Additive identity

Formal definition

Let N be a group that is closed under the operation of addition, denoted +. An additive identity for N, denoted e, is an element in N such that for any element n in N, $${\displaystyle e+n=n=n+e.}$$

Further examples

• In a group, the additive identity is the identity element of the group, is often denoted 0, and is unique (see below for proof).
• A ring or field is a group under the operation of addition and thus these also have a unique additive identity 0. This is defined to be different from the multiplicative identity 1 if the ring (or field) has more than one element. If the additive identity and the multiplicative identity are the same, then the ring is trivial (proved below).
• In the ring M_{m × n}(R) of m-by-n matrices over a ring R, the additive identity is the zero matrix, denoted O or 0, and is the m-by-n matrix whose entries consist entirely of the identity element 0 in R. For example, in the 2×2 matrices over the integers ⁠${\displaystyle {\mathrm{M}}_2(\mathbb{Z})}$⁠ the additive identity is $${\displaystyle 0=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]}$$
• In the quaternions, 0 is the additive identity.
• In the ring of functions from ⁠${\displaystyle \mathbb{R}\to \mathbb{R}}$⁠, the function mapping every number to 0 is the additive identity.
• In the additive group of vectors in ⁠${\displaystyle {\mathbb{R}}^n,}$⁠ the origin or zero vector is the additive identity.

Properties

The additive identity is unique in a group

Let (G, +) be a group and let 0 and 0' in G both denote additive identities, so for any g in G, $${\displaystyle 0+g=g=g+0,0^{\prime }+g=g=g+0^{\prime }.}$$

It then follows from the above that $${\displaystyle 0^{\prime }=0^{\prime }+0=0^{\prime }+0=0.}$$

The additive identity annihilates ring elements

In a system with a multiplication operation that distributes over addition, the additive identity is a multiplicative absorbing element, meaning that for any s in S, s · 0 = 0. This follows because:

$${\displaystyle \begin{array}{cc}s\cdot 0& =s\cdot (0+0)=s\cdot 0+s\cdot 0\\ \Rightarrow s\cdot 0& =s\cdot 0-s\cdot 0\\ \Rightarrow s\cdot 0& =0.\end{array}}$$

The additive and multiplicative identities are different in a non-trivial ring

Let R be a ring and suppose that the additive identity 0 and the multiplicative identity 1 are equal, i.e. 0 = 1. Let r be any element of R. Then $${\displaystyle r=r\times 1=r\times 0=0}$$ proving that R is trivial, i.e. R = {0}. The contrapositive, that if R is non-trivial then 0 is not equal to 1, is therefore shown.