Proofs of elementary ring properties

The following proofs of elementary ring properties use only the axioms that define a mathematical ring:

Basics

Multiplication by zero

Theorem:${\displaystyle 0\cdot a=a\cdot 0=0}$

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${\displaystyle 0\cdot a=(0+0)\cdot a=(0\cdot a)+(0\cdot a)}$

By subtracting (i.e. adding the additive inverse of) ${\displaystyle 0\cdot a}$ on both sides of the equation, we get the desired result. The proof that ${\displaystyle a\cdot 0=0}$ is similar.

Unique identity element per binary operation

Theorem: The identity element e for a binary operation (addition or multiplication) of a ring is unique.

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If there is another identity element ${\displaystyle e'}$ for the binary operation, then ${\displaystyle e'a=ae'=a}$, and when ${\displaystyle a=e}$, ${\displaystyle e'e=ee'=e=e'}$ where ${\displaystyle ab}$ is the binary operation on ring elements ${\displaystyle a}$ and ${\displaystyle b}$.

Unique additive inverse element

Theorem:- a as the additive inverse element for a is unique.

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If there is another inverse element ${\displaystyle -a'}$ for ${\displaystyle a}$, then ${\displaystyle -a=-a+0=-a+a-a'=0-a'=-a'}$.

Unique multiplicative inverse element

Theorem:a⁻¹ as the multiplicative inverse element for a is unique.

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If there is another inverse element ${\displaystyle a^{-1'}}$ for ${\displaystyle a}$, then ${\displaystyle a^{-1}=a^{-1}\times 1=a^{-1}\times a\times a^{-1'}=1\times a^{-1'}=a^{-1'}}$.

Zero ring

Theorem: A ring ${\displaystyle (R,+,\cdot )}$ is the zero ring (that is, consists of precisely one element) if and only if ${\displaystyle 0=1}$.

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Suppose that ${\displaystyle 1=0}$. Let ${\displaystyle a}$ be any element in ${\displaystyle R}$; then ${\displaystyle a=a\cdot 1=a\cdot 0=0}$. Therefore, ${\displaystyle (R,+,\cdot )}$ is the zero ring. Conversely, if ${\displaystyle (R,+,\cdot )}$ is the zero ring, it must contain precisely one element by its definition. Therefore, ${\displaystyle 0}$ and ${\displaystyle 1}$ is the same element, i.e. ${\displaystyle 0=1}$.

Multiplication by negative one

Theorem:${\displaystyle (-1)a=-a}$

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${\displaystyle (-1)\cdot a+a=(-1)\cdot a+1\cdot a=((-1)+1)\cdot a=0\cdot a=0}$

Therefore ${\displaystyle (-1)\cdot a=(-1)\cdot a+0=(-1)\cdot a+(a+(-a))=((-1)\cdot a+a)+(-a)=0+(-a)=(-a)}$.

Multiplication by additive inverse

Theorem:${\displaystyle (-a)\cdot b=a\cdot (-b)=-(ab)}$

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To prove that the first expression equals the second one, ${\displaystyle (-a)\cdot b=((-1)\cdot a)\cdot b=(a\cdot (-1))\cdot b=a\cdot ((-1)\cdot b)=a(-b).}$

To prove that the first expression equals the third one, ${\displaystyle (-a)\cdot b=((-1)\cdot a)\cdot b=(-1)\cdot (a\cdot b).}$

A pseudo-ring does not necessarily have a multiplicative identity element. To prove that the first expression equals the third one without assuming the existence of a multiplicative identity, we show that ${\displaystyle (-a)\cdot b}$ is indeed the inverse of ${\displaystyle (a\cdot b)}$ by showing that adding them up results in the additive identity element,

${\displaystyle (a\cdot b)+(-a)\cdot b=(a-a)\cdot b=0\cdot b=0}$.