The Form of Multi-additive Symmetric Functions

In this paper we show that every n-additive symmetric function, between quite general structures like abelian groups and semigroups, can be factorized into a composition of additive function with the product of additive functions. We also show that every two n-additive functions, defined on a product of groups, with equal counterimages of the positive half-line must be proportional.


Introduction
Multi-additive symmetric functions play an important role in the theory of functional equations: they naturally generalize additive functions, their diagonalizations are regarded as generalizations of monomial functions, and they appear frequently as solutions (or components of solutions) of functional equations. The general construction of multi-additive symmetric functions is well known (see eg. [1]). Some known facts about multi-additive symmetric functions can be found eg. in [3,5,7,8,11]. The main result of this paper is the characterization of multi-additive symmetric functions in the form of an additive function of the product of additive functions.
The structure of the paper is as follows. In the first section we recall some well known definitions and theorems from the theory of groups and semigroups but also state and prove a few results concerning extensions of a multi-additive symmetric functions defined on a semigroup. The section ends with a simple, but important lemma connecting notions of an algebraical and linear independence. The second section contains main results of the paper -we show

Multi-additive Symmetric Function on Groups and Semigroups
We start by recalling here some notions and results from the theory of groups and semigroups. Unless stated otherwise we refer to [4,Appendix A]) and [2]. Definition 1. An abelian group G is torsion-free if every element except the identity has the infinite order.

Definition 2.
A subset L of a group G is said to be independent if it does not contain the identity 0 and for all distinct x 1 , . . . , x k ∈ L we have n1,...,n k ∈Z (n 1 x 1 + · · · + n k x k = 0 =⇒ n 1 x 1 = · · · = n k x k = 0) .
Let G be an abelian group, A be the family of all independent sets L in G consisting only of elements whose order is infinite or a power of a prime and such that L is maximal with respect to these properties. Similarly, let A 0 [A p ] be the family of independent sets L 0 [L p ] in G consisting of elements whose order is infinite [a power of p] and such that L 0 [L p ] is maximal with respect to these properties; p denotes any prime integer. The cardinal number of any set in A [resp. A 0 , A p ] is called the rank of G [resp. the torsion-free rank, the p-rank] and is denoted by r(G) [resp. r 0 (G), r p (G)] (all the sets in A [resp. A 0 , A p ] have the same cardinal number).
Let p be a prime number. The Prüfer p-group is the unique p-group in which every element has p different p-th roots. Alternatively we can write Theorem 3. Let S be a cancellative abelian semigroup. Then there exists an abelian group G such that S ≤ G and G = S − S.
Theorem 4. Let (S, +) be an abelian semigroup. Then a relation ∼ given by is the equivalence relation, S/ ∼ with the operation + : is the cancellative abelian semigroup and the function κ : S → S/ ∼ given by is the semigroup homomorphism. Moreover, if S is divisible, then S/ ∼ and G are divisible, where G is an abelian group such that S/ ∼ ≤ G and G = Proof. It is easy to see that relation ∼ is reflexive and symmetric. Let x, y, z ∈ S, x ∼ y, y ∼ z. Then there exist u, v ∈ S such that x + u = y + u and y + v = z + v. Now we notice that which shows that x ∼ z, and hence relation ∼ is transitive. We proceed to show that operation + defined by (2) is well-defined. Let x 1 , x 2 , y 1 , y 2 ∈ S be such that x 1 ∼ x 2 and y 1 ∼ y 2 . Then there exist u, v ∈ S such that x 1 + u = x 2 + u and y 1 + v = y 2 + v. Hence we have It is also easily seen that S/ ∼ with the operation + given by (2) is the abelian semigroup and the function κ given by (3)  Assume that S is divisible, and let G be an abelian group such that G = S/ ∼ − S/ ∼ . Let x ∈ S, m ∈ N. Then there exist y ∈ S such that x = my, and thus κ(x) = κ(my) = mκ(y), hence S/ ∼ is divisible. Let further x ∈ G, m ∈ N. Then x = κ(u) − κ(v) for some u, v ∈ S, and moreover there exist y, z ∈ S such that u = my and v = mz. It follows that and hence G is divisible.
For semigroups we define a cardinal number with a meaning similar to the rank of a group. Definition 6. Let S be an abelian semigroup. The cardinal number r 0 (S) is equal to the torsion-free rank r 0 (G), where G is an abelian group such that After these preparations we may now pass to multi-additive functions. By Perm(n) we denote the set of all bijections of the set {1, . . . , n}.

Definition 7.
Let S be an abelian semigroup, H be an abelian group. The function A n : S n → H is called n-additive symmetric if for all x, y, x 1 , . . . , x n ∈ S and σ ∈ Perm(n). If A n is an n-additive symmetric, Such diagonalizations will be called generalized monomials of degree n.
Let us quote here the theorem of Székelyhidi about extending n-additive symmetric functions from subgroup to the whole group, that we use later. For our further purposes we also need a similar result on extending n-additive symmetric functions from semigroup S to group G generated by S in the sense of Theorem 3, which is stated and proved below.
Theorem 5 ([13, Theorem 2]). Let G be an abelian group, D be a divisible abelian group, n ∈ N. Furthermore, let H be a subgroup of G, and A n : H n → D an n-additive symmetric function. Then A n can be extended to an n-additive symmetric mapping of G n into D, that is, there exists an n-additive symmetric function A n : G n → D such that Theorem 6. Let S be a cancellative abelian semigroup, H be an abelian group, n ∈ N. Furthermore, let A n : S n → H be an n-additive symmetric function. Then, for an abelian group G such that S ≤ G and G = S − S, the function A n can be uniquely extended to an n-additive symmetric mapping of G n into H, that is, there exists an n-additive symmetric function A n : Proof. We define A n : G n → H by the formula where g 1 , . . . , g n , h 1 , . . . , h n ∈ S and We have and thus We now observe that By the induction we obtain which shows that A n is well-defined. It is immediate that A n is n-additive symmetric.
On the other hand, if A n is an extension of A n , then equation (4) holds which implies the uniqueness of A n . The following elementary result will be indispensable in the proof of one of the corollaries given in the next section. (1), H be an abelian group. Furthermore, let n ∈ N and A n : S n → H be an n-additive symmetric function. Then A n : (S/ ∼ ) n → H given by the formula

Theorem 7. Let S be an abelian semigroup, relation ∼ be defined by
is the well-defined n-additive symmetric function.
Proof . Let g 1 , . . . , g n , h 1 , . . . , h n ∈ S be such that [ We prove by the induction that for all y k , . . . , y n ∈ S, 1 ≤ k ≤ n. For k = 1 and y 2 , . . . , y n ∈ S we have Assume that (5)  which ends the proof of (5). In particular, we have hence the function A n is well-defined. We see at once that A n is n-additive symmetric.
In the last part of this section we will state a simple, yet crucial observation concerning the notions of algebraical and linear independence. We start with some facts from the theory of fields. Let E be a subfield of a field F . The set S ⊂ F is said to be algebraically independent over E if f (t 1 , . . . , t n ) = 0 for all non-zero polynomials f ∈ E[X 1 , . . . , X n ] and all distinct t 1 , . . . , t n ∈ S, n ∈ N. It is known that every extension F/E has a maximal algebraically independent subset which is called a transcendence base for F/E. The cardinality of each transcendence base for F/E is the same and it is denoted by trdeg(F/E). Let K be a field of characteristic 0 and let T ⊂ K be an algebraically independent set over the field Q of rational numbers. Define the sets The following statement is elementary and its proof is left to the reader. Lemma 1. For every n ∈ N, the set T n is linearly independent over Q and its elements are determined uniquely up to the order of multiplication, i.e.: n∈N x1,...,xn∈T y1,...,yn∈T In particular, T n can be extended to the basis of K over Q.

Main Results
We are now ready to present main results of the paper. The first theorem presents a characterization of n-additive symmetric functions as an additive function of a product of additive functions. The following corollaries presents the same factorization in a slightly different settings.

Theorem 8.
Let G be a divisible abelian group, H be an abelian group, n ∈ N, A n : G n → H an n-additive symmetric function. Furthermore, if n = 1, then let G or H be torsion-free. Then for every field K of characteristic 0 and trdeg(K/Q) ≥ r 0 (G) there exist additive maps φ : Proof. In view of Theorem 1 we have (up to isomorphism) Indeed, when n = 1 and G is torsion-free, then G 0 = {0}. When n = 1 and H is torsion-free we have that for x ∈ G of the finite order, the order of A 1 (x) divides the order of x, hence we obtain A 1 (x) = 0. If n ≥ 2, x 1 , . . . , x n ∈ G and kx 1 = 0 for some k ∈ N, then since G is divisible, there exists y 2 ∈ G such that x 2 = ky 2 . Hence We have trdeg(K/Q) ≥ r 0 (G 1 ) = r 0 (G) = |I|, hence there exists an algebraically independent set T ⊂ K over Q of the cardinality |I|. Let f : I → T be a bijection. We will denote by 1 i an element of G 1 such that it is equal to 1 on the i-th coordinate and equal to 0 on all remaining coordinates. We now define the function φ : G → K by the formula It is easily seen that φ is additive and ker φ = G 0 . In view of Lemma 1 there exists a basis H of K/Q such that T n ⊂ H. We define the map ψ 0 : H → H by the formula This function is well-defined. Indeed, if x = φ(g 1 )·. . .·φ(g n ) = φ(h 1 )·. . .·φ(h n ) for some g 1 , . . . , g n , h 1 , . . . , h n ∈ G, then in view of Lemma 1 there exists σ ∈ Perm(n) such that i∈{1,...,n} φ(h i ) = φ(g σ(i) ), which give us h i = g σ(i) + z i , for i ∈ {1, . . . , n}, where z 1 , . . . , z n ∈ G 0 , and thus A n (g 1 , . . . , g n ) = A n (h 1 , . . . , h n ). Obviously, we can uniquely extend ψ 0 to the additive map ψ : K → H. Finally, we show that Eq. (6) holds. Let x 1 , . . . , x n ∈ G. Then and hence we have A n (q 1,i1 1 i1 , . . . , q n,in 1 in ) which ends the proof. Proof. When G is divisible, then it is exactly Theorem 8. Assume that H is divisible. In view of Theorem 2 we can extend the group G to the divisible group G 1 such that r 0 (G 1 ) = r 0 (G). Next, applying Theorem 5 we can extend A n to the n-additive symmetric map A n : G n 1 → H. By Theorem 8 for every field K of characteristic 0 and trdeg(K/Q) ≥ r 0 (G 1 ) = r 0 (G) there exist additive maps φ 0 : which ends the proof. Proof. Let G be an abelian group such that S/ ∼ ≤ G, G = S/ ∼ − S/ ∼ . We combine Theorems 7, 4 and 6 to get the n-additive symmetric map A n : G n → H such that A n (κ(x 1 ), . . . , κ(x n )) = A n (x 1 , . . . , x n ), x 1 , . . . , x n ∈ S.
In view of Corollary 1 for every field K of characteristic 0 and trdeg(K/Q) ≥ r 0 (S) there exist additive maps φ 0 : G → K, ψ : K → H such that Let φ : S → K be given by φ = φ 0 • κ. Then which ends the proof.
At the end of this section we are obliged to notice that the assumptions of Theorem 8 are certainly fulfilled in the real numbers world, that is when G = H = K = R, hence we have the following result.

Examples and Auxiliary Results
Let us first notice that in Theorem 8, in the case when H = K, n ≥ 2, we cannot write A n in the form A suitable example is given in [9]. Moreover, in the case when G = K, n ≥ 2, we are not able to write A n in the form as the following example shows.
Example 1. Let n ≥ 2, G = H = R, and take a discontinuous additive function a : R → R. Consider an n-additive symmetric function and suppose that (7) holds for some additive function ψ : R → R. Notice that diagonalization A * n (x) = a(x) 2 x n−2 , x ∈ R, is discontinuous and A * n (x) ≥ 0 for x ≥ 0. From (7) it follows that A * n (x) = ψ(x n ), hence ψ is also discontinuous and positive on the positive real axis, which is a contradiction to well known property of additive functions.
The following example shows that the field K in Theorem 8 cannot be too small. Example 2. Take G = Q 2 and H = Q. Let a, b ∈ Q be such that a < 0 < b and define A 2 : G 2 → H by A 2 ((q 1 ; q 2 ), (r 1 ; r 2 )) = aq 1 r 1 + bq 2 r 2 , q 1 , q 2 , r 1 , r 2 ∈ Q.
Let us also notice that Theorem 8 does not hold for n = 1 and G, H which are divisible but not torsion-free.
Dealing with generalized polynomial functions, we often assume that the domain or the codomain is uniquely divisible by n!. It is not enough for Corollary 1 to hold. Example 4. Let n ∈ N, G = {m/(n!) k : m ∈ Z, k ∈ N}. It is easily seen that G (with the usual addition) is an abelian group uniquely divisible by n!. Let A : G → G be an identity map, which obviously is additive. Suppose that there exist a field K of characteristic 0 and additive functions φ : G → K, ψ : K → G such that A = ψ • φ. Let x ∈ K. Then ψ(x) = m/(n!) k for some m ∈ Z and k ∈ N. Hence (n! + 1) j ψ(x/(n! + 1) j ) = ψ(x) = m (n!) k , j ∈ N, and thus we have ψ(x/(n! + 1) j ) = m/(n!) k /(n! + 1) j for j ∈ N. Since G is not divisible by (n! + 1), then m = 0, which means that ψ(x) = 0. We conclude that A = ψ • φ = 0, which give us a contradiction.
Finally we can apply the results given above to generalized polynomial functions.