Additive map

Short description: Z-module homomorphism

In algebra, an additive map, [math]\displaystyle{ Z }[/math]-linear map or additive function is a function [math]\displaystyle{ f }[/math] that preserves the addition operation:^[1] [math]\displaystyle{ f(x + y) = f(x) + f(y) }[/math] for every pair of elements [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] in the domain of [math]\displaystyle{ f. }[/math] For example, any linear map is additive. When the domain is the real numbers, this is Cauchy's functional equation. For a specific case of this definition, see additive polynomial.

More formally, an additive map is a [math]\displaystyle{ \Z }[/math]-module homomorphism. Since an abelian group is a [math]\displaystyle{ \Z }[/math]-module, it may be defined as a group homomorphism between abelian groups.

A map [math]\displaystyle{ V \times W \to X }[/math] that is additive in each of two arguments separately is called a bi-additive map or a [math]\displaystyle{ \Z }[/math]-bilinear map.^[2]

Examples

Typical examples include maps between rings, vector spaces, or modules that preserve the additive group. An additive map does not necessarily preserve any other structure of the object; for example, the product operation of a ring.

If [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math] are additive maps, then the map [math]\displaystyle{ f + g }[/math] (defined pointwise) is additive.

Properties

Definition of scalar multiplication by an integer

Suppose that [math]\displaystyle{ X }[/math] is an additive group with identity element [math]\displaystyle{ 0 }[/math] and that the inverse of [math]\displaystyle{ x \in X }[/math] is denoted by [math]\displaystyle{ -x. }[/math] For any [math]\displaystyle{ x \in X }[/math] and integer [math]\displaystyle{ n \in \Z, }[/math] let: [math]\displaystyle{ n x := \left\{ \begin{alignat}{9} & &&0 && && &&~~~~ && &&~\text{ when } n = 0, \\ & &&x &&+ \cdots + &&x &&~~~~ \text{(} n &&\text{ summands) } &&~\text{ when } n \gt 0, \\ & (-&&x) &&+ \cdots + (-&&x) &&~~~~ \text{(} |n| &&\text{ summands) } &&~\text{ when } n \lt 0, \\ \end{alignat} \right. }[/math] Thus [math]\displaystyle{ (-1) x = - x }[/math] and it can be shown that for all integers [math]\displaystyle{ m, n \in \Z }[/math] and all [math]\displaystyle{ x \in X, }[/math] [math]\displaystyle{ (m + n) x = m x + n x }[/math] and [math]\displaystyle{ - (n x) = (-n) x = n (-x). }[/math] This definition of scalar multiplication makes the cyclic subgroup [math]\displaystyle{ \Z x }[/math] of [math]\displaystyle{ X }[/math] into a left [math]\displaystyle{ \Z }[/math]-module; if [math]\displaystyle{ X }[/math] is commutative, then it also makes [math]\displaystyle{ X }[/math] into a left [math]\displaystyle{ \Z }[/math]-module.

Homogeneity over the integers

If [math]\displaystyle{ f : X \to Y }[/math] is an additive map between additive groups then [math]\displaystyle{ f(0) = 0 }[/math] and for all [math]\displaystyle{ x \in X, }[/math] [math]\displaystyle{ f(-x) = - f(x) }[/math] (where negation denotes the additive inverse) and^{[proof 1]} [math]\displaystyle{ f(n x) = n f(x) \quad \text{ for all } n \in \Z. }[/math] Consequently, [math]\displaystyle{ f(x - y) = f(x) - f(y) }[/math] for all [math]\displaystyle{ x, y \in X }[/math] (where by definition, [math]\displaystyle{ x - y := x + (-y) }[/math]).

In other words, every additive map is homogeneous over the integers. Consequently, every additive map between abelian groups is a homomorphism of [math]\displaystyle{ \Z }[/math]-modules.

Homomorphism of [math]\displaystyle{ \Q }[/math]-modules

If the additive abelian groups [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] are also a unital modules over the rationals [math]\displaystyle{ \Q }[/math] (such as real or complex vector spaces) then an additive map [math]\displaystyle{ f : X \to Y }[/math] satisfies:^{[proof 2]} [math]\displaystyle{ f(q x) = q f(x) \quad \text{ for all } q \in \Q \text{ and } x \in X. }[/math] In other words, every additive map is homogeneous over the rational numbers. Consequently, every additive maps between unital [math]\displaystyle{ \Q }[/math]-modules is a homomorphism of [math]\displaystyle{ \Q }[/math]-modules.

Despite being homogeneous over [math]\displaystyle{ \Q, }[/math] as described in the article on Cauchy's functional equation, even when [math]\displaystyle{ X = Y = \R, }[/math] it is nevertheless still possible for the additive function [math]\displaystyle{ f : \R \to \R }[/math] to not be homogeneous over the real numbers; said differently, there exist additive maps [math]\displaystyle{ f : \R \to \R }[/math] that are not of the form [math]\displaystyle{ f(x) = s_0 x }[/math] for some constant [math]\displaystyle{ s_0 \in \R. }[/math] In particular, there exist additive maps that are not linear maps.

Notes

↑ Leslie Hogben (2013), Handbook of Linear Algebra (3 ed.), CRC Press, pp. 30–8, ISBN 9781498785600
↑ N. Bourbaki (1989), Algebra Chapters 1–3, Springer, p. 243

Proofs

↑ [math]\displaystyle{ f(0) = f(0 + 0) = f(0) + f(0) }[/math] so adding [math]\displaystyle{ -f(0) }[/math] to both sides proves that [math]\displaystyle{ f(0) = 0. }[/math] If [math]\displaystyle{ x \in X }[/math] then [math]\displaystyle{ 0 = f(0) = f(x + (-x)) = f(x) + f(-x) }[/math] so that [math]\displaystyle{ f(-x) = - f(x) }[/math] where by definition, [math]\displaystyle{ (-1) f(x) := - f(x). }[/math] Induction shows that if [math]\displaystyle{ n \in \N }[/math] is positive then [math]\displaystyle{ f(n x) = n f(x) }[/math] and that the additive inverse of [math]\displaystyle{ n f(x) }[/math] is [math]\displaystyle{ n (- f(x)), }[/math] which implies that [math]\displaystyle{ f((-n) x) = f(n (-x)) = n f(-x) = n (- f(x)) = -(n f(x)) = (-n) f(x) }[/math] (this shows that [math]\displaystyle{ f(n x) = n f(x) }[/math] holds for [math]\displaystyle{ n \lt 0 }[/math]). [math]\displaystyle{ \blacksquare }[/math]
↑ Let [math]\displaystyle{ x \in X }[/math] and [math]\displaystyle{ q = \frac{m}{n} \in \Q }[/math] where [math]\displaystyle{ m, n \in \Z }[/math] and [math]\displaystyle{ n \gt 0. }[/math] Let [math]\displaystyle{ y := \frac{1}{n} x. }[/math] Then [math]\displaystyle{ n y = n \left(\frac{1}{n} x\right) = \left(n \frac{1}{n}\right) x = (1) x = x, }[/math] which implies [math]\displaystyle{ f(x) = f(n y) = n f(y) = n f\left(\frac{1}{n} x\right) }[/math] so that multiplying both sides by [math]\displaystyle{ \frac{1}{n} }[/math] proves that [math]\displaystyle{ f\left(\frac{1}{n} x\right) = \frac{1}{n} f(x). }[/math] Consequently, [math]\displaystyle{ f(q x) = f\left(\frac{m}{n} x\right) = m f\left(\frac{1}{n} x\right) = m\left(\frac{1}{n} f(x)\right) = q f(x). }[/math] [math]\displaystyle{ \blacksquare }[/math]

References

Roger C. Lyndon; Paul E. Schupp (2001), Combinatorial Group Theory, Springer

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[1] Leslie Hogben (2013), Handbook of Linear Algebra (3 ed.), CRC Press, pp. 30–8, ISBN 9781498785600

[2] N. Bourbaki (1989), Algebra Chapters 1–3, Springer, p. 243

[3] [math]\displaystyle{ f(0) = f(0 + 0) = f(0) + f(0) }[/math] so adding [math]\displaystyle{ -f(0) }[/math] to both sides proves that [math]\displaystyle{ f(0) = 0. }[/math] If [math]\displaystyle{ x \in X }[/math] then [math]\displaystyle{ 0 = f(0) = f(x + (-x)) = f(x) + f(-x) }[/math] so that [math]\displaystyle{ f(-x) = - f(x) }[/math] where by definition, [math]\displaystyle{ (-1) f(x) := - f(x). }[/math] Induction shows that if [math]\displaystyle{ n \in \N }[/math] is positive then [math]\displaystyle{ f(n x) = n f(x) }[/math] and that the additive inverse of [math]\displaystyle{ n f(x) }[/math] is [math]\displaystyle{ n (- f(x)), }[/math] which implies that [math]\displaystyle{ f((-n) x) = f(n (-x)) = n f(-x) = n (- f(x)) = -(n f(x)) = (-n) f(x) }[/math] (this shows that [math]\displaystyle{ f(n x) = n f(x) }[/math] holds for [math]\displaystyle{ n \lt 0 }[/math]). [math]\displaystyle{ \blacksquare }[/math]

[4] Let [math]\displaystyle{ x \in X }[/math] and [math]\displaystyle{ q = \frac{m}{n} \in \Q }[/math] where [math]\displaystyle{ m, n \in \Z }[/math] and [math]\displaystyle{ n \gt 0. }[/math] Let [math]\displaystyle{ y := \frac{1}{n} x. }[/math] Then [math]\displaystyle{ n y = n \left(\frac{1}{n} x\right) = \left(n \frac{1}{n}\right) x = (1) x = x, }[/math] which implies [math]\displaystyle{ f(x) = f(n y) = n f(y) = n f\left(\frac{1}{n} x\right) }[/math] so that multiplying both sides by [math]\displaystyle{ \frac{1}{n} }[/math] proves that [math]\displaystyle{ f\left(\frac{1}{n} x\right) = \frac{1}{n} f(x). }[/math] Consequently, [math]\displaystyle{ f(q x) = f\left(\frac{m}{n} x\right) = m f\left(\frac{1}{n} x\right) = m\left(\frac{1}{n} f(x)\right) = q f(x). }[/math] [math]\displaystyle{ \blacksquare }[/math]

[1]

[2]

[proof 1]

[proof 2]

Additive map

Contents

Examples

Properties

See also

Notes

References