additive inverse of a sum in a ring

Let $R$ be a ring with elements $a,b\in R$. Suppose we want to find the inverse of the element $(a+b)\in R$. (Note that we call the element $(a+b)$ the sum of $a$ and $b$.) So we want the unique element $c\in R$ so that $(a+b)+c=0$. Actually, let’s put $c=(-a)+(-b)$ where $(-a)\in R$ is the additive inverse of $a$ and $(-b)\in R$ is the additive inverse of $b$. Because addition in the ring is both associative and commutative we see that

$(a+b)+((-a)+(-b))$ $=$ $(a+(-a))+(b+(-b))$

$=$ $0+0=0$

since $(-a)\in R$ is the additive inverse of $a$ and $(-b)\in R$ is the additive inverse of $b$. Since additive inverses are unique this means that the additive inverse of $(a+b)$ must be $(-a)+(-b)$. We write this as

$$-(a+b)=(-a)+(-b).$$

It is important to note that we cannot just distribute the minus sign across the sum because this would imply that $-1\in R$ which is not the case if our ring is not with unity.

Title	additive inverse of a sum in a ring
Canonical name	AdditiveInverseOfASumInARing
Date of creation	2013-03-22 15:45:02
Last modified on	2013-03-22 15:45:02
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	11
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 16B70