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: $0\cdot a=a\cdot 0=0$

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$0\cdot a=(0+0)\cdot a=(0\cdot a)+(0\cdot a)$

By subtracting (i.e. adding the additive inverse of) $0\cdot a$ on both sides of the equation, we get the desired result. The proof that $a\cdot 0=0$ 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 $e^{\prime }$ for the binary operation, then $e^{\prime }a=a{e}^{\prime }=a$, and when $a=e$, $e^{\prime }e=e{e}^{\prime }=e=e^{\prime }$ where $ab$ is the binary operation on ring elements $a$ and $b$.

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 $-a^{\prime }$ for $a$, then $-a=-a+0=-a+a-a^{\prime }=0-a^{\prime }=-a^{\prime }$.

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 $a^{-1^{\prime }}$ for $a$, then $a^{-1}=a^{-1}\times 1=a^{-1}\times a\times a^{-1^{\prime }}=1\times a^{-1^{\prime }}=a^{-1^{\prime }}$.

Zero ring

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

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

Multiplication by negative one

Theorem: $(-1)a=-a$

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$(-1)\cdot a+a=(-1)\cdot a+1\cdot a=((-1)+1)\cdot a=0\cdot a=0$

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

Multiplication by additive inverse

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

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

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

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 $(-a)\cdot b$ is indeed the inverse of $(a\cdot b)$ by showing that adding them up results in the additive identity element,

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

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