(x)=\frac{3}{2}a(1)^{2}a(x)^{2}-\frac{1}{2}a(1)^{3}a(x^{2})\\\
=\frac{3}{2}a(1)^{2}a(1)^{2}\varphi(x)^{2}-\frac{1}{2}a(1)^{3}a(1)\varphi(x^{2})=a(1)^{4}\varphi(x)^{2}\qquad\left(x\in\mathbb{F}\right).$

∎

######  Remark 12.

Using the ideas of the results of the third section, the case

$f(x^{n})=a(x)^{2n}\qquad\left(x\in\mathbb{F}\right)$

where $n$ is a fixed positive integer, can be reduced to the above studied
case.

######  Remark 13.

Observe that during the proof of Lemmas 3, 5, 6 the fact that we considered
complex-valued mappings, was not used at all. We remark that these statements
as well as their proofs are exactly the same for functions defined on a field
$\mathbb{F}$ and mapping to another field $\mathbb{K}$.

The situation is slightly different in case of Lemma 4, since in that case
derivations are involved. Nevertheless, Lemma 4 also holds true (with an
unchanged proof) for mappings defined on $\mathbb{F}$ and having values in
$\mathbb{K}$, where $\mathbb{F}\subset\mathbb{K}$ are fields.

######  Remark 14.

We would also like to clarify why we considered only complex-valued functions.
Obviously, the investigated problems are meaningful in a much more general
setting. The importance of this condition lies in our method. Namely, in each
case we showed firstly that the involved mappings
$f\colon\mathbb{F}\to\mathbb{C}$ are exponential polynomials of the
multiplicative group $\mathbb{F}^{\times}$. This enabled us to describe the
unknown function completely. This method however relies on the notion of
exponential polynomials and the theory of this notion is well-developed only
for complex-valued mappings. Maybe with a different approach the general case
can also be handled.

The assumption that the field $\mathbb{F}$ has to have zero characteristic is
caused by the limitations of the Polarization formula, since $n$-additive
functions are uniquely determined by their diagonalizations only if the
characteristic of the domain is large enough or zero.

######  Open Problem 5.

Let $\mathbb{F}$ and $\mathbb{K}$ be fields, $n$ be a positive integer,
$P\in\mathbb{F}[x]$ and $Q\in\mathbb{K}[x]$ be polynomials. Determine those
generalized monomials $f,g\colon\mathbb{F}\to\mathbb{K}$ of degree at most $n$
that also fulfill equation

$f(P(x))=Q(f(x))$

for each $x\in\mathbb{F}$.

A particularly interesting case of this problem is when at least one of the
fields $\mathbb{F}$ and $\mathbb{K}$ has finite characteristic.

######  Acknowledgment.

The author would like to thank Professor José María Almira (University of
Murcia, Spain) and Professor Rezső Lovas (University of Debrecen, Hungary) for
their generous help and valuable comments that improved the quality of the
paper.

Project No. K134191 has been implemented with the support provided by the
National Research, Development and Innovation Fund of Hungary, financed under
the K_20 funding scheme. The research of the author has partially been carried
out with the help of the Project 2019-2.1.11-TÉT-2019-00049, which has been
implemented with the support provided from the National Research, Development
and Innovation Fund of Hungary, financed under the TÉT funding scheme.

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