additive map

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 $f$ and $g$ are additive maps, then the map $f\; +\; g$ (defined pointwise) is additive.

Definition of scalar multiplication by an integer

Suppose that $X$ is an additive group with identity element $0$ and that the inverse of $x\; \backslash in\; X$ is denoted by $-x.$ For any $x\; \backslash in\; X$ and integer $n\; \backslash in\; \backslash Z,$ let:

$$n\; x\; :=\; \backslash left\backslash \{$$

\begin{alignat}{9}

& &&0 && && &&~~~~ && &&~\text{ when } n = 0, \\

& &&x &&+ \cdots + &&x &&~~~~ \text{(} n &&\text{ summands) } &&~\text{ when } n > 0, \\

& (-&&x) &&+ \cdots + (-&&x) &&~~~~ \text{(} |n| &&\text{ summands) } &&~\text{ when } n < 0, \\

\end{alignat}

\right.

Thus $(-1)\; x\; =\; -\; x$ and it can be shown that for all integers $m,\; n\; \backslash in\; \backslash Z$ and all $x\; \backslash in\; X,$ $(m\; +\; n)\; x\; =\; m\; x\; +\; n\; x$ and $-\; (n\; x)\; =\; (-n)\; x\; =\; n\; (-x).$

This definition of scalar multiplication makes the cyclic subgroup $\backslash Z\; x$ of $X$ into a left $\backslash Z$-module; if $X$ is commutative, then it also makes $X$ into a left $\backslash Z$-module.

Homogeneity over the integers

If $f\; :\; X\; \backslash to\; Y$ is an additive map between additive groups then $f(0)\; =\; 0$ and for all $x\; \backslash in\; X,$ $f(-x)\; =\; -\; f(x)$ (where negation denotes the additive inverse) and$f(0)\; =\; f(0\; +\; 0)\; =\; f(0)\; +\; f(0)$ so adding $-f(0)$ to both sides proves that $f(0)\; =\; 0.$ If $x\; \backslash in\; X$ then $0\; =\; f(0)\; =\; f(x\; +\; (-x))\; =\; f(x)\; +\; f(-x)$ so that $f(-x)\; =\; -\; f(x)$ where by definition, $(-1)\; f(x)\; :=\; -\; f(x).$ Induction shows that if $n\; \backslash in\; \backslash N$ is positive then $f(n\; x)\; =\; n\; f(x)$ and that the additive inverse of $n\; f(x)$ is $n\; (-\; f(x)),$ which implies that $f((-n)\; x)\; =\; f(n\; (-x))\; =\; n\; f(-x)\; =\; n\; (-\; f(x))\; =\; -(n\; f(x))\; =\; (-n)\; f(x)$ (this shows that $f(n\; x)\; =\; n\; f(x)$ holds for ). $\backslash blacksquare$

$$f(n\; x)\; =\; n\; f(x)\; \backslash quad\; \backslash text\{\; for\; all\; \}\; n\; \backslash in\; \backslash Z.$$

Consequently, $f(x\; -\; y)\; =\; f(x)\; -\; f(y)$ for all $x,\; y\; \backslash in\; X$ (where by definition, $x\; -\; y\; :=\; x\; +\; (-y)$).

In other words, every additive map is homogeneous over the integers. Consequently, every additive map between abelian groups is a homomorphism of $\backslash Z$-modules.

Homomorphism of $\backslash Q$-modules

If the additive abelian groups $X$ and $Y$ are also a unital modules over the rationals $\backslash Q$ (such as real or complex vector spaces) then an additive map $f\; :\; X\; \backslash to\; Y$ satisfies:Let $x\; \backslash in\; X$ and $q\; =\; \backslash frac\{m\}\{n\}\; \backslash in\; \backslash Q$ where $m,\; n\; \backslash in\; \backslash Z$ and $n\; >\; 0.$ Let $y\; :=\; \backslash frac\{1\}\{n\}\; x.$ Then $n\; y\; =\; n\; \backslash left(\backslash frac\{1\}\{n\}\; x\backslash right)\; =\; \backslash left(n\; \backslash frac\{1\}\{n\}\backslash right)\; x\; =\; (1)\; x\; =\; x,$ which implies $f(x)\; =\; f(n\; y)\; =\; n\; f(y)\; =\; n\; f\backslash left(\backslash frac\{1\}\{n\}\; x\backslash right)$ so that multiplying both sides by $\backslash frac\{1\}\{n\}$ proves that $f\backslash left(\backslash frac\{1\}\{n\}\; x\backslash right)\; =\; \backslash frac\{1\}\{n\}\; f(x).$ Consequently, $f(q\; x)\; =\; f\backslash left(\backslash frac\{m\}\{n\}\; x\backslash right)\; =\; m\; f\backslash left(\backslash frac\{1\}\{n\}\; x\backslash right)\; =\; m\backslash left(\backslash frac\{1\}\{n\}\; f(x)\backslash right)\; =\; q\; f(x).$ $\backslash blacksquare$

$$f(q\; x)\; =\; q\; f(x)\; \backslash quad\; \backslash text\{\; for\; all\; \}\; q\; \backslash in\; \backslash Q\; \backslash text\{\; and\; \}\; x\; \backslash in\; X.$$

In other words, every additive map is homogeneous over the rational numbers. Consequently, every additive maps between unital $\backslash Q$-modules is a homomorphism of $\backslash Q$-modules.

Despite being homogeneous over $\backslash Q,$ as described in the article on Cauchy's functional equation, even when $X\; =\; Y\; =\; \backslash R,$ it is nevertheless still possible for the additive function $f\; :\; \backslash R\; \backslash to\; \backslash R$ to {{em|not}} be homogeneous over the real numbers; said differently, there exist additive maps $f\; :\; \backslash R\; \backslash to\; \backslash R$ that are {{em|not}} of the form $f(x)\; =\; s\_0\; x$ for some constant $s\_0\; \backslash in\; \backslash R.$

In particular, there exist additive maps that are not linear maps.

• {{annotated link|Antilinear map}}

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Proofs

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• {{citation|author1=Roger C. Lyndon|author2=Paul E. Schupp|title=Combinatorial Group Theory|publisher=Springer|year=2001}}

Category:Ring theory

Category:Morphisms

Category:Types of functions