Additive map

In algebra, an additive map, $Z$ -linear map or additive function is a function $f$ that preserves the addition operation:^[1] $f(x+y)=f(x)+f(y)$ for every pair of elements $x$ and $y$ in the domain of $f.$ 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 $\mathbb {Z}$ -module homomorphism. Since an abelian group is a $\mathbb {Z}$ -module, it may be defined as a group homomorphism between abelian groups.

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

Properties

Definition of scalar multiplication by an integer

Suppose that $X$ is an additive group with identity element $0$ and that the inverse of $x\in X$ is denoted by $-x.$ For any $x\in X$ and integer $n\in \mathbb {Z} ,$ let: ${\begin{alignedat}{9}nx&:=&&0&&&&&&~~~~&&&&~{\text{ when }}n=0,\\nx&:=&&x&&+\cdots +&&x&&~~~~{\text{(}}n&&{\text{-summands) }}&&~{\text{ when }}n>0,\\nx&:=(-&&x)&&+\cdots +(-&&x)&&~~~~{\text{(}}|n|&&{\text{-summands) }}&&~{\text{ when }}n<0,\\\end{alignedat}}$ Thus $(-1)x=-x$ and it can be shown that for all integers $m,n\in \mathbb {Z}$ and all $x\in X,$ $(m+n)x=mx+nx$ and $-(nx)=(-n)x=n(-x).$ This definition of scalar multiplication makes the cyclic subgroup $\mathbb {Z} x$ of $X$ into a left $\mathbb {Z}$ -module; if $X$ is commutative, then it also makes $X$ into a left $\mathbb {Z}$ -module. The next result states that every additive map is homogeneous over the integers; that is, every additive maps between abelian groups is a homomorphism of $\mathbb {Z}$ -modules.

Homogeneity over the integers

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

Homomorphism of $\mathbb {Q}$ -modules

If the additive abelian groups $X$ and $Y$ are also a unital modules over the rationals $\mathbb {Q}$ (such as real or complex vector space) then an additive map $f:X\to Y$ satisfies:^{[proof 2]} $f(qx)=qf(x)\quad {\text{ for all }}q\in \mathbb {Q} {\text{ and }}x\in X.$

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

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

References

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

[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] $f(0)=f(0+0)=f(0)+f(0)$ so adding $-f(0)$ to both sides proves that $f(0)=0.$ If $x\in X$ 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 $n\in \mathbb {N}$ is positive then $f(nx)=nf(x)$ and that the additive inverse of $nf(x)$ is $n(-f(x)),$ which implies that $f((-n)x)=f(n(-x))=nf(-x)=n(-f(x))=-(nf(x))=(-n)f(x)$ (this shows that $f(nx)=nf(x)$ holds for $n<0$ ). $\blacksquare$

[4] Let $x\in X$ and $q={\frac {m}{n}}\in \mathbb {Q}$ where $m,n\in \mathbb {Z}$ and $n>0.$ Let $y:={\frac {1}{n}}x.$ Then $ny=n\left({\frac {1}{n}}x\right)=\left(n{\frac {1}{n}}\right)x=(1)x=x,$ which implies $f(x)=f(ny)=nf(y)=nf\left({\frac {1}{n}}x\right)$ so that multiplying both sides by ${\frac {1}{n}}$ proves that $f\left({\frac {1}{n}}x\right)={\frac {1}{n}}f(x).$ Consequently, $f(qx)=f\left({\frac {m}{n}}x\right)=mf\left({\frac {1}{n}}x\right)=m\left({\frac {1}{n}}f(x)\right)=qf(x).$ $\blacksquare$

[1]

[2]

[proof 1]

[proof 2]

Examples

Properties

See also

Notes

References