In algebra, an additive map, ${\displaystyle Z}$-linear map or additive function is a function ${\displaystyle f}$ that preserves the addition operation:[1]

${\displaystyle f(x+y)=f(x)+f(y)}$
for every pair of elements ${\displaystyle x}$ and ${\displaystyle y}$ in the domain of ${\displaystyle f.}$ 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 ${\displaystyle \mathbb {Z} }$-module homomorphism. Since an abelian group is a ${\displaystyle \mathbb {Z} }$-module, it may be defined as a group homomorphism between abelian groups.

A map ${\displaystyle V\times W\to X}$ that is additive in each of two arguments separately is called a bi-additive map or a ${\displaystyle \mathbb {Z} }$-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 ${\displaystyle f}$ and ${\displaystyle g}$ are additive maps, then the map ${\displaystyle f+g}$ (defined pointwise) is additive.

## Properties

Definition of scalar multiplication by an integer

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

{\displaystyle nx:=\left\{{\begin{alignedat}{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{alignedat}}\right.}
Thus ${\displaystyle (-1)x=-x}$ and it can be shown that for all integers ${\displaystyle m,n\in \mathbb {Z} }$ and all ${\displaystyle x\in X,}$ ${\displaystyle (m+n)x=mx+nx}$ and ${\displaystyle -(nx)=(-n)x=n(-x).}$ This definition of scalar multiplication makes the cyclic subgroup ${\displaystyle \mathbb {Z} x}$ of ${\displaystyle X}$ into a left ${\displaystyle \mathbb {Z} }$-module; if ${\displaystyle X}$ is commutative, then it also makes ${\displaystyle X}$ into a left ${\displaystyle \mathbb {Z} }$-module.

Homogeneity over the integers

If ${\displaystyle f:X\to Y}$ is an additive map between additive groups then ${\displaystyle f(0)=0}$ and for all ${\displaystyle x\in X,}$ ${\displaystyle f(-x)=-f(x)}$ (where negation denotes the additive inverse) and[proof 1]

${\displaystyle f(nx)=nf(x)\quad {\text{ for all }}n\in \mathbb {Z} .}$
Consequently, ${\displaystyle f(x-y)=f(x)-f(y)}$ for all ${\displaystyle x,y\in X}$ (where by definition, ${\displaystyle x-y:=x+(-y)}$).

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

Homomorphism of ${\displaystyle \mathbb {Q} }$-modules

If the additive abelian groups ${\displaystyle X}$ and ${\displaystyle Y}$ are also a unital modules over the rationals ${\displaystyle \mathbb {Q} }$ (such as real or complex vector spaces) then an additive map ${\displaystyle f:X\to Y}$ satisfies:[proof 2]

${\displaystyle f(qx)=qf(x)\quad {\text{ for all }}q\in \mathbb {Q} {\text{ and }}x\in X.}$
In other words, every additive map is homogeneous over the rational numbers. Consequently, every additive maps between unital ${\displaystyle \mathbb {Q} }$-modules is a homomorphism of ${\displaystyle \mathbb {Q} }$-modules.

Despite being homogeneous over ${\displaystyle \mathbb {Q} ,}$ as described in the article on Cauchy's functional equation, even when ${\displaystyle X=Y=\mathbb {R} ,}$ it is nevertheless still possible for the additive function ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ to not be homogeneous over the real numbers; said differently, there exist additive maps ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ that are not of the form ${\displaystyle f(x)=s_{0}x}$ for some constant ${\displaystyle s_{0}\in \mathbb {R} .}$ In particular, there exist additive maps that are not linear maps.

1. ^ ${\displaystyle f(0)=f(0+0)=f(0)+f(0)}$ so adding ${\displaystyle -f(0)}$ to both sides proves that ${\displaystyle f(0)=0.}$ If ${\displaystyle x\in X}$ then ${\displaystyle 0=f(0)=f(x+(-x))=f(x)+f(-x)}$ so that ${\displaystyle f(-x)=-f(x)}$ where by definition, ${\displaystyle (-1)f(x):=-f(x).}$ Induction shows that if ${\displaystyle n\in \mathbb {N} }$ is positive then ${\displaystyle f(nx)=nf(x)}$ and that the additive inverse of ${\displaystyle nf(x)}$ is ${\displaystyle n(-f(x)),}$ which implies that ${\displaystyle f((-n)x)=f(n(-x))=nf(-x)=n(-f(x))=-(nf(x))=(-n)f(x)}$ (this shows that ${\displaystyle f(nx)=nf(x)}$ holds for ${\displaystyle n<0}$). ${\displaystyle \blacksquare }$
2. ^ Let ${\displaystyle x\in X}$ and ${\displaystyle q={\frac {m}{n}}\in \mathbb {Q} }$ where ${\displaystyle m,n\in \mathbb {Z} }$ and ${\displaystyle n>0.}$ Let ${\displaystyle y:={\frac {1}{n}}x.}$ Then ${\displaystyle ny=n\left({\frac {1}{n}}x\right)=\left(n{\frac {1}{n}}\right)x=(1)x=x,}$ which implies ${\displaystyle f(x)=f(ny)=nf(y)=nf\left({\frac {1}{n}}x\right)}$ so that multiplying both sides by ${\displaystyle {\frac {1}{n}}}$ proves that ${\displaystyle f\left({\frac {1}{n}}x\right)={\frac {1}{n}}f(x).}$ Consequently, ${\displaystyle f(qx)=f\left({\frac {m}{n}}x\right)=mf\left({\frac {1}{n}}x\right)=m\left({\frac {1}{n}}f(x)\right)=qf(x).}$ ${\displaystyle \blacksquare }$