Additive map

Z-module homomorphism

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

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

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

Properties

Definition of scalar multiplication by an integer

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

n x := { 0            when  n = 0 , x + ⋯ + x         ( n  summands)     when  n > 0 , ( − x ) + ⋯ + ( − x )         ( | n |  summands)     when  n < 0 , {\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 ( − 1 ) x = − x {\displaystyle (-1)x=-x} and it can be shown that for all integers m , n ∈ Z {\displaystyle m,n\in \mathbb {Z} } and all x ∈ X , {\displaystyle x\in X,} ( m + n ) x = m x + n x {\displaystyle (m+n)x=mx+nx} and − ( n x ) = ( − n ) x = n ( − x ) . {\displaystyle -(nx)=(-n)x=n(-x).} This definition of scalar multiplication makes the cyclic subgroup Z x {\displaystyle \mathbb {Z} x} of X {\displaystyle X} into a left Z {\displaystyle \mathbb {Z} } -module; if X {\displaystyle X} is commutative, then it also makes X {\displaystyle X} into a left Z {\displaystyle \mathbb {Z} } -module.

Homogeneity over the integers

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

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

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

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

f ( q x ) = q f ( x )  for all  q ∈ Q  and  x ∈ X . {\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 Q {\displaystyle \mathbb {Q} } -modules is a homomorphism of Q {\displaystyle \mathbb {Q} } -modules.

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

See also

Notes

  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

Proofs

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

References