Leibniz-Additive Functions on UFD’s

Viachaslau I. Murashka Faculty of Mathematics and Technologies of Programming Francisk Skorina Gomel State University Gomel 246019 Belarus

                   mvimath@yandex.ru

Andrey D. Goncharenko and Irina N. Goncharenko State Educational Establishment “Gymnasium 71” Gomel 246036 Belarus

             goncharenkoandrey8@gmail.com
                    ira nika@tut.by

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≥1} and solutions of the equation f(x) = αx. As corollaries we obtain results about the arithmetic derivative and partial arithmetic derivatives.

1 Introduction

Recall [2, 12, 13] that the arithmetic derivative is a function D : ℕ → ℕ, such that 1. D(p) = 1 for all prime p;

2. D satisfies the Leibniz rule: D(mn) = D(m)n + D(n)m.

It is known that the arithmetic derivative is not a linear function:

D(2 + 3) = D(5) = 1 and D(2) + D(3) = 2.

One can define the arithmetic derivative for a rational number [13]. Since D is not a linear function, it is difficult to solve even equations D(x) = 2a and D(D(x)) = 1. Ufnarovski and Åhlander [13] 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-Marcén [4].

Note that the partial arithmetic derivative D_{p}(n) and the arithmetic subderivative D_{S}(n) of an integer number satisfy the Leibniz rule [11].

Also the analogues of arithmetic derivative were considered over different sets of elements [13]. Haukkanen et al. [7] discussed the arithmetic derivative on non-unique factorization domains. Emmons et al. [3] described all functions f : ℤ_{n} → ℤ_{n} that satisfy the Leibniz rule. In particular, all values of such functions are divisors of zero. Kovic̆ [10] constructed functions defined on Gaussian rationals that satisfy the Leibniz rule.

Let h be a completely multiplicative function. According to Merikoski et al. [9, 11] a function f : ℕ → 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 ∈ 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.).

2 Preliminary results

The notation and terminology agree with the book [1]. We refer the reader to this book for the results on ring theory. Through ℕ, ℙ, ℤ, ℤ_{+}, ℚ, and ℚ_{+} we denote here the sets of natural, prime, integer, non-negative integer, rational, and non-negative rational numbers respectively.

Let Φ be an integral domain [1, p. 368]. Recall that the field of fractions Frac(Φ) of Φ is the set \{\frac{a}{b} | a ∈ Φ, b ∈ Φ \ \{0\} with \frac{a}{b} = \frac{c}{d} if ∃k ≠ 0 such that a = kc and b = kd\} with two operations:

\frac{a}{b} + \frac{c}{d} = \frac{a · d + c · b}{b · d} and \frac{a}{b} · \frac{c}{d} = \frac{a · c}{b · d} ∀a, b, c, d ∈ Φ and b, d ≠ 0.

Note that Frac(ℤ) = ℚ and Frac(ℤ[i]) = ℚ(i).

If p is a prime in a UFD, then all elements associated with it are also primes. Let ℙ(Φ) be some maximal by inclusion set of non-associated primes in Φ. 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 Φ has the following unique factorization:

y = u · ∏_{p∈ℙ(Φ)} p^{v_{p}(y)}

where u is a unit, v_{p}(y) ∈ ℤ_{+}, and v_{p}(y) ≠ 0 for finitely many p ∈ ℙ(Φ).

Recall that the characteristic of an integral domain Φ 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 Φ 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 Φ be a UFD such that its group of units G is finitely generated. Then

G ≃ A_{1} × A_{2} × ⋯ × A_{n} × B_{1} × B_{2} × ⋯ × B_{k}

with n, k ≥ 0, A_{i} ≃ ℤ, and B_{j} ≃ ℤ_{n_{j}} where n_{j} is a power of a prime for all j. Let A(Φ) be the set containing exactly one generating element a_{i} of each subgroup A_{i} and B(Φ) 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 Φ. If the characteristic of Φ is equal to 0, then B(Φ) is empty. Define R(Φ) = A(Φ) ∪ B(Φ).

If the characteristic of Φ is equal to 0, then all its units u can be uniquely represented

u = u_{0} · ∏_{i=1}^{n} a_{i}^{γ_{i}}

with γ_{i} ∈ ℤ and u_{0} is an element of a finite order.

If the characteristic of Φ is equal to p ≠ 0, then all its units u can be uniquely represented

u = u_{0} · ∏_{i=1}^{n} a_{i}^{γ_{i}} · ∏_{j=1}^{m} b_{j}^{δ_{j}}

with γ_{i} ∈ ℤ, δ_{j} ∈ {0, 1, …, n_{j} − 1}, and u_{0} is an element of a finite order t with gcd(t, p) = 1.

Let ℙ^{∗}(Φ) = ℙ(Φ) ∪ R(Φ). So the factorization of an element q ≠ 0 in Φ can be written uniquely in the following form

q = u_{0} · ∏_{r∈R(Φ)} r^{v_{r}(q)} · ∏_{p∈ℙ(Φ)} p^{v_{p}(q)} = u_{0} · ∏_{p∈ℙ^{∗}(Φ)} p^{v_{p}(q)}.

This means that every element q ≠ 0 of Frac(Φ) can be written uniquely in the following form

q = u_{0} · ∏_{p∈ℙ^{∗}(Φ)} p^{v_{p}(q)}

where v_{p}(q) ∈ ℤ for all p ∈ ℙ^{∗}(Φ) and v_{p}(q) ≠ 0 only for a finite number of p ∈ ℙ^{∗}(Φ).

Let m, n ∈ Frac(Φ) \ {0}. It is easy to check that v_{p}(m · n) = v_{p}(m) + v_{p}(n) for any p ∈ ℙ(Φ). Note that v_{a_{i}}(m · n) = v_{a_{i}}(m) + v_{a_{i}}(n). If characteristic of Φ is p, then v_{b_{i}}(m · n) ≡ v_{b_{i}}(m) + v_{b_{i}}(n) (mod n_{i}), where n_{i} = p^{α} for some α. This means that v_{b_{i}}(m · n) ≡ v_{b_{i}}(m) + v_{b_{i}}(n) (mod p). All these mean that v_{r}(m · n) = v_{r}(m) + v_{r}(n) in Frac(Φ).

A Gaussian integer is a complex number a + bi with a, b ∈ 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 ℤ[i] is the cyclic group of order 4. So R(ℤ[i]) = ∅. We can chose ℙ^{∗}(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 ∈ ℕ and a^{2}+b^{2} is a natural prime of the form 4k+1. Then ℙ^{∗}(ℤ[i]) = {1+i, 3, 1+2i, 1−2i, 7, 2+3i, 2−3i, 11, …}.

An Eisenstein integer is a complex number a + bω with a, b ∈ ℤ and ω = \frac{−1+i\sqrt{3}}{2} is a solution of the equation ω^{2} + ω + 1 = 0. Note that Eisenstein integers form a UFD. Any Eisenstein prime divides some integer prime, with 3 = (1+2ω)^{2} where 1+2ω is an Eisenstein prime. An integer prime p ≡ 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 ℤ[ω] is the cyclic group of order 6. So R(ℤ[ω]) = ∅. We can chose ℙ^{∗}(ℤ[ω]) in the following way: an integer prime p ≡ 2 (mod 3); 1 + 2ω and numbers of the form a + bω and a + bω^{2} with a ∈ ℕ, |a| < |b| and a^{2} − ab + b^{2} is a natural prime number p ≡ 1 (mod 3). Then ℙ^{∗}(ℤ[ω]) = {2, 1 + 2ω, 5, 1 + 3ω, 1 + 3ω^{2}, 11, …}.

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 + bω) = a^{2} − ab + b^{2}.

Recall that ℤ[\sqrt{2}] = \{a + b\sqrt{2} | a, b ∈ ℤ\} is a UFD. Note that its group of units is generated by \{−1, \sqrt{2} + 1\} and is isomorphic to ℤ_{2} × ℤ. So in this case we can choose R(ℤ[\sqrt{2}]) = \{\sqrt{2} + 1\}.

3 Main results

The following well-known function classes play an important role in our paper.

Definition 1. Let K be a ring. A function h : K → K is called

(1) Completely multiplicative, if h(1) = 1 and h(mn) = h(m)h(n) holds for all m, n ∈ K.

(2) Completely additive, if h(mn) = h(m) + h(n) holds for all m, n ∈ K.

Definition 2 ([11]). Let K be a ring and h : K → K be a completely multiplicative function. A function f : K → K is called L-additive with respect to h if f(mn) = f(m)h(n)+f(n)h(m) holds for all m, n ∈ K.

If h(n) ≡ 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 L-additive.

Theorem 3. Let Φ be an integral domain and h, f : Φ → Φ be completely multiplicative and L-additive with respect to h functions. Then there exist unique functions h̄, f̄: Frac(Φ) → Frac(Φ), such that h̄ is completely multiplicative, f̄ is L-additive with respect to h̄, and h(x) = h̄(x) and f(x) = f̄(x) ∀x ∈ Φ iff h(x) ≠ 0 ∀x ∈ Φ \ {0}.

Let q : Frac(Φ) → Frac(Φ) be completely multiplicative. In these terms a function V : Frac(Φ) → Frac(Φ) is L-additive with respect to q where V (x) = \frac{f̄(x)·q(x)}{h̄(x)} for x ≠ 0 and V (0) = 0.

Proof. Assume that h(x) ≠ 0 ∀x ∈ Φ \ {0}. Then for all x ∈ Frac(Φ) there exist m, n ∈ Φ with x = \frac{m}{n}, n ≠ 0. Let h̄(x) = \frac{h(m)}{h(n)} and f̄(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 ≠ 0 such that m_{1} = km_{2} and n_{1} = kn_{2}. The following equality shows that h̄ is well-defined:

h̄ (\frac{m_{1}}{n_{1}} ) = \frac{h(m_{1})}{h(n_{1})} = \frac{h(km_{2})}{h(kn_{2})} = \frac{h(k) · h(m_{2})}{h(k) · h(n_{2})} = h̄ (\frac{m_{2}}{n_{2}} ) .

Let x = x_{1} · 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} · m_{2} · k and n = n_{1} · n_{2} · k. The following equality shows that h̄ is completely multiplicative:

h̄(x_{1}) · h̄(x_{2}) = \frac{h(m_{1})}{h(n_{1})} · \frac{h(m_{2})}{h(n_{2})} · \frac{h(k)}{h(k)} = \frac{h(k · m_{1} · m_{2})}{h(k · n_{1} · n_{2})} = \frac{h(m)}{h(n)} = h̄(x).

Let ḡ be a completely multiplicative extension of h. Then

1 = ḡ(1) = ḡ(x · x^{−1}) = ḡ(x) · ḡ(x^{−1}).

Therefore ḡ(x^{−1}) = \frac{1}{ḡ(x)} = \frac{1}{h(x)} ∀x ∈ Φ \ \{0\}. The following shows that h̄ is unique:

ḡ(x) = ḡ (\frac{m}{n}) = ḡ(m) · ḡ(n^{−1}) = \frac{h(m)}{h(n)} = h̄(x).

The following equality shows that f̄ is well-defined:

f̄(\frac{km}{kn}) = \frac{f(km) · h(kn) − f(kn) · h(km)}{h^{2}(kn)}

= \frac{(f(k)h(m) + f(m)h(k)) · h(k)h(n) − (f(k)h(n) + f(n)h(k)) · 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)} = f̄(x).

Let x = x_{1} · 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} · m_{2} · k and n = n_{1} · n_{2} · k. The following equality shows that f̄ is L-additive with respect to h̄:

h̄(x_{1})f̄(x_{2}) + h̄(x_{2})f̄(x_{1}) = h̄ (\frac{m_{1}}{n_{1}} ) f̄(\frac{m_{2}}{n_{2}} ) + h̄ (\frac{m_{2}}{n_{2}} ) f̄(\frac{m_{1}}{n_{1}} )

= \frac{h(m_{1})}{h(n_{1})} · (\frac{f(m_{2})h(n_{2}) − f(n_{2})h(m_{2})}{h^{2}(n_{2})}) + \frac{h(m_{2})}{h(n_{2})} · (\frac{f(m_{1})h(n_{1}) − f(n_{1})h(m_{1})}{h^{2}(n_{1})})

= \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}) · 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}) · h^{2}(n_{2})}

= \frac{h(n_{1}n_{2})(h(m_{1})f(m_{2}) + h(m_{2})f(m_{1}))}{h^{2}(n_{1}n_{2})} − \frac{h(m_{1}m_{2})(h(n_{1})f(n_{2}) + h(n_{2})f(n_{1}))}{h^{2}(n_{1}n_{2})}

= \frac{h(n_{1}n_{2})f(m_{1}m_{2}) − h(m_{1}m_{2})f(n_{1}n_{2})}{h^{2}(n_{1}n_{2})} = f̄(\frac{m_{1} · m_{2}}{n_{1} · n_{2}} ) = f̄(\frac{k · m_{1} · m_{2}}{k · n_{1} · n_{2}} ) = f̄(x).

Let ḡ be an L-additive with respect to h̄ extension of f̄:

0 = ḡ(1) = ḡ(xx^{−1}) = ḡ(x)h̄(x^{−1}) + ḡ(x^{−1})h̄(x) = \frac{f(x)}{h(x)} + ḡ(x^{−1})h(x).

Thus ḡ(x^{−1}) = −\frac{f(x)}{h^{2}(x)}∀x ∈ Φ \ \{0\} and

ḡ (\frac{m}{n}) = ḡ(m)h̄ (\frac{1}{n})+ḡ (\frac{1}{n}) h̄(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)} = f̄(\frac{m}{n}) .

Therefore the function f̄ is unique.

Let us prove the converse statement. Assume that f and h can be extended to Frac(Φ). Suppose that there exists x ∈ Φ \ {0} with h(x) = 0. Then x^{−1} ∈ Frac(Φ). Therefore 1 = h̄(x · x^{−1}) = h̄(x) · h̄(x^{−1}) = 0. This is a contradiction.

Let us prove that V (x) is L-additive with respect to q. Let m, n ∈ Frac(Φ). Assume that m, n ≠ 0. Then

V (mn) = \frac{f̄(mn) · q(mn)}{h̄(mn)} = \frac{(f̄(m)h̄(n) + f̄(n)h̄(m)) q(mn)}{h̄(m) · h̄(n)}

= \frac{f̄(m)h̄(n)q(mn)}{h̄(m)h̄(n)} + \frac{f̄(n)h̄(m)q(mn)}{h̄(m)h̄(n)} = \frac{f̄(m)q(m)q(n)}{h̄(m)} + \frac{f̄(n)q(n)q(m)}{h̄(n)}

= V (m)q(n) + V (n)q(m).

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.

Example 4. Let

h(n) = \begin{cases}1, & n = 1; \\ 0, & n ≠ 1; \end{cases} and f(n) = \begin{cases}1, & n ∈ ℙ; \\ 0, & n ∉ ℙ. \end{cases}

Note that f is L-additive with respect to h and we cannot extend f and h from ℤ to ℚ.

Corollary 5. A function f : Frac(Φ) → Frac(Φ) is L-additive iff f(0) = 0 and \frac{f(x)}{x} is completely additive.

Proof. Since every completely additive function is L-additive with respect to h(x) ≡ 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 3.

Corollary 6 ([6, Theorem 3.1]). A function D_{S}^{f}(n) is L-additive iff ∑_{p∈S} \frac{f_{p}(n)}{p} is completely additive where D_{S}^{f}(n) is as defined in [6].

Proposition 7. Let Φ be an integral domain and f, g : Frac(Φ) → Frac(Φ) be L-additive functions. Then

1. h(x) = α · f(x) is L-additive.

2. h(x) = f(x) + g(x) is L-additive.

3. f (\frac{a}{b}) = \frac{f(a)b−f(b)a}{b^{2}}.

4. f(a^{n}) = n · a^{n−1} · f(a), ∀n ∈ ℤ.

Proof. 1. Let m, n ∈ Frac(Φ). Then f(mn) = f(m)n + f(n)m. So h(mn) = α · (f(n)m + f(m)n) = (α · f(n))m + (α · f(m))n = h(n)m + h(m)n. Hence h(x) = α · f(x) is L-additive.

2. Let m, n ∈ Frac(Φ). 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) is L-additive.

3. From f(1) = 0 it follows that f(1) = f (n · \frac{1}{n}) = \frac{f(n)}{n} + f (\frac{1}{n}) n = 0. So f (\frac{1}{n}) = −\frac{f(n)}{n^{2}}. Thus f (\frac{a}{b}) = \frac{f(a)}{b} + f (\frac{1}{b}) a = \frac{f(a)b−f(b)a}{b^{2}}.

4. Let us prove this statement by induction. Note that 0 = f(1) = f(a^{0}) = 0a^{−1}f(a) and f(a) = 1a^{0}f(a). Assume that the statement holds for n ∈ ℕ. Let us prove this statement for n + 1 : f(a^{n+1}) = f(a · 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} ·a^{−n}) = f(a^{n})a^{−n} + a^{n}f(a^{−n}) = \frac{nf(a)}{a} + a^{n}f(a^{−n}) and therefore f(a^{−n}) = −n · a^{−(n+1)}f(a).

Let S ⊆ ℙ^{∗}(Φ). We shall call the function

D_{S}(x) = x · ∑_{p∈S} \frac{v_{p}(x)}{p}

an arithmetic subderivative in Φ. If |S| = 1, then we shall call it a partial arithmetic derivative and if S = ℙ^{∗}(Φ), then we shall call it the arithmetic derivative in Φ. These functions for integers were studied, for example, by Merikoski et al. [11]. Note that the function D_{S} is also referred to as “arithmetic type derivative” [5].

Corollary 8. Let Φ be a UFD such that its group of units is finitely generated and S ⊆ ℙ^{∗}(Φ). Then D_{S}(x) is L-additive.

Proof. Recall that v_{p}(x) is completely additive. Then x · v_{p}(x) is a L-additive function by Corollary 5. Since v_{p}(x) ≠ 0 only for finite number of p ∈ ℙ^{∗}(Φ), D_{S}(x) = x · ∑_{p∈S} \frac{v_{p}(x)}{p} is L-additive by 1 and 2 of Proposition 7.

Theorem 9. Let Φ be a UFD such that its group of units is finitely generated. A function g : Frac(Φ) \ {0} → Frac(Φ) is completely additive iff

g(n) = ∑_{p∈ℙ^{∗}(Φ)} (v_{p}(n)g(p)) ∀n ∈ Frac(Φ).

Proof. Let g be a completely additive function. Note that for all n ∈ Frac(Φ) holds

n = u_{0} · ∏_{p∈ℙ^{∗}(Φ)} p^{v_{p}(n)},

where u_{0} is a unit of Φ of a finite order k (if char(Frac(Φ)) = p ≠ 0, then gcd(k, p) = 1). Note that g(1 · 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 Φ is an integral domain, we see that g(u_{0}) = 0.

Since g is completely additive, it is clear that g(x^{k}) = k · g(x) for all k ∈ ℕ. We proved that g(x^{0}) = g(1) = 0 = 0 · g(x). From 0 = g(1) = g(x^{k} · x^{−k}) = g(x^{k}) + g(x^{−k}) it follows that g(x^{−k}) = (−k) · g(x). Thus g(x^{k}) = k · g(x) for all k ∈ ℤ. Then

g(n) = g (u · ∏_{\begin{matrix}p∈ℙ^{∗}(Φ), \\ v_{p}(n)≠0 \end{matrix}} p^{v_{p}(n)}) = g(u) + ∑_{\begin{matrix}p∈ℙ^{∗}(Φ), \\ v_{p}(n)≠0 \end{matrix}}(v_{p}(n)g(p)) = ∑_{p∈ℙ^{∗}(Φ)}(v_{p}(n)g(p)).

To prove the converse let

g(n) = ∑_{p∈ℙ^{∗}(Φ)}(v_{p}(n)g(p)).

Then

g(m · n) = ∑_{p∈ℙ^{∗}(Φ)}(v_{p}(m · n)g(p)) = ∑_{\begin{matrix}p∈ℙ^{∗}(Φ) \\ v_{p}(n)≠0 or v_{p}(m)≠0 \end{matrix}}(v_{p}(m · n)g(p))

= ∑_{\begin{matrix}p∈ℙ^{∗}(Φ) \\ v_{p}(n)≠0 or v_{p}(m)≠0 \end{matrix}}((v_{p}(n) + v_{p}(m))g(p))

= ∑_{\begin{matrix}p∈ℙ^{∗}(Φ) \\ v_{p}(n)≠0 \end{matrix}}(v_{p}(n)g(p)) + ∑_{\begin{matrix}p∈ℙ^{∗}(Φ) \\ v_{p}(m)≠0 \end{matrix}}(v_{p}(m)g(p)) = g(n) + g(m).

This means that g is completely additive.

Corollary 10. Let Φ be a UFD such that its group of units is finitely generated and f : Frac(Φ) → Frac(Φ) be an L-additive function. Then

f(x) = x · ∑_{p∈ℙ^{∗}(Φ), v_{p}(x)≠0} \frac{f(p)v_{p}(x)}{p}.

Corollary 11 ([13, Theorem 1]). Let D be the arithmetic derivative of rational numbers. Then

D(x) = x · ∑_{p∈ℙ, v_{p}(x)≠0} \frac{v_{p}(x)}{p} = D_{ℙ}(x).

Corollary 12. Let F be a finite field and f : F → F be an L-additive function. Then f(x) = 0 for all x ∈ F.

Proof. Since F is a finite field, there is a prime p such that the characteristic of F is p and |F | = p^{α}. It is well-known that the group of units of F is a cyclic group of order |F | − 1. Hence R(F) = ∅. Since every non-zero element in F has the inverse, ℙ(F) = ∅. Thus ℙ^{∗}(F) = ∅. Now f(x) = 0 for all x ∈ F by Corollary 10.

Example 13. Let us show that the arithmetic derivatives in integers, Gaussian integers, and Eisenstein integers of the same number can be different. For example,

D_{ℙ}(6) = 5, D_{ℙ(ℤ[i])}(6) = 8 − 6i, and D_{ℙ(ℤ[ω])}(6) = \frac{15+6ω}{1+2ω} = 3 − 4\sqrt{3}i.

Definition 14. We shall say that a UFD Γ of characteristic 0 satisfies (∗) if its group of units is finitely generated and for any prime p ∈ Γ there is m ∈ ℤ with p | m.

In Definition 14 by ℤ we mean the smallest subring of Γ which contains 1. Note that ℤ[i] and ℤ[ω] satisfy (∗), but ℤ[x] does not satisfy (∗).

Lemma 15. Let Γ satisfy (∗) and f : Frac(Γ) → Frac(Γ). If f(x) ∈ Γ for all x ∈ Γ, then the denominator of \frac{f(p)}{p} is equal to 1 or p for all p ∈ ℙ^{∗}(Γ).

Proof. Note that f(p) ∈ Γ for all prime p ∈ ℙ(Γ). 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 ∈ R(Γ), then \frac{f(p)}{p} ∈ Γ. Therefore its denominator is 1.

Let f^{(0)}(x) = x and f^{(i)}(x) = f(f^{(i−1)}(x)) for every i ∈ ℕ.

Theorem 16. Let Γ satisfy (∗), p ∈ ℙ(Γ) and f : Frac(Γ) → Frac(Γ) be an L-additive function with f(x) ∈ Γ for all x ∈ Γ. Then

v_{p}(f^{(k)}(x)) ≥ max{n ∈ ℤ | n ≤ v_{p}(x) and p | n} ∀x ∈ Γ.

Proof. From Corollary 10 it follows that

f(x) = x · (\frac{v_{p}(x)f(p)}{p} + ∑_{p_{i}∈ℙ^{∗}(Γ), p_{i}≠p, v_{p_{i}}(x)≠0} \frac{v_{p_{i}}(x)f(p_{i})}{p_{i}} ) =

= x · (\frac{v_{p}(x)f(p)}{p} + \frac{A}{B} ) = x · (\frac{v_{p}(x)f(p)B + Ap}{p · B}) .

Since ℤ ⊆ Γ, we may assume that A, B ∈ Γ and gcd(B, p) = 1 by Lemma 15. Note that v_{p}(x)f(p)B +Ap ∈ Γ. Therefore v_{p}(f(x)) ≥ v_{p}(x)−1 and if p | v_{p}(x), then v_{p}(f(x)) ≥ v_{p}(x).

Since Γ satisfies (∗), we see that {n ∈ ℤ | n ≤ v_{p}(x) and p | n} ≠ ∅. This is a set of integers bounded from above. Hence it has the greatest element β = max{n ∈ ℤ | n ≤ v_{p}(x) and p | n}.

Suppose that v_{p}(f^{(k)}(x)) < β for some k. Note that v_{p}(f^{(m)}(x)) ∈ ℤ and v_{p}(f^{(m+1)}(x)) − v_{p}(f^{(m)}(x)) ≥ −1 for all m. Hence there is n with v_{p}(f^{(n)}(x)) = β and v_{p}(f^{(n+1)}(x)) = β − 1. Since p | β, we have the contradiction.

Corollary 17. Let f ∈ {D_{ℙ^{∗}(Γ)}, D_{p}, D_{S}} and x ∈ Frac(Γ). Then there are finitely many different denominators of numbers in the sequence (f^{(k)}(x))_{k≥1}.

Definition 18. A function W : Frac(Γ) → ℚ_{+} is called norm-like if

0) W (x) = 0 ⇔ x = 0.

1) W (a) · W (b) = W (ab) for all a, b ∈ Frac(Γ).

2) W (x) ∈ ℕ ∪ {0} for all x ∈ Γ.

3) W (x) = n has a finite number of solutions in Γ for all n ∈ ℕ.

Lemma 19. The absolute value of rational number |·| : Frac(ℤ) → ℚ_{+}, the norm of Gaussian rational N_{1} : Frac(ℤ[i]) → ℚ_{+} and the norm of Eisenstein rational N_{2} : Frac(ℤ[ω]) → ℚ_{+} are norm-like functions.

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 ℤ[i]. Let x = a + bi. Then N_{1}(x) = a^{2} + b^{2} = n with a, b ∈ ℤ. So 0 ≤ |a| ≤ \sqrt{n} and 0 ≤ |b| ≤ \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 ℤ[ω]. Let y = c + dω. Then N_{2}(y) = a^{2} + b^{2} − ab = m with a, b ∈ 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 ≤ |v| < \sqrt{m} and 0 ≤ |k| ≤ \sqrt{\frac{m}{3}}, we see that N_{2}(y) = m has a finite number of solutions.

Theorem 20. Let Γ satisfy (∗), g : Frac(Γ) → Frac(Γ) be an L-additive function with g(x) ∈ Γ for all x ∈ Γ and W be a norm-like function. If x ∈ Frac(Γ), then either W (g^{(k)}(x)) → +∞ or the sequence (g^{(k)}(x))_{k≥1} is periodic starting from some k.

Proof. Let x ∈ Frac(Γ). Assume that g^{(k)}(x) = 0 for some k. Then g^{(n)}(x) = 0 for all n ≥ k. Hence Theorem 20 for such x is proved. Now assume that g^{(k)}(x) ≠ 0 for all k. Note that if a, b ≠ 0, then

W (a) = W (b · \frac{a}{b}) = W (b) · W (\frac{a}{b}) ⇒ W (\frac{a}{b}) = \frac{W (a)}{W (b)}.

Let p_{1}^{α_{1}} · p_{2}^{α_{2}} ⋯ p_{n}^{α_{n}} be the denominator of x. Then by Theorem 16 there are β_{1}, β_{2}, …, β_{n} and A_{k} ∈ Γ such that g^{(k)}(x) can be written as _{\frac{A_{k}}{p_{1}^{β_{1}}·p_{2}^{β_{2}}···p_{n}^{βn}}} (this fraction is not necessary

irreducible).

Suppose that W (g^{(k)}(x)) ↛ +∞. This means that

∃A : ∀M ∃m > M : W (g^{(m)}(x)) ≤ A.

The last inequality is equivalent to W (A_{m}) ≤ A · W (p_{1}^{β_{1}} · p_{2}^{β_{2}} ⋯ p_{n}^{β_{n}}).

Note that B = ⌈A⌉ · W (p_{1}^{β_{1}} · p_{2}^{β_{2}} ⋯ p_{n}^{β_{n}}) ∈ ℕ and there is a finite number of y ∈ Γ with W (y) ≤ B by 3) of Definition 18. 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·(m_{2}−m_{1}))}(x) = g^{(m_{1})}(x) and m_{2} − m_{1} = T is a period of g^{(k)}(x).

Corollary 21. Let f ∈ {D_{ℙ^{∗}(Γ)}, D_{p}, D_{S}}.

(a) If x ∈ ℚ, then either |f^{(k)}(x)| → +∞ or (f^{(k)}(x))_{k≥1} is periodic starting from some natural number k.

(b) If x ∈ ℚ(i), then either N_{1}(f^{(k)}(x)) → +∞ or (f^{(k)}(x))_{k≥1} is periodic starting from some natural number k.

(c) If x ∈ ℚ(ω), then either N_{2}(f^{(k)}(x)) → +∞ or (f^{(k)}(x))_{k≥1} is periodic starting from some natural number k.

Theorem 22. Let Φ be a UFD such that its group of units is finitely generated, α ∈ Frac(Φ) and f : Frac(Φ) → Frac(Φ) be an L-additive function.

1. Let x_{0} be a non-zero solution of

f(x) = αx  (1)

Then x = y · x_{0} for any solution x of (1) where y is a solution of f(y) = 0.

2. Let S = {p ∈ ℙ^{∗}(Φ) | f(p) ≠ 0}. If v_{p}(x) = 0 for all p ∈ S, then f(x) = 0.

3. There exists a bijection between non-zero solutions l of (1) with P = {p | v_{p}(l) ≠ 0} ⊆ S and integer solutions {v_{p}(x) | p ∈ ℙ^{∗}(Φ)} of

α = ∑_{p∈ℙ^{∗}(Φ)} (v_{p}(x) · \frac{f(p)}{p}) .

4. Assume that Frac(Φ) = ℚ. Let \frac{f(p)}{p} = \frac{c_{p}}{z_{p}} be an irreducible fraction for all p ∈ S. The

equation (1) has a solution x iff β = α · δ ∈ ℤ is divisible by gcd (\frac{δf(p)}{p} | p ∈ P ) where P = {p ∈ ℙ | v_{p}(x) ≠ 0} and δ = lcm(z_{p} | p ∈ P).

Proof. 1. Let x = x_{0} · y, with f(y) = 0. Since f is an L-additive function, we see that f(x) = f(y · x_{0}) = f(y)x_{0} + f(x_{0})y = f(x_{0})y = (αx_{0})y = αx. Hence x is a solution of (1).

Assume now that x_{1} is a non-zero solution of (1). Let us show that x_{1} = y · x_{0} with f(y) = 0. Let y = \frac{x_{1}}{x_{0}} . Then f(y) = f (\frac{x_{1}}{x_{0}} ) = \frac{f(x_{1})x_{0}−f(x_{0})x_{1}}{x_{0}^{2}} = \frac{αx_{1}x_{0}−αx_{0}x_{1}}{x_{0}^{2}} = 0.

2. By Corollary 10 we have that

f(x) = x · ∑_{p∈ℙ^{∗}(Φ), v_{p}(x)≠0} \frac{f(p)v_{p}(x)}{p} = 0.

3. By Corollary 10 we have that

f(l) = α · l = l · ∑_{p∈ℙ^{∗}(Φ), v_{p}(l)≠0} \frac{f(p)v_{p}(l)}{p} ⇔ α = ∑_{p∈ℙ^{∗}(Φ), v_{p}(l)≠0} v_{p}(l) · \frac{f(p)}{p}.

4. By 3) we have that

α = \frac{f(x)}{x} = ∑_{p∈P} (v_{p}(x)\frac{f(p)}{p}) = ∑_{p∈P} \frac{v_{p}(x)c_{p}}{z_{p}} .

Note that

β = α · δ = ∑_{p∈P} (\frac{c_{p}δ}{z_{p}} · v_{p}(x)) = ∑_{p∈P} (\frac{f(p)δ}{p} · v_{p}(x)) ∈ ℤ.

This equation is a linear Diophantine equation. It is well-known that this equation has a solution iff β is divisible by gcd (\frac{δf(p)}{p} | p ∈ P ). Then we know all v_{p}(x). Hence we know ±x.

Corollary 23 ([8, Theorem 3]). Let p ∈ ℙ and α ∈ ℚ. The equation D_{p}(x) = αx has a nontrivial solution iff αp ∈ ℤ. Then all nontrivial solutions are of the form x = cp^{αp}, where p |̸ c ∈ ℚ \ {0}. Conversely, all numbers of this from are nontrivial solutions.

Corollary 24 ([13, Theorem 18]). Let α = \frac{a}{b} be a rational number with gcd(a, b) = 1, b > 0. Then the equation D(x) = αx has non-zero rational solutions iff b is a product of different primes or b = 1.

Theorem 25. Let a field F be a finite algebraic extension of ℚ. If f : F → F is an L-additive linear function, then f(x) ≡ 0.

Proof. According to Artin’s theorem on primitive elements, every finite algebraic extension of ℚ is simple, i.e., there exists α ∈ F with F = ℚ(α). Let g be a minimal polynomial of α over 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) + ⋯ + f(y)}_{n} = n · f(y) and f(ny) = f(n)y + f(y)n.

Therefore f(n)y = 0. Hence f(n) = 0 for all n ∈ ℕ. Note that f(−n) = −f(n) = 0. Thus f(x) = 0 for all x ∈ ℤ.

Let g(x) = a_{n}x^{n} + ⋯ + a_{1}x + a_{0}. Then

0 = f(0) = f(g(α)) = ∑_{i=0}^{n} f(a_{i}α^{i}) = ∑_{i=0}^{n} a_{i}f(α^{i}) = ∑_{i=1}^{n} a_{i}(iα^{i−1})f(α) = g′(α)f(α).

Since g′(α) ≠ 0, we see that f(α) = 0. Therefore f(x) = 0 for every x ∈ ℤ[α]. It is easy to see that for every x ∈ F there are m ∈ ℤ[α] and n ∈ ℤ with x = \frac{m}{n}. So f(x) = 0 by 3 of Proposition 7. Thus f(x) = 0 for all x ∈ ℚ(α) = F.

4 Final remarks

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

Question 26. Let J be a factorization domain which is not UFD. Are there any non-zero L-additive functions over J?

Haukkanen et al. [7] discussed some ideas about this question.

According to Theorem 25 there are no L-additive linear non-zero functions over any finite extension of ℚ.

Question 27. Describe all UFD J such that there are no L-additive linear non-zero functions over J.

By Corollary 12 there are no non-zero L-additive functions over a finite field. According to Emmons et al. [3] every value of an L-additive function over ℤ_{n} is a divisor of zero.

Question 28. Let f be an L-additive function over a finite ring. Is it true that all values of f are divisors of zero?

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2010 Mathematics Subject Classification. Primary 11A25; Secondary 11A51, 11R27. Keywords: Leibniz-additive function, arithmetic derivative, completely multiplicative function, completely additive function, unique factorization domain (UFD), field of fractions.

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

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

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