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\begin{center}
\vskip 1cm{\LARGE\bf Leibniz-Additive Functions on UFD's
}
\vskip 1cm
\large
Viachaslau I. Murashka\\
Faculty of Mathematics and Technologies of Programming\\
Francisk Skorina Gomel State University\\
Gomel 246019 \\
Belarus\\
\href{mailto:mvimath@yandex.ru}{\tt mvimath@yandex.ru}\\
\ \\
Andrey D. Goncharenko and Irina N. Goncharenko\\
State Educational Establishment ``Gymnasium 71''\\
Gomel 246036\\
Belarus\\
\href{mailto:goncharenkoandrey8@gmail.com}{\tt goncharenkoandrey8@gmail.com}\\
\href{mailto:ira\_nika@tut.by}{\tt ira\_nika@tut.by}
\end{center}

\vskip .2 in

\begin{abstract}
Recall that an arithmetic function $f$ is called an L-additive function with
respect to a completely multiplicative function $h$ if $f(mn
)=f( m )h( n )+f( n )h( m
)$ holds for all $m$ and $n$. We study L-additive functions in the fields
of fractions of unique factorization domains (UFD). In particular, we
describe all L-additive functions over given UFD such that these functions
can be extended to its field of fractions. We find the exact formula for an
L-additive function in the terms of prime elements. For a given L-additive
function $f( x )$ we study the properties of the sequence $(f^{( k
)}( x ))_{k\geq 1}$ and solutions of the equation $f( x
)=\alpha x$. As corollaries we obtain results about the arithmetic
derivative and partial arithmetic derivatives.
\end{abstract}

\section{Introduction}
Recall \cite{1,11,2} that the arithmetic derivative is a function $D: \mathbb{N}\to \mathbb{N}$, such that
\begin{enumerate}
    \item $D(p)=1$ for all prime $p$;
    \item $D$ satisfies the Leibniz rule: $D(mn)=D(m)n+D(n)m$.
\end{enumerate}
It is known that the arithmetic derivative is not a linear function:
$$
D(2+3)=D(5)=1 \text{ and } D(2)+D(3)=2.
$$
One can define the arithmetic derivative for a rational number \cite{2}. Since $D$ is not a linear function, it is difficult to solve even equations $D(x)=2a$ and $D(D(x))=1$. Ufnarovski and {\AA}hlander \cite{2} showed that the first equation has a solution for any natural $a$ if Goldbach's conjecture is true and the second equation has infinite number of solutions if twin prime conjecture is true. Equations of the form $D(x)=ax+1$ are connected to the conjecture about Guiga numbers as was shown by Grau and Oller-Marc\'en \cite{3}.

Note that the partial arithmetic derivative $D_p(n)$ and the arithmetic subderivative $D_S(n)$ of an integer number satisfy the Leibniz rule \cite{4}.

Also the analogues of arithmetic derivative were considered over different
sets of elements \cite{2}. Haukkanen et al.\ \cite{5} discussed the arithmetic derivative on non-unique factorization domains. Emmons et al.\ \cite{6} described all functions $f: \mathbb{Z}_n \to \mathbb{Z}_n$ that satisfy the Leibniz rule.
In particular, all values of such functions are divisors of zero.
Kovi\u c \cite{6.5} constructed functions defined on Gaussian rationals that satisfy the Leibniz rule.

Let $h$ be a completely multiplicative function.
According to Merikoski et al.\ \cite{12, 4} a function $f: \mathbb{N} \to \mathbb{N}$ is called L-additive with
respect to $h$ if $f(mn)=f(m)h(n)+f(n)h(m)$ holds for all $m, n \in \mathbb{N}$.

The aim of this paper is to study L-additive functions in the fields of fractions over unique factorization domains (Gaussian integers, Eisenstein integers and etc.).

\section{Preliminary results}

The notation and terminology agree with the book \cite{8}. We refer the
reader to this book for the results on ring theory.
Through $\mathbb{N},\,\mathbb{P},\, \mathbb{Z},\,\mathbb{Z}_+,\,\mathbb{Q}$, and $\mathbb{Q}_+$ we denote here the sets of natural, prime, integer, non-negative integer, rational, and non-negative rational numbers respectively.

Let $\Phi$ be an integral domain \cite[p.\ 368]{8}. Recall that the field of fractions $
\Frac(\Phi)$ of $\Phi$ is the set $\{\frac{a}{b} \mid a \in \Phi, b \in \Phi \setminus\{0\}$ with $\frac{a}{b}=\frac{c}{d}$ if $\exists k\neq 0$ such that $a=kc$ and $b=kd\}$ with two operations:
$$
\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d+c\cdot b}{b\cdot d}\,\, \text{and} \hspace{2mm} \frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d} \hspace{2mm} \forall a,b,c,d \in \Phi \,\, \text{and}\,\, b,d\neq 0.
$$
Note that $\Frac(\mathbb{Z})=\mathbb{Q}$ and $\Frac(\mathbb{Z}[i])=\mathbb{Q}(i)$.

If $p$ is a prime in a UFD, then all elements associated with
it are also primes. Let $\mathbb{P}(\Phi)$ be some maximal by inclusion set
of non-associated primes in $\Phi$. Note that a UFD can have
infinite number of maximal by inclusion sets of non-associated primes. Now
the unique factorization means that every non-zero element of $\Phi$ has the following unique factorization:
$$
y=u\cdot \prod_{p\in \mathbb{P}(\Phi)}p^{v_p (y) }
$$
where $u$ is a unit, $v_p(y)\in \mathbb{Z}_+$, and $v_p(y)\neq 0$ for finitely many $p\in \mathbb{P}(\Phi)$.

Recall that the characteristic of an integral domain $\Phi$ is defined to be the smallest number of times one must use the ring's multiplicative identity in a sum to get the additive identity. If this sum never reaches the additive identity, then $\Phi$ is said to have characteristic zero.
Note that if the characteristic of an integral
domain is not equal to $0$, then it is a prime. Also note that the
characteristics of an integral domain and its field of fractions coincide.

Let $\Phi$ be a UFD such that its group of units $G$ is finitely generated. Then
$$
G \simeq A_1\times A_2\times \dots \times A_n\times B_1\times B_2\times \dots \times B_k
$$
with $n, k \geq 0$, $A_i \simeq \mathbb{Z}$, and $B_j \simeq \mathbb{Z}_{n_j}$ where $n_j$ is a power of a prime for all $j$. Let $A(\Phi)$ be the set containing exactly one generating
element $a_i$ of each subgroup $A_i$ and $B(\Phi)$ be the set containing exactly one generating
element $b_j$ of each subgroup $B_j$ such that its order is divisible by the characteristic of $\Phi$. If the characteristic of $\Phi$ is equal to $0$, then $B(\Phi)$ is empty. Define $R(\Phi)=A(\Phi)\cup B(\Phi)$.

If the characteristic of $\Phi$ is equal to $0$, then all its units $u$ can be uniquely represented
$$
u=u_0\cdot \prod_{i=1}^n a_i^{\gamma_i}
$$
with $\gamma_i \in \mathbb{Z}$ and $u_0$ is an element of a finite order.

If the characteristic of $\Phi$ is equal to $p\neq 0$, then all
its units $u$ can be uniquely represented
$$
u=u_0\cdot \prod_{i=1}^n a_i^{\gamma_i} \cdot \prod_{j=1}^m b_j^{\delta_j}
$$
with $\gamma_i \in \mathbb{Z}, \delta_j \in \{0, 1, \ldots , n_j-1\}$, and $u_0$ is an element of a finite order
$t$ with $\gcd(t,p)=1$.

Let $\mathbb{P}^*(\Phi)=\mathbb{P}(\Phi)\cup R(\Phi)$. So the factorization of an element $q\neq 0$ in $\Phi$ can be written uniquely in the following form
$$
q=u_0 \cdot \prod_{r\in R(\Phi)}r^{v_r(q)}\cdot \prod_{p\in \mathbb{P}(\Phi)}p^{v_p(q)}=u_0 \cdot \prod_{p\in \mathbb{P}^*(\Phi)}p^{v_p(q)}.
$$
This means that every element $q\neq 0$ of $\Frac(\Phi)$ can be written uniquely in the following form
$$
q=u_0 \cdot \prod_{p\in \mathbb{P}^*(\Phi)}p^{v_p(q)}
$$
where $v_p(q)\in \mathbb{Z}$ for all $p\in \mathbb{P}^*(\Phi)$ and $v_p(q)\neq 0$ only for a finite number of $p\in \mathbb{P}^*(\Phi)$.

Let $m, n\in \Frac(\Phi)\setminus\{0\}$. It is easy to check that $v_p(m\cdot n)=v_p(m)+v_p(n)$ for any
$p\in \mathbb{P}(\Phi)$. Note that $v_{a_i}(m\cdot n)=v_{a_i}(m)+v_{a_i}(n)$. If characteristic of
$\Phi$ is $p$, then $v_{b_i}(m\cdot n)\equiv v_{b_i}(m)+v_{b_i}(n)$ (mod $n_i$), where
$n_i=p^\alpha$ for some $\alpha$. This means that $v_{b_i}(m\cdot n)\equiv v_{b_i}(m)+v_{b_i}(n)$ (mod $p$). All these mean that $v_r(m\cdot n)=v_r(m)+v_r(n)$ in $\Frac(\Phi)$.

A \textit{Gaussian integer} is a complex number $a+bi$ with $a,b\in \mathbb{Z}$ and $i^2=-1$. Note that Gaussian integers form a UFD. Any Gaussian prime divides some integer prime wherein in Gaussian integers $2=-i(1+i)^2$ with $1+i$ is a prime; every integer
prime of the form $4k+1$ is a product of two non-associated
Gaussian primes of the form $a+bi$ and $a-bi$; every
integer prime of the form $4k+3$ is a Gaussian prime. Note that
the group of units of $\mathbb{Z}[i]$ is the cyclic group of order $4$. So
$R(\mathbb{Z}[i])=\emptyset$. We can chose $\mathbb{P}^*(\mathbb{Z}[i])$ in the following way:
natural primes of the form $4k+3$; $1+i$; numbers
of the form $a+bi$ and $a-bi$ with $a<b,\,a,b\in \mathbb{N}$
and $a^2+b^2$ is a natural prime of the form $4k+1$. Then $\mathbb{P}^*(\mathbb{Z}[i])=\{1+i,3,1+2i,1-2i,7,2+3i,2-3i,11,\ldots \}$.

An \textit{Eisenstein integer} is a complex number $a+b\omega$ with $a,b\in \mathbb{Z}$ and $\omega=\frac{-1+i\sqrt{3}}{2}$ is a solution of the equation $\omega^2+\omega+1=0$. Note that Eisenstein integers form a
UFD. Any Eisenstein prime divides some integer prime, with $3=(1+2\omega)^2$
where $1+2\omega$ is an Eisenstein prime. An integer prime $p\equiv 2$ (mod $3$) is an Eisenstein prime. The remaining integer primes are the
products of two non-associated Eisenstein primes. Note that the group of units
of $\mathbb{Z}[\omega]$ is the cyclic group of order $6$. So $R(\mathbb{Z}[\omega])=\emptyset$. We can chose $\mathbb{P}^*(\mathbb{Z}[\omega])$ in the following way: an
integer prime $p\equiv 2$ (mod $3$); $1+2\omega$ and numbers of the
form $a+b\omega$ and $a+b\omega^2$ with $a\in \mathbb{N}, |a|<|b|$ and
$a^2-ab+b^2$ is a natural prime number $p\equiv 1$ (mod $3$). Then $\mathbb{P}^* (\mathbb{Z}[\omega])=\{2,1+2\omega,5,1+3\omega,1+3\omega^2,11,\ldots\}$.

Note that the norm of a Gaussian integer is $N_1(a+bi)=a^2+b^2$.
The norm of an Eisenstein integer is $N_2(a+b\omega)=a^2-ab+b^2$.

Recall that $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}\, |\, a,b\in \mathbb{Z}\}$ is a UFD. Note that its group of units is generated by $\{-1, \sqrt{2}+1\}$ and is isomorphic to $\mathbb{Z}_2\times \mathbb{Z}$. So in this case we can choose $R(\mathbb{Z}[\sqrt{2}])=\{\sqrt{2}+1\}$.
\section{Main results}
The following well-known function classes play an important role in our paper.

\begin{definition} Let $K$ be a ring. A function $h:K\to K$ is called

\begin{enumerate}
    \item [(1)] \textit{Completely multiplicative}, if $h(1)=1$ and $h(mn)=h(m)h(n)$ holds for all $m,n\in K$.
    \item [(2)] \textit{Completely additive}, if $h(mn)=h(m)+h(n)$ holds for all $m,n\in K$.
\end{enumerate}
\end{definition}
\begin{definition}[\cite{4}] Let $K$ be a ring and $h:K\to K$ be a completely multiplicative function. A function $f:K\to K$ is called \textit{L-additive with respect to} $h$ if $f(mn)=f(m)h(n)+f(n)h(m)$ holds for all $m,n\in K$.
\end{definition}
If $h(n)\equiv 1$, then every L-additive function with respect to $h$ is completely additive. Every L-additive function with respect to
$h(x)=x$ will be called \textit{L-additive}.

\begin{theorem} \label{thm1} Let $\Phi$ be an integral domain and $h,f:\Phi \to \Phi$ be completely multiplicative and L-additive with respect to  $h$ functions. Then there exist unique functions $\Bar{h},\Bar{f}:
\Frac(\Phi)\to \Frac(\Phi)$, such that $\Bar{h}$ is completely multiplicative, $\Bar{f}$ is L-additive with respect to $\Bar{h}$, and $h(x)=\Bar{h}(x)$ and $f(x)=\Bar{f}(x)\, \forall x\in \Phi$ iff $h(x)\neq 0\, \forall x \in \Phi \setminus \{0\}$.

Let $q:\Frac(\Phi)\to \Frac(\Phi)$ be completely multiplicative. In these terms a function $V:\Frac(\Phi)\to \Frac(\Phi)$ is L-additive with respect to $q$ where $V(x)=\frac{\Bar{f}(x)\cdot q(x)}{\Bar{h}(x)}$ for $x\neq 0$ and $V(0)=0$.
\end{theorem}
\begin{proof} Assume that $h(x)\neq 0\, \forall x\in \Phi \setminus \{0\}$. Then for all $x\in \Frac(\Phi)$ there exist $m,n \in \Phi$ with $x=\frac{m}{n}, n\neq 0$. Let
$\Bar{h}(x)=\frac{h(m)}{h(n)}$ and $\Bar{f}(x)=\frac{f(m)h(n)-f(n)h(m)}{h^2(n)}$.

Let $x=\frac{m_1}{n_1}=\frac{m_2}{n_2}$. Then there is $k\neq 0$ such that $m_1=km_2$ and $n_1=kn_2$. The following equality shows that $\Bar{h}$ {is well-defined:}
$$
\Bar{h}\left(\frac{m_1}{n_1}\right)=\frac{h(m_1)}{h(n_1)}=\frac{h(km_2)}{h(kn_2)}=\frac{h(k)\cdot h(m_2)}{h(k)\cdot h(n_2)}=\Bar{h}\left(\frac{m_2}{n_2}\right).
$$
Let $x=x_1\cdot x_2,\, x=\frac{m}{n},\, x_1=\frac{m_1}{n_1}$ and $x_2=\frac{m_2}{n_2}$. Then $m=m_1\cdot m_2 \cdot k$ and $n=n_1\cdot n_2\cdot k$. The
following equality shows that $\Bar{h}$ {is completely multiplicative:}
$$
\Bar{h}(x_1)\cdot \Bar{h}(x_2)=\frac{h(m_1)}{h(n_1)}\cdot \frac{h(m_2)}{h(n_2)}\cdot \frac{h(k)}{h(k)}=\frac{h(k\cdot m_1 \cdot m_2)}{h(k\cdot n_1 \cdot n_2)}=\frac{h(m)}{h(n)}=\Bar{h}(x).
$$
Let $\Bar{g}$ be a completely multiplicative extension of $h$. Then
$$
1=\Bar{g}(1)=\Bar{g}(x\cdot x^{-1})=\Bar{g}(x)\cdot \Bar{g}(x^{-1}).
$$
Therefore $\Bar{g}(x^{-1})=\frac{1}{\Bar{g}(x)}=\frac{1}{h(x)}\, \forall x \in \Phi \setminus \{0\}$. The following shows that $\Bar{h}$ {is unique}:
$$
\Bar{g}(x)=\Bar{g}\left(\frac{m}{n}\right)=\Bar{g}(m)\cdot \Bar{g}(n^{-1})=\frac{h(m)}{h(n)}=\Bar{h}(x).
$$

The following equality shows that $\Bar{f}$ {is well-defined:}

\begin{align*}
\Bar{f}\left(\frac{km}{kn} \right)&=\frac{f(km)\cdot h(kn)-f(kn)\cdot h(km)}{h^2(kn)}\\&=\frac{(f(k)h(m)+f(m)h(k))\cdot h(k)h(n)-(f(k)h(n)+f(n)h(k))\cdot h(k)h(m)}{h^2(kn)}
\\&=\frac{f(k)h(m)h(k)h(n)+f(m)h^2(k)h(n)-f(k)h(n)h(k)h(m)-f(n)h^2(k) h(m)}{h^2(k)h^2(n)}\\&=\frac{h^2(k)(f(m)h(n)-f(n)h(m))}{h^2(k) h^2(n)}=\frac{f(m)h(n)-f(n)h(m)}{h^2(n)}=\Bar{f}(x).
\end{align*}
Let $x=x_1\cdot x_2, \, x=\frac{m}{n},\, x_1=\frac{m_1}{n_1}$, and $x_2=\frac{m_2}{n_2}$. Then $m=m_1\cdot m_2\cdot k$ and $n=n_1\cdot n_2\cdot k$. The following equality shows that $\Bar{f}$ is {L-additive with respect to} $\Bar{h}$:
\begin{align*}
&\Bar{h}(x_1)\Bar{f}(x_2)+\Bar{h}(x_2)\Bar{f}(x_1)=\Bar{h}\left(\frac{m_1}{n_1}\right)\Bar{f}\left(\frac{m_2}{n_2}\right)+\Bar{h}\left(\frac{m_2}{n_2}\right)\Bar{f}\left(\frac{m_1}{n_1}\right)
\\&=\frac{h(m_1)}{h(n_1)}\cdot \left(\frac{f(m_2)h(n_2)-f(n_2)h(m_2)}{h^2(n_2)}\right)+\frac{h(m_2)}{h(n_2)}\cdot \left(\frac{f(m_1)h(n_1)-f(n_1)h(m_1)}{h^2(n_1)}\right)
\\&=\frac{f(m_2)h(m_1)h(n_2)h(n_1)-f(n_2)h(n_1)h(m_1)h(m_2)}{h^2(n_1)\cdot h^2(n_2)}
\\&+\frac{f(m_1)h(m_2)h(n_2)h(n_1)-f(n_1)h(n_2)h(m_1)h(m_2)}{h^2(n_1)\cdot h^2(n_2)}
\end{align*}
\begin{align*}&=\frac{h(n_1n_2)(h(m_1)f(m_2)+h(m_2)f(m_1))}{h^2(n_1n_2)}-\frac{h(m_1m_2)(h(n_1)f(n_2)+h(n_2)f(n_1))}{h^2(n_1n_2)}\\&=\frac{h(n_1n_2)f(m_1m_2)-h(m_1m_2)f(n_1n_2)}{h^2(n_1n_2)}=\Bar{f}\left(\frac{m_1\cdot m_2}{n_1\cdot n_2}\right)=\Bar{f}\left(\frac{k\cdot m_1\cdot m_2}{k\cdot n_1\cdot n_2}\right)=\Bar{f}(x).
\end{align*}

Let $\Bar{g}$ be an L-additive with respect to $\Bar{h}$ extension
of $\Bar{f}$:
$$
0=\Bar{g}(1)=\Bar{g}(x x^{-1})=\Bar{g}(x)\Bar{h}(x^{-1})+\Bar{g}(x^{-1})\Bar{h}(x)=\frac{f(x)}{h(x)}+\Bar{g}(x^{-1})h(x).
$$
Thus $\Bar{g}(x^{-1})=-\frac{f(x)}{h^2(x)} \forall x\in\Phi \setminus \{0\} \text{ and}$
$$
 \Bar{g}\left(\frac{m}{n}\right)=\Bar{g}(m)\Bar{h}\left(\frac{1}{n}\right)+\Bar{g}\left(\frac{1}{n}\right)\Bar{h}(m)=\frac{f(m)}{h(n)}-\frac{f(n)h(m)}{h^2(n)}=\frac{f(m)h(n)-f(n)h(m)}{h^2(n)}=\Bar{f}\left(\frac{m}{n}\right).
$$
Therefore the function $\Bar{f}$ {is unique}.

{Let us prove the converse statement}. Assume that $f$ and $h$ can be extended to $\Frac(\Phi)$. Suppose that there exists $x\in \Phi  \setminus \{0\}$ with $h(x)=0$. Then $x^{-1}\in \Frac(\Phi)$.
Therefore $1=\Bar{h}(x\cdot x^{-1})=\Bar{h}(x)\cdot \Bar{h}(x^{-1})=0$. {This is a contradiction}.

{Let us prove that $V(x)$ is L-additive with respect
to $q$.} Let $m,n\in \Frac(\Phi)$. Assume that $m,n\neq 0$. Then
\begin{align*}
V(mn)&=\frac{\Bar{f}(mn)\cdot q(mn)}{\Bar{h}(mn)}=\frac{\left(\Bar{f}(m)\Bar{h}(n)+\Bar{f}(n)\Bar{h}(m)\right)q(mn)}{\Bar{h}(m)\cdot \Bar{h}(n)}\\&=\frac{\Bar{f}(m)\Bar{h}(n)q(mn)}{\Bar{h}(m)\Bar{h}(n)}+\frac{\Bar{f}(n)\Bar{h}(m)q(mn)}{\Bar{h}(m)\Bar{h}(n)}=\frac{\Bar{f}(m)q(m)q(n)}{\Bar{h}(m)}+\frac{\Bar{f}(n)q(n)q(m)}{\Bar{h}(n)}\\&=V(m)q(n)+V(n)q(m).
\end{align*}
Assume now that $m=0$. Note that $q(0)=0$. Then
$$
V(mn)=V(0)=0=0q(n)+V(n)0=V(m)q(n)+V(n)q(m).
$$
The case $n=0$ is the same.
Hence $V(x)$ {is L-additive with respect to} $q$. \end{proof}

\begin{example}
Let
\begin{displaymath}
h(n)=\begin{cases} 
  1,& n=1;\\
  0,& n\neq 1;
\end{cases} 
\quad  \text{and} \quad
f(n)=\begin{cases}
 1,& n\in \mathbb{P};\\ 
 0,& n \notin \mathbb{P}.
\end{cases}
\end{displaymath} 
Note that $f$ is L-additive with respect to $h$ and we cannot extend $f$ and $h$ from $\mathbb{Z}$ to $\mathbb{Q}$.
\end{example}

\begin{corollary} \label{cr1.1} A function $f:\Frac(\Phi)\to \Frac(\Phi)$ is L-additive iff $f(0)=0$ and $\frac{f(x)}{x}$ is completely additive.
\end{corollary}
\begin{proof}
Since every completely additive function is L-additive with respect to $h(x)\equiv 1$ and every L-additive function is L-additive with respect to $h(x)=x$, the statement of corollary directly follows from the last statement of Theorem \ref{thm1}.
\end{proof}

\begin{corollary}[{\cite[Theorem 3.1]{7}}] \label{cr1.2}
A function $D_S^f(n)$ is L-additive iff $\sum_{p\in S}\frac{f_p(n)}{p}$ is completely additive where $D_S^f(n)$ is as defined in \cite{7}.
\end{corollary}

\begin{proposition} \label{pr1}
Let $\Phi$ be an integral domain and $f,\, g:\Frac(\Phi)\to \Frac(\Phi)$ be L-additive functions. Then
\begin{enumerate}
    \item $h(x)=\alpha \cdot f(x)$ is L-additive.
    \item $h(x)=f(x)+g(x)$ is L-additive.
    \item $f\left(\frac{a}{b}\right)=\frac{f(a)b-f(b)a}{b^2}$.
    \item $f(a^n)=n\cdot a^{n-1}\cdot f(a),\, \forall n\in \mathbb{Z}$.
\end{enumerate}
\end{proposition}

\begin{proof}
\begin{enumerate}
    \item Let $m,n \in \Frac(\Phi)$. Then $f(mn)=f(m)n+f(n)m$.
So $h(mn)=\alpha \cdot (f(n)m+f(m)n)=(\alpha \cdot f(n))m+(\alpha \cdot f(m))n=h(n)m+h(m)n$. Hence $h(x)=\alpha \cdot f(x)$ is \textit{L-additive}.

    \item Let $m,n \in \Frac(\Phi)$. Then $f(mn)=f(m)n+f(n)m$ and $g(mn)=g(m)n+g(n)m$. Now $h(mn)=f(m)n+g(m)n+f(n)m+g(n)m=n(f(m)+g(m))+m(f(n)+g(n))=h(m)n+h(n)m$. So $h(x)=f(x)+g(x)$ \textit{is L-additive.}

    \item From $f(1)=0$ it follows that $f(1)=f\left(n\cdot \frac{1}{n}\right)=\frac{f(n)}{n}+f\left(\frac{1}{n}\right)n=0$. So $f\left(\frac{1}{n}\right)=-\frac{f(n)}{n^2}$. Thus $f\left(\frac{a}{b}\right)=\frac{f(a)}{b}+f\left(\frac{1}{b}\right)a=\frac{f(a)b-f(b)a}{b^2}$.

    \item Let us prove this statement by induction. Note that $0=f(1)=f(a^0)=0a^{-1}f(a)$ and $f(a)=1a^0f(a)$. Assume that the statement holds for $n\in \mathbb{N}$. Let us prove this statement for $n+1:\, f(a^{n+1})=f(a\cdot a^n)=f(a)a^n+f(a^n)a=f(a)a^n+nf(a)a^{n-1}a=(n+1)f(a)a^n$. Now we prove this statement for a negative $n:\, 0=f(1)=f(a^n\cdot a^{-n})=f(a^n)a^{-n}+a^nf(a^{-n})=\frac{nf(a)}{a}+a^nf(a^{-n})$ and therefore $f(a^{-n})=-n\cdot a^{-(n+1)}f(a)$.
\end{enumerate}
\end{proof}

Let $S\subseteq \mathbb{P}^*(\Phi)$. We shall call the function
$$
D_S(x)=x\cdot \sum_{p\in S} \frac{v_p(x)}{p}
$$
\textit{an arithmetic subderivative in} $\Phi$. If $|S|=1$, then we shall call it \textit{a partial arithmetic derivative} and if
$S=\mathbb{P}^*(\Phi)$, then we shall call it the \textit{arithmetic derivative in} $\Phi$. These functions for integers were studied, for example, by Merikoski et al.\ \cite{4}. Note that the function $D_S$ is also referred to as ``arithmetic type derivative'' \cite{13}.

\begin{corollary} \label{cr1.3}
Let $\Phi$ be a UFD such  that  its  group  of  units  is  finitely  generated and $S \subseteq \mathbb{P}^*(\Phi)$. Then $D_S(x)$ is L-additive.
\end{corollary}
\begin{proof}
Recall that $v_p(x)$ is completely additive. Then
$x\cdot v_p(x)$ is a L-additive function by Corollary \ref{cr1.1}. Since $v_p(x)\neq 0$ only for finite number of $p\in \mathbb{P}^*(\Phi)$,
$D_S(x)=x\cdot \sum_{p\in S}\frac{v_p(x)}{p}$ is L-additive by 1 and 2 of Proposition \ref{pr1}.
\end{proof}

\begin{theorem} \label{thm2}
Let $\Phi$ be a UFD such that its group of units is finitely generated. A function $g:\Frac(\Phi) \setminus \{0\}\to \Frac(\Phi)$ is completely additive iff
$$
g(n)=\sum_{p\in \mathbb{P}^*(\Phi)}\left(v_p(n)g(p)\right)\, \forall n\in \Frac(\Phi).
$$
\end{theorem}

\begin{proof} Let $g$ be a completely additive function. Note that for all $n\in \Frac(\Phi)$ holds
$$
n=u_0\cdot \prod_{p\in \mathbb{P}^*(\Phi)}p^{v_p(n)},
$$
where $u_0$ is a unit of $\Phi$ of a finite order $k$ (if $\text{char}(\Frac(\Phi))=p\neq 0$, then $\gcd(k,p)=1$). Note that $g(1\cdot 1)=g(1)+g(1)$. Therefore $g(1)=0$. Hence
$$
g(1)=g(u_0^k)=kg(u_0)=0.
$$
Since $\gcd(k,p)=1$ and $\Phi$ is an integral domain, we see that $g(u_0)=0$.


Since $g$ is completely additive, it is clear that $g(x^k)=k\cdot g(x)$ for all $k\in \mathbb{N}$. We proved that $g(x^0)=g(1)=0=0\cdot g(x)$. From $0=g(1)=g(x^k\cdot x^{-k})=g(x^k)+g(x^{-k})$ it follows that $g(x^{-k})=(-k)\cdot g(x)$.
Thus $g(x^k)=k\cdot g(x)$ for all $k\in \mathbb{Z}$. Then
$$
    g(n)=g \left(u\cdot \prod_{\substack{p\in \mathbb{P}^*(\Phi),\\ v_p (n)\neq 0}} p^{v_p(n)} \right)=g(u)+\sum_{\substack{p\in \mathbb{P}^*(\Phi),\\ v_p (n)\neq 0}}(v_p(n)g(p))=\sum_{p\in \mathbb{P}^*(\Phi)}(v_p(n)g(p)).
$$
To prove the converse let
$$
    g(n)=\sum_{p\in \mathbb{P}^*(\Phi)}(v_p(n)g(p)).
$$
Then
\begin{align*}
    g(m\cdot n)&=\sum_{p\in \mathbb{P}^*(\Phi)}(v_p(m\cdot n)g(p))=\sum_{\substack{p\in \mathbb{P}^*(\Phi)\\ v_p(n)\neq 0 \,\text{or}\, v_p(m)\neq 0}}(v_p(m\cdot n)g(p)) \\ &= \sum_{\substack{p\in \mathbb{P}^*(\Phi)\\ v_p(n)\neq 0 \, \text{or}\, v_p(m)\neq 0}}((v_p(n)+v_p(m))g(p)) \\ &=\sum_{\substack{p\in \mathbb{P}^*(\Phi)\\ v_p(n)\neq 0}}(v_p(n)g(p))+\sum_{\substack{p\in \mathbb{P}^*(\Phi)\\ v_p(m)\neq 0}}(v_p(m)g(p))=g(n)+g(m).
\end{align*}
This means that $g$ is completely additive. \end{proof}

\begin{corollary} \label{cr2.1}
Let $\Phi$ be a UFD such that its group of units is finitely generated and $f:\Frac(\Phi)\to\,\Frac(\Phi)$ be an L-additive function. Then
$$
f(x)=x\cdot \sum_{p\in \mathbb{P}^*(\Phi),\, v_p(x)\neq 0}\frac{f(p)v_p(x)}{p}.
$$
\end{corollary}
\begin{corollary}[{\cite[Theorem 1]{2}}] \label{cr2.2}
Let $D$ be the arithmetic derivative of rational numbers. Then
$$
D(x)=x\cdot \sum_{p\in \mathbb{P},\, v_p(x)\neq 0}\frac{v_p(x)}{p}=D_{\mathbb{P}}(x).
$$
\end{corollary}
\begin{corollary} \label{cr2.3}
Let $F$ be a finite field and $f:F\to F$ be an L-additive function. Then $f(x)=0$ for all $x\in F$.
\end{corollary}
\begin{proof}
Since $F$ is a finite field, there is a
prime $p$ such that the characteristic of $F$ is $p$ and $|F|=p^{\alpha}$. It is well-known
that the group of units of $F$ is a cyclic group of order
$|F|-1$. Hence $R(F)=\emptyset$. Since every non-zero element
in $F$ has the inverse, $\mathbb{P}(F)=\emptyset$. Thus $\mathbb{P}^*(F)=\emptyset$. Now $f(x)=0$ for all $x\in F$ by Corollary \ref{cr2.1}.
\end{proof}
\begin{example}
Let us show that the arithmetic derivatives in integers, Gaussian integers, and Eisenstein integers of the same number can be different. For example, \center $D_{\mathbb{P}}(6)=5,\, D_{\mathbb{P}(\mathbb{Z}[i])}(6)=8-6i$, and $D_{\mathbb{P}(\mathbb{Z}[\omega])}(6)=\frac{15+6\omega}{1+2\omega}=3-4\sqrt{3}i$.
\end{example}

\begin{definition} \label{df3}
We shall say that a UFD $\Gamma$ of characteristic $0$ \textit{satisfies} $(*)$ if its group of units is finitely generated and for any prime $p\in \Gamma$ there is $m\in \mathbb{Z}$ with $p\mid m$.
\end{definition}

In Definition \ref{df3} by $\mathbb{Z}$ we mean the smallest subring of $\Gamma$ which contains $1$. Note that $\mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$ satisfy $(*)$, but $\mathbb{Z}[x]$ does not satisfy $(*)$.

\begin{lemma} \label{lm1}
Let $\Gamma$ satisfy $(*)$ and $f: \Frac(\Gamma) \to \Frac(\Gamma)$. If $f(x)\in \Gamma$ for all $x \in \Gamma$, then the denominator of
$\frac{f(p)}{p}$ is equal to $1$ or $p$ for all $p\in \mathbb{P}^*(\Gamma)$.
\end{lemma}
\begin{proof}
Note that $f(p)\in \Gamma$ for all prime $p \in \mathbb{P}(\Gamma)$. Then we have either $f(p)$ is divisible by $p$ and the denominator of $\frac{f(p)}{p}$ is $1$ or $f(p)$ is not divisible by $p$ and the denominator of $\frac{f(p)}{p}$ is $p$. If $p\in R(\Gamma)$, then $\frac{f(p)}{p}\in \Gamma$. Therefore its denominator is 1.
\end{proof}
Let $f^{(0)}(x)=x$ and $f^{(i)}(x)=f(f^{(i-1)}(x))$ for every $i \in \mathbb{N}$.
\begin{theorem} \label{thm3}
Let $\Gamma$ satisfy $(*), p\in \mathbb{P}(\Gamma)$ and $f:\Frac(\Gamma)\to \Frac(\Gamma)$ be an L-additive function with $f(x)\in \Gamma$ for all $x \in \Gamma$. Then
$$
v_p(f^{(k)}(x)) \geq \max\{n\in \mathbb{Z}\,|\, n \le v_p(x) \, \mathrm{and} \, p\mid n \} \, \forall x \in \Gamma.
$$
\end{theorem}
\begin{proof}
From Corollary \ref{cr2.1} it follows that
\begin{align*}
    f(x)=x \cdot \left(\frac{v_p(x)f(p)}{p}+\sum_{p_i\in \mathbb{P}^*(\Gamma),\, p_i\neq p,\, v_{p_i}(x)\neq 0}\frac{v_{p_i}(x)f(p_i)}{p_i}\right)=\\=x\cdot \left(\frac{v_p(x)f(p)}{p}+\frac{A}{B}\right)=x\cdot \left(\frac{v_p(x)f(p)B+Ap}{p\cdot B}\right).
\end{align*}

\
Since $\mathbb{Z}\subseteq \Gamma$, we may assume that $A,\, B\in \Gamma$ and $\gcd(B,p)=1$ by Lemma \ref{lm1}. Note that $v_p(x)f(p)B+Ap\in \Gamma$. Therefore $v_p(f(x))\geq v_p(x)-1$ and if $p \mid v_p(x)$, then $v_p(f(x))\geq v_p(x)$.

Since $\Gamma$ satisfies $(*)$, we see that $\{n\in \mathbb{Z}\, |\, n \le v_p(x)\, \text{and} \, p\mid n\}\neq \emptyset$. This is a set of integers bounded from above. Hence it has the greatest element $\beta=\max \{n\in \mathbb{Z} \mid n \le v_p(x)\,\text{and}\, p\mid n\}$.

Suppose that $v_p(f^{(k)}(x))<\beta$ for some $k$.
Note that $v_p(f^{(m)}(x))\in \mathbb{Z}$ and $v_p(f^{(m+1)}(x))-v_p(f^{(m)}(x))\geq -1$ for all $m$. Hence there is $n$ with $v_p(f^{(n)}(x))=\beta$ and $v_p(f^{(n+1)}(x))=\beta-1$. Since $p\mid \beta$, we have the contradiction. \end{proof}

\begin{corollary} \label{cr3.1}
Let $f\in \{D_{\mathbb{P}^*(\Gamma)},\, D_{{p}},\, D_S\}$ and $x\in \Frac(\Gamma)$. Then there are finitely many different denominators of numbers in the sequence $(f^{(k)}(x))_{k\geq 1}$.
\end{corollary}
\begin{definition} \label{df4}
A function $W:\Frac(\Gamma)\to \mathbb{Q}_+$ is called \textit{norm-like} if

\begin{enumerate}
    \item [0)] $\,\, W(x)=0\,\Leftrightarrow x=0$.
    \item [1)] $\,\, W(a)\cdot W(b)=W(ab)$ for all $a,b\in \Frac(\Gamma)$.
    \item [2)] $\,\, W(x)\in \mathbb{N} \cup \{0\}$ for all $x \in \Gamma$.
    \item [3)] $\,\, W(x)=n$ has a finite number of solutions in $\Gamma$ for all $n \in \mathbb{N}$.
\end{enumerate}
\end{definition}

\begin{lemma} \label{lm2} The absolute value of rational number $|\cdot|:\Frac(\mathbb{Z})\to \mathbb{Q}_+$, the norm of Gaussian rational $N_1:\Frac(\mathbb{Z}[i])\to \mathbb{Q}_+$ and the norm of Eisenstein rational $N_2:\Frac(\mathbb{Z}[\omega])\to \mathbb{Q}_+$ are norm-like functions.
\end{lemma}
\begin{proof} Obviously, all these functions have properties $0-2$ and the absolute value of rational number has property $3$.

Let us prove that the equation $N_1(x)=n$ has a finite number of solutions
in $\mathbb{Z}[i]$. Let $x=a+bi$. Then $N_1(x)=a^2+b^2=n$ with $a,b \in \mathbb{Z}$. So $0\le |a| \le \sqrt{n}$ and $0\le |b| \le \sqrt{n}$. Therefore $N_1(x)=n$ has a finite
number of solutions.

Let us prove that the equation $N_2(y)=n$ has a finite number of solutions in $\mathbb{Z}[\omega]$. Let $y=c+d\omega$.
Then $N_2(y)=a^2+b^2-ab=m$ with $a,b\in \mathbb{Z}$. So $(a-b)^2+ab=m$. Let $a=v+k$ and $b=v-k$. Therefore $(v+k-v+k)^2+v^2-k^2=3k^2+v^2=m$. Note that the denominators of $v$ and $k$ can be equal to $1$ or $2$. Since $0 \le |v| <\sqrt{m}$ and $0\le |k| \le \sqrt{\frac{m}{3}}$, we see that $N_2(y)=m$ has a finite number of solutions.
\end{proof}

\begin{theorem} \label{thm4}
Let $\Gamma$ satisfy $(*),\, g:\Frac(\Gamma)\to \Frac(\Gamma)$ be an L-additive function with $g(x)\in \Gamma$ for all $x \in \Gamma$ and $W$ be a norm-like function. If $x\in \Frac(\Gamma)$, then either $W(g^{(k)}(x))\to +\infty$ or the sequence $(g^{(k)}(x))_{k\geq 1}$ is periodic starting from some $k$.
\end{theorem}

\begin{proof} Let $x\in \Frac(\Gamma)$. Assume that $g^{(k)}(x)=0$ for some $k$. Then $g^{(n)}(x)=0$ for all $n\geq k$. Hence Theorem \ref{thm4} for such $x$ is proved. Now assume that $g^{(k)}(x)\neq 0$ for all $k$. Note that if $a,b\neq 0$, then
$$
W(a)=W\left(b\cdot \frac{a}{b}\right)=W(b)\cdot W\left(\frac{a}{b}\right) \Rightarrow W\left(\frac{a}{b}\right)=\frac{W(a)}{W(b)}.
$$
Let $p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ be the denominator of $x$. Then by Theorem \ref{thm3} there are $\beta_1,\, \beta_2, \ldots, \beta_n$ and $A_k \in \Gamma$ such that $g^{(k)}(x)$ can be written as $\frac{A_k}{p_1^{\beta_1}\cdot p_2^{\beta_2} \cdots p_n^{\beta_n}}$ (this fraction is not necessary irreducible).

Suppose that $W(g^{(k)}(x))\not \to +\infty$. This means that
$$
\exists A:\, \forall M\, \exists m > M:W(g^{(m)}(x))\le A.
$$
The last inequality is equivalent to $W(A_m)\le A\cdot W(p_1^{\beta_1}\cdot p_2^{\beta_2} \cdots  p_n^{\beta_n})$.

Note that $B=\lceil A \rceil \cdot W(p_1^{\beta_1}\cdot p_2^{\beta_2} \cdots  p_n^{\beta_n})\in \mathbb{N}$ and there is a finite number of $y\in \Gamma$ with $W(y) \le B$ by 3) of Definition \ref{df4}. Therefore for infinitely many $m$ the number $W(A_m)$ belongs to
the same finite set. Hence for infinitely many $m$ all $W(A_m)$ are equal by Pigeonhole principle. So there are $m_1$ and $m_2$ with $A_{m_1}=A_{m_2}$. Now $g^{(m_1)}(x)=g^{(m_2)}(x)$ where $m_2>m_1$. Thus $g^{(m_1+i\cdot (m_2-m_1))}(x)=g^{(m_1)}(x)$ and $m_2-m_1=T$ is a period of $g^{(k)}(x)$. \end{proof}
\begin{corollary} \label{cr4.1}
Let $f\in \{D_{\mathbb{P}^*(\Gamma)},\, D_{p},\, D_S\}$.

\begin{enumerate}
    \item [(a)] If $x\in \mathbb{Q}$, then either $|f^{(k)}(x)|\to +\infty$ or $(f^{(k)}(x))_{k\geq 1}$ is periodic starting from some natural number $k$.
    \item [(b)] If $x\in \mathbb{Q}(i)$, then either $N_1(f^{(k)}(x))\to + \infty$ or $(f^{(k)}(x))_{k\geq 1}$ is periodic starting from some natural number $k$.
    \item [(c)] If $x\in \mathbb{Q}(\omega)$, then either $N_2(f^{(k)}(x))\to + \infty$ or $(f^{(k)}(x))_{k\geq 1}$ is periodic starting from some natural number $k$.
\end{enumerate}
\end{corollary}
\begin{theorem} \label{thm5}
Let $\Phi$ be a UFD such  that  its  group  of  units  is  finitely  generated, $\alpha \in \Frac(\Phi)$ and $f:\Frac(\Phi)\to \Frac(\Phi)$ be an L-additive function.

\begin{enumerate}
    \item Let $x_0$ be a non-zero solution of \begin{equation} \label{f(x)}
    f(x)=\alpha x
\end{equation}
Then $x=y\cdot x_0$ for any solution $x$ of \eqref{f(x)} where $y$ is a solution of $f(y)=0$.
    \item Let $S=\{p\in \mathbb{P}^*(\Phi)\,|\, f(p)\neq 0\}$. If $v_p(x)=0$ for all $p\in S$, then $f(x)=0$.
    \item There exists a bijection between non-zero solutions $l$ of \eqref{f(x)} with $P=\{p\mid v_p(l)\neq 0\}\subseteq S$ and integer solutions $\{v_p(x)\mid p\in \mathbb{P}^*(\Phi)\}$ of
$$
\alpha = \sum_{p\in \mathbb{P}^*(\Phi)}\left(v_p(x)\cdot \frac{f(p)}{p}\right).
$$
    \item Assume that $\Frac(\Phi)=\mathbb{Q}$. Let $\frac{f(p)}{p}=\frac{c_p}{z_p}$ be an irreducible fraction for all $p\in S$. The equation \eqref{f(x)} has a solution $x$ iff $\beta =\alpha \cdot \delta \in \mathbb{Z}$ is divisible by $\gcd\left(\frac{\delta f(p)}{p}\, |\, p\in P\right)$ where $P=\{p\in \mathbb{P}\, |\, v_p(x)\neq 0\}$ and $\delta = \lcm(z_p\,|\,p\in P)$.
\end{enumerate}
\end{theorem}

\begin{proof}
\begin{enumerate}
    \item Let $x=x_0\cdot y$, with $f(y)=0$.
Since $f$ is an L-additive function, we see that $f(x)=f(y\cdot x_0)=f(y)x_0+f(x_0)y=f(x_0)y=(\alpha x_0)y=\alpha x$. Hence $x$ is a  solution of \eqref{f(x)}.

Assume now that $x_1$ is a non-zero solution of \eqref{f(x)}.
Let us show that $x_1=y\cdot x_0$ with $f(y)=0$. Let $y=\frac{x_1}{x_0}$. Then $f(y)=f\left(\frac{x_1}{x_0}\right)=\frac{f(x_1)x_0-f(x_0)x_1}{x_0^2}=\frac{\alpha x_1x_0-\alpha x_0 x_1}{x_0^2}=0$.
    \item By Corollary \ref{cr2.1} we have that
$$
f(x)=x\cdot \sum_{p\in \mathbb{P}^*(\Phi),\, v_p(x)\neq 0}\frac{f(p)v_p(x)}{p}=0.
$$
    \item By Corollary \ref{cr2.1} we have that
$$
f(l)=\alpha \cdot l=l\cdot \sum_{p\in \mathbb{P}^*(\Phi),\, v_p(l)\neq 0}\frac{f(p)v_p(l)}{p}\Leftrightarrow \alpha = \sum_{p\in \mathbb{P}^*(\Phi),\, v_p(l)\neq 0}v_p(l)\cdot \frac{f(p)}{p}.
$$
    \item By \textit{3)} we have that
$$
\alpha=\frac{f(x)}{x}=\sum_{p\in P}\left(v_p(x)\frac{f(p)}{p}\right)=\sum_{p\in P}\frac{v_p(x)c_p}{z_p}.
$$
Note that
$$
\beta=\alpha \cdot \delta =\sum_{p\in P}\left(\frac{c_p\delta}{z_p}\cdot v_p(x)\right)=\sum_{p\in P}\left(\frac{f(p)\delta}{p}\cdot v_p(x)\right)\in \mathbb{Z}.
$$
This equation is a linear Diophantine equation. It is well-known that this equation has a solution iff $\beta$ is divisible by $\gcd\left(\frac{\delta f(p)}{p}\,|\,p\in P\right)$. Then we know all $v_p(x)$. Hence we know $\pm x$.
\end{enumerate}
\end{proof}

\begin{corollary}[{\cite[Theorem 3]{10}}] \label{cr5.1}
Let $p\in \mathbb{P}$ and $\alpha \in \mathbb{Q}$. The equation $D_p(x)=\alpha x$ has a nontrivial solution iff $\alpha p\in \mathbb{Z}$. Then all nontrivial solutions are of the form $x=cp^{\alpha p}$, where $p\not|\,c\in \mathbb{Q}\setminus\{0\}$. Conversely, all numbers of this from are nontrivial solutions.
\end{corollary}

\begin{corollary}[{\cite[Theorem 18]{2}}] \label{cr5.2}
Let $\alpha =\frac{a}{b}$ be a rational number with $\gcd(a,b)=1,\, b>0$. Then the equation $D(x)=\alpha x$ has non-zero rational solutions iff $b$ is a product of different primes or $b=1$.
\end{corollary}
\begin{theorem} \label{thm6}
Let a field $F$ be a finite algebraic extension of $\mathbb{Q}$. If $f:F\to F$
is an L-additive linear function, then $f(x)\equiv 0$.
\end{theorem}
\begin{proof}
According to Artin's theorem on primitive elements, every finite algebraic extension of $\mathbb{Q}$ is simple, i.e., there exists $\alpha \in F$ with $F=\mathbb{Q}(\alpha)$. Let $g$ be a minimal polynomial of $\alpha$ over $\mathbb{Q}$. We may assume that all coefficients of $g$ are integer. It is known that $g$ does not have multiple
zeros.

Since $f(xy)=f(x)y+f(y)x$ and $f(x+y)=f(x)+f(y)$, we see that
$$
f(ny)=\underbrace{f(y)+f(y)+\dots+f(y)}_{n}=n\cdot f(y)\,\, \text{and} \,\, f(ny)=f(n)y+f(y)n.
$$
Therefore $f(n)y=0$. Hence $f(n)=0$ for all $n\in \mathbb{N}$. Note that $f(-n)=-f(n)=0$. Thus $f(x)=0$ for all $x\in \mathbb{Z}$.

Let $g(x)=a_n x^n +\dots+a_1 x+a_0$. Then
$$
0=f(0)=f(g(\alpha))=\sum_{i=0}^n f(a_i \alpha ^i)=
\sum_{i=0}^n a_i f(\alpha ^i)=\sum_{i=1}^n a_i (i \alpha^{i-1})f(\alpha)=g'(\alpha) f(\alpha).
$$

Since $g'(\alpha)\neq 0$, we see that $f(\alpha)=0$. Therefore $f(x)=0$ for every $x\in \mathbb{Z}[\alpha]$. It is easy to see that for every $x\in F$ there are $m\in \mathbb{Z}[\alpha]$ and $n\in \mathbb{Z}$ with $x=\frac{m}{n}$. So $f(x)=0$ by $3$ of Proposition \ref{pr1}. Thus $f(x)=0$ for all $x\in \mathbb{Q}(\alpha)=F$.
\end{proof}

\section{Final remarks}

In this paper we studied L-additive functions over unique factorization
domains. So it is natural to ask the following question.

\begin{question}
Let $J$ be a factorization domain which is not UFD. Are there any non-zero L-additive functions over $J$?
\end{question}
Haukkanen et al.\ \cite{5} discussed some ideas about this question.

According to Theorem \ref{thm6} there are no L-additive linear non-zero functions over any finite extension of $\mathbb{Q}$.

\begin{question}
Describe all UFD $J$ such that there are no L-additive linear non-zero functions over $J$.
\end{question}
By Corollary \ref{cr2.3} there are no non-zero L-additive functions over a finite field. According to Emmons et al.\ \cite{6} every value of an L-additive function over $\mathbb{Z}_n$ is a divisor of zero.

\begin{question}
Let $f$ be an L-additive function over a finite ring. Is it true that all values of $f$ are divisors of zero?
\end{question}

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\noindent 2010 \textit{Mathematics Subject Classification.} Primary 11A25; Secondary 11A51, 11R27.

\noindent \textit{Keywords:} Leibniz-additive function, arithmetic derivative,
completely multiplicative function, completely additive function, unique
factorization domain (UFD), field of fractions.

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\noindent (Concerned with sequences \seqnum{A000040} and \seqnum{A003415}.)

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\noindent
Received June 12 2020;
revised versions received  August 22 2020; August 25 2020.
Published in {\it Journal of Integer Sequences}, October 17 2020.

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